1 00:00:00,040 --> 00:00:02,460 The following content is provided under a Creative 2 00:00:02,460 --> 00:00:03,870 Commons license. 3 00:00:03,870 --> 00:00:06,910 Your support will help MIT OpenCourseWare continue to 4 00:00:06,910 --> 00:00:10,560 offer high quality educational resources for free. 5 00:00:10,560 --> 00:00:13,460 To make a donation or view additional materials from 6 00:00:13,460 --> 00:00:17,390 hundreds of MIT courses, visit MIT OpenCourseWare at 7 00:00:17,390 --> 00:00:18,640 ocw.mit.edu. 8 00:00:22,790 --> 00:00:24,630 PROFESSOR: So last time we started talking 9 00:00:24,630 --> 00:00:26,420 about random processes. 10 00:00:26,420 --> 00:00:30,710 A random process is a random experiment that 11 00:00:30,710 --> 00:00:32,570 evolves over time. 12 00:00:32,570 --> 00:00:35,110 And conceptually, it's important to realize that it's 13 00:00:35,110 --> 00:00:37,760 a single probabilistic experiment 14 00:00:37,760 --> 00:00:39,250 that has many stages. 15 00:00:39,250 --> 00:00:42,440 Actually, it has an infinite number of stages. 16 00:00:42,440 --> 00:00:47,060 And we discussed the simplest random process there is, the 17 00:00:47,060 --> 00:00:50,270 Bernoulli process, which is nothing but the sequence of 18 00:00:50,270 --> 00:00:51,570 Bernoulli trials-- 19 00:00:51,570 --> 00:00:54,580 an infinite sequence of Bernoulli trials. 20 00:00:54,580 --> 00:00:58,250 For example, flipping a coin over and over. 21 00:00:58,250 --> 00:01:01,640 Once we understand what's going on with that process, 22 00:01:01,640 --> 00:01:05,519 then what we want is to move into a continuous time version 23 00:01:05,519 --> 00:01:06,750 of the Bernoulli process. 24 00:01:06,750 --> 00:01:09,420 And this is what we will call the Poisson process. 25 00:01:09,420 --> 00:01:11,970 And for the Poisson process, we're going to do exactly the 26 00:01:11,970 --> 00:01:14,300 same things that we did for the Bernoulli process. 27 00:01:14,300 --> 00:01:18,210 That is, talk about the number of arrivals during a given 28 00:01:18,210 --> 00:01:21,400 time period, and talk also about the time between 29 00:01:21,400 --> 00:01:24,160 consecutive arrivals, and for the distribution of 30 00:01:24,160 --> 00:01:27,660 inter-arrival times. 31 00:01:27,660 --> 00:01:30,660 So let's start with a quick review of what we 32 00:01:30,660 --> 00:01:32,680 discussed last time. 33 00:01:32,680 --> 00:01:35,500 First, a note about language. 34 00:01:35,500 --> 00:01:38,100 If you think of coin tosses, we then talk 35 00:01:38,100 --> 00:01:40,660 about heads and tails. 36 00:01:40,660 --> 00:01:43,430 If you think of these as a sequence of trials, you can 37 00:01:43,430 --> 00:01:47,145 talk about successes and failures. 38 00:01:47,145 --> 00:01:50,312 The language that we will be using will be more the 39 00:01:50,312 --> 00:01:51,800 language of arrivals. 40 00:01:51,800 --> 00:01:56,020 That is, if in a given slot you have a success, you say 41 00:01:56,020 --> 00:01:57,930 that something arrived. 42 00:01:57,930 --> 00:02:00,470 If you have a failure, nothing arrived. 43 00:02:00,470 --> 00:02:03,090 And that language is a little more convenient and more 44 00:02:03,090 --> 00:02:06,820 natural, especially when we talk about continuous time-- 45 00:02:06,820 --> 00:02:10,250 to talk about arrivals instead of successes. 46 00:02:10,250 --> 00:02:12,640 But in any case, for the Bernoulli process let's keep, 47 00:02:12,640 --> 00:02:14,990 for a little bit, the language of successes. 48 00:02:14,990 --> 00:02:18,530 Whereas working in discrete time, we have time slots. 49 00:02:18,530 --> 00:02:20,870 During each time slot, we have an 50 00:02:20,870 --> 00:02:22,810 independent Bernoulli trial. 51 00:02:22,810 --> 00:02:25,750 There is probability p of having a success. 52 00:02:25,750 --> 00:02:29,400 Different slots are independent of each other. 53 00:02:29,400 --> 00:02:33,190 And this probability p is the same for any given time slot. 54 00:02:33,190 --> 00:02:36,540 So for this process we will discuss the one random 55 00:02:36,540 --> 00:02:39,240 variable of interest, which is the following. 56 00:02:39,240 --> 00:02:43,060 If we have n time slots, or n trials, how many 57 00:02:43,060 --> 00:02:44,600 arrivals will there be? 58 00:02:44,600 --> 00:02:46,990 Or how many successes will there be? 59 00:02:46,990 --> 00:02:50,910 Well, this is just given by the binomial PMF. 60 00:02:50,910 --> 00:02:54,780 Number of successes in n trials is a random variable 61 00:02:54,780 --> 00:02:58,420 that has a binomial PMF, and we know what this is. 62 00:02:58,420 --> 00:03:01,150 Then we talked about inter-arrival times. 63 00:03:01,150 --> 00:03:04,490 The time until the first arrival happens has a 64 00:03:04,490 --> 00:03:07,890 geometric distribution. 65 00:03:07,890 --> 00:03:11,770 And we have seen that from some time ago. 66 00:03:11,770 --> 00:03:14,890 Now if you start thinking about the time until k 67 00:03:14,890 --> 00:03:20,040 arrivals happen, and we denote that by Yk, this is the time 68 00:03:20,040 --> 00:03:22,190 until the first arrival happens. 69 00:03:22,190 --> 00:03:24,860 And then after the first arrival happens, you have to 70 00:03:24,860 --> 00:03:27,000 wait some time until the second arrival 71 00:03:27,000 --> 00:03:28,730 happens, and so on. 72 00:03:28,730 --> 00:03:32,650 And then the time from the (k -1)th arrival, until 73 00:03:32,650 --> 00:03:34,820 arrival number k. 74 00:03:34,820 --> 00:03:37,680 The important thing to realize here is that because the 75 00:03:37,680 --> 00:03:41,830 process has a memorylessness property, once the first 76 00:03:41,830 --> 00:03:45,330 arrival comes, it's as if we're starting from scratch 77 00:03:45,330 --> 00:03:48,030 and we will be flipping our coins until the 78 00:03:48,030 --> 00:03:49,560 next arrival comes. 79 00:03:49,560 --> 00:03:52,350 So the time it will take until the next arrival comes will 80 00:03:52,350 --> 00:03:54,810 also be a geometric random variable. 81 00:03:54,810 --> 00:03:57,790 And because different slots are independent, whatever 82 00:03:57,790 --> 00:04:00,580 happens after the first arrival is independent from 83 00:04:00,580 --> 00:04:02,640 whatever happened before. 84 00:04:02,640 --> 00:04:06,250 So T1 and T2 will be independent random variables. 85 00:04:06,250 --> 00:04:08,940 And similarly, all the way up to Tk. 86 00:04:08,940 --> 00:04:13,460 So the time until the k-th arrival is a sum of 87 00:04:13,460 --> 00:04:17,640 independent geometric random variables, with the same 88 00:04:17,640 --> 00:04:19,459 parameter p. 89 00:04:19,459 --> 00:04:23,010 And we saw last time that we can find the probability 90 00:04:23,010 --> 00:04:25,850 distribution of Yk. 91 00:04:25,850 --> 00:04:30,880 The probability that Yk takes a value of t is equal to-- 92 00:04:30,880 --> 00:04:36,110 there's this combinatorial factor here, and then you get 93 00:04:36,110 --> 00:04:41,590 p to the k, (1-p) to the (t-k), and this formula is 94 00:04:41,590 --> 00:04:48,020 true for t equal to k, k+1, and so on. 95 00:04:48,020 --> 00:04:49,950 And this distribution has a name. 96 00:04:49,950 --> 00:04:51,660 It's called the Pascal PMF. 97 00:04:54,500 --> 00:04:57,080 So this is all there is to know about 98 00:04:57,080 --> 00:04:59,130 the Bernoulli process. 99 00:04:59,130 --> 00:05:02,880 One important comment is to realize what exactly this 100 00:05:02,880 --> 00:05:05,540 memorylessness property is saying. 101 00:05:05,540 --> 00:05:07,450 So I discussed it a little bit last time. 102 00:05:07,450 --> 00:05:10,490 Let me reiterate it. 103 00:05:10,490 --> 00:05:13,460 So we have a Bernoulli process, which is a sequence 104 00:05:13,460 --> 00:05:15,230 of Bernoulli trials. 105 00:05:15,230 --> 00:05:17,990 And these are (0,1) random variables that 106 00:05:17,990 --> 00:05:20,320 keep going on forever. 107 00:05:20,320 --> 00:05:27,400 So someone is watching this movie of Bernoulli trials B_t. 108 00:05:27,400 --> 00:05:31,540 And at some point, they say they think, or something 109 00:05:31,540 --> 00:05:33,990 interesting has happened, why don't you come 110 00:05:33,990 --> 00:05:36,300 in and start watching? 111 00:05:36,300 --> 00:05:39,760 So at some time t, they tell you to come 112 00:05:39,760 --> 00:05:41,300 in and start watching. 113 00:05:41,300 --> 00:05:44,780 So what you will see once you come in will 114 00:05:44,780 --> 00:05:48,320 be this future trials. 115 00:05:48,320 --> 00:05:52,610 So actually what you will see is a random process, whose 116 00:05:52,610 --> 00:05:57,170 first random variable is going to be the first one that you 117 00:05:57,170 --> 00:05:59,340 see, B_(t +1). 118 00:05:59,340 --> 00:06:03,040 The second one is going to be this, and so on. 119 00:06:03,040 --> 00:06:06,850 So this is the process that's seen by the person who's asked 120 00:06:06,850 --> 00:06:10,360 to come in and start watching at that time. 121 00:06:10,360 --> 00:06:15,220 And the claim is that this process is itself a Bernoulli 122 00:06:15,220 --> 00:06:20,380 process, provided that the person who calls you into the 123 00:06:20,380 --> 00:06:23,740 room does not look into the future. 124 00:06:23,740 --> 00:06:27,100 The person who calls you into the room decides to call you 125 00:06:27,100 --> 00:06:31,180 in only on the basis of what they have seen so far. 126 00:06:31,180 --> 00:06:33,860 So for example, who calls you into the room might have a 127 00:06:33,860 --> 00:06:39,600 rule that says, as soon as I see a sequence of 3 heads, I 128 00:06:39,600 --> 00:06:43,730 ask the other person to come in. 129 00:06:43,730 --> 00:06:46,390 So if they use that particular rule, it means that when 130 00:06:46,390 --> 00:06:49,900 you're called in, the previous 3 were heads. 131 00:06:49,900 --> 00:06:53,190 But this doesn't give you any information about the future. 132 00:06:53,190 --> 00:06:55,260 And so the future ones will be just 133 00:06:55,260 --> 00:06:57,160 independent Bernoulli trials. 134 00:06:57,160 --> 00:07:00,370 If on the other hand, the person who calls you in has 135 00:07:00,370 --> 00:07:04,180 seen the movie before and they use a rule, such as, for 136 00:07:04,180 --> 00:07:09,710 example, I call you in just before 3 heads show up for the 137 00:07:09,710 --> 00:07:10,820 first time. 138 00:07:10,820 --> 00:07:13,850 So the person calls you in based on knowledge that these 139 00:07:13,850 --> 00:07:15,430 two would be three heads. 140 00:07:15,430 --> 00:07:17,490 If they have such foresight-- 141 00:07:17,490 --> 00:07:19,820 if they can look into the future-- 142 00:07:19,820 --> 00:07:25,050 then X1, X2, X3, they're certain to be three heads, so 143 00:07:25,050 --> 00:07:27,190 they do not correspond to random 144 00:07:27,190 --> 00:07:29,680 independent Bernoulli trials. 145 00:07:29,680 --> 00:07:33,660 So to rephrase this, the process is memoryless. 146 00:07:33,660 --> 00:07:38,460 It does not matter what has happened in the past. 147 00:07:38,460 --> 00:07:42,090 And that's true even if you are called into the room and 148 00:07:42,090 --> 00:07:45,700 start watching at a random time, as long as that random 149 00:07:45,700 --> 00:07:50,950 time is determined in a causal way on the basis of what has 150 00:07:50,950 --> 00:07:52,340 happened so far. 151 00:07:52,340 --> 00:07:55,660 So you are called into the room in a causal manner, just 152 00:07:55,660 --> 00:07:57,630 based on what's happened so far. 153 00:07:57,630 --> 00:08:00,030 What you're going to see starting from that time will 154 00:08:00,030 --> 00:08:03,240 still be a sequence of independent Bernoulli trials. 155 00:08:03,240 --> 00:08:06,560 And this is the argument that we used here, essentially, to 156 00:08:06,560 --> 00:08:09,190 argue that this T2 is an independent random 157 00:08:09,190 --> 00:08:11,030 variable from T1. 158 00:08:11,030 --> 00:08:14,530 So a person is watching the movie, sees the first success. 159 00:08:17,260 --> 00:08:19,810 And on the basis of what they have seen-- 160 00:08:19,810 --> 00:08:21,620 they have just seen the first success-- 161 00:08:21,620 --> 00:08:23,590 they ask you to come in. 162 00:08:23,590 --> 00:08:24,350 You come in. 163 00:08:24,350 --> 00:08:27,260 What you're going to see is a sequence of Bernoulli trials. 164 00:08:27,260 --> 00:08:32,260 And you wait this long until the next success comes in. 165 00:08:32,260 --> 00:08:35,390 What you see is a Bernoulli process, as if the process was 166 00:08:35,390 --> 00:08:37,299 just starting right now. 167 00:08:37,299 --> 00:08:40,830 And that convinces us that this should be a geometric 168 00:08:40,830 --> 00:08:43,200 random variable of the same kind as this one, as 169 00:08:43,200 --> 00:08:47,080 independent from what happened before. 170 00:08:47,080 --> 00:08:47,370 All right. 171 00:08:47,370 --> 00:08:49,610 So this is pretty much all there is to know about the 172 00:08:49,610 --> 00:08:50,650 Bernoulli process. 173 00:08:50,650 --> 00:08:52,860 Plus the two things that we did at the end of the last 174 00:08:52,860 --> 00:08:55,640 lecture where we merge two independent Bernoulli 175 00:08:55,640 --> 00:08:58,070 processes, we get a Bernoulli process. 176 00:08:58,070 --> 00:09:01,240 If we have a Bernoulli process and we split it by flipping a 177 00:09:01,240 --> 00:09:05,450 coin and sending things one way or the other, then we get 178 00:09:05,450 --> 00:09:07,790 two separate Bernoulli processes. 179 00:09:07,790 --> 00:09:10,590 And we see that all of these carry over to 180 00:09:10,590 --> 00:09:11,690 the continuous time. 181 00:09:11,690 --> 00:09:14,700 And our task for today is basically to work these 182 00:09:14,700 --> 00:09:18,000 continuous time variations. 183 00:09:18,000 --> 00:09:21,440 So the Poisson process is a continuous time version of the 184 00:09:21,440 --> 00:09:23,480 Bernoulli process. 185 00:09:23,480 --> 00:09:25,250 Here's the motivation for considering 186 00:09:25,250 --> 00:09:26,930 it a Bernoulli process. 187 00:09:26,930 --> 00:09:29,850 So you have that person whose job is to sit outside 188 00:09:29,850 --> 00:09:32,120 the door of a bank. 189 00:09:32,120 --> 00:09:38,280 And they have this long sheet, and for every one second slot, 190 00:09:38,280 --> 00:09:42,560 they mark an X if a person came in, or they mark 191 00:09:42,560 --> 00:09:45,760 something else if no one came in during that slot. 192 00:09:45,760 --> 00:09:48,500 Now the bank manager is a really scientifically trained 193 00:09:48,500 --> 00:09:50,530 person and wants very accurate results. 194 00:09:50,530 --> 00:09:53,380 So they tell you, don't use one second slots, use 195 00:09:53,380 --> 00:09:54,400 milliseconds slots. 196 00:09:54,400 --> 00:09:57,160 So you have all those slots and you keep filling if 197 00:09:57,160 --> 00:09:59,950 someone arrived or not during that slot. 198 00:09:59,950 --> 00:10:01,870 Well then you come up with an idea. 199 00:10:01,870 --> 00:10:06,760 Why use millisecond slots and keep putting crosses or zero's 200 00:10:06,760 --> 00:10:08,150 into each slot? 201 00:10:08,150 --> 00:10:12,380 It's much simpler if I just record the exact times when 202 00:10:12,380 --> 00:10:14,010 people came in. 203 00:10:14,010 --> 00:10:16,380 So time is continuous. 204 00:10:16,380 --> 00:10:20,340 I don't keep doing something at every time slot. 205 00:10:20,340 --> 00:10:24,440 But instead of the time axis, I mark the times at which 206 00:10:24,440 --> 00:10:26,370 customers arrive. 207 00:10:26,370 --> 00:10:28,620 So there's no real need for slots. 208 00:10:28,620 --> 00:10:32,370 The only information that you want is when did we have 209 00:10:32,370 --> 00:10:34,130 arrivals of people. 210 00:10:34,130 --> 00:10:37,830 And we want to now model a process of this kind happening 211 00:10:37,830 --> 00:10:41,880 in continuous time, that has the same flavor, however, as 212 00:10:41,880 --> 00:10:44,210 the Bernoulli process. 213 00:10:44,210 --> 00:10:48,170 So that's the model we want to develop. 214 00:10:48,170 --> 00:10:48,550 OK. 215 00:10:48,550 --> 00:10:52,880 So what are the properties that we're going to have? 216 00:10:52,880 --> 00:10:57,190 First, we're going to assume that intervals over the same 217 00:10:57,190 --> 00:11:01,340 length behave probabilistically in an 218 00:11:01,340 --> 00:11:04,500 identical fashion. 219 00:11:04,500 --> 00:11:06,740 So what does that mean? 220 00:11:06,740 --> 00:11:09,640 Think of an interval of some given length. 221 00:11:09,640 --> 00:11:12,450 During the interval of that length, there's going to be a 222 00:11:12,450 --> 00:11:14,750 random number of arrivals. 223 00:11:14,750 --> 00:11:17,140 And that random number of arrivals is going to have a 224 00:11:17,140 --> 00:11:19,050 probability distribution. 225 00:11:19,050 --> 00:11:20,630 So that probability distribution-- 226 00:11:20,630 --> 00:11:24,820 let's denote it by this notation. 227 00:11:24,820 --> 00:11:29,410 We fix t, we fix the duration. 228 00:11:29,410 --> 00:11:31,650 So this is fixed. 229 00:11:31,650 --> 00:11:34,180 And we look at the different k's. 230 00:11:34,180 --> 00:11:37,190 The probability of having 0 arrivals, the probability of 1 231 00:11:37,190 --> 00:11:40,100 arrival, the probability of 2 arrivals, and so on. 232 00:11:40,100 --> 00:11:42,630 So this thing is essentially a PMF. 233 00:11:42,630 --> 00:11:46,920 So it should have the property that the sum over all k's of 234 00:11:46,920 --> 00:11:49,890 this P_(k, tau) should be equal to 1. 235 00:11:52,620 --> 00:11:57,490 Now, hidden inside this notation is an assumption of 236 00:11:57,490 --> 00:11:59,550 time homogeneity. 237 00:11:59,550 --> 00:12:03,520 That is, this probability distribution for the number of 238 00:12:03,520 --> 00:12:09,020 arrivals only depends on the length of the interval, but 239 00:12:09,020 --> 00:12:13,230 not the exact location of the interval on the time axis. 240 00:12:13,230 --> 00:12:18,810 That is, if I take an interval of length tau, and I ask about 241 00:12:18,810 --> 00:12:21,200 the number of arrivals in this interval. 242 00:12:21,200 --> 00:12:24,990 And I take another interval of length tau, and I ask about 243 00:12:24,990 --> 00:12:27,790 the number of arrivals during that interval. 244 00:12:27,790 --> 00:12:31,040 Number of arrivals here, and number of arrivals there have 245 00:12:31,040 --> 00:12:34,400 the same probability distribution, which is 246 00:12:34,400 --> 00:12:36,990 denoted this way. 247 00:12:36,990 --> 00:12:41,650 So the statistical behavior of arrivals here is the same as 248 00:12:41,650 --> 00:12:44,310 the statistical behavioral of arrivals there. 249 00:12:44,310 --> 00:12:46,820 What's the relation with the Bernoulli process? 250 00:12:46,820 --> 00:12:48,510 It's very much like the assumption-- 251 00:12:48,510 --> 00:12:49,710 the Bernoulli process-- 252 00:12:49,710 --> 00:12:52,530 that in different slots, we have the same 253 00:12:52,530 --> 00:12:54,450 probability of success. 254 00:12:54,450 --> 00:12:56,770 Every slot looks probabilistically 255 00:12:56,770 --> 00:12:58,400 as any other slot. 256 00:12:58,400 --> 00:13:02,760 So similarly here, any interval of length tau looks 257 00:13:02,760 --> 00:13:06,790 probabilistically as any other interval of length tau. 258 00:13:06,790 --> 00:13:09,710 And the number of arrivals during that interval is a 259 00:13:09,710 --> 00:13:12,350 random variable described by these probabilities. 260 00:13:12,350 --> 00:13:15,570 Number of arrivals here is a random variable described by 261 00:13:15,570 --> 00:13:18,270 these same probabilities. 262 00:13:18,270 --> 00:13:19,900 So that's our first assumption. 263 00:13:19,900 --> 00:13:21,030 Then what else? 264 00:13:21,030 --> 00:13:23,340 In the Bernoulli process we had the assumption that 265 00:13:23,340 --> 00:13:27,710 different time slots were independent of each other. 266 00:13:27,710 --> 00:13:32,970 Here we do not have time slots, but we can still think 267 00:13:32,970 --> 00:13:37,490 in a similar way and impose the following assumption, that 268 00:13:37,490 --> 00:13:40,970 these joint time intervals are statistically independent. 269 00:13:40,970 --> 00:13:42,640 What does that mean? 270 00:13:42,640 --> 00:13:45,530 Does a random number of arrivals during this interval, 271 00:13:45,530 --> 00:13:48,270 and the random number of arrivals during this interval, 272 00:13:48,270 --> 00:13:49,830 and the random number of 273 00:13:49,830 --> 00:13:51,730 arrivals during this interval-- 274 00:13:51,730 --> 00:13:55,110 so these are three different random variables-- 275 00:13:55,110 --> 00:13:58,900 these three random variables are independent of each other. 276 00:13:58,900 --> 00:14:02,170 How many arrivals we got here is independent from how many 277 00:14:02,170 --> 00:14:04,040 arrivals we got there. 278 00:14:04,040 --> 00:14:07,410 So this is similar to saying that different time slots were 279 00:14:07,410 --> 00:14:08,110 independent. 280 00:14:08,110 --> 00:14:10,020 That's what we did in discrete time. 281 00:14:10,020 --> 00:14:13,190 The continuous time analog is this independence assumption. 282 00:14:13,190 --> 00:14:16,400 So for example, in particular, number of arrivals here is 283 00:14:16,400 --> 00:14:19,440 independent from the number of arrivals there. 284 00:14:19,440 --> 00:14:22,980 So these are two basic assumptions about the process. 285 00:14:25,620 --> 00:14:30,270 Now in order to write down a formula, eventually, about 286 00:14:30,270 --> 00:14:33,050 this probability distribution-- 287 00:14:33,050 --> 00:14:36,860 which is our next objective, we would like to say something 288 00:14:36,860 --> 00:14:38,790 specific about this distribution 289 00:14:38,790 --> 00:14:40,310 of number of arrivals-- 290 00:14:40,310 --> 00:14:43,980 we need to add a little more structure into the problem. 291 00:14:43,980 --> 00:14:47,380 And we're going to make the following assumption. 292 00:14:47,380 --> 00:14:51,140 If we look at the time interval of length delta-- 293 00:14:51,140 --> 00:14:54,090 and delta now is supposed to be a small number, so a 294 00:14:54,090 --> 00:14:55,900 picture like this-- 295 00:14:55,900 --> 00:15:00,790 during a very small time interval, there is a 296 00:15:00,790 --> 00:15:06,140 probability that we get exactly one arrival, which is 297 00:15:06,140 --> 00:15:07,750 lambda times delta. 298 00:15:07,750 --> 00:15:10,630 Delta is the length of the interval and lambda is a 299 00:15:10,630 --> 00:15:14,510 proportionality factor, which is sort of the intensity of 300 00:15:14,510 --> 00:15:16,190 the arrival process. 301 00:15:16,190 --> 00:15:21,000 Bigger lambda means that a little interval is more likely 302 00:15:21,000 --> 00:15:24,140 to get an arrival. 303 00:15:24,140 --> 00:15:25,595 So there's a probability lambda 304 00:15:25,595 --> 00:15:27,570 times delta of 1 arrival. 305 00:15:27,570 --> 00:15:31,740 The remaining probability goes to 0 arrivals. 306 00:15:31,740 --> 00:15:35,560 And when delta is small, the probability of 2 arrivals can 307 00:15:35,560 --> 00:15:39,660 be approximated by 0. 308 00:15:39,660 --> 00:15:42,460 So this is a description of what happens during 309 00:15:42,460 --> 00:15:45,240 a small, tiny slot. 310 00:15:45,240 --> 00:15:48,200 Now this is something that's supposed to be true in some 311 00:15:48,200 --> 00:15:51,530 limiting sense, when delta is very small. 312 00:15:51,530 --> 00:15:56,070 So the exact version of this statement would be that this 313 00:15:56,070 --> 00:16:02,050 is an equality, plus order of delta squared terms. 314 00:16:02,050 --> 00:16:04,060 So this is an approximate equality. 315 00:16:04,060 --> 00:16:07,850 And what approximation means is that in the limit of small 316 00:16:07,850 --> 00:16:11,970 deltas, the dominant terms-- 317 00:16:11,970 --> 00:16:15,900 the constant and the first order term are given by this. 318 00:16:15,900 --> 00:16:19,760 Now when delta is very small, second order terms in delta do 319 00:16:19,760 --> 00:16:21,130 not matter. 320 00:16:21,130 --> 00:16:24,380 They are small compared to first order terms. 321 00:16:24,380 --> 00:16:26,190 So we ignore this. 322 00:16:26,190 --> 00:16:30,280 So you can either think in terms of an exact relation, 323 00:16:30,280 --> 00:16:34,640 which is the probabilities are given by this, plus delta 324 00:16:34,640 --> 00:16:35,990 squared terms. 325 00:16:35,990 --> 00:16:38,730 Or if you want to be a little more loose, you just write 326 00:16:38,730 --> 00:16:41,000 here, as an approximate equality. 327 00:16:41,000 --> 00:16:44,080 And the understanding is that this equality holds-- 328 00:16:44,080 --> 00:16:50,850 approximately becomes more and more correct as 329 00:16:50,850 --> 00:16:53,410 delta goes to 0. 330 00:16:53,410 --> 00:16:57,250 So another version of that statement would be that if you 331 00:16:57,250 --> 00:17:03,280 take the limit as delta goes to 0, of p, the probability of 332 00:17:03,280 --> 00:17:06,829 having 1 arrival in an interval of length delta, 333 00:17:06,829 --> 00:17:10,300 divided by delta, this is equal to lambda. 334 00:17:10,300 --> 00:17:16,010 So that would be one version of an exact statement of what 335 00:17:16,010 --> 00:17:19,250 we are assuming here. 336 00:17:19,250 --> 00:17:22,750 So this lambda, we call it the arrival rate, or the intensity 337 00:17:22,750 --> 00:17:23,930 of the process. 338 00:17:23,930 --> 00:17:27,349 And clearly, if you double lambda, then a little interval 339 00:17:27,349 --> 00:17:29,340 is likely -- 340 00:17:29,340 --> 00:17:31,630 you expect to get -- 341 00:17:31,630 --> 00:17:34,200 the probability of obtaining an arrival during that 342 00:17:34,200 --> 00:17:35,740 interval has doubled. 343 00:17:35,740 --> 00:17:40,400 So in some sense we have twice as intense arrival process. 344 00:17:40,400 --> 00:17:46,470 If you look at the number of arrivals during delta 345 00:17:46,470 --> 00:17:54,490 interval, what is the expected value of that random variable? 346 00:17:54,490 --> 00:18:00,100 Well with probability lambda delta we get 1 arrival. 347 00:18:00,100 --> 00:18:01,240 And with the remaining 348 00:18:01,240 --> 00:18:03,610 probability, we get 0 arrivals. 349 00:18:03,610 --> 00:18:06,680 So it's just lambda times delta. 350 00:18:06,680 --> 00:18:10,640 So expected number of arrivals during a little interval is 351 00:18:10,640 --> 00:18:12,170 lambda times delta. 352 00:18:12,170 --> 00:18:15,460 So expected number of arrivals is proportional to lambda, and 353 00:18:15,460 --> 00:18:19,050 that's again why we call lambda the arrival rate. 354 00:18:19,050 --> 00:18:22,820 If you send delta to the denominator in this equality, 355 00:18:22,820 --> 00:18:26,380 it tells you that lambda is the expected number of 356 00:18:26,380 --> 00:18:30,000 arrivals per unit time. 357 00:18:30,000 --> 00:18:37,010 So the arrival rate is expected number of arrivals 358 00:18:37,010 --> 00:18:38,750 per unit time. 359 00:18:38,750 --> 00:18:42,580 And again, that justifies why we call lambda the intensity 360 00:18:42,580 --> 00:18:43,830 of this process. 361 00:18:46,316 --> 00:18:46,760 All right. 362 00:18:46,760 --> 00:18:49,680 So where are we now? 363 00:18:49,680 --> 00:18:53,740 For the Bernoulli process, the number of arrivals during a 364 00:18:53,740 --> 00:19:00,210 given interval of length n had the PMF that we knew it was 365 00:19:00,210 --> 00:19:01,545 the binomial PMF. 366 00:19:04,190 --> 00:19:07,530 What is the formula for the corresponding PMF for the 367 00:19:07,530 --> 00:19:09,190 continuous time process? 368 00:19:09,190 --> 00:19:12,570 Somehow we would like to use our assumptions and come up 369 00:19:12,570 --> 00:19:16,100 with the formula for this quantity. 370 00:19:16,100 --> 00:19:19,110 So this tells us about the distribution of number of 371 00:19:19,110 --> 00:19:23,750 arrivals during an interval of some general length. 372 00:19:23,750 --> 00:19:27,690 We have made assumptions about the number of arrivals during 373 00:19:27,690 --> 00:19:30,410 an interval of small length. 374 00:19:30,410 --> 00:19:34,350 An interval of big length is composed of many intervals of 375 00:19:34,350 --> 00:19:37,830 small length, so maybe this is the way to go. 376 00:19:37,830 --> 00:19:43,120 Take a big interval, and split it into many intervals of 377 00:19:43,120 --> 00:19:44,970 small length. 378 00:19:44,970 --> 00:19:48,410 So we have here our time axis. 379 00:19:48,410 --> 00:19:51,480 And we have an interval of length tau. 380 00:19:51,480 --> 00:19:55,240 And I'm going to split it into lots of little intervals of 381 00:19:55,240 --> 00:19:56,580 length delta. 382 00:19:56,580 --> 00:19:59,000 So how many intervals are we going to have? 383 00:19:59,000 --> 00:20:03,060 The number of intervals is going to be the total time, 384 00:20:03,060 --> 00:20:04,620 divided by delta. 385 00:20:07,520 --> 00:20:12,960 Now what happens during each one of these little intervals? 386 00:20:12,960 --> 00:20:22,380 As long as the intervals are small, what you have is that 387 00:20:22,380 --> 00:20:24,240 during an interval, you're going to have 388 00:20:24,240 --> 00:20:27,220 either 0 or 1 arrival. 389 00:20:27,220 --> 00:20:29,940 The probability of more than 1 arrival during a little 390 00:20:29,940 --> 00:20:31,950 interval is negligible. 391 00:20:31,950 --> 00:20:35,380 So with this picture, you have essentially a Bernoulli 392 00:20:35,380 --> 00:20:39,970 process that consists of so many trials. 393 00:20:39,970 --> 00:20:43,160 And during each one of those trials, we have a probability 394 00:20:43,160 --> 00:20:46,730 of success, which is lambda times delta. 395 00:20:51,845 --> 00:20:54,330 Different little intervals here are 396 00:20:54,330 --> 00:20:56,140 independent of each other. 397 00:20:56,140 --> 00:20:58,670 That's one of our assumptions, that these joint time 398 00:20:58,670 --> 00:21:00,380 intervals are independent. 399 00:21:00,380 --> 00:21:05,590 So approximately, what we have is a Bernoulli process. 400 00:21:05,590 --> 00:21:06,980 We have independence. 401 00:21:06,980 --> 00:21:09,250 We have the number of slots of interest. 402 00:21:09,250 --> 00:21:11,450 And during each one of the slots we have a certain 403 00:21:11,450 --> 00:21:13,530 probability of success. 404 00:21:13,530 --> 00:21:17,300 So if we think of this as another good approximation of 405 00:21:17,300 --> 00:21:18,870 the Poisson process-- 406 00:21:18,870 --> 00:21:21,090 with the approximation becoming more and more 407 00:21:21,090 --> 00:21:23,595 accurate as delta goes to 0 -- 408 00:21:23,595 --> 00:21:28,150 what we should do would be to take the formula for the PMF 409 00:21:28,150 --> 00:21:32,320 of number of arrivals in a Bernoulli process, and then 410 00:21:32,320 --> 00:21:37,230 take the limit as delta goes to 0. 411 00:21:37,230 --> 00:21:45,260 So in the Bernoulli process, the probability of k arrivals 412 00:21:45,260 --> 00:21:52,730 is n choose k, and then you have p to the k. 413 00:21:52,730 --> 00:21:57,610 Now in our case, we have here lambda times delta, delta is 414 00:21:57,610 --> 00:21:59,340 tau over n. 415 00:22:02,190 --> 00:22:08,410 Delta is tau over n, so p is lambda times tau divided by n. 416 00:22:08,410 --> 00:22:11,010 So here's our p -- 417 00:22:11,010 --> 00:22:13,690 Lambda tau over n -- 418 00:22:13,690 --> 00:22:22,760 to the power k, and then times one minus this-- this is our 419 00:22:22,760 --> 00:22:24,540 one minus p-- 420 00:22:24,540 --> 00:22:25,790 to the power n-k. 421 00:22:30,010 --> 00:22:35,730 So this is the exact formula for the Bernoulli process. 422 00:22:35,730 --> 00:22:39,830 For the Poisson process, what we do is we take that formula 423 00:22:39,830 --> 00:22:43,360 and we let delta go to 0. 424 00:22:43,360 --> 00:22:48,150 As delta goes to 0, n goes to infinity. 425 00:22:48,150 --> 00:22:51,280 So that's the limit that we're taking. 426 00:22:51,280 --> 00:22:55,580 On the other hand, this expression lambda times tau-- 427 00:22:59,730 --> 00:23:03,740 lambda times tau, what is it going to be? 428 00:23:03,740 --> 00:23:06,860 Lambda times tau is equal to n times p. 429 00:23:09,900 --> 00:23:11,990 n times p, is that what I want? 430 00:23:21,300 --> 00:23:22,550 No, let's see. 431 00:23:26,600 --> 00:23:28,110 Lambda tau is np. 432 00:23:28,110 --> 00:23:29,990 Yeah. 433 00:23:29,990 --> 00:23:32,060 So lambda tau is np. 434 00:23:53,370 --> 00:23:54,030 All right. 435 00:23:54,030 --> 00:23:59,320 So we have this relation, lambda tau equals np. 436 00:23:59,320 --> 00:24:03,070 These two numbers being equal kind of makes sense. np is the 437 00:24:03,070 --> 00:24:05,890 expected number of successes you're going to get in the 438 00:24:05,890 --> 00:24:07,750 Bernoulli process. 439 00:24:07,750 --> 00:24:08,780 Lambda tau-- 440 00:24:08,780 --> 00:24:11,800 since lambda is the arrival rate and you have a total time 441 00:24:11,800 --> 00:24:15,710 of tau, lambda tau you can think of it as the number of 442 00:24:15,710 --> 00:24:19,750 expected arrivals in the Bernoulli process. 443 00:24:19,750 --> 00:24:22,000 We're doing a Bernoulli approximation 444 00:24:22,000 --> 00:24:23,250 to the Poisson process. 445 00:24:23,250 --> 00:24:26,150 We take the formula for the Bernoulli, and now take the 446 00:24:26,150 --> 00:24:30,060 limit as n goes to infinity. 447 00:24:30,060 --> 00:24:35,330 Now lambda tau over n is equal to p, so it's clear what this 448 00:24:35,330 --> 00:24:37,040 term is going to give us. 449 00:24:37,040 --> 00:24:39,695 This is just p to the power k. 450 00:24:48,820 --> 00:24:53,230 It will actually take a little more work than that. 451 00:24:53,230 --> 00:24:58,210 Now I'm not going to do the algebra, but I'm just telling 452 00:24:58,210 --> 00:25:03,390 you that one can take the limit in this formula here, as 453 00:25:03,390 --> 00:25:05,000 n goes to infinity. 454 00:25:05,000 --> 00:25:08,860 And that will give you another formula, the final formula for 455 00:25:08,860 --> 00:25:10,320 the Poisson PMF. 456 00:25:10,320 --> 00:25:13,400 One thing to notice is that here you have something like 1 457 00:25:13,400 --> 00:25:17,460 minus a constant over n, to the power n. 458 00:25:17,460 --> 00:25:21,630 And you may recall from calculus a formula of this 459 00:25:21,630 --> 00:25:26,540 kind, that this converges to e to the minus c. 460 00:25:26,540 --> 00:25:29,560 If you remember that formula from calculus, then you will 461 00:25:29,560 --> 00:25:32,750 expect that here, in the limit, you are going to get 462 00:25:32,750 --> 00:25:36,520 something like an e to the minus lambda tau. 463 00:25:36,520 --> 00:25:39,180 So indeed, we will get such a term. 464 00:25:39,180 --> 00:25:42,230 There is some work that needs to be done to find the limit 465 00:25:42,230 --> 00:25:45,880 of this expression, times that expression. 466 00:25:45,880 --> 00:25:48,690 The algebra is not hard, it's in the text. 467 00:25:48,690 --> 00:25:51,340 Let's not spend more time doing this. 468 00:25:51,340 --> 00:25:53,820 But let me just give you the formula of what 469 00:25:53,820 --> 00:25:55,620 comes at the end. 470 00:25:55,620 --> 00:25:59,720 And the formula that comes at the end is of this form. 471 00:25:59,720 --> 00:26:03,710 So what matters here is not so much the specific algebra that 472 00:26:03,710 --> 00:26:07,690 you will do to go from this formula to that one. 473 00:26:07,690 --> 00:26:09,370 It's kind of straightforward. 474 00:26:09,370 --> 00:26:14,535 What's important is the idea that the Poisson process, by 475 00:26:14,535 --> 00:26:19,710 definition, can be approximated by a Bernoulli 476 00:26:19,710 --> 00:26:25,040 process in which we have a very large number of slots-- 477 00:26:25,040 --> 00:26:27,970 n goes to infinity. 478 00:26:27,970 --> 00:26:32,600 Whereas we have a very small probability of success during 479 00:26:32,600 --> 00:26:34,420 each time slot. 480 00:26:34,420 --> 00:26:38,640 So a large number of slots, but tiny probability of 481 00:26:38,640 --> 00:26:40,480 success during each slot. 482 00:26:40,480 --> 00:26:42,370 And we take the limit as the slots 483 00:26:42,370 --> 00:26:44,680 become smaller and smaller. 484 00:26:44,680 --> 00:26:47,170 So with this approximation we end up with 485 00:26:47,170 --> 00:26:49,030 this particular formula. 486 00:26:49,030 --> 00:26:51,890 And this is the so-called Poisson PMF. 487 00:26:51,890 --> 00:26:53,930 Now this function P here -- 488 00:26:53,930 --> 00:26:55,190 has two arguments. 489 00:26:55,190 --> 00:26:58,320 The important thing to realize is that when you think of this 490 00:26:58,320 --> 00:27:02,900 as a PMF, you fix t to tau. 491 00:27:02,900 --> 00:27:06,010 And for a fixed tau, now this is a PMF. 492 00:27:06,010 --> 00:27:11,260 As I said before, the sum over k has to be equal to 1. 493 00:27:11,260 --> 00:27:15,510 So for a given tau, these probabilities add up to 1. 494 00:27:15,510 --> 00:27:20,590 The formula is moderately messy, but not too messy. 495 00:27:20,590 --> 00:27:24,570 One can work with it without too much pain. 496 00:27:24,570 --> 00:27:28,460 And what's the mean and variance of this PMF? 497 00:27:28,460 --> 00:27:31,560 Well what's the expected number of arrivals? 498 00:27:31,560 --> 00:27:35,680 If you think of this Bernoulli analogy, we know that the 499 00:27:35,680 --> 00:27:37,940 expected number of arrivals in the Bernoulli 500 00:27:37,940 --> 00:27:41,250 process is n times p. 501 00:27:41,250 --> 00:27:44,500 In the approximation that we're using in these 502 00:27:44,500 --> 00:27:48,170 procedure, n times p is the same as lambda tau. 503 00:27:48,170 --> 00:27:52,490 And that's why we get lambda tau to be the expected number 504 00:27:52,490 --> 00:27:53,480 of arrivals. 505 00:27:53,480 --> 00:27:56,140 Here I'm using t instead of tau. 506 00:27:56,140 --> 00:28:01,670 The expected number of arrivals is lambda t. 507 00:28:01,670 --> 00:28:05,450 So if you double the time, you expect to get 508 00:28:05,450 --> 00:28:07,290 twice as many arrivals. 509 00:28:07,290 --> 00:28:10,960 If you double the arrival rate, you expect to get twice 510 00:28:10,960 --> 00:28:12,760 as many arrivals. 511 00:28:12,760 --> 00:28:15,290 How about the formula for the variance? 512 00:28:15,290 --> 00:28:18,930 The variance of the Bernoulli process is np, 513 00:28:18,930 --> 00:28:22,720 times one minus p. 514 00:28:22,720 --> 00:28:25,800 What does this go to in the limit? 515 00:28:25,800 --> 00:28:31,170 In the limit that we're taking, as delta goes to zero, 516 00:28:31,170 --> 00:28:33,550 then p also goes to zero. 517 00:28:33,550 --> 00:28:37,270 The probability of success in any given slot goes to zero. 518 00:28:37,270 --> 00:28:39,700 So this term becomes insignificant. 519 00:28:39,700 --> 00:28:45,980 So this becomes n times p, which is again lambda t, or 520 00:28:45,980 --> 00:28:47,710 lambda tau. 521 00:28:47,710 --> 00:28:50,960 So the variance, instead of having this more complicated 522 00:28:50,960 --> 00:28:54,290 formula of the variance is the Bernoulli process, here it 523 00:28:54,290 --> 00:28:56,840 gets simplified and it's lambda t. 524 00:28:56,840 --> 00:29:00,580 So interestingly, the variance in the Poisson process is 525 00:29:00,580 --> 00:29:03,200 exactly the same as the expected value. 526 00:29:03,200 --> 00:29:06,360 So you can look at this as just some interesting 527 00:29:06,360 --> 00:29:07,780 coincidence. 528 00:29:07,780 --> 00:29:10,260 So now we're going to take this formula and 529 00:29:10,260 --> 00:29:11,370 see how to use it. 530 00:29:11,370 --> 00:29:14,060 First we're going to do a completely trivial, 531 00:29:14,060 --> 00:29:16,560 straightforward example. 532 00:29:16,560 --> 00:29:24,630 So 15 years ago when that example was made, email was 533 00:29:24,630 --> 00:29:27,230 coming at a rate of five messages per hour. 534 00:29:27,230 --> 00:29:30,660 I wish that was the case today. 535 00:29:30,660 --> 00:29:38,450 And now emails that are coming in, let's say during the day-- 536 00:29:38,450 --> 00:29:41,750 the arrival rates of emails are probably different in 537 00:29:41,750 --> 00:29:42,960 different times of the day. 538 00:29:42,960 --> 00:29:46,840 But if you fix a time slot, let's say 1:00 to 2:00 in the 539 00:29:46,840 --> 00:29:49,370 afternoon, there's probably a constant rate. 540 00:29:49,370 --> 00:29:53,050 And email arrivals are reasonably well modeled by a 541 00:29:53,050 --> 00:29:54,790 Poisson process. 542 00:29:54,790 --> 00:29:58,220 Speaking of modeling, it's not just email arrivals. 543 00:29:58,220 --> 00:30:02,290 Whenever arrivals happen in a completely random way, without 544 00:30:02,290 --> 00:30:05,370 any additional structure, the Poisson process is a good 545 00:30:05,370 --> 00:30:07,010 model of these arrivals. 546 00:30:07,010 --> 00:30:10,200 So the times at which car accidents will happen, that's 547 00:30:10,200 --> 00:30:11,450 a Poisson processes. 548 00:30:15,530 --> 00:30:19,550 If you have a very, very weak light source that's shooting 549 00:30:19,550 --> 00:30:24,290 out photons, just one at a time, the times at which these 550 00:30:24,290 --> 00:30:27,240 photons will go out is well modeled again 551 00:30:27,240 --> 00:30:28,670 by a Poisson process. 552 00:30:28,670 --> 00:30:30,540 So it's completely random. 553 00:30:30,540 --> 00:30:35,230 Or if you have a radioactive material where one atom at a 554 00:30:35,230 --> 00:30:43,720 time changes at random times. 555 00:30:43,720 --> 00:30:45,920 So it's a very slow radioactive decay. 556 00:30:45,920 --> 00:30:48,900 The time at which these alpha particles, or whatever we get 557 00:30:48,900 --> 00:30:51,580 emitted, again is going to be described 558 00:30:51,580 --> 00:30:53,200 by a Poisson process. 559 00:30:53,200 --> 00:30:58,220 So if you have arrivals, or emissions, that happen at 560 00:30:58,220 --> 00:31:02,660 completely random times, and once in a while you get an 561 00:31:02,660 --> 00:31:07,500 arrival or an event, then the Poisson process is a very good 562 00:31:07,500 --> 00:31:10,070 model for these events. 563 00:31:10,070 --> 00:31:12,200 So back to emails. 564 00:31:12,200 --> 00:31:16,350 Get them at a rate of five messages per day, per hour. 565 00:31:16,350 --> 00:31:19,420 In 30 minutes this is half an hour. 566 00:31:19,420 --> 00:31:23,770 So what we have is that lambda t, total 567 00:31:23,770 --> 00:31:26,520 number of arrivals is-- 568 00:31:26,520 --> 00:31:29,020 the expected number of arrivals is-- 569 00:31:29,020 --> 00:31:33,810 lambda is five, t is one-half, if we talk about hours. 570 00:31:33,810 --> 00:31:36,480 So lambda t is two to the 0.5. 571 00:31:36,480 --> 00:31:41,220 The probability of no new messages is the probability of 572 00:31:41,220 --> 00:31:48,560 zero, in time interval of length t, which, in our case, 573 00:31:48,560 --> 00:31:51,790 is one-half. 574 00:31:51,790 --> 00:31:55,510 And then we look back into the formula from the previous 575 00:31:55,510 --> 00:31:59,550 slide, and the probability of zero arrivals is lambda t to 576 00:31:59,550 --> 00:32:03,770 the power zero, divided by zero factorial, and then an e 577 00:32:03,770 --> 00:32:05,450 to the lambda t. 578 00:32:05,450 --> 00:32:07,840 And you plug in the numbers that we have. 579 00:32:07,840 --> 00:32:10,380 Lambda t to the zero power is one. 580 00:32:10,380 --> 00:32:12,040 Zero factorial is one. 581 00:32:12,040 --> 00:32:15,500 So we're left with e to the minus 2.5. 582 00:32:15,500 --> 00:32:18,860 And that number is 0.08. 583 00:32:18,860 --> 00:32:22,090 Similarly, you can ask for the probability that you get 584 00:32:22,090 --> 00:32:24,850 exactly one message in half an hour. 585 00:32:24,850 --> 00:32:27,420 And that would be-- the probability of one message in 586 00:32:27,420 --> 00:32:28,680 one-half an hour-- 587 00:32:28,680 --> 00:32:32,590 is going to be lambda t to the first power, divided by 1 588 00:32:32,590 --> 00:32:38,230 factorial, e to the minus lambda t, which-- 589 00:32:38,230 --> 00:32:41,900 as we now get the extra lambda t factor-- is going to be 2.5, 590 00:32:41,900 --> 00:32:43,650 e to the minus 2.5. 591 00:32:43,650 --> 00:32:46,930 And the numerical answer is 0.20. 592 00:32:46,930 --> 00:32:50,450 So this is how you use the PMF formula for the Poisson 593 00:32:50,450 --> 00:32:55,540 distribution that we had in the previous slide. 594 00:32:55,540 --> 00:32:55,890 All right. 595 00:32:55,890 --> 00:33:00,010 So this was all about the distribution of 596 00:33:00,010 --> 00:33:01,780 the number of arrivals. 597 00:33:01,780 --> 00:33:03,350 What else did we do last time? 598 00:33:03,350 --> 00:33:08,250 Last time we also talked about the time it takes until the 599 00:33:08,250 --> 00:33:09,500 k-th arrival. 600 00:33:12,390 --> 00:33:12,790 OK. 601 00:33:12,790 --> 00:33:16,020 So let's try to figure out something about this 602 00:33:16,020 --> 00:33:18,260 particular distribution. 603 00:33:18,260 --> 00:33:21,180 We can derive the distribution of the time of the k-th 604 00:33:21,180 --> 00:33:24,730 arrival by using the exact same argument 605 00:33:24,730 --> 00:33:27,360 as we did last time. 606 00:33:27,360 --> 00:33:31,650 So now the time of the k-th arrival is a 607 00:33:31,650 --> 00:33:33,830 continuous random variable. 608 00:33:33,830 --> 00:33:36,160 So it has a PDF. 609 00:33:36,160 --> 00:33:38,430 Since we are in continuous time, arrivals can 610 00:33:38,430 --> 00:33:39,900 happen at any time. 611 00:33:39,900 --> 00:33:42,310 So Yk is a continuous random variable. 612 00:33:45,200 --> 00:33:48,160 But now let's think of a time interval of 613 00:33:48,160 --> 00:33:49,410 length little delta. 614 00:33:52,370 --> 00:33:58,620 And use our usual interpretation of PDFs. 615 00:33:58,620 --> 00:34:03,180 The PDF of a random variable evaluated at a certain time 616 00:34:03,180 --> 00:34:08,010 times delta, this is the probability that the Yk falls 617 00:34:08,010 --> 00:34:09,514 in this little interval. 618 00:34:13,460 --> 00:34:16,639 So as I've said before, this is the best way of thinking 619 00:34:16,639 --> 00:34:18,420 about PDFs. 620 00:34:18,420 --> 00:34:22,179 PDFs give you probabilities of little intervals. 621 00:34:22,179 --> 00:34:25,540 So now let's try to calculate this probability. 622 00:34:25,540 --> 00:34:29,880 For the k-th arrival to happen inside this little interval, 623 00:34:29,880 --> 00:34:31,550 we need two things. 624 00:34:31,550 --> 00:34:35,790 We need an arrival to happen in this interval, and we need 625 00:34:35,790 --> 00:34:41,530 k minus one arrivals to happen during that interval. 626 00:34:41,530 --> 00:34:41,880 OK. 627 00:34:41,880 --> 00:34:45,469 You'll tell me, but it's possible that we might have 628 00:34:45,469 --> 00:34:50,130 the k minus one arrival happen here, and the k-th arrival to 629 00:34:50,130 --> 00:34:51,219 happen here. 630 00:34:51,219 --> 00:34:53,050 In principle, that's possible. 631 00:34:53,050 --> 00:34:56,139 But in the limit, when we take delta very small, the 632 00:34:56,139 --> 00:34:59,850 probability of having two arrivals in the same little 633 00:34:59,850 --> 00:35:01,830 slot is negligible. 634 00:35:01,830 --> 00:35:06,870 So assuming that no two arrivals can happen in the 635 00:35:06,870 --> 00:35:10,940 same mini slot, then for the k-th one to happen here, we 636 00:35:10,940 --> 00:35:15,710 must have k minus one during this interval. 637 00:35:15,710 --> 00:35:20,210 Now because we have assumed that these joint intervals are 638 00:35:20,210 --> 00:35:23,900 independent of each other, this breaks down into the 639 00:35:23,900 --> 00:35:33,070 probability that we have exactly k minus one arrivals, 640 00:35:33,070 --> 00:35:37,600 during the interval from zero to t, times the probability of 641 00:35:37,600 --> 00:35:41,410 exactly one arrival during that little interval, which is 642 00:35:41,410 --> 00:35:43,420 lambda delta. 643 00:35:43,420 --> 00:35:51,010 We do have a formula for this from the previous slide, which 644 00:35:51,010 --> 00:35:59,340 is lambda t, to the k minus 1, over k minus one factorial, 645 00:35:59,340 --> 00:36:07,190 times e to minus lambda t. 646 00:36:07,190 --> 00:36:09,070 And then lambda times delta. 647 00:36:14,910 --> 00:36:16,160 Did I miss something? 648 00:36:24,680 --> 00:36:26,310 Yeah, OK. 649 00:36:26,310 --> 00:36:26,970 All right. 650 00:36:26,970 --> 00:36:30,220 And now you cancel this delta with that delta. 651 00:36:30,220 --> 00:36:36,820 And that gives us a formula for the PDF of the time until 652 00:36:36,820 --> 00:36:39,170 the k-th arrival. 653 00:36:39,170 --> 00:36:43,290 This PDF, of course, depends on the number k. 654 00:36:43,290 --> 00:36:46,850 The first arrival is going to happen somewhere in 655 00:36:46,850 --> 00:36:48,040 this range of time. 656 00:36:48,040 --> 00:36:50,140 So this is the PDF that it has. 657 00:36:50,140 --> 00:36:53,170 The second arrival, of course, is going to happen later. 658 00:36:53,170 --> 00:36:54,860 And the PDF is this. 659 00:36:54,860 --> 00:36:57,880 So it's more likely to happen around these times. 660 00:36:57,880 --> 00:37:01,410 The third arrival has this PDF, so it's more likely to 661 00:37:01,410 --> 00:37:03,690 happen around those times. 662 00:37:03,690 --> 00:37:08,020 And if you were to take k equal to 100, 663 00:37:08,020 --> 00:37:10,470 you might get a PDF-- 664 00:37:10,470 --> 00:37:13,260 it's extremely unlikely that the k-th arrival happens in 665 00:37:13,260 --> 00:37:18,060 the beginning, and it might happen somewhere down there, 666 00:37:18,060 --> 00:37:20,010 far into the future. 667 00:37:20,010 --> 00:37:22,230 So depending on which particular arrival we're 668 00:37:22,230 --> 00:37:25,510 talking about, it has a different probability 669 00:37:25,510 --> 00:37:26,350 distribution. 670 00:37:26,350 --> 00:37:30,340 The time of the 100th arrival, of course, is expected to be a 671 00:37:30,340 --> 00:37:34,100 lot larger than the time of the first arrival. 672 00:37:34,100 --> 00:37:38,550 Incidentally, the time of the first arrival has a PDF whose 673 00:37:38,550 --> 00:37:40,160 form is quite simple. 674 00:37:40,160 --> 00:37:43,850 If you let k equal to one here, this term disappears. 675 00:37:43,850 --> 00:37:46,310 That term becomes a one. 676 00:37:46,310 --> 00:37:49,880 You're left with just lambda, e to the minus lambda. 677 00:37:49,880 --> 00:37:53,210 And you recognize it, it's the exponential distribution. 678 00:37:53,210 --> 00:37:57,210 So the time until the first arrival in a Poisson process 679 00:37:57,210 --> 00:38:00,160 is an exponential distribution. 680 00:38:00,160 --> 00:38:02,150 What was the time of the first arrival in 681 00:38:02,150 --> 00:38:03,970 the Bernoulli process? 682 00:38:03,970 --> 00:38:07,060 It was a geometric distribution. 683 00:38:07,060 --> 00:38:11,170 Well, not coincidentally, these two look quite a bit 684 00:38:11,170 --> 00:38:13,030 like the other. 685 00:38:13,030 --> 00:38:17,980 A geometric distribution has this kind of shape. 686 00:38:17,980 --> 00:38:21,900 The exponential distribution has that kind of shape. 687 00:38:21,900 --> 00:38:25,560 The geometric is just a discrete version of the 688 00:38:25,560 --> 00:38:27,090 exponential. 689 00:38:27,090 --> 00:38:29,860 In the Bernoulli case, we are in discrete time. 690 00:38:29,860 --> 00:38:32,540 We have a PMF for the time of the first 691 00:38:32,540 --> 00:38:35,080 arrival, which is geometric. 692 00:38:35,080 --> 00:38:38,540 In the Poisson case, what we get is the limit of the 693 00:38:38,540 --> 00:38:41,560 geometric as you let those lines become closer and 694 00:38:41,560 --> 00:38:46,480 closer, which gives you the exponential distribution. 695 00:38:46,480 --> 00:38:50,430 Now the Poisson process shares all the memorylessness 696 00:38:50,430 --> 00:38:52,870 properties of the Bernoulli process. 697 00:38:52,870 --> 00:38:56,750 And the way one can argue is just in terms of this picture. 698 00:38:56,750 --> 00:39:00,250 Since the Poisson process is the limit of Bernoulli 699 00:39:00,250 --> 00:39:03,570 processes, whatever qualitative processes you have 700 00:39:03,570 --> 00:39:07,340 in the Bernoulli process remain valid 701 00:39:07,340 --> 00:39:08,360 for the Poisson process. 702 00:39:08,360 --> 00:39:11,470 In particular we have this memorylessness property. 703 00:39:11,470 --> 00:39:15,120 You let the Poisson process run for some time, and then 704 00:39:15,120 --> 00:39:16,600 you start watching it. 705 00:39:16,600 --> 00:39:18,520 What ever happened in the past has no 706 00:39:18,520 --> 00:39:20,220 bearing about the future. 707 00:39:20,220 --> 00:39:23,150 Starting from right now, what's going to happen in the 708 00:39:23,150 --> 00:39:27,330 future is described again by a Poisson process, in the sense 709 00:39:27,330 --> 00:39:30,530 that during every little slot of length delta, there's going 710 00:39:30,530 --> 00:39:33,790 to be a probability of lambda delta of having an arrival. 711 00:39:33,790 --> 00:39:36,590 And that probably lambda delta is the same-- is 712 00:39:36,590 --> 00:39:38,070 always lambda delta-- 713 00:39:38,070 --> 00:39:41,270 no matter what happened in the past of the process. 714 00:39:41,270 --> 00:39:47,040 And in particular, we could use this argument to say that 715 00:39:47,040 --> 00:39:50,460 the time until the k-th arrival is the time that it 716 00:39:50,460 --> 00:39:53,720 takes for the first arrival to happen. 717 00:39:53,720 --> 00:39:56,380 OK, let me do it for k equal to two. 718 00:39:56,380 --> 00:39:59,630 And then after the first arrival happens, you wait a 719 00:39:59,630 --> 00:40:02,600 certain amount of time until the second arrival happens. 720 00:40:02,600 --> 00:40:06,410 Now once the first arrival happened, that's in the past. 721 00:40:06,410 --> 00:40:07,400 You start watching. 722 00:40:07,400 --> 00:40:10,690 From now on you have mini slots of length delta, each 723 00:40:10,690 --> 00:40:13,230 one having a probability of success lambda delta. 724 00:40:13,230 --> 00:40:16,230 It's as if we started the Poisson process from scratch. 725 00:40:16,230 --> 00:40:19,310 So starting from that time, the time until the next 726 00:40:19,310 --> 00:40:22,840 arrival is going to be again an exponential distribution, 727 00:40:22,840 --> 00:40:26,010 which doesn't care about what happened in the past, how long 728 00:40:26,010 --> 00:40:28,000 it took you for the first arrival. 729 00:40:28,000 --> 00:40:33,410 So these two random variables are going to be independent 730 00:40:33,410 --> 00:40:38,140 and exponential, with the same parameter lambda. 731 00:40:38,140 --> 00:40:42,570 So among other things, what we have done here is we have 732 00:40:42,570 --> 00:40:48,130 essentially derived the PDF of the sum of k independent 733 00:40:48,130 --> 00:40:49,320 exponentials. 734 00:40:49,320 --> 00:40:53,990 The time of the k-th arrival is the sum of k 735 00:40:53,990 --> 00:40:56,230 inter-arrival times. 736 00:40:56,230 --> 00:40:59,380 The inter-arrival times are all independent of each other 737 00:40:59,380 --> 00:41:01,450 because of memorylessness. 738 00:41:01,450 --> 00:41:04,245 And they all have the same exponential distribution. 739 00:41:07,130 --> 00:41:08,980 And by the way, this gives you a way to 740 00:41:08,980 --> 00:41:11,080 simulate the Poisson process. 741 00:41:11,080 --> 00:41:14,070 If you wanted to simulate it on your computer, you would 742 00:41:14,070 --> 00:41:20,140 have one option to break time into tiny, tiny slots. 743 00:41:20,140 --> 00:41:24,030 And for every tiny slot, use your random number generator 744 00:41:24,030 --> 00:41:27,520 to decide whether there was an arrival or not. 745 00:41:27,520 --> 00:41:29,810 To get it very accurate, you would have to 746 00:41:29,810 --> 00:41:32,090 use tiny, tiny slots. 747 00:41:32,090 --> 00:41:35,280 So that would be a lot of computation. 748 00:41:35,280 --> 00:41:38,530 The more clever way of simulating the Poisson process 749 00:41:38,530 --> 00:41:42,320 is you use your random number generator to generate a sample 750 00:41:42,320 --> 00:41:45,280 from an exponential distribution and call that 751 00:41:45,280 --> 00:41:47,240 your first arrival time. 752 00:41:47,240 --> 00:41:50,050 Then go back to the random number generator, generate 753 00:41:50,050 --> 00:41:53,040 another independent sample, again from the same 754 00:41:53,040 --> 00:41:54,780 exponential distribution. 755 00:41:54,780 --> 00:41:58,490 That's the time between the first and the second arrival, 756 00:41:58,490 --> 00:42:01,390 and you keep going that way. 757 00:42:01,390 --> 00:42:03,260 So as a sort of a quick summary, 758 00:42:03,260 --> 00:42:04,910 this is the big picture. 759 00:42:04,910 --> 00:42:08,630 This table doesn't tell you anything new. 760 00:42:08,630 --> 00:42:12,230 But it's good to have it as a reference, and to look at it, 761 00:42:12,230 --> 00:42:14,740 and to make sure you understand what all the 762 00:42:14,740 --> 00:42:16,300 different boxes are. 763 00:42:16,300 --> 00:42:18,930 Basically the Bernoulli process runs in discrete time. 764 00:42:18,930 --> 00:42:20,960 The Poisson process runs in continuous time. 765 00:42:20,960 --> 00:42:25,140 There's an analogy of arrival rates, p per trial, or 766 00:42:25,140 --> 00:42:27,270 intensity per unit time. 767 00:42:27,270 --> 00:42:32,190 We did derive, or sketched the derivation for the PMF of the 768 00:42:32,190 --> 00:42:33,610 number of arrivals. 769 00:42:33,610 --> 00:42:37,810 And the Poisson distribution, which is the distribution that 770 00:42:37,810 --> 00:42:40,450 we get, this Pk of t. 771 00:42:40,450 --> 00:42:44,220 Pk and t is the limit of the binomial when we take the 772 00:42:44,220 --> 00:42:49,710 limit in this particular way, as delta goes to zero, and n 773 00:42:49,710 --> 00:42:51,220 goes to infinity. 774 00:42:51,220 --> 00:42:54,270 The geometric becomes an exponential in the limit. 775 00:42:54,270 --> 00:42:56,960 And the distribution of the time of the k-th arrival-- 776 00:42:56,960 --> 00:42:59,600 we had a closed form formula last time for 777 00:42:59,600 --> 00:43:01,050 the Bernoulli process. 778 00:43:01,050 --> 00:43:03,930 We got the closed form formula this time 779 00:43:03,930 --> 00:43:05,230 for the Poisson process. 780 00:43:05,230 --> 00:43:08,940 And we actually used exactly the same argument to get these 781 00:43:08,940 --> 00:43:12,320 two closed form formulas. 782 00:43:12,320 --> 00:43:12,650 All right. 783 00:43:12,650 --> 00:43:18,280 So now let's talk about adding or merging Poisson processes. 784 00:43:18,280 --> 00:43:21,060 And there's two statements that we can make here. 785 00:43:21,060 --> 00:43:25,970 One has to do with adding Poisson random variables, just 786 00:43:25,970 --> 00:43:26,820 random variables. 787 00:43:26,820 --> 00:43:28,290 There's another statement about 788 00:43:28,290 --> 00:43:30,770 adding Poisson processes. 789 00:43:30,770 --> 00:43:34,540 And the second is a bigger statement than the first. 790 00:43:34,540 --> 00:43:36,140 But this is a warm up. 791 00:43:36,140 --> 00:43:39,140 Let's work with the first statement. 792 00:43:39,140 --> 00:43:42,460 So the claim is that the sum of independent Poisson random 793 00:43:42,460 --> 00:43:45,340 variables is Poisson. 794 00:43:45,340 --> 00:43:45,990 OK. 795 00:43:45,990 --> 00:43:50,490 So suppose that we have a Poisson process with rate-- 796 00:43:50,490 --> 00:43:51,760 just for simplicity-- 797 00:43:51,760 --> 00:43:53,170 lambda one. 798 00:43:53,170 --> 00:43:56,240 And I take the interval from zero to two. 799 00:43:56,240 --> 00:44:00,620 And that take then the interval from two until five. 800 00:44:00,620 --> 00:44:03,720 The number of arrivals during this interval-- 801 00:44:03,720 --> 00:44:06,730 let's call it n from zero to two-- 802 00:44:06,730 --> 00:44:13,920 is going to be a Poisson random variable, with 803 00:44:13,920 --> 00:44:18,240 parameter, or with mean, two. 804 00:44:18,240 --> 00:44:24,340 The number of arrivals during this interval is n from time 805 00:44:24,340 --> 00:44:26,340 two until five. 806 00:44:26,340 --> 00:44:31,120 This is again a Poisson random variable with mean equal to 807 00:44:31,120 --> 00:44:34,690 three, because the arrival rate is 1 and the duration of 808 00:44:34,690 --> 00:44:36,990 the interval is three. 809 00:44:36,990 --> 00:44:41,320 These two random variables are independent. 810 00:44:41,320 --> 00:44:43,760 They obey the Poisson distribution 811 00:44:43,760 --> 00:44:45,640 that we derived before. 812 00:44:45,640 --> 00:44:50,930 If you add them, what you get is the number of arrivals 813 00:44:50,930 --> 00:44:53,850 during the interval from zero to five. 814 00:44:53,850 --> 00:44:56,290 Now what kind of distribution does this 815 00:44:56,290 --> 00:44:57,910 random variable have? 816 00:44:57,910 --> 00:45:00,760 Well this is the number of arrivals over an interval of a 817 00:45:00,760 --> 00:45:03,600 certain length in a Poisson process. 818 00:45:03,600 --> 00:45:08,580 Therefore, this is also Poisson with mean five. 819 00:45:16,520 --> 00:45:19,040 Because for the Poisson process we know that this 820 00:45:19,040 --> 00:45:23,300 number of arrivals is Poisson, this is Poisson, but also the 821 00:45:23,300 --> 00:45:26,610 number of overall arrivals is also Poisson. 822 00:45:26,610 --> 00:45:30,040 This establishes that the sum of a Poisson plus a Poisson 823 00:45:30,040 --> 00:45:32,200 random variable gives us another 824 00:45:32,200 --> 00:45:33,630 Poisson random variable. 825 00:45:33,630 --> 00:45:37,110 So adding Poisson random variables gives us a Poisson 826 00:45:37,110 --> 00:45:38,720 random variable. 827 00:45:38,720 --> 00:45:42,660 But now I'm going to make a more general statement that 828 00:45:42,660 --> 00:45:44,940 it's not just number of arrivals during 829 00:45:44,940 --> 00:45:46,415 a fixed time interval-- 830 00:45:50,420 --> 00:45:53,240 it's not just numbers of arrivals for given time 831 00:45:53,240 --> 00:45:54,260 intervals-- 832 00:45:54,260 --> 00:45:57,770 but rather if you take two different Poisson processes 833 00:45:57,770 --> 00:46:02,330 and add them up, the process itself is Poisson in the sense 834 00:46:02,330 --> 00:46:05,930 that this process is going to satisfy all the assumptions of 835 00:46:05,930 --> 00:46:07,510 a Poisson process. 836 00:46:07,510 --> 00:46:11,060 So the story is that you have a red bulb that flashes at 837 00:46:11,060 --> 00:46:13,350 random times at the rate of lambda one. 838 00:46:13,350 --> 00:46:14,980 It's a Poisson process. 839 00:46:14,980 --> 00:46:19,080 You have an independent process where a green bulb 840 00:46:19,080 --> 00:46:21,230 flashes at random times. 841 00:46:21,230 --> 00:46:24,800 And you happen to be color blind, so you just see when 842 00:46:24,800 --> 00:46:26,630 something is flashing. 843 00:46:26,630 --> 00:46:29,920 So these two are assumed to be independent Poisson processes. 844 00:46:29,920 --> 00:46:34,968 What can we say about the process that you observe? 845 00:46:34,968 --> 00:46:40,250 So in the processes that you observe, if you take a typical 846 00:46:40,250 --> 00:46:45,170 time interval of length little delta, what can happen during 847 00:46:45,170 --> 00:46:48,380 that little time interval? 848 00:46:48,380 --> 00:46:55,280 The red process may have something flashing. 849 00:46:55,280 --> 00:46:56,815 So red flashes. 850 00:46:59,850 --> 00:47:01,580 Or the red does not. 851 00:47:06,610 --> 00:47:12,170 And for the other bulb, the green bulb, there's two 852 00:47:12,170 --> 00:47:13,020 possibilities. 853 00:47:13,020 --> 00:47:17,910 The green one flashes. 854 00:47:17,910 --> 00:47:20,900 And the other possibility is that the green does not. 855 00:47:24,990 --> 00:47:25,330 OK. 856 00:47:25,330 --> 00:47:29,070 So there's four possibilities about what can happen during a 857 00:47:29,070 --> 00:47:31,170 little slot. 858 00:47:31,170 --> 00:47:36,080 The probability that the red one flashes and the green one 859 00:47:36,080 --> 00:47:39,750 flashes, what is this probability? 860 00:47:39,750 --> 00:47:43,510 It's lambda one delta that the first one flashes, and lambda 861 00:47:43,510 --> 00:47:47,290 two delta that the second one does. 862 00:47:47,290 --> 00:47:50,280 I'm multiplying probabilities here because I'm making the 863 00:47:50,280 --> 00:47:52,645 assumption that the two processes are independent. 864 00:47:55,330 --> 00:47:57,330 OK. 865 00:47:57,330 --> 00:48:00,130 Now the probability that the red one flashes 866 00:48:00,130 --> 00:48:01,440 is lambda one delta. 867 00:48:01,440 --> 00:48:08,210 But the green one doesn't is one, minus lambda two delta. 868 00:48:08,210 --> 00:48:12,840 Here the probability would be that the red one does not, 869 00:48:12,840 --> 00:48:16,400 times the probability that the green one does. 870 00:48:16,400 --> 00:48:20,750 And then here we have the probability that none of them 871 00:48:20,750 --> 00:48:26,790 flash, which is whatever is left. 872 00:48:26,790 --> 00:48:29,600 But it's one minus lambda one delta, times one 873 00:48:29,600 --> 00:48:33,160 minus lambda two delta. 874 00:48:33,160 --> 00:48:36,920 Now we're thinking about delta as small. 875 00:48:36,920 --> 00:48:43,260 So think of the case where delta goes to zero, but in a 876 00:48:43,260 --> 00:48:49,840 way that we keep the first order terms. 877 00:48:49,840 --> 00:48:54,070 We keep the delta terms, but we throw away the delta 878 00:48:54,070 --> 00:48:55,020 squared terms. 879 00:48:55,020 --> 00:48:58,170 Delta squared terms are much smaller than the delta terms 880 00:48:58,170 --> 00:49:00,260 when delta becomes small. 881 00:49:00,260 --> 00:49:01,920 If we do that-- 882 00:49:01,920 --> 00:49:05,650 if we only keep the order of delta terms-- 883 00:49:05,650 --> 00:49:07,940 this term effectively disappears. 884 00:49:07,940 --> 00:49:09,110 This is delta squared. 885 00:49:09,110 --> 00:49:11,550 So we make it zero. 886 00:49:11,550 --> 00:49:14,550 So the probability of having simultaneously a red and a 887 00:49:14,550 --> 00:49:17,940 green flash during a little interval is negligible. 888 00:49:17,940 --> 00:49:20,150 What do we get here? 889 00:49:20,150 --> 00:49:23,200 Lambda delta times one survives, but 890 00:49:23,200 --> 00:49:24,910 this times that doesn't. 891 00:49:24,910 --> 00:49:28,820 So we can throw that away. 892 00:49:28,820 --> 00:49:32,190 So the approximation that we get is lambda one delta. 893 00:49:32,190 --> 00:49:34,010 Similarly here, this goes away. 894 00:49:34,010 --> 00:49:36,420 We're left with a lambda two delta. 895 00:49:36,420 --> 00:49:42,140 And this is whatever remains, whatever is left. 896 00:49:42,140 --> 00:49:45,000 So what do we have? 897 00:49:45,000 --> 00:49:51,400 That there is a probability of seeing a flash, either a red 898 00:49:51,400 --> 00:49:54,360 or a green, which is lambda one delta, 899 00:49:54,360 --> 00:49:57,020 plus lambda two delta. 900 00:49:57,020 --> 00:50:03,590 So if we take a little interval of length delta here, 901 00:50:03,590 --> 00:50:11,780 it's going to see an arrival with probability approximately 902 00:50:11,780 --> 00:50:15,100 lambda one, plus lambda two, delta. 903 00:50:15,100 --> 00:50:20,940 So every slot in this merged process has an arrival 904 00:50:20,940 --> 00:50:25,780 probability with a rate which is the sum of the rates of 905 00:50:25,780 --> 00:50:27,600 these two processes. 906 00:50:27,600 --> 00:50:29,640 So this is one part of the definition 907 00:50:29,640 --> 00:50:31,680 of the Poisson process. 908 00:50:31,680 --> 00:50:34,890 There's a few more things that one would need to verify. 909 00:50:34,890 --> 00:50:37,980 Namely, that intervals of the same length have the same 910 00:50:37,980 --> 00:50:41,000 probability distribution and that different slots are 911 00:50:41,000 --> 00:50:42,710 independent of each other. 912 00:50:42,710 --> 00:50:50,780 This can be argued by starting from here because different 913 00:50:50,780 --> 00:50:53,620 intervals in this process are independent from each other. 914 00:50:53,620 --> 00:50:56,900 Different intervals here are independent from each other. 915 00:50:56,900 --> 00:50:59,900 It's not hard to argue that different intervals in the 916 00:50:59,900 --> 00:51:03,580 merged process will also be independent of each other. 917 00:51:03,580 --> 00:51:06,480 So the conclusion that comes at the end is that this 918 00:51:06,480 --> 00:51:10,130 process is a Poisson process, with a total rate which is 919 00:51:10,130 --> 00:51:13,210 equal to the sum of the rate of the two processes. 920 00:51:13,210 --> 00:51:17,010 And now if I tell you that an arrival happened in the merged 921 00:51:17,010 --> 00:51:20,530 process at a certain time, how likely is it that 922 00:51:20,530 --> 00:51:23,470 it came from here? 923 00:51:23,470 --> 00:51:24,950 How likely is it? 924 00:51:24,950 --> 00:51:26,980 We go to this picture. 925 00:51:26,980 --> 00:51:30,140 Given that an arrival occurred-- 926 00:51:30,140 --> 00:51:36,050 which is the event that this or that happened-- 927 00:51:36,050 --> 00:51:39,330 what is the probability that it came from the first 928 00:51:39,330 --> 00:51:42,060 process, the red one? 929 00:51:42,060 --> 00:51:45,190 Well it's the probability of this divided by the 930 00:51:45,190 --> 00:51:48,030 probability of this, times that. 931 00:51:48,030 --> 00:51:52,760 Given that this event occurred, you want to find the 932 00:51:52,760 --> 00:51:56,560 conditional probability of that sub event. 933 00:51:56,560 --> 00:51:58,960 So we're asking the question, out of the total probability 934 00:51:58,960 --> 00:52:00,660 of these two, what fraction of that 935 00:52:00,660 --> 00:52:02,790 probability is assigned here? 936 00:52:02,790 --> 00:52:05,300 And this is lambda one delta, after we 937 00:52:05,300 --> 00:52:07,040 ignore the other terms. 938 00:52:07,040 --> 00:52:09,170 This is lambda two delta. 939 00:52:09,170 --> 00:52:15,040 So that fraction is going to be lambda one, over lambda one 940 00:52:15,040 --> 00:52:16,770 plus lambda two. 941 00:52:16,770 --> 00:52:17,640 What does this tell you? 942 00:52:17,640 --> 00:52:21,820 If lambda one and lambda two are equal, given that I saw an 943 00:52:21,820 --> 00:52:25,580 arrival here, it's equally likely to be red or green. 944 00:52:25,580 --> 00:52:29,716 But if the reds have a much higher arrival rate, when I 945 00:52:29,716 --> 00:52:32,700 see an arrival here, it's more likely this 946 00:52:32,700 --> 00:52:34,050 number will be large. 947 00:52:34,050 --> 00:52:38,390 So it's more likely to have come from the red process. 948 00:52:38,390 --> 00:52:40,830 OK so we'll continue with this story and do some 949 00:52:40,830 --> 00:52:42,080 applications next time.