If Z|Y ~ Bin(p, y) and Y ~ Poisson(L) then Z ~ Poisson(p*L)?

Question

I checked whether this question was answered before but because of notation, it's hard to see. I am a reading a paper that defines the following two RVs $$ z \mid y \sim Binomial(\pi, y) \\ y \sim Poisson(\lambda) $$ then concludes (by integration and Bayes Rule) that $$ z \sim Poisson(\pi \cdot \lambda) \\ y - z \sim Poisson( (1 - \pi)\cdot \lambda) $$

I tried working it out on paper but since I am not a trained statistician, I am not sure where I am going wrong. If I want to derive $z \sim Poisson(\pi\lambda)$, then I use the conditional probability i.e. $$ p(z) = \int_y p(z, y) \,\, dy $$ where $p(z, y)$ is the joint probability. Expanding this out, I have $$p(z) = \int_y {y\choose z} \pi^z (1 - \pi)^{(y - z)} \frac{e^\lambda \lambda^y}{y!} \, \, dy $$ but I am guessing I have to get to the following equation somehow $$p(z) = \frac{e^{\pi\lambda} (\pi\lambda)^{z}}{z!}$$ but I was not able to manipulate the integral above to get this form. Not sure if that's even possible.

Answer 1

This follows from some fairly standard distribution theory. Define $Y_1 \sim \text{Poisson}(\pi \lambda)$ and $Y_2 \sim \text{Poisson}((1-\pi) \lambda)$ independently, and let $Y = Y_1 + Y_2$ and $Z = Y_1$. Then the following facts are quickly derived:

• $Y \sim \text{Poisson}(\lambda)$ (can be checked by computing the moment generating function).

• $[Z \mid Y = y] \sim \text{Binomial}(\pi, y)$ because, using independence,

$$ f(z \mid y) = \frac{\Pr(Y_1 = z, Y_2 = y - z)}{\Pr(Y = y)} = \binom{y}{z} \pi^z (1 - \pi)^{y-z} $$

• It is true by definition that $Y - Z = Y_2 \sim \text{Poisson}((1-\pi) \lambda)$ and that $Z = Y_1 \sim \text{Poisson}(\pi \lambda)$, which are the results you wanted.

Hence, there exist $Z$ and $Y$ with the properties you want, but since the joint distribution is uniquely characterized by your conditions for $(Z,Y)$ it follows that this is true for all $Z$ and $Y$ satisfying your conditions.

Answer 2

Its a bit of algebra, but here is my try

The expression of the density after you pull out the terms not involving $y$ are

$$ p(z) = \pi^z \exp(\lambda) \sum_{z \leq y} \binom{y}{z} (1-\pi)^{y-z} \dfrac{\lambda^y}{y!}$$

The $y!$ cancels from the binomial coefficient

$$ = \pi^z \dfrac{\exp(\lambda)}{z!} \sum_{{z \leq y}} (1-\pi)^{y-z} \dfrac{\lambda^y}{(y-z)!}$$

And since the index is only for $0\leq y-z$, then let $k=y-z$

$$ = \pi^z \dfrac{\exp(\lambda)}{z!} \sum_{k} (1-\pi)^{k} \dfrac{\lambda^{k+z}}{(k)!}$$

Simplifying more

$$ = (\lambda\pi)^z \dfrac{\exp(\lambda)}{z!} \sum_{k} (1-\pi)^{k} \dfrac{\lambda^{k}}{(k)!}$$

You'll notice the sum is the expression for $\exp(\lambda - \lambda \pi)$

And so we wind up with

$$ p(z) = (\lambda \pi)^z \dfrac{\exp(-\pi\lambda)}{z!}$$

Which I believe means

$$z \sim \operatorname{Poisson}(\pi \lambda)$$

Answer 3

To get $E(Z)$ and $Var(Z),$ this can be viewed as a random sum of random variables. In particular, $Z$ is a sum of a random number $Y$ of Bernoulli random variables each with success probability $\pi.$

Here is a histogram of 100,000 simulated realizations of $Z,$ using $\lambda = 20, \pi = 0.4$ along with exact probabilities (centers of red circles) for $\mathsf{Pois}(8).$

set.seed(2020)
lam = 20;  pp = 0.4
y = rpois(10^5, lam)
z = rbinom(10^5, y, pp)
summary(z)
   Min. 1st Qu.  Median    Mean 3rd Qu.    Max. 
  0.000   6.000   8.000   8.001  10.000  22.000 

mx = max(z);  cutp = (-1:mx)+.5
hdr = "Histogram of Simulated Z with Density of POIS(8)"
hist(z, prob=T, br=cutp, col="skyblue2", main=hdr)
 points(0:mx, dpois(0:mx, pp*lam), col="red")

Notes: (1) @aleshing is correct that, on account of discreteness, the integral should be treated as a sum.

(2) In R code: Cannot use pi for $\pi$ because it is a reserved constant in R. If y happens to return $0,$ rbinom is programmed to return $0.$

(3) In case it is of interest: UNL course handout on random sum of random variables.