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Here's an exercise in An Intermediate Course in Probability by Gut.

The probability generating function (pgf) of a nonnegative integer-valued random variable $Y$ is $g_Y(t)=Et^Y$, and the pgf of a sum $S_N$ of a random number $N$ of i.i.d. random variables $X_1,X_2,\ldots$ is $g_{S_N}(t)=g_N(g_X(t))$.

Exercise 6.2 Consider a game of roulette and [someone] who bets one dollar until number $13$ appears, with the roulette wheel having numbers $0,\ldots,36$. Then [they bet] one dollar the same number of times on number $36$. Find the pgf of [their] loss in the second round. Try also finding the pgf for [their] overall loss.

Here's my attempt. Let $N\in \text{Fs}(1/37)$ be the number of bets on number $13$ (here $\text{Fs}(1/37)$ is the geometric distribution that models the first success), and let $Y_1,Y_2,\ldots$ be the losses in the bets on number $36$. Thus $$Y_k=\begin{cases} 1,&\text{if number 36 does not appear}\\ -35&(\text{i.e.} -36+1)\quad\text{otherwise},\end{cases}$$and $Y_1,Y_2,\ldots$ are independent with $P(Y_k=1)=36/37$ and $P(Y_k=-35)=1/37$ (note that a negative loss is a gain).

So the player's loss in the second round equals $X=Y_1+\ldots +Y_N$. We know the generating function of $X$ is $g_X(t)=g_N(g_Y(t))$. Computing the generating function of the geometric distribution is straightforward; it is $\frac{tp}{1-(1-p)t}$ where $p=1/37$. But what is the pgf of $Y$? $Y$ seems to be both positive and negative, and hence I'm confused if it even has a pgf.

The total loss is $X+N-36$, but from the first part I am unsure how to find the pgf of this random variable.

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The number of spins until the first $13$ appears follows a Geometric distribution with $p = 1/37$, i.e. the probability that the first $13$ appears on spin $n$ is $p (1-p)^n$ for $n \ge 1$. So the PGF for the number of spins is $$g_N(t) = \sum_{n=1}^{\infty} p(1-p)^n t^n= \frac{pt}{1-(1-p)t}$$ where we have applied the formula for the sum of an infinite geometric series.

If $X$ is the loss on a single spin in the second part, then $X=0$ with probability $p$ and $X=1$ with probability $1-p$ (regardless of the number of the spin). So the PGF of $X$ is $$g_X(t) = p + (1-p)t$$

Applying the theorem quoted in the OP, the PGF of the total loss is $$g_N(g_X (t)) = \frac{p(p + (1-p)t)}{1-(1-p)(p + (1-p)t)}$$

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  • $\begingroup$ Thanks! I don't see how the expression for $g_N(g_X (t))$ can be true. The expected total loss is $E(X+N-36)=2$. This follows from the fact that $$EX=EN\cdot EY=37\cdot\left(1\cdot\frac{36}{37}-35\cdot\frac1{37}\right)=1,$$and $EN=37$. The left-hand derivative of a PGF at $t=1$ should give the expected value. Differentiating your expression with $p=1/37$ and evaluating at $t=1$, WolframAlpha yields $36$. Something's not right, but I don't know what. $\endgroup$
    – psie
    Commented Jul 27 at 16:48
  • $\begingroup$ Maybe it is supposed to be that way. Would you know how to find the PGF for his overall loss? I'm unsure which random variable describes the overall loss. $\endgroup$
    – psie
    Commented Jul 27 at 17:41
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    $\begingroup$ @pse The expected value of the loss is the expected value of the number of spins times the expected value of the loss on one spin: $37 \times (36/37) = 36$. The PGF above is the PGF of the overall loss. Frankly, I don't understand what the exercise statement means by "the loss in the second round", unless by that is meant the overall loss. So maybe I don't understand the problem. $\endgroup$
    – awkward
    Commented Jul 27 at 20:54

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