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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 $P(Z=z)$.
The exponential distribution is a special case of the gamma distribution, so you have a Poisson-gamma compound-distribution (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)
[1] 0.6667809
> var(xx)
[1] 1.1111
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 $Z$, $$P(Z=z)=\int_{y=0}^\infty P(X(5y)=z|Y=y)f_Y (y)dy$$
