1 00:00:00,740 --> 00:00:03,350 In this segment, we go through a quick review 2 00:00:03,350 --> 00:00:05,700 of a few properties of the Bernoulli process 3 00:00:05,700 --> 00:00:07,420 that we already know. 4 00:00:07,420 --> 00:00:11,410 We start by thinking about the number of successes or arrivals 5 00:00:11,410 --> 00:00:13,740 in the first n time slots. 6 00:00:13,740 --> 00:00:16,270 This is the following quantity. 7 00:00:16,270 --> 00:00:20,300 At each time we add a 0 or a 1, depending on 8 00:00:20,300 --> 00:00:22,650 whether we've had a success or not, 9 00:00:22,650 --> 00:00:25,150 then by adding those numbers, we get the total number 10 00:00:25,150 --> 00:00:26,690 of successes. 11 00:00:26,690 --> 00:00:31,080 Now we already know that the number of successes in n trials 12 00:00:31,080 --> 00:00:34,410 obeys a binomial distribution, so the probability 13 00:00:34,410 --> 00:00:40,320 of having k successes is given by the binomial probabilities. 14 00:00:40,320 --> 00:00:46,380 And this is a formula that holds for k equal to 0 up to n, 15 00:00:46,380 --> 00:00:49,790 which are the possible numbers for the random variable S. 16 00:00:49,790 --> 00:00:52,740 For this random variable, we know the expected value. 17 00:00:52,740 --> 00:00:54,500 It's n times p. 18 00:00:54,500 --> 00:00:57,660 And we also know its variance, which is n times 19 00:00:57,660 --> 00:01:01,410 p times 1 minus p. 20 00:01:01,410 --> 00:01:03,340 Another random variable of interest 21 00:01:03,340 --> 00:01:07,240 is the time until the first success or arrival. 22 00:01:07,240 --> 00:01:10,430 So this is defined to be the smallest 23 00:01:10,430 --> 00:01:17,005 i for which the random variable Xi is equal to 1. 24 00:01:19,670 --> 00:01:21,900 We have done this calculation in the past. 25 00:01:21,900 --> 00:01:27,550 The probability that the first success appears at time k is 26 00:01:27,550 --> 00:01:34,520 the same as the probability that the first k minus 1 trials 27 00:01:34,520 --> 00:01:37,850 resulted in 0's. 28 00:01:37,850 --> 00:01:41,530 And then, the k-th trial resulted in a 1. 29 00:01:41,530 --> 00:01:46,620 And so the probability of this is 1 minus p, 30 00:01:46,620 --> 00:01:51,070 the probability of 0, and we have k minus 1 of them, 31 00:01:51,070 --> 00:01:54,870 times p, the probability that the next trial gives us 32 00:01:54,870 --> 00:01:56,080 a success. 33 00:01:56,080 --> 00:02:02,300 And this formula is valid for k being 1, 2, and so on, 34 00:02:02,300 --> 00:02:04,850 which is the range of possible values 35 00:02:04,850 --> 00:02:07,160 of this random variable T1. 36 00:02:07,160 --> 00:02:10,020 This is the familiar geometric distribution 37 00:02:10,020 --> 00:02:13,300 that we have dealt with on several occasions. 38 00:02:13,300 --> 00:02:16,930 And in particular, we know the expected value and the variance 39 00:02:16,930 --> 00:02:20,520 of the geometric random variable.