A game of roulette and generating functions

[psie]

Homework Statement

Consider a game of roulette and Charlie who bets one dollar until number $13$ appears, with the roulette wheel having numbers $0,\ldots,36$. Then he bets one dollar the same number of times on number $36$. Find the generating function of his loss in the second round. Try also finding the generating function for his overall loss.

Relevant Equations

The probability generating function (pgf) of a sum $S_N$ of random number $N$ of i.i.d. random variables $X_1,X_2,\ldots$ is $g_{S_N}(t)=g_N(g_X(t))$.

Here's my attempt. So, 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 Charlie'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 generating function of $Y$?

I have not yet been able to think about the total loss part of the problem.

 

[psie]

Maybe I didn't do it right. As before, let $N\in\text{Ge}(p)$, with pdf $f(n)=(1-p)^{n-1}p$ for $n\geq 1$ and $p=1/37$, be the number of bets on $13$. If $$Y_k=\begin{cases} 1,&\text{if number 36 does not appear}\\ 0&\text{otherwise}\end{cases},$$ then $P(Y_k=1)=1-p$ and $P(Y_k=0)=p$. Then the total loss in the second round is $X=Y_1+\ldots +Y_N$. Applying the formula for $g_X(t)=g_N(g_Y(t))$, with $$g_N(t)=\frac{tp}{1-(1-p)t},\quad g_Y(t)=p+(1-p)t,$$we have that the pgf of his loss in the second round is $$g_N(g_X (t)) = \frac{p(p + (1-p)t)}{1-(1-p)(p + (1-p)t)}.$$ I am unsure about the pgf for his overall loss though and would be really grateful for some help. I am not sure what the random variable is that describes his overall loss.

 

[psie]

My educated guess is that the overall loss is $X+N$. What complicates things is that I don't think they are independent. In what follows, it'll be clearer if we write $X=S_N$. The pgf is (I think) $$Et^{S_N+N}=E\left(E\left(t^{S_N+N}\mid N\right)\right).$$ Now, $E\left(t^{S_N+N}\mid N\right)$ is the r.v. $h(N)$, where $$\begin{align*}h(n)&=E\left(t^{S_N+N}\mid N=n\right) \\ &=E\left(t^{S_n+n}\mid N=n\right) \\ &=E\left(t^{S_n+n}\right) \\ &=t^nEt^{S_n} \\ &=t^n (Et^X)^n \\ &=t^n(g_X(t))^n.\end{align*}$$ So the pgf of the overall loss, i.e. of $X+N$, is $$Eh(N)=E(t\cdot g_X(t))^N=g_N(t\cdot g_X(t)).$$ The formula at least makes sense if we write $S_N+N = \sum_{n=1}^N (X_n+1)$.

 

[psie]

I realize I abused the notation here in #3 compared to #2. So the pgf of $X+N$, where $X=Y_1+\ldots+Y_N$, is $$g_N(t\cdot g_Y(t)),$$ and not $g_N(t\cdot g_{S_N}(t))$. Sorry.