## Chicken and Egg (Probability) Problem

The following problem comes from a stats course taught by Joe Blitzstein in Fall ’09–my freshman year of college. Suppose a chicken lays $N$ eggs, where $N$ is Poisson distributed with parameter $\lambda > 0$. Given that the chicken has layed $N$ eggs, each egg hatches independently of the other eggs with probability $p \in (0,1)$. Mathematically speaking, if $X$ denotes the number of hatched eggs, the random variable $X | N$ is binomial distributed with parameters $N,p$. What is the distribution of $X$?

We use conditional probability to tackle this problem. Specifically, we condition on the number of eggs $N$ the chicken lays.

$\displaystyle \begin{array}{lcl}\mathbb{P}(X = k)&=&\sum_{n=0}^{\infty}\mathbb{P}(X = k| N = n)\mathbb{P}(N = n)\\&=&\sum_{n=k}^{\infty}\mathbb{P}(X=k|N=n)\mathbb{P}(N=n)\\&=&\sum_{n=k}^{\infty}{n \choose k}p^{k}(1-p)^{n-k}\dfrac{e^{-\lambda}\lambda^{n}}{n!}\\&=&\dfrac{e^{-\lambda}p^{k}\lambda^{k}}{k!}\sum_{n=k}^{\infty}\dfrac{(1-p)^{n-k}\lambda^{n-k}}{(n-k)!}\end{array}$

The infinite series in the last expression should be recognizable to the reader as the Taylor expansion of $e^{\lambda(1-p)}$. Substituting this in, we obtain

$\displaystyle \mathbb{P}(X=k) = \left(\dfrac{e^{-\lambda}p^{k}\lambda^{k}}{k!}\right)e^{\lambda(1-p)}=\dfrac{e^{-\lambda p}(p\lambda)^{k}}{k!}$

But this last expression is the probability mass function (PMF) of a Poisson random variable with parameter $\lambda p$ evaluated at $k$.

Consider now the number of eggs which don’t hatch, which is the random variable $Y := N-X$? At first glance, it seems that $X$ and $Y$ are highly correlated. When I first encountered this problem, I thought this was indeed the case. However, it turns out that $X$ and $Y$ are independent. To see this, we show that the joint distribution of $X$ and $Y$ factors into the product of the marginal distributions. A completely analogous argument to that above shows that the marginal distribution for $Y$ is Poisson with parameter $\lambda(1-p)$. Conditioning on $N$, we have that

$\displaystyle\begin{array}{lcl}\mathbb{P}(X=k,Y=j) &=&\sum_{n=0}^{\infty}\mathbb{P}(X=k,Y=j \mid N=n)\mathbb{P}(N=n)\end{array}$

Observe that

$\displaystyle\mathbb{P}(X=k,Y=j\mid N=n)=\begin{cases} {{k+j}\choose k}p^{k}(1-p)^{j}&{n=k+j}\\ 0&{n\neq k+j}\end{cases}$

so that

$\displaystyle\begin{array}{lcl}\mathbb{P}(X=k,Y=j)=\mathbb{P}(X=k,Y=j\mid N=k+j)\mathbb{P}(N=k+j)&=&{{k+j}\choose k}p^{k}(1-p)^{j}\mathbb{P}(N=k+j)\\&=&{{k+j}\choose k}p^{k}(1-p)^{j}\left(\dfrac{e^{-\lambda}\lambda^{k+j}}{(k+j)!}\right)\\&=&\left(\dfrac{e^{-\lambda p}(p\lambda)^{k}}{k!}\right)\left(\dfrac{e^{-\lambda(1-p)}((1-p)\lambda)^{j}}{j!}\right)\\&=&\mathbb{P}(X=k)\mathbb{P}(Y=j)\end{array}$

What if instead of taking the probability $p$ that an egg hatches to be a known constant, we treat $p$ as a random variable? I know nothing about chicken egg hatching probabilities, so I’m not sure what would be a good choice of distribution for $p$. Any thoughts?

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