Sum of geometric and Poisson distribution Suppose I have $X \sim \mathrm{Geom}(p)$ and $Y=\mathrm{Pois}(\lambda)$.
I want to create $Z = X + Y$, where the $X$ begins at $0$ rather than $1$.
Is this possible? Then I would calculate the mean and variance.
 A: Are $X$ and $Y$ independent?
For any random variables, $E[X+Y]=E[X]+E[Y]$; and for any independent random variables, $\text{Var}(X+Y)=\text{Var}(X)+\text{Var}(Y)$
A: Something that I thought of to compute the pmf, then the expectation, is:
$Pr(X=k)= p(1-k)^{k}$
$Pr(Y=k)=\frac{\lambda^k e^{\lambda}}{k!}$
Then:
$Pr(Z=k)=p(1-p)^{k} \frac{\lambda^k e^{\lambda}}{k!}$
Finally, the expected value of $Z$ would be
$E[Z] = \sum_{k=0}^{\infty}k  p(1-p)^k \frac{\lambda^k e^{\lambda}}{k!} $
$E[Z] = p \sum_{k=0}^{\infty}k (1-p)^k \frac{\lambda^k e^{\lambda}}{k!} $
Now we split into $k = 0$, and then remaining $k$:
$E[Z] = p\{0 \frac{\lbrack(1-p)\lambda\rbrack^0 e^{-\lambda}}{0!}\} + p\sum_{k=1}^{\infty}\frac{\lbrack(1-p)\lambda\rbrack^k e^{-\lambda}}{(k-1)!}$
$E[Z] = p \sum_{k=1}^{\infty}\frac{\lbrack(1-p)\lambda\rbrack^k e^{-\lambda}}{(k-1)!}$
$E[Z] = e^{-\lambda} p\lbrack(1-p)\lambda\rbrack  \sum_{k=1}^{\infty}\frac{\lbrack(1-p)\lambda\rbrack^{k-1}}{(k-1)!}$
Let $j=k-1$
$E[Z] = e^{-\lambda} p\lbrack(1-p)\lambda\rbrack \sum_{j=0}^{\infty} \frac{\lbrack(1-p)\lambda\rbrack^j e^{-\lambda}}{j!}$
$E[Z] = e^{-\lambda} p\lbrack(1-p)\lambda\rbrack e^{(1-p)\lambda}$
$E[Z] =  p\lbrack(1-p)\lambda\rbrack e^{-\lambda p}$
Then the variance is straightforward.
Thanks!
