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 $P(Z=z)$.

  • 2
    $\begingroup$ 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. $\endgroup$ – COOLSerdash Aug 13 '20 at 20:52

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


The mean and variance we derived above match the simulations, too:

> mean(xx)
[1] 0.6667809
> var(xx)
[1] 1.1111
  • 1
    $\begingroup$ 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. $\endgroup$ – whuber Aug 13 '20 at 21:38
  • $\begingroup$ @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. $\endgroup$ – Stephan Kolassa Aug 14 '20 at 6:34
  • $\begingroup$ @whuber: again, you are completely right. I have edited my answer. $\endgroup$ – 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 $Z$, $$P(Z=z)=\int_{y=0}^\infty P(X(5y)=z|Y=y)f_Y (y)dy$$


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.