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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