I'm working with Bradley Efron (2010): Large Scale Inference and my question concerns the proof of Lemma 2.3.

Here we have $z_i \sim F_0$ with probability $\pi_0$, $z_i \sim F_1$ with probability $\pi_1$ and $N \sim Poi(\eta)$. The $z_i$ are independent of each other and $i$ runs from $1$ to $N$.

$N_0(\mathcal{Z}) := \vert \{ i: z_i \sim F_0, z_i \in \mathcal{Z} \} \vert $

$N_1(\mathcal{Z}) := \vert \{ i: z_i \sim F_1, z_i \in \mathcal{Z} \} \vert $

$N_+(\mathcal{Z}) := N_0(\mathcal{Z})+ N_1(\mathcal{Z}) $

Due to "well known properties of the Poisson distribution" it follows that

  1. $N_0(\mathcal{Z}) \sim Poi(\eta \pi_0 F_0(\mathcal{Z}))$ independently of $N_1(\mathcal{Z}) \sim Poi(\eta \pi_1 F_1(\mathcal{Z}))$

as well as

  1. $N_0(\mathcal{Z}) \vert N_+(\mathcal{Z}) \sim Bin(N_+(\mathcal{Z}), \pi_0 F_0(\mathcal{Z}) / F(\mathcal{Z}) )$

With $F = \pi_0 F_0 + \pi_1 F_1$ being the "mixed distribution".

Can someone help me out with those well known properties needed to conclude 1. and 2. ?

  • $\begingroup$ Hint:Look up infinite divisibility of the Poisson. $\endgroup$
    – user237392
    Aug 26, 2015 at 14:50

1 Answer 1


Ha! Nice question, I also had the exact same problem while I was reading Efron's book and it took me some time to figure it out. I guess thus it is by no chance that I have previously posted answers to prove theorems which will help answer your question..

First of all, one important ingredient you forgot to mention in your question is that $\pi_0 + \pi_1 = 1$.

Let's first consider one of the $z_i$ individually. I think it is easier to think in terms of the generating two-groups model, rather than the mixture, i.e. for each $z_i$ let's also introduce a random variable $H_i$, such that:

$$ \begin{aligned} &H_i \sim \text{Bernoulli}(\pi_1) \\ &z_i | H_i = 1 \sim F_1 \\ &z_i | H_i = 0 \sim F_0 \end{aligned} $$

Now notice that:

$$ \Pr[H_i = 0, z_i \in \mathcal{Z}] = \Pr[H_i = 0]\Pr[z_i \in \mathcal{Z} \vert H_i=0]=\pi_0F_0(\mathcal{Z}) $$

This means that $$\mathbf{1}_{\{H_i = 0, z_i \in \mathcal{Z}\}} \sim \text{Bernoulli}(\pi_0F_0(\mathcal{Z}))$$ Since the $(z_i, H_i)_i$ are independent, also the $(\mathbf{1}_{\{H_i = 0, z_i \in \mathcal{Z}\}})_i$ will be independent. Thus it follows that for $n$ fixed:

$$\sum_{i=1}^n \mathbf{1}_{\{H_i = 0, z_i \in \mathcal{Z}\}} \sim \text{Binomial}(n,\pi_0F_0(\mathcal{Z}))$$

Since $N_0(\mathcal{Z}) = \sum_{i=1}^N \mathbf{1}_{\{H_i = 0, z_i \in \mathcal{Z}\}}$, we get our 0-th result (before the 2 your mention), which is a result conditional on $N$:

$$N_0(\mathcal{Z}) \big\vert N \sim \text{Binomial}(N,\pi_0F_0(\mathcal{Z}))$$

Similarly we could get:

$$N_1(\mathcal{Z}) \big\vert N \sim \text{Binomial}(N,\pi_1F_1(\mathcal{Z}))$$

Now we actually need to make the above results a bit stronger: For this let's also define $\widetilde{N}_0(\mathcal{Z}) = \sum_{i=1}^N \mathbf{1}_{\{H_i = 0, z_i \notin \mathcal{Z}\}}$, $\widetilde{N}_1(\mathcal{Z}) = \sum_{i=1}^N \mathbf{1}_{\{H_i = 1, z_i \notin \mathcal{Z}\}}$.

By definition it holds that

$$\widetilde{N}_0(\mathcal{Z}) + \widetilde{N}_1(\mathcal{Z}) + N_0(\mathcal{Z}) + N_1(\mathcal{Z}) = N$$

Using the same argument as above (i.e. considering what happens at a single $i$ and then summing up independent random variables and finally conditioning), we can now show that:

$$ \left(\widetilde{N}_0(\mathcal{Z}), \widetilde{N}_1(\mathcal{Z}), N_0(\mathcal{Z}),N_1(\mathcal{Z}\right)\big\vert N \sim \text{Multinomial}(N, \pi_0(1-F_0(\mathcal{Z}), \pi_1(1-F_1(\mathcal{Z}), \pi_0 F_0(\mathcal{Z}), \pi_1 F_1(\mathcal{Z})) $$

Now because $N \sim \text{Pois}(\eta)$ we can apply the theorem, which I previously proved here. This immediately yields your first result.

The second result now follows from the first result by applying the theorem I proved here.


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