Mean and variance of the sum of cgf gamma and poisson distribution

Question

Suppose that we have the sum of two cumulant generating function: $\log e^{m(e^t-1)} + \log(1-dt)^{-c}$, and we wish to find the expectation and variance without differentiation.

I realize that $\log e^{m(e^t-1)} $ is the cgf of a Poisson distribution with parameter $m$, and $\log(1-dt)^{-c}$ is the cgf of a Gamma distribution with parameters $c,d$.

My attempt to find the mean and variance: Suppose that $X$~$Poisson(m)$ and $Y$~$Gamma(c,d)$. Then: $$E(X+Y) = E(X) + E(Y) = m + cd$$

I am not quite sure with how to compute the variance, as it involves covariance term between the two variables. I was told that the answer should be $m+cd^2$, but if that is the case how do I show that $Cov(X,Y)=0$?

Answer 1

Note: if $M_X(t)$ and $M_Y(t)$ are the moment generating functions of $X$ and $Y$, then the product $$M_X (t) M_Y(t) = M_W(t)$$ is the MGF of their sum $W = X+Y$, where $X$ and $Y$ are independent. Note they do not need to be identically distributed for this to hold.

Since the cumulant generating function is the logarithm of the MGF, we see that if $X$ and $Y$ have CGFs $$C_X(t) = \log M_X(t) = \log e^{m(e^t - 1)}, \quad C_Y(t) = \log M_Y(t) = \log (1-dt)^{-c},$$ then $W = X+Y$ has CGF $$C_W(t) = C_X(t) + C_Y(t),$$ and we conclude that $$\operatorname{E}[W] = \operatorname{E}[X] + \operatorname{E}[Y]$$ by linearity, and $$\operatorname{E}[W^2] = \operatorname{E}[X^2] + \operatorname{E}[Y^2] + 2\operatorname{E}[X]\operatorname{E}[Y].$$ The covariance is zero because the sum of the CGFs corresponds to a random variable that is, by construction, the sum of independent random variables.