2
$\begingroup$

I'm working on the following problem

Each time you flip a certain coin, heads appears with probability $p$. Suppose that you flip the coin a random number of $N$ times, where $N$ has the Poisson distribution with parameter $\lambda$ and is independent of the outcomes of the flips. Find the distributions of the numbers $X$ and $Y$ of the resulting heads and tails, respectively, and show that $X$ and $Y$ are independent.

What I tried, is conditioning on the value of $N$: \begin{eqnarray} \mathbb{P}(X=x) & = & \sum_{k=0}^{\infty}\mathbb{P}(X=x \ | \ N=k)\mathbb{P}(N=k)\\ & = & \sum_{k=0}^{\infty}\binom{k}{x}p^x(1-p)^{k-x}\frac{\lambda^ke^{-\lambda}}{k!}\\ & = & \sum_{k=x}^{\infty}\binom{k}{x}p^x(1-p)^{k-x}\frac{\lambda^ke^{-\lambda}}{k!}.\\ \end{eqnarray} Similarly, for $Y$ i found $$\mathbb{P}(y=y)=\sum_{k=y}^{\infty}\binom{k}{y}p^{k-y}(1-p)^y\frac{\lambda^ke^{-\lambda}}{k!}.$$ I tried to work this out but I didn't seem to go anywhere. The answer should be that $X \sim Pois(\lambda p)$ and because of symmetry we would have $Y \sim Pois(\lambda (1-p))$.

Can anyone provide some help about how to from where I came to $X \sim Pois(\lambda p)$? Thanks in advance.

$\endgroup$
4

1 Answer 1

4
$\begingroup$

I would suggest to use moment-generating functions (MGF): simpler, faster proof. Namely, you have, for $t\in\mathbb{R}$, $$\begin{align} \mathbb{E} e^{tX} &= \mathbb{E}[ \mathbb{E}[ e^{tX} \mid N ] ] \stackrel{(\dagger)}{=} \mathbb{E}[ (1-p+pe^{t})^N ]\\ &= \mathbb{E}[ e^{N\ln(1-p+pe^{t})} ] \stackrel{(\ddagger)}{=} \exp(\lambda(e^{\ln(1-p+pe^{t})}-1))\\ &= \exp(\lambda((1-p+pe^{t})-1))\\ &= \exp(\lambda p(e^{t}-1)) \end{align}$$ where $(\dagger)$ uses the expression of the MGF of a Binomial distribution with parameters $N$ and $p$, and $(\ddagger)$ that of the MGF of a Poisson distribution with parameter $\lambda$ (applied to the argument $t'\stackrel{\rm def}{=}\ln(1-p+pe^{t})$).

At the end, you get that, for every $t\in\mathbb{R}$, $$ \mathbb{E} e^{tX} = \exp(\lambda p(e^{t}-1)) \tag{$\ast$} $$ which is the MGF of a Poisson distribution with parameter $\lambda p$. As the MGF characterizes the distribution (when it exists), we have the result.


However, if you want to finish your computation: here how it goes. I assume $p\neq 1$, otherwise the answer is trivial. \begin{align} \mathbb{P}\{X=n\} &= \sum_{k=n}^\infty \binom{k}{n}p^n(1-p)^{k-n} \frac{\lambda^k e^{-\lambda}}{k!}\\ &= e^{-\lambda}\frac{p^n}{(1-p)^n}\sum_{k=n}^\infty \binom{k}{n}(1-p)^{k} \frac{\lambda^k}{k!}\\ &= e^{-\lambda}\frac{p^n}{(1-p)^n}\sum_{k=n}^\infty \frac{k!}{n!(k-n)!}(1-p)^{k} \frac{\lambda^k}{k!}\\ &= e^{-\lambda}\frac{p^n}{n!(1-p)^n}\sum_{k=n}^\infty \frac{1}{(k-n)!}(1-p)^{k} \lambda^k\\ &= e^{-\lambda}\frac{p^n}{n!(1-p)^n}\sum_{\ell=0}^\infty \frac{1}{\ell!}(1-p)^{\ell+n} \lambda^{\ell+n}\\ &= e^{-\lambda}\frac{(\lambda p)^n}{n!}\sum_{\ell=0}^\infty \frac{(1-p)^{\ell} \lambda^{\ell}}{\ell!}\\ &= e^{-\lambda}\frac{(\lambda p)^n}{n!}e^{\lambda(1-p)} = \boxed{e^{-\lambda p}\frac{(\lambda p)^n}{n!}} \end{align} and you get the probability mass function of a Poisson r.v. with parameter $\lambda p$, as desired.

$\endgroup$
8
  • $\begingroup$ Thanks for this answer, but I have never seen moment-generating functions before. Would you know a more elementary approach? $\endgroup$ Nov 6, 2017 at 19:37
  • 1
    $\begingroup$ Yes. I can update my answer to show how to finish your computation. $\endgroup$
    – Clement C.
    Nov 6, 2017 at 19:40
  • $\begingroup$ @VáclavMordvinov Updated. $\endgroup$
    – Clement C.
    Nov 6, 2017 at 19:46
  • 1
    $\begingroup$ What you call the MGF is usually referred to as the characteristic function being related to the MGF by $\phi_X(t)=M_X(it)$, where $\phi$ is the CF and $M$ the MGF (of some RV). Or perhaps it was a typo, as you start out with $E(e^{itX})$ but end with the expression for the MGFβ€”without any $i$. $\endgroup$ Nov 6, 2017 at 19:54
  • 2
    $\begingroup$ @LoveTooNap29 I know what the CF is, and its relation to the MGF. I used the MGF for simplicity here, to avoid taking the N-power of a complex number. It'd have made things less nice. $\endgroup$
    – Clement C.
    Nov 6, 2017 at 19:56

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .