1 00:00:00,000 --> 00:00:00,130 2 00:00:00,130 --> 00:00:02,770 Hi, in this problem, we're going to look at competing 3 00:00:02,770 --> 00:00:04,080 exponential. 4 00:00:04,080 --> 00:00:07,500 So we have three exponential random variables, X with 5 00:00:07,500 --> 00:00:11,510 parameter lambda, Y with parameters mu, and Z with 6 00:00:11,510 --> 00:00:13,090 parameter nu. 7 00:00:13,090 --> 00:00:16,920 And we want to calculate some probability. 8 00:00:16,920 --> 00:00:18,550 And the probability that we want to calculate is the 9 00:00:18,550 --> 00:00:25,150 probability that X is less than Y is less than Z. 10 00:00:25,150 --> 00:00:31,630 Now we can reinterpret this as 3 plus Poisson processes. 11 00:00:31,630 --> 00:00:34,900 Because the link between exponentials and Poisson 12 00:00:34,900 --> 00:00:37,720 processes is that the inter-arrival time of Poisson 13 00:00:37,720 --> 00:00:40,080 processes are exponentially distributed. 14 00:00:40,080 --> 00:00:44,200 So you can think of X being the time until the first 15 00:00:44,200 --> 00:00:48,760 arrival in a Poisson process with parameter lambda. 16 00:00:48,760 --> 00:00:52,670 And same thing for Y is the first arrival time of a 17 00:00:52,670 --> 00:00:54,420 Poisson process with parameter mu. 18 00:00:54,420 --> 00:00:56,730 The same thing for Z and nu. 19 00:00:56,730 --> 00:01:02,530 And so in that interpretation, X less than Y less than Z, you 20 00:01:02,530 --> 00:01:06,380 could interpret as a race, meaning that X finishes first 21 00:01:06,380 --> 00:01:09,880 followed by Y and then Z comes in last. 22 00:01:09,880 --> 00:01:12,370 So with that interpretation, let's see if we can calculate 23 00:01:12,370 --> 00:01:14,620 what this probability is. 24 00:01:14,620 --> 00:01:22,190 We can rewrite this probability as a combination 25 00:01:22,190 --> 00:01:23,410 of two things occurring. 26 00:01:23,410 --> 00:01:33,010 One is that X is less than the minimum of Y and Z. And then 27 00:01:33,010 --> 00:01:38,020 the other is that Y is less than Z. 28 00:01:38,020 --> 00:01:39,850 So what is this first event mean? 29 00:01:39,850 --> 00:01:42,970 This first event means that X comes in first. 30 00:01:42,970 --> 00:01:46,060 And it doesn't matter whether Y comes in second 31 00:01:46,060 --> 00:01:47,500 or Z comes in second. 32 00:01:47,500 --> 00:01:51,170 So we first say that X has to come in first which means it 33 00:01:51,170 --> 00:01:57,130 has to beat the better of Y and Z. And then that combined 34 00:01:57,130 --> 00:02:01,350 with the fact that Y does better than Z is the same 35 00:02:01,350 --> 00:02:03,350 thing as saying that X is first, Y is 36 00:02:03,350 --> 00:02:05,630 second, and Z is third. 37 00:02:05,630 --> 00:02:09,610 And now, let's try to argue using Poisson processes, that 38 00:02:09,610 --> 00:02:12,300 these two events are actually independent. 39 00:02:12,300 --> 00:02:20,120 So this event occurring means that X is smaller than Y and 40 00:02:20,120 --> 00:02:25,930 Z. So let's take these Poisson processes, and because these 41 00:02:25,930 --> 00:02:29,360 random variables are assumed to be independent, these are 42 00:02:29,360 --> 00:02:31,980 independent Poisson processes. 43 00:02:31,980 --> 00:02:33,560 So we can merge them. 44 00:02:33,560 --> 00:02:35,490 So let's merge these two. 45 00:02:35,490 --> 00:02:38,270 And we'll get a Poisson process that has 46 00:02:38,270 --> 00:02:43,640 rate mu plus nu. 47 00:02:43,640 --> 00:02:49,820 And we can also merge this first one and that one. 48 00:02:49,820 --> 00:02:53,410 And we'll get another Poisson process with predator lambda 49 00:02:53,410 --> 00:02:58,220 plus mu plus nu. 50 00:02:58,220 --> 00:03:02,060 So in that context, what does it mean that X is less than 51 00:03:02,060 --> 00:03:03,590 the minimum Y and Z? 52 00:03:03,590 --> 00:03:09,470 It just means that in this merged process, the first 53 00:03:09,470 --> 00:03:16,490 arrival came from the X process. 54 00:03:16,490 --> 00:03:20,090 In that case, if that's true, then X is less than minimum Y 55 00:03:20,090 --> 00:03:25,270 and Z. Well, let's say that event does occur that the 56 00:03:25,270 --> 00:03:29,430 first arrival is from the X process. 57 00:03:29,430 --> 00:03:35,690 Now we're interested in what the order of the second two 58 00:03:35,690 --> 00:03:36,270 arrivals are. 59 00:03:36,270 --> 00:03:37,650 Is it Y first and then Z? 60 00:03:37,650 --> 00:03:39,070 Or Z first and then Y? 61 00:03:39,070 --> 00:03:40,770 Well, it doesn't matter because of 62 00:03:40,770 --> 00:03:42,270 the fresh start property. 63 00:03:42,270 --> 00:03:48,040 Because after this arrival comes, and say it is from the 64 00:03:48,040 --> 00:03:52,140 X process, the Poisson processes start anew, and 65 00:03:52,140 --> 00:03:54,330 they're still independent. 66 00:03:54,330 --> 00:03:57,670 And so what happens after that is independent of what 67 00:03:57,670 --> 00:03:59,560 happened here, when X arrived. 68 00:03:59,560 --> 00:04:04,160 And so whether Y came first followed by Z, or Z came first 69 00:04:04,160 --> 00:04:07,340 followed by Y is independent of what happened here. 70 00:04:07,340 --> 00:04:10,710 And so because of that, these two events are independent, 71 00:04:10,710 --> 00:04:12,830 and so when we have the probability of the 72 00:04:12,830 --> 00:04:14,980 intersection of two independent events, we can 73 00:04:14,980 --> 00:04:18,419 write that as the product of those two probabilities. 74 00:04:18,419 --> 00:04:29,870 75 00:04:29,870 --> 00:04:33,790 Now, what is the probability of this first event? 76 00:04:33,790 --> 00:04:36,220 The probability that X is less than the minimum Y and Z? 77 00:04:36,220 --> 00:04:39,440 Well, we just said that that corresponds to the first 78 00:04:39,440 --> 00:04:44,160 arrival of this merge process being from the X process. 79 00:04:44,160 --> 00:04:48,950 Well, that probability is lambda over lambda 80 00:04:48,950 --> 00:04:51,650 plus mu plus nu. 81 00:04:51,650 --> 00:04:55,953 So it's equal to this ratio where the process that you're 82 00:04:55,953 --> 00:04:58,530 interested in, its rate comes in the numerator. 83 00:04:58,530 --> 00:05:05,010 And then the merged rate is on the denominator, 84 00:05:05,010 --> 00:05:06,300 And what about the second one? 85 00:05:06,300 --> 00:05:09,780 What's the probability that Y is less than Z? 86 00:05:09,780 --> 00:05:13,300 Well, let's go now to this merge process where we merged 87 00:05:13,300 --> 00:05:17,140 just the Y and Z processes and see which one comes first. 88 00:05:17,140 --> 00:05:20,170 Well, in that case what we want to know is in this merge 89 00:05:20,170 --> 00:05:23,830 process, what is the probability that the first 90 00:05:23,830 --> 00:05:26,030 arrival came from the Y process? 91 00:05:26,030 --> 00:05:29,880 Well, analogously, that probability is going to mu 92 00:05:29,880 --> 00:05:34,850 over mu plus nu. 93 00:05:34,850 --> 00:05:37,170 And that gives us our answer. 94 00:05:37,170 --> 00:05:42,120 And so we see that what looked like a pretty complex 95 00:05:42,120 --> 00:05:44,830 calculation when we reinterpreted it in terms of 96 00:05:44,830 --> 00:05:49,310 Poisson processes, it becomes relatively easy to solve. 97 00:05:49,310 --> 00:05:53,290 But this still seems like a complicated expression. 98 00:05:53,290 --> 00:05:54,820 So let's try to check to see whether it 99 00:05:54,820 --> 00:05:56,000 actually makes sense. 100 00:05:56,000 --> 00:05:59,260 So one way to do that is to look at a specific example of 101 00:05:59,260 --> 00:06:01,640 the choice of lambda, mu, and nu, and see if it actually 102 00:06:01,640 --> 00:06:03,190 makes sense. 103 00:06:03,190 --> 00:06:10,250 So one example is suppose that all three of these parameters 104 00:06:10,250 --> 00:06:11,650 are the same. 105 00:06:11,650 --> 00:06:14,070 Well, if they're all the same then this probability, the 106 00:06:14,070 --> 00:06:18,980 first part becomes 1 over 3, 1/3. 107 00:06:18,980 --> 00:06:22,230 And the second one is 1 over 2. 108 00:06:22,230 --> 00:06:26,350 And so if all three parameters are the same, probability 109 00:06:26,350 --> 00:06:27,760 becomes 1/6. 110 00:06:27,760 --> 00:06:29,790 And let's see if that makes sense. 111 00:06:29,790 --> 00:06:33,700 If all three parameters are the same, that means that 112 00:06:33,700 --> 00:06:38,340 these rates, these arrival rates are all the same. 113 00:06:38,340 --> 00:06:44,690 And what that means is that any three ordering of these 114 00:06:44,690 --> 00:06:47,700 three arrivals is as likely as any other ordering. 115 00:06:47,700 --> 00:06:50,840 And what we're interested in is the probability of one's 116 00:06:50,840 --> 00:06:54,630 particular ordering happening which is X first then Y then 117 00:06:54,630 --> 00:06:55,470 Z. 118 00:06:55,470 --> 00:06:59,180 But if everything is symmetric then any of the orderings is 119 00:06:59,180 --> 00:07:01,020 as likely as any other one. 120 00:07:01,020 --> 00:07:02,580 And how many orderings are there? 121 00:07:02,580 --> 00:07:06,180 Well, there's three choices for who comes in first. 122 00:07:06,180 --> 00:07:07,300 Two for who comes in second. 123 00:07:07,300 --> 00:07:09,760 And one for who comes in last. 124 00:07:09,760 --> 00:07:12,760 So there's a total of six possible orders in which these 125 00:07:12,760 --> 00:07:15,460 three contestants, if you think of it that way, could 126 00:07:15,460 --> 00:07:16,890 finish this race. 127 00:07:16,890 --> 00:07:19,480 And out of none of those, we want the probability that one 128 00:07:19,480 --> 00:07:21,230 of those outcomes happens. 129 00:07:21,230 --> 00:07:23,090 And so the probability should be 1/6. 130 00:07:23,090 --> 00:07:25,210 And that's what our formula tells us. 131 00:07:25,210 --> 00:07:30,720 So as I said, in this problem, we saw how to reinterpret 132 00:07:30,720 --> 00:07:34,020 exponentials in the context of Poisson processes that helped 133 00:07:34,020 --> 00:07:35,360 us solve a-- 134 00:07:35,360 --> 00:07:36,610