If numbers of heads and tails are independent, then number of tosses $N \sim \mathrm{Poisson}$

Question

A fair coin is tossed a random number $N$ of times, giving a total $X$ of heads and $Y=N-X$ tails. Show that if $X$ and $Y$ are independent and the generating function $G_N(s)$ of $N$ exists for $s$ in a neighbourhood of $s=1$, then $N$ is Poisson distributed.

In other exercises, I showed the converse , which was relatively easy. I'm working through my probability book myself, but my book does not provide an answer.

I came this far:

Because we have a fair coin, each coin toss follows a $\mathrm{Bernoulli}(\frac12)$ distribution, thus having generating function $G_{X_i}(s)=\frac12+\frac12 s$ for each $i \in \{1,2,\ldots, n\}$. Using the random sum formula I found $G_N(s)=(G_N(\frac12 +\frac12 s))^2$, for the probability generating function for the random sum, because $N=X+Y$, because each coin toss yields either a heads or a tails.

The book gives as a hint: use $H(s)=G_N(1-s)$. However, I have no idea how to solve this one. This is also supposed to be one of the most difficult exercises so I'm just curious how this one has to be solved. Any ideas?

Answer 1

Let $N$ denote the total number of tossing coins, $X$ denote the total number of heads, and $Y$ denote the total number of tails. Then we have $X \sim B(N, p), 0 < p < 1, q = 1-p$ and $N = X + Y$.

The moment generating function of $X$ is \begin{equation} \begin{aligned} G_{X}(s) &= E(s^{X}) \\ &= E \{ E (s^{X} \mid N)\} \\ &= E \left[\sum_{x=0}^{N} \left\{s^{x} \binom{N}{x} p^{x} (1-p)^{N-x} \right\}\right] \\ &= E \left[\sum_{x=0}^{N} \left\{\binom{N}{x} (ps)^{x} (1-p)^{N-x} \right\}\right] \\ &= E \{ (ps + 1-p)^{N}\} \\ &= E \{ (ps + q)^{N}\} = G_{N}(ps + q). \qquad (1) \end{aligned} \end{equation}

By symmetry of $X$ and $Y$, we have \begin{equation} G_{Y}(t) = G_{N}(p + qt). \qquad (2) \end{equation}

We want to leverage the independence between $X$ and $Y$. By using the moment generating function, this is equivalent to $E(s^{X}t^{Y}) = E(s^{X})E(t^{Y})$.

Thus we transform $E(s^{X}t^{Y})$ as follows. \begin{equation} \begin{aligned} E(s^{X}t^{Y}) &= E(s^{X}t^{N-X}) \\ &= E\left\{\left(\frac{s}{t}\right)^{X}t^{N}\right\} \\ &= E \left[ E \left\{\left(\frac{s}{t}\right)^{X}t^{N} \mid N\right\} \right]\\ &= E \left[t^{N} E \left\{\left(\frac{s}{t}\right)^{X} \mid N\right\} \right]\\ &= E \left\{t^{N} \left[\sum_{x=0}^{N}\left\{\left(\frac{s}{t}\right)^{x}\binom{N}{x}p^{x}(1-p)^{N-x} \right\} \right]\right\}\\ &= E \left\{t^{N} \left[\sum_{x=0}^{N}\left\{\binom{N}{x}\left(\frac{ps}{t}\right)^{x} (1-p)^{N-x} \right\} \right]\right\}\\ &= E \left[t^{N} \left\{\left(\frac{ps}{t}\right) + 1-p \right\}^{N} \right]\\ &= E \left[t^{N} \left\{\left(\frac{ps}{t}\right) + q \right\}^{N} \right]\\ &= E \left[\left(ps + qt \right)^{N} \right]\\ &= G_{N}(ps + qt). \qquad (3) \end{aligned} \end{equation}

Combining (1) (2) (3), we get the following equality: \begin{equation} G_{N}(ps + qt) = G_{N}(ps + q)G_{N}(p + qt). \end{equation}

Let $H(s) = G_{N}(1-s)$, and we get \begin{equation}\nonumber H(1 - ps - qt) = H(1 - ps - q)H(1 - p - qt). \end{equation}

That is, \begin{equation} H\{p(1-s) + q(1-t)\} = H\{p(1-s)\}H\{q(1-t)\}. \end{equation}

Since all continuous solutions of the equation $f(x+y) = f(x)f(y)$ take the form $f(x) = e^{\lambda x}$ for some $\lambda$, we have $H(s) = e^{\lambda s}$ for some $\lambda$. Therefore, $G_{N}(s) = H(1-s) = e^{\lambda (1-s)}$ for some $\lambda$.

Note that the moment generating function of $N$ coincides with the moment generating function of Poisson distribution with parameter $-\lambda$, which finishes the proof.

Answer 2

I should start by saying that I don't have much experience with generating functions, but it looks to me like what you have there is simply a functional equation to solve: $G_n(s) = (G_N(1/2+s/2))^2$. The hint from the book is to introduce a new function to simplify that: let $H(s) = G_N(1-s)$ and our equation becomes $H(s) = [H(s/2)]^2$.

Now if $G_N(s)$ exists around $s = 1$, then $G_N(1) = 1$ and so $H(0) = 1$. Note also that $H(s)$ is non-negative whenever $H(s/2)$ exists, and that in fact it must be positive, because if it was $0$, then so are $H(s/2), H(s/4), \dots, H(s/2^n)$ which would make a discontinuity at $H(0) = 1$. Then, introduce yet another function $F(s) = \ln H(s)$ to simplify even further; now we have $F(s) = 2F(s/2)$ with $F(0) = 0$. Obviously any function of the form $F(s) = \lambda s$ satisfies that, so you just need to show that it must be of that form.

Consider a sequence of values $s, s/2, s/4, \dots, s/2^n, \dots$ which approaches $0$. We have $$ {F(s)\over s} = {F(s/2)\over s/2} = \dots = {F(s/2^n)\over s/2^n} = \dots $$ Can you see where this is going? If we suppose that $F(s)/ s \neq F(s')/s'$ again we would get a discontinuity. Can you take it from here to get back to what $G_N(s)$ must be?