Let $U_1$ and $U_2$ be independent Skellam random variables, i.e. $U_1=X_1-X_2$ and $U_2=X_3-X_4$, where $X_i \sim Pois(\lambda_i)$. Let $a, b, c, d \in R_+$. We know that $U_1+U_2$ be Skellam random variable.
- Would $S=aX_1-bX_2+cX_2-dX_3$ be also Skellam random variable?
- What would be $E|S|^4?$
Best Answer
1) In general does not hold. Note that the MGF of $X \sim \text{Poisson}(\lambda)$ is
$$ M_X(t) = \exp\left\{\lambda(e^t - 1)\right\} $$
and the MGF of $Y \sim \text{Skellam}(\mu_1,\mu_2)$ is
$$ M_Y(t) = \exp\left\{\mu_1 e^t + \mu_2 e^{-t} - (\mu_1 + \mu_2)\right\}$$
The MGF of $S = aX_1 - bX_2 + cX_3 - dX_4$ is
$$ \begin{align} M_S(t) &= E\left[e^{(aX_1 - bX_2 + cX_3 - dX_4)t}\right] \\ &= E\left[e^{X_1(at)}e^{X_2(-bt)}e^{X_3(ct)}e^{X_4(-dt)}\right] \\ &= E\left[e^{X_1(at)}\right] E\left[e^{X_2(-bt)}\right] E\left[e^{X_3(ct)}\right] E\left[e^{X_4(-dt)}\right] \\ &= M_{X_1}(at)M_{X_2}(-bt)M_{X_3}(ct)M_{X_4}(-dt) \\ &= \exp\left\{\lambda_1(e^{at} - 1)\right\} \exp\left\{\lambda_2(e^{-bt} - 1)\right\} \exp\left\{\lambda_3(e^{ct} - 1)\right\} \exp\left\{\lambda_4(e^{-dt} - 1)\right\} \\ &= \exp\left\{\lambda_1 e^{at} + \lambda_2 e^{-bt} + \lambda_3 e^{ct} + \lambda_4 e^{-dt} - (\lambda_1 + \lambda_2 + \lambda_3 + \lambda_4)\right\} \end{align}$$
So we see that we need to have $a = b = c = d = 1$ (or one of the $\{a, c\}$ and/or $\{b, d\}$ can be $0$) in order for $S$ to have a Skellam distribution.
2) You may direct expand $S^4$ to calculate, or differentiate $M_S(t)$ $4$ times.
$$ \begin{align} & E\left[|S|^4\right] \\ =~& E\left[S^4\right]\\ =~& \left. \frac {d^4} {dt^4} M_S(t) \right|_{t = 0} \\ =~ & (a^4\lambda_1 + b^4\lambda_2 + c^4\lambda_3 + d^4\lambda_4) \\ & + 4(a^3\lambda_1 - b^3\lambda_2 + c^3\lambda_3 - d^3\lambda_4) (a\lambda_1 - b\lambda_2 + c\lambda_3 - d\lambda_4) \\ & + 3(a^2\lambda_1 + b^2\lambda_2 + c^2\lambda_3 + d^2\lambda_4)^2 \\ & + 6(a^2\lambda_1 + b^2\lambda_2 + c^2\lambda_3 + d^2\lambda_4) (a\lambda_1 - b\lambda_2 + c\lambda_3 - d\lambda_4)^2 \\ & + (a\lambda_1 - b\lambda_2 + c\lambda_3 - d\lambda_4)^4 \end{align} $$