1 00:00:00,000 --> 00:00:00,430 2 00:00:00,430 --> 00:00:00,860 Hi. 3 00:00:00,860 --> 00:00:03,820 In this problem, we have an absent-minded professor who 4 00:00:03,820 --> 00:00:06,860 will inadvertently give us some practice with exponential 5 00:00:06,860 --> 00:00:08,460 random variables. 6 00:00:08,460 --> 00:00:13,630 So the professor has made two appointments with two students 7 00:00:13,630 --> 00:00:17,480 and inadvertently made them at the same time. 8 00:00:17,480 --> 00:00:21,390 And what we do is we model the duration of these appointments 9 00:00:21,390 --> 00:00:23,340 with an exponential random variable. 10 00:00:23,340 --> 00:00:26,760 So remember, an exponential random variable is a 11 00:00:26,760 --> 00:00:29,850 continuous random variable that takes on non-negative 12 00:00:29,850 --> 00:00:35,980 values, and it's parametrized by a rate parameter, lambda. 13 00:00:35,980 --> 00:00:39,030 And the exponential random variable is often used to 14 00:00:39,030 --> 00:00:42,670 model durations of time-- so time until something happens, 15 00:00:42,670 --> 00:00:46,310 so for example, in this case, time until the student leaves 16 00:00:46,310 --> 00:00:47,420 or the appointment is over. 17 00:00:47,420 --> 00:00:51,180 Or sometimes you will also use it to be as a model of time 18 00:00:51,180 --> 00:00:52,430 until something fails. 19 00:00:52,430 --> 00:00:54,600 20 00:00:54,600 --> 00:00:59,010 And one thing that will be useful is the CDF of this 21 00:00:59,010 --> 00:01:00,240 exponential random variable. 22 00:01:00,240 --> 00:01:02,280 So the probability that it's less than or equal to some 23 00:01:02,280 --> 00:01:07,290 value, little t, is equal to 1 minus e to the minus lambda t. 24 00:01:07,290 --> 00:01:10,970 So this is, of course, valid only when t is non-negative. 25 00:01:10,970 --> 00:01:15,750 The other useful property is that the expected value of an 26 00:01:15,750 --> 00:01:17,790 exponential random variable is just 1 over 27 00:01:17,790 --> 00:01:20,280 the parameter lambda. 28 00:01:20,280 --> 00:01:22,390 And the last thing that we'll use specifically in this 29 00:01:22,390 --> 00:01:24,450 problem is the memory list property of 30 00:01:24,450 --> 00:01:25,410 exponential random variables. 31 00:01:25,410 --> 00:01:30,490 And so recall that that just means that if you pop in the 32 00:01:30,490 --> 00:01:36,700 middle of an exponential random variable, the 33 00:01:36,700 --> 00:01:40,350 distribution for that random variable going forward from 34 00:01:40,350 --> 00:01:43,345 the point where you popped in is exactly the same as if it 35 00:01:43,345 --> 00:01:44,400 had just started over. 36 00:01:44,400 --> 00:01:46,430 So that's why we call it the memory list property. 37 00:01:46,430 --> 00:01:49,250 Basically, the past doesn't really matter, and going 38 00:01:49,250 --> 00:01:52,670 forward from whatever point that you observe it at, it 39 00:01:52,670 --> 00:01:58,310 looks as if it had just started over afresh. 40 00:01:58,310 --> 00:02:02,350 And last thing we'll use, which is a review of a concept 41 00:02:02,350 --> 00:02:06,780 from earlier, is total expectation. 42 00:02:06,780 --> 00:02:08,889 So let's actually model this problem 43 00:02:08,889 --> 00:02:10,979 with two random variables. 44 00:02:10,979 --> 00:02:17,080 Let's let T1 be the time that the first student takes in the 45 00:02:17,080 --> 00:02:20,760 appointment and T2 be the time that the second student takes. 46 00:02:20,760 --> 00:02:23,160 And what we're told in the problem is that they're both 47 00:02:23,160 --> 00:02:26,000 exponential with mean 30 minutes. 48 00:02:26,000 --> 00:02:29,240 So remember the mean being 30 minutes means that the lambda 49 00:02:29,240 --> 00:02:30,640 is 1 over the mean. 50 00:02:30,640 --> 00:02:34,260 And so the lambda in this case would be 1/30. 51 00:02:34,260 --> 00:02:35,770 And importantly, we're also told that they are 52 00:02:35,770 --> 00:02:36,710 independent. 53 00:02:36,710 --> 00:02:40,160 So how long the first person takes is independent of how 54 00:02:40,160 --> 00:02:41,960 long the second person takes. 55 00:02:41,960 --> 00:02:46,560 So the first student arrives on time and take some random 56 00:02:46,560 --> 00:02:48,110 amount of time, T1. 57 00:02:48,110 --> 00:02:51,330 The second student arrives exactly five minutes late. 58 00:02:51,330 --> 00:02:55,170 And whatever the second person meets with the professor, that 59 00:02:55,170 --> 00:02:58,310 student will then take some random amount of time, T2. 60 00:02:58,310 --> 00:03:01,810 What we're asked to do is find the expected time between when 61 00:03:01,810 --> 00:03:03,270 the first student arrives-- 62 00:03:03,270 --> 00:03:05,370 so we can just call that time 0-- 63 00:03:05,370 --> 00:03:08,920 and when the second student leaves. 64 00:03:08,920 --> 00:03:12,580 Now you may say, well we're dealing with expectations, so 65 00:03:12,580 --> 00:03:13,890 it's easier. 66 00:03:13,890 --> 00:03:19,640 And in this case, it probably is just the expectation of how 67 00:03:19,640 --> 00:03:21,540 long the first student takes plus the expectation of how 68 00:03:21,540 --> 00:03:22,540 long the second student takes. 69 00:03:22,540 --> 00:03:26,450 So it should be about 60 minutes or exactly 60 minutes. 70 00:03:26,450 --> 00:03:28,940 Now, why is that not exactly right? 71 00:03:28,940 --> 00:03:36,450 It's because there is a small wrinkle, that the students may 72 00:03:36,450 --> 00:03:38,660 not go exactly back to back. 73 00:03:38,660 --> 00:03:41,970 So let's actually draw out a time frame of what might 74 00:03:41,970 --> 00:03:44,640 actually happen. 75 00:03:44,640 --> 00:03:47,240 So here's time 0, when the first student arrives. 76 00:03:47,240 --> 00:03:52,270 And the first will go for some amount of time and leave. 77 00:03:52,270 --> 00:03:54,390 And now let's consider two scenarios. 78 00:03:54,390 --> 00:04:00,570 One scenario is that the first student takes more than five 79 00:04:00,570 --> 00:04:01,560 minutes to complete. 80 00:04:01,560 --> 00:04:10,120 Well then the second student will have arrived at 5 minutes 81 00:04:10,120 --> 00:04:14,210 and then will be already waiting whenever this first 82 00:04:14,210 --> 00:04:14,810 student leaves. 83 00:04:14,810 --> 00:04:16,180 So then the second student will 84 00:04:16,180 --> 00:04:18,310 immediately pick up and continue. 85 00:04:18,310 --> 00:04:22,740 And in that case, we do have two exponentials back to back. 86 00:04:22,740 --> 00:04:25,370 But there could be another situation. 87 00:04:25,370 --> 00:04:32,110 Suppose that the first student didn't take very long at all 88 00:04:32,110 --> 00:04:37,260 and finished within five minutes, in which case the 89 00:04:37,260 --> 00:04:38,490 second student hasn't arrived yet. 90 00:04:38,490 --> 00:04:42,530 So this professor is idle in between here. 91 00:04:42,530 --> 00:04:45,970 And so we actually don't necessarily have two of them 92 00:04:45,970 --> 00:04:47,190 going back to back. 93 00:04:47,190 --> 00:04:50,260 So there's an empty period in between that we have to 94 00:04:50,260 --> 00:04:52,270 account for. 95 00:04:52,270 --> 00:04:57,730 So with that in mind, we see that we have two scenarios. 96 00:04:57,730 --> 00:04:59,640 And so what does that beg to use? 97 00:04:59,640 --> 00:05:02,520 Well, we can split them up into the two scenarios and 98 00:05:02,520 --> 00:05:05,140 then calculate expectations with each one and then use 99 00:05:05,140 --> 00:05:09,280 total expectation to find the overall 100 00:05:09,280 --> 00:05:12,090 expected length of time. 101 00:05:12,090 --> 00:05:15,350 OK, so let's begin with the first scenario. 102 00:05:15,350 --> 00:05:19,690 103 00:05:19,690 --> 00:05:24,450 The first scenario is that, let's say, the first student 104 00:05:24,450 --> 00:05:27,710 finished within five minutes. 105 00:05:27,710 --> 00:05:30,840 So what does that mean in terms of the definitions that 106 00:05:30,840 --> 00:05:31,870 we've used? 107 00:05:31,870 --> 00:05:37,050 That means T1 is less than or equal to 5. 108 00:05:37,050 --> 00:05:40,460 So if the first student took less than five minutes, then 109 00:05:40,460 --> 00:05:41,840 what happens? 110 00:05:41,840 --> 00:05:46,070 Then we know that the amount of time that 111 00:05:46,070 --> 00:05:47,320 you'd need to take-- 112 00:05:47,320 --> 00:05:51,510 113 00:05:51,510 --> 00:05:52,600 let's call that something else. 114 00:05:52,600 --> 00:05:58,190 Let's call that X. So X is the random variable that we're 115 00:05:58,190 --> 00:06:01,580 interested in, the time between when the first student 116 00:06:01,580 --> 00:06:04,140 comes and the second student leaves. 117 00:06:04,140 --> 00:06:05,950 This is the value that we want to find. 118 00:06:05,950 --> 00:06:10,440 Well we know that we're guaranteed that there will be 119 00:06:10,440 --> 00:06:12,720 a five-minute interval. 120 00:06:12,720 --> 00:06:17,260 So first student will come, and then the second person 121 00:06:17,260 --> 00:06:19,070 will take over. 122 00:06:19,070 --> 00:06:31,340 So we're guaranteed that the first five minutes will be the 123 00:06:31,340 --> 00:06:34,380 difference between when time starts and when the second 124 00:06:34,380 --> 00:06:35,240 student arrives. 125 00:06:35,240 --> 00:06:39,670 And then, after that, it's just however long the second 126 00:06:39,670 --> 00:06:44,320 student takes, which is just the expected value of T2. 127 00:06:44,320 --> 00:06:47,770 And T2 is a exponential random variable with mean 30. 128 00:06:47,770 --> 00:06:51,325 So in this case, it's just 35. 129 00:06:51,325 --> 00:06:55,800 130 00:06:55,800 --> 00:06:58,410 So the first student doesn't take very long. 131 00:06:58,410 --> 00:07:02,320 Then we just get the five minutes, that little buffer, 132 00:07:02,320 --> 00:07:04,170 plus however long the second student takes, which, on 133 00:07:04,170 --> 00:07:06,640 average, is 30 minutes. 134 00:07:06,640 --> 00:07:09,260 Now what is the probability of this happening? 135 00:07:09,260 --> 00:07:12,820 The probability of this happening is the probability 136 00:07:12,820 --> 00:07:16,080 that the first student takes less than five minutes. 137 00:07:16,080 --> 00:07:21,980 And here is where we use the CDF that we wrote out earlier. 138 00:07:21,980 --> 00:07:30,650 It's going to be 1 minus e to the minus lambda t. 139 00:07:30,650 --> 00:07:35,550 So in this case, t is five and lambda is 1/30. 140 00:07:35,550 --> 00:07:40,440 So it's minus 5/30 is the probability. 141 00:07:40,440 --> 00:07:43,580 All right, now let's consider the second case. 142 00:07:43,580 --> 00:07:47,690 The second case is that the first student actually takes 143 00:07:47,690 --> 00:07:50,360 longer than five minutes. 144 00:07:50,360 --> 00:07:54,350 OK, so what happens in that case? 145 00:07:54,350 --> 00:07:54,880 Here's five minutes. 146 00:07:54,880 --> 00:07:58,300 The first student came to five minutes. 147 00:07:58,300 --> 00:08:00,665 The second student arrived, and the first 148 00:08:00,665 --> 00:08:02,640 student is still going. 149 00:08:02,640 --> 00:08:06,320 So he goes for some amount of time. 150 00:08:06,320 --> 00:08:10,310 And then whenever he finishes, the second student continues. 151 00:08:10,310 --> 00:08:13,120 So now the question is, what is the total amount of 152 00:08:13,120 --> 00:08:14,960 time in this case? 153 00:08:14,960 --> 00:08:21,220 Well, you can think of it as using 154 00:08:21,220 --> 00:08:22,150 the memory list property. 155 00:08:22,150 --> 00:08:24,020 This is where it comes in. 156 00:08:24,020 --> 00:08:27,550 So the first five minutes, we know that it was already taken 157 00:08:27,550 --> 00:08:31,390 because we're considering the second scenario, which we're 158 00:08:31,390 --> 00:08:34,860 given that T1 is greater than 5. 159 00:08:34,860 --> 00:08:38,190 And so the question now is, if we know that, how much longer 160 00:08:38,190 --> 00:08:39,650 does it take? 161 00:08:39,650 --> 00:08:41,799 How much longer past the five-minute mark does the 162 00:08:41,799 --> 00:08:43,049 first student take? 163 00:08:43,049 --> 00:08:46,570 And by the member list property, we know that it's as 164 00:08:46,570 --> 00:08:49,210 if the first student started over. 165 00:08:49,210 --> 00:08:51,800 So there was no memory of the first five minutes, and it's 166 00:08:51,800 --> 00:08:55,310 as if the first student just arrived also at the 167 00:08:55,310 --> 00:08:57,000 five-minute mark and met with the professor. 168 00:08:57,000 --> 00:09:02,950 So past the five-minute mark, it's as if you have a new 169 00:09:02,950 --> 00:09:06,300 exponential random variable, still with mean 30. 170 00:09:06,300 --> 00:09:12,490 And so what we get is that, in this case, you get the 171 00:09:12,490 --> 00:09:18,760 guaranteed five minutes, and then you get the memory list 172 00:09:18,760 --> 00:09:25,860 continuation of the first student's appointment. 173 00:09:25,860 --> 00:09:28,460 So you get another 30 minutes on average because of the 174 00:09:28,460 --> 00:09:29,640 memory list property. 175 00:09:29,640 --> 00:09:32,470 And then whenever the first student finally does finish 176 00:09:32,470 --> 00:09:35,950 up, the second student will immediately take over because 177 00:09:35,950 --> 00:09:37,080 he has already arrived. 178 00:09:37,080 --> 00:09:38,580 It's past the five-minute mark. 179 00:09:38,580 --> 00:09:41,520 And then that second student will take, again, on average, 180 00:09:41,520 --> 00:09:42,530 30 more minutes. 181 00:09:42,530 --> 00:09:47,130 So what you get is, in this case, the appointment lasts 65 182 00:09:47,130 --> 00:09:49,540 minutes on average. 183 00:09:49,540 --> 00:09:51,100 Now what is the probability of this case? 184 00:09:51,100 --> 00:09:55,410 The probability of this case is the probability that T1 is 185 00:09:55,410 --> 00:09:57,150 greater than 5. 186 00:09:57,150 --> 00:10:00,720 And now we know that that is just the complement of this, 1 187 00:10:00,720 --> 00:10:01,520 minus that. 188 00:10:01,520 --> 00:10:04,050 So it's just e to the minus 5/30. 189 00:10:04,050 --> 00:10:06,840 190 00:10:06,840 --> 00:10:10,490 So now we have both scenarios. 191 00:10:10,490 --> 00:10:12,850 We have the probabilities of each scenario, and we have the 192 00:10:12,850 --> 00:10:15,980 expectation under each scenario. 193 00:10:15,980 --> 00:10:18,810 Now all that remains now is to combine them using total 194 00:10:18,810 --> 00:10:20,060 expectation. 195 00:10:20,060 --> 00:10:26,990 196 00:10:26,990 --> 00:10:36,880 So I really should have written expectation of X given 197 00:10:36,880 --> 00:10:38,920 T1 is less than or equal to 5 here. 198 00:10:38,920 --> 00:10:42,630 And this is expectation of X given that T1 is 199 00:10:42,630 --> 00:10:44,840 greater than 5. 200 00:10:44,840 --> 00:10:49,840 So expectation of X overall is the probability that T1 is 201 00:10:49,840 --> 00:10:54,250 less than or equal to 5 times the expectation of X given 202 00:10:54,250 --> 00:10:59,190 that T1 is less than or equal to 5 plus the probability that 203 00:10:59,190 --> 00:11:04,790 T1 is greater than 5 times the expectation of X given that T1 204 00:11:04,790 --> 00:11:07,640 is greater than 5. 205 00:11:07,640 --> 00:11:10,440 And we have all four of these pieces here. 206 00:11:10,440 --> 00:11:23,220 so it's 35 times 1 minus e to the minus 5/30 plus 65 times e 207 00:11:23,220 --> 00:11:26,650 to the minus 5 5/30. 208 00:11:26,650 --> 00:11:31,600 And it turns out that this is approximately 209 00:11:31,600 --> 00:11:39,790 equal to 60.394 minutes. 210 00:11:39,790 --> 00:11:42,610 All right, so what have we found? 211 00:11:42,610 --> 00:11:47,430 We found that the original guess that we had, if we just 212 00:11:47,430 --> 00:11:50,360 had two meetings back to back, was on average it would take 213 00:11:50,360 --> 00:11:51,310 60 minutes. 214 00:11:51,310 --> 00:11:55,150 It turns out that, because of the way that things are set 215 00:11:55,150 --> 00:11:58,890 up, because of the five minute thing, it actually takes a 216 00:11:58,890 --> 00:12:00,780 little longer than 60 minutes on average. 217 00:12:00,780 --> 00:12:01,700 And why is that? 218 00:12:01,700 --> 00:12:08,630 It's because the five sometimes adds an extra 219 00:12:08,630 --> 00:12:13,930 buffer, adds a little bit of extra amount, because it would 220 00:12:13,930 --> 00:12:18,820 have been shorter in this scenario because, if the both 221 00:12:18,820 --> 00:12:21,030 students had arrived on time, then the second student would 222 00:12:21,030 --> 00:12:23,570 have been able to pick up right here immediately. 223 00:12:23,570 --> 00:12:26,180 And so both appointments would have ended sooner. 224 00:12:26,180 --> 00:12:29,880 But because the second student didn't arrive until five 225 00:12:29,880 --> 00:12:32,490 minutes, there was some empty space that was wasted. 226 00:12:32,490 --> 00:12:36,140 And that's where you get you the little bit of extra time. 227 00:12:36,140 --> 00:12:39,760 So this is a nice problem just to get some more exercise with 228 00:12:39,760 --> 00:12:42,590 exponential random variables and also nicely illustrates 229 00:12:42,590 --> 00:12:46,560 the memory list property, which was a key points in 230 00:12:46,560 --> 00:12:47,680 order to solve this. 231 00:12:47,680 --> 00:12:52,170 And it also is nice because we get to review a useful tool 232 00:12:52,170 --> 00:12:56,350 that we've been using all course long, which is to split 233 00:12:56,350 --> 00:13:00,010 things into different scenarios and then solve the 234 00:13:00,010 --> 00:13:02,770 simpler problems and then combine them up, for example 235 00:13:02,770 --> 00:13:05,490 using total expectation. 236 00:13:05,490 --> 00:13:08,120 So I hope that was helpful, and see you next time. 237 00:13:08,120 --> 00:13:10,334