1 00:00:00,000 --> 00:00:00,950 2 00:00:00,950 --> 00:00:01,530 Hi. 3 00:00:01,530 --> 00:00:04,059 In this problem, we're going to look at random incidence 4 00:00:04,059 --> 00:00:05,910 under Erlang arrivals. 5 00:00:05,910 --> 00:00:08,230 First, let's parse what that means. 6 00:00:08,230 --> 00:00:12,500 In a Poisson process, remember, the time between 7 00:00:12,500 --> 00:00:15,680 arrivals, or the inter-arrival time, is distributed as an 8 00:00:15,680 --> 00:00:17,710 exponential random variable. 9 00:00:17,710 --> 00:00:22,240 And random incidence for a Poisson process refers to the 10 00:00:22,240 --> 00:00:25,690 somewhat surprising result that when you consider a 11 00:00:25,690 --> 00:00:29,700 specific time, say, T-star, then the length of the 12 00:00:29,700 --> 00:00:33,650 inter-arrival interval that contains that time T-star is 13 00:00:33,650 --> 00:00:39,500 not distributed as an exponential random variable. 14 00:00:39,500 --> 00:00:42,510 It's actually distributed as an Erlang random variable of 15 00:00:42,510 --> 00:00:47,520 order 2 or it's distributed as a sum of two exponential 16 00:00:47,520 --> 00:00:48,530 random variables. 17 00:00:48,530 --> 00:00:50,670 And the reason for that is that it 18 00:00:50,670 --> 00:00:52,080 comprises of two parts. 19 00:00:52,080 --> 00:00:58,180 One is the time since the last arrival until T-star, which is 20 00:00:58,180 --> 00:01:02,590 exponentially distributed, and the time from T-star until the 21 00:01:02,590 --> 00:01:06,170 next arrival, which is also exponentially distributed. 22 00:01:06,170 --> 00:01:08,990 So that brings us to a review of what Erlang random 23 00:01:08,990 --> 00:01:10,190 variables are. 24 00:01:10,190 --> 00:01:14,550 An Erlang random variable of order k is just the sum of k 25 00:01:14,550 --> 00:01:17,170 independent and identically distributed 26 00:01:17,170 --> 00:01:18,760 exponential random variables. 27 00:01:18,760 --> 00:01:22,840 So to be more specific, if Ti is an exponential random 28 00:01:22,840 --> 00:01:27,630 variable with parameter lambda, then if you take kiid 29 00:01:27,630 --> 00:01:33,140 copies of Ti and add them up, and call that Yk, then Yk is 30 00:01:33,140 --> 00:01:36,330 an Erlang random variable of order k. 31 00:01:36,330 --> 00:01:41,620 And one other fact is that the mean of Yk, the mean of an 32 00:01:41,620 --> 00:01:45,940 Erlang random variable of order k, is just k, the order, 33 00:01:45,940 --> 00:01:49,880 over lambda, which is the rate of the underlying exponential 34 00:01:49,880 --> 00:01:51,960 random variables. 35 00:01:51,960 --> 00:01:55,395 So as a consequence, if you have an Erlang random variable 36 00:01:55,395 --> 00:01:58,980 of order two and that random variable also has a mean of 37 00:01:58,980 --> 00:02:02,530 two over lambda, we can interpret that random variable 38 00:02:02,530 --> 00:02:05,620 as just being the sum of two exponential random variables. 39 00:02:05,620 --> 00:02:08,750 2 iid exponential random variables, T1 and T2, where 40 00:02:08,750 --> 00:02:14,300 each one takes exponential with the rate in lambda. 41 00:02:14,300 --> 00:02:17,920 OK, so in this problem now, we're dealing with the random 42 00:02:17,920 --> 00:02:21,170 incidence not under Poisson processes, but under something 43 00:02:21,170 --> 00:02:26,570 else, which we call here an Erlang process with Erlang 44 00:02:26,570 --> 00:02:28,180 arrival times. 45 00:02:28,180 --> 00:02:31,730 So to be more specific, what we're saying is that, instead 46 00:02:31,730 --> 00:02:35,520 of inter-arrival time being exponentially distributed, in 47 00:02:35,520 --> 00:02:38,260 this process, and inter-arrival time is actually 48 00:02:38,260 --> 00:02:41,880 distributed as an Erlang random variable of order 2 49 00:02:41,880 --> 00:02:43,800 with mean 2 over lambda. 50 00:02:43,800 --> 00:02:48,740 So to be explicit, this is no longer a Poisson process. 51 00:02:48,740 --> 00:02:52,590 It's some other process because the inter-arrival 52 00:02:52,590 --> 00:02:54,620 times are not exponential. 53 00:02:54,620 --> 00:02:57,730 So let's make use of this fact that we talked about earlier 54 00:02:57,730 --> 00:03:00,770 because now we know that the inter-arrival times of this 55 00:03:00,770 --> 00:03:04,220 Erlang process are Erlang order 2 56 00:03:04,220 --> 00:03:05,450 with mean 2 over lambda. 57 00:03:05,450 --> 00:03:08,540 But we know that that can just be re-interpreted as a sum of 58 00:03:08,540 --> 00:03:13,320 two simple exponentials, each with parameter lambda. 59 00:03:13,320 --> 00:03:21,860 So let's just draw another picture and imagine that for 60 00:03:21,860 --> 00:03:25,470 each of these arrivals, so say we have three sample arrivals 61 00:03:25,470 --> 00:03:30,470 in this Erlang process, we can fill in, kind of, the gaps 62 00:03:30,470 --> 00:03:38,060 between these with additional arrivals. 63 00:03:38,060 --> 00:03:48,980 And then think of each one of these times as all being 64 00:03:48,980 --> 00:03:54,050 exponential with parameter lambda. 65 00:03:54,050 --> 00:03:58,040 So this is a valid interpretation because when we 66 00:03:58,040 --> 00:04:08,440 connect these, these inter-arrival times correspond 67 00:04:08,440 --> 00:04:11,590 to the combination of two inter-arrival times, which we 68 00:04:11,590 --> 00:04:15,490 know we can split that into just two exponentials. 69 00:04:15,490 --> 00:04:19,060 So each one of these is an exponential random variable. 70 00:04:19,060 --> 00:04:23,980 And when you combine them, you get an Erlang order of 2. 71 00:04:23,980 --> 00:04:27,300 But the nice thing about this is that if we look at this 72 00:04:27,300 --> 00:04:36,060 diagram, it actually is just exactly a Poisson process with 73 00:04:36,060 --> 00:04:41,240 a rate lambda because now, what we're 74 00:04:41,240 --> 00:04:42,690 dealing with are exactly-- 75 00:04:42,690 --> 00:04:45,930 the inter-arrival times are now exactly 76 00:04:45,930 --> 00:04:47,460 exponential random variables. 77 00:04:47,460 --> 00:04:52,770 And so this is in fact, now, just a simple Poisson process. 78 00:04:52,770 --> 00:04:58,770 And we can also just think of it as we take the Poisson 79 00:04:58,770 --> 00:05:02,400 process, and take every other arrival, say, all the 80 00:05:02,400 --> 00:05:07,650 even-numbered arrivals, and make those corresponds to be 81 00:05:07,650 --> 00:05:11,440 arrivals in the Erlang process. 82 00:05:11,440 --> 00:05:18,810 OK, so now let's think about some specific time T-star. 83 00:05:18,810 --> 00:05:29,340 We want to know what is the distribution of the length of 84 00:05:29,340 --> 00:05:32,760 this to be specific inter-arrival interval that 85 00:05:32,760 --> 00:05:35,910 T-star is in. 86 00:05:35,910 --> 00:05:41,290 Well, what we can do is take it down to the level of this 87 00:05:41,290 --> 00:05:43,500 Poisson process and look at it from there. 88 00:05:43,500 --> 00:05:49,860 Well, we do that because, for a Poisson process, we know 89 00:05:49,860 --> 00:05:51,910 about random incidence for Poisson processes. 90 00:05:51,910 --> 00:05:55,190 And we know how to deal with Poisson processes. 91 00:05:55,190 --> 00:05:57,660 So let's think about this now. 92 00:05:57,660 --> 00:05:59,890 Well, T-star is here. 93 00:05:59,890 --> 00:06:03,580 And what we know from random incidence for a Poisson 94 00:06:03,580 --> 00:06:09,380 processes is that the length of this inter-arrival interval 95 00:06:09,380 --> 00:06:14,160 for the Poisson process, we know that this is an 96 00:06:14,160 --> 00:06:17,180 exponential plus an exponential. 97 00:06:17,180 --> 00:06:25,780 So combined, this is Erlang order 2. 98 00:06:25,780 --> 00:06:30,100 99 00:06:30,100 --> 00:06:34,070 But that only covers from here to here. 100 00:06:34,070 --> 00:06:36,580 And what we want is actually from here to there. 101 00:06:36,580 --> 00:06:42,730 Well now, we tack on an extra exponential because we know 102 00:06:42,730 --> 00:06:44,930 that the inter-arrival times-- 103 00:06:44,930 --> 00:06:46,580 the time between this arrival and that arrival in the 104 00:06:46,580 --> 00:06:49,670 Poisson process is just another exponential. 105 00:06:49,670 --> 00:06:52,720 And now all of these are in [INAUDIBLE] time intervals. 106 00:06:52,720 --> 00:06:54,100 And they're all independent. 107 00:06:54,100 --> 00:06:59,010 And so the time of this inter-arrival interval in the 108 00:06:59,010 --> 00:07:02,200 Erlang process is just going to be the sum of three 109 00:07:02,200 --> 00:07:04,690 independent exponentials within the 110 00:07:04,690 --> 00:07:06,670 underlying Poisson process. 111 00:07:06,670 --> 00:07:10,140 And so to answer here is actually, it's going to be an 112 00:07:10,140 --> 00:07:13,790 Erlang of order 3. 113 00:07:13,790 --> 00:07:16,960 114 00:07:16,960 --> 00:07:20,040 Now this is one possible scenario for 115 00:07:20,040 --> 00:07:21,610 how this might occur. 116 00:07:21,610 --> 00:07:27,830 Another scenario is actually that T-star is somewhere else. 117 00:07:27,830 --> 00:07:29,760 So let's draw this again. 118 00:07:29,760 --> 00:07:34,300 119 00:07:34,300 --> 00:07:45,880 And suppose now, in this case, T-star landed between an even 120 00:07:45,880 --> 00:07:49,160 numbered arrival in the Poisson process and an odd 121 00:07:49,160 --> 00:07:50,880 numbered arrival. 122 00:07:50,880 --> 00:07:53,750 Now it could also arrive between an odd numbered and an 123 00:07:53,750 --> 00:07:55,560 even numbered arrival. 124 00:07:55,560 --> 00:08:02,170 So it could be that T-star is actually here. 125 00:08:02,170 --> 00:08:04,690 Well, but in this case, it's actually more or less the same 126 00:08:04,690 --> 00:08:11,110 thing because now what we want is the length of this entire 127 00:08:11,110 --> 00:08:15,100 inter-arrival interval, which, in the Poisson world, we can 128 00:08:15,100 --> 00:08:18,490 break it down into random incidence within this 129 00:08:18,490 --> 00:08:21,150 interval, this inter-arrival interval, which is two 130 00:08:21,150 --> 00:08:30,070 exponentials, or an Erlang of 2, plus this interval, which 131 00:08:30,070 --> 00:08:32,020 is just a standard inter-arrival time, which is 132 00:08:32,020 --> 00:08:33,360 another exponential. 133 00:08:33,360 --> 00:08:36,970 So in this case as well, we have the sum of three 134 00:08:36,970 --> 00:08:38,900 independent exponential random variables. 135 00:08:38,900 --> 00:08:43,820 And so, in either case, we have that the inter-arrival 136 00:08:43,820 --> 00:08:48,250 time in the Erlang process is an Erlang of order 3. 137 00:08:48,250 --> 00:08:52,080 And so the final answer is, in fact, that the inter-arrival 138 00:08:52,080 --> 00:08:56,600 for random incidence under these Erlang-type arrivals is 139 00:08:56,600 --> 00:08:59,030 an Erlang of order 3. 140 00:08:59,030 --> 00:09:04,040 OK, so in this problem we looked at the random incidence 141 00:09:04,040 --> 00:09:07,290 under a different type of an arrival process, not Poisson, 142 00:09:07,290 --> 00:09:09,270 but with Erlang random variables. 143 00:09:09,270 --> 00:09:11,600 But we used the insight that Erlang really can be 144 00:09:11,600 --> 00:09:16,110 re-interpreted as the sum of independent and identically 145 00:09:16,110 --> 00:09:18,155 distributed exponential random variables. 146 00:09:18,155 --> 00:09:25,110 And exponential random variables can be viewed as one 147 00:09:25,110 --> 00:09:27,990 way of interpreting and viewing a Poisson process. 148 00:09:27,990 --> 00:09:32,460 And so by going through those steps, we were able to use 149 00:09:32,460 --> 00:09:34,330 what we know about random incidence under Poisson 150 00:09:34,330 --> 00:09:37,300 processes to help us solve this problem of random 151 00:09:37,300 --> 00:09:39,030 incidence its Erlang arrivals. 152 00:09:39,030 --> 00:09:40,510 So I hope that was helpful. 153 00:09:40,510 --> 00:09:41,760 And I'll see you next time. 154 00:09:41,760 --> 00:09:44,334