Question

7. A probability generating function encodes the probabilities of different events happening with a gener- ating function. For example, consider tossing a coin until the first time you get a heads. Then, let pn be the probability that you had to flip the coin n times until you got a heads for the first time. The probability generating function for this problem would be f(z) = Σ Pnz". 膨:1 (a) If your coin has a probability p of landing heads on any particular coin toss, find the value of pa in terms of p. (b) Find a closed form for the probability generating function for this coin (c) What is f(1)? Give a non-algebraic reason for why this is true. (d) The average number of times you need to flip the coin in order to get a heads would be given by n Pn Compute this value using your generating function from part (b)

Answer 1

(a)

For getting heads for the first time on nth toss, we have got all tails on (n-1) tosses.

probability of tails = 1-p and all tosses are independent of each other, then

$p_n = (1-p)^{n-1}p$

(b)

$f(x)= \sum_{n=1}^{\infty } p_n x^n = \sum_{n=1}^{\infty } p(1-p)^{n-1} x^n = px\sum_{n=1}^{\infty } ((1-p)x)^{n-1}$

$= px\sum_{n=0}^{\infty } ((1-p)x)^{n}$

$= px \frac{1}{1 - (1-p)x} = \frac{px}{1 - x + px}$ (By sum of geometric series with common ratio = (1-p)x and first term = 1}

(c)

$f(1)= \frac{p*1}{1 - 1 + p*1} = p/p = 1$

This is true because f(1) is sum of all probabilities for possible values of n which is always equal to 1.

(d)

By definition of probability generating function, average number of times need to flip the coin is,

$f'(1)= \frac{\mathrm{d} }{\mathrm{d} x}_{x=1} \frac{px}{1 - x + px} = \frac{\mathrm{d} }{\mathrm{d} x}_{x=1} \frac{p(1-x+px)- px(-1 + p)}{(1 - x + px)^2}$

$= \frac{p(1-1+p*1)- p*1(-1 + p)}{(1 - 1 + p*1)^2} = \frac{p^2 + p - p^2}{p^2} = \frac{1}{p}$