Question 5 [18 marks] a) Two gamblers toss an unbiased coin...

Question 5 [18 marks] a) Two gamblers toss an unbiased coin in each round of the game. Let Yj be the difference between the number of heads (0 or 1) tossed by the first player and the number of heads (0 or 1) tossed by the second player in the j-th round of the game, j = 1,...,n. Then {Yj, 1 ? j ? n} is a sequence of independent random variables such that Yj = { 1 with probability 1/4, 0 with probability 1/2, -1 with probability 1/4; and Sn = ?_{j=1}^{n} Yj. i) Show that the probability generating function (pgf) of Sn, for z ? 0, GSn(z) = (z + 1)^{2n} / (4z)^n. Hence compute P(Sn = 0). ii) Suppose that the gamblers play 5 rounds of this game. Find the probability that the gamblers score the same number of heads.

Transcript

00:01 So in this question we have a y j which is equal to one with probability a quarter, zero with probability a half, and minus one with probability a quarter.

00:16 So what does that mean? well that means that the probability that y j is equal to y is one half times one minus y squared over two, because then when y equals zero we get a half and when r equals plus or minus one we get a quarter.

00:31 And this is valid for y in the set minus one, zero, or one.

00:37 And just to be clear y j is equal to the number, the difference between the number of heads tossed by the first player and the number of heads tossed by the second player in the jth round.

00:52 So it's the number of heads player one minus the number of heads tossed by player two in round j.

01:15 So that's what it is.

01:18 Now s n is the sum from one to n of y j.

01:23 So let's get the probability generating function of s n.

01:28 G s n of z is the expected value of z to the s n, which is the expected value of z to the sum from one to n of y j.

01:39 But since the y's are independent, we're told in the question this is the product from one to n of the expected value of z to the y j.

01:50 So this is the product from one to n of, now we're going to have y j is minus one with probability of quarter.

02:02 So we have one quarter one over z plus one half z to the zero plus one quarter times z to the one.

02:13 Now this doesn't depend on j anymore, so now we can just raise it to the power of n.

02:17 So let's pull a one over four z out to the power of n, and then we'll have one plus two z plus z squared, which is also raised to the power of n.

02:28 But that's just one plus z squared to the power of n, which is one plus z to the two n over four z to the n.

02:38 So there's our p g f, g s n of z.

02:43 Now we have to note that this has a pole of order n.

02:51 So that means that the probability that s n is equal to k is going to be the k plus nth derivative.

03:00 So one over k plus n factorial times the k plus nth derivative of z to the n g s n z.

03:16 And we have to insert the z to the n so that we don't have negative powers of z in our expansion, and then we evaluate this at z equals zero.

03:26 And that's why we need to have no negative poles, so that we can evaluate at z equals zero and get a finite answer...

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