# Poisson Distribution with Exponential Parameter

If we have $$X(k)\sim Pois(2k)$$ and $$Y \sim Exp(15)$$ and $$Z=X(5Y)$$. How can we determine $$E(Z)$$, $$Var(Z)$$ and $$P(Z = z)$$.

So far I'm thinking \begin{align*} E(Z) &= E(X(5Y)) \\ &= E(Pois(10Y)) \\ &= E(10Y) \\ &= 10E(Y) \\ &= \frac{10}{15} \end{align*}

Similarly for Variance: \begin{align*} Var(Z) &= E(X(5Y)) \\ &= 100Var(Y) \\ &= \frac{100}{225} \end{align*}

I am not sure if this line of reasoning is correct and any guidance is appreciated. Also not too sure how to approach $$P(Z=z)$$.

• If my reasoning is correct, $Z$ follows a Geometric($3/5$) distribution with mean $2/3$ and variance $10/9$. This is in good agreement with some quick simulations. – COOLSerdash Aug 13 '20 at 20:52

## 3 Answers

The exponential distribution is a special case of the gamma distribution, so you have a Poisson-gamma (also known, confusingly, as a "mixture"). The resulting distribution is a negative binomial one - more specifically, a geometric distribution.

Specifically, you have $$Z\sim\text{Pois}(\lambda)$$, where $$10\lambda\sim\text{Exp}(15)$$ - so $$\lambda\sim\text{Exp}(\frac{15}{10})=\text{Exp}(\frac{3}{2})$$ (per Wikipedia), which is $$\Gamma(1,\frac{3}{2})$$ in the shape-rate parameterization. The Wikipedia entry for the negbin as a Poisson-gamma mixture then gives the parameters of the resulting negbin as $$r=1$$ and $$\frac{1-p}{p}=\frac{3}{2}$$, or $$p=\frac{2}{5}$$. Finally, Wikipedia again gives us the mean, variance and PMF:

\begin{align*} \mu &= \frac{pr}{1-p} = \frac{2/5}{1-2/5} = \frac{2}{3} \\ \sigma^2 &= \frac{pr}{(1-p)^2} =\frac{2/5}{(1-2/5)^2} = \frac{10}{9} \\ P(Z=z) &= {z+r-1\choose z}p^z(1-p)^r = (1-p)p^z. \end{align*}

(Note that there is a little confusion at Wikipedia for the PMF, with $$p$$ and $$1-p$$ switching places between the box at the top and the section on the Poisson-gamma mixture. The formula here is the correct one and is taken from the Poisson-gamma mixture section.)

As COOLSerdash writes, we recognize this as a geometric distribution, which is also noted at the negbin Wikipedia page under "Related distributions" as the special case for $$r=1$$.

I like confirming calculations like these with simulations. (Actually, that is how I found the confusion for the PMF at the Wikipedia page.) Things seem to work out fine. R code:

rate <- 15
n_sims <- 1e7
set.seed(1) # for reproducibility
yy <- rexp(n_sims,rate=15)
xx <- rpois(n_sims,5*2*yy)

hh <- hist(xx,breaks=seq(-0.5,max(xx)+0.5),col="grey",freq=FALSE,las=1)
pp <- 2/5
lines(hh$$mids,pp^hh$$mids*(1-pp),type="o",pch=19,col="red") The mean and variance we derived above match the simulations, too:

> mean(xx)
 0.6667809
> var(xx)
 1.1111

• It would be kind of you to explain or illustrate the calculations you are referring to, because I suspect they are quite different from those attempted in the question. – whuber Aug 13 '20 at 21:38
• @whuber: you are completely right. It was late yesterday, and I'd rather post a short answer than none at all. I'll try to get around to elaborating later today or in the next days. – Stephan Kolassa Aug 14 '20 at 6:34
• @whuber: again, you are completely right. I have edited my answer. – Stephan Kolassa Aug 14 '20 at 7:05

The variance calculation is incorrect. You must use the law of total variance:

$$\operatorname{Var}[Z] = \operatorname{E}[\operatorname{Var}[Z \mid Y]] + \operatorname{Var}[\operatorname{E}[Z \mid Y]].$$ The conditional variance and conditional expectation are equal since $$Z \mid Y$$ is Poisson: $$\operatorname{Var}[Z \mid Y] = \operatorname{E}[Z \mid Y] = 10Y.$$ Then $$\operatorname{Var}[Z] = \operatorname{E}[10Y] + \operatorname{Var}[10Y] = \frac{10}{15} + \frac{10^2}{15^2} = \frac{10}{9}.$$

To compute the PMF of $$Z$$, we note $$\Pr[Z = z] = \int_{y=0}^\infty \Pr[Z = z \mid Y = y] f_Y(y) \, dy = \int_{y=0}^\infty e^{-10y} \frac{(10y)^z}{z!} 15 e^{-15y} \, dy.$$ The remainder of the computation I leave as an exercise.

As a matter of principle, I would like to point out that I find the choice of notation in this question to be detestable. I would have written the model as such: $$Y \sim \operatorname{Exponential}(15), \\ Z \mid Y \sim \operatorname{Poisson}(10Y),$$ and ignored $$X$$ entirely.

Hint: $$E(X(5Y))= E(E(X(5y)|Y=y))=E(10Y)$$ and $$V(X(5Y))\\ = V(E(X(5y)|Y=y))+E(V(X(5y)|Y=y)) \\= V(10Y)+E(10Y)$$ Now calculate it using your form of exponential distribution. And your distribution of $$Z$$, $$P(Z=z)=\int_{y=0}^\infty P(X(5y)=z|Y=y)f_Y (y)dy$$