I have the following problem:
Use moment generating functions to determine whether $3X + Y$ is Poisson if $X$ and $Y$ are i.i.d. Pois($\lambda$).
Hint: If $X \sim$ Pois($\lambda$), then its moment generating function is
$$M_X(t) = e^{\lambda(e^t – 1)}, \ \ \text{for} \ t \in \mathbb{R}$$
My solution is as follows:
$$\begin{align} M_{3X + Y}(t) &= E[e^{(3X + Y)t}] \ \ \text{(By the definition of moment generating function.)} \\ &= E[e^{3Xt}] E[e^{Yt}] \ \ \text{(Since $X$ and $Y$ are independent.)} \\ &= M_X(3t) M_Y(t) \\ &= [e^{\lambda(e^{3t} – 1)}][e^{\lambda(e^t – 1)}] \ \ \text{(Since $X$ and $Y$ are Poisson random variables.)} \\ &= e^{\lambda(e^{3t} + e^t – 2)} \end{align}$$
We want to determine whether $3X + Y$ is Poisson. I am told that, if $3X + Y$ is Poisson, then there must exist some $\mu$ such that
$$ e^{\lambda(e^{3t} + e^t – 2)} = e^{\mu(e^t – 1)} \ \ \text{for all $t \in \mathbb{R}$} $$
But don't we need this to be true for all $\mu$? After all, otherwise we could just take the case $\lambda = \mu = 0$ and say that $3X + Y$ is Poisson, even though, according to the solution, it isn't.
So what am I misunderstanding here? How am I supposed to show a counterexample?
I would greatly appreciate it if people could please take the time to clarify this.
Best Answer
If we require $$e^{\lambda(e^{3t} + e^t - 2)} = e^{\mu(e^t - 1)} \tag{1}$$ to hold for all $t \in \mathbb R$, then let this is equivalent to showing that there exists some constant $\mu > 0$ with respect to $s$ (but which may depend on $\lambda$ in some nontrivial manner) such that $$\lambda(s^3 + s - 2) = \mu(s - 1)$$ for all $s > 0$, or that $$\lambda s^3 + (\lambda - \mu)s - 2\lambda + \mu = 0. \tag{2}$$ So if this equation must hold for all positive $s$, choose particular values for $s$ and show the resulting system has no valid solution; e.g., $s = 2$ implies $8\lambda - \mu = 0$, but $s = 3$ implies $28\lambda - 2\mu = 0$, and together, these would require $\lambda = \mu = 0$, which is not allowed. The fact that equation $(2)$ can be trivially satisfied for some values of $s$ (e.g., $s = 1$), is insufficient, because the requirement is that there is some choice of $\mu > 0$ for a given $\lambda > 0$ such that the cubic is identically zero for all positive $s$.
Addendum. It is illustrative to see how the same approach can be applied to a situation in which we do obtain a distribution from the same parametric family. Consider $X \sim \operatorname{Poisson}(\lambda_1)$, $Y \sim \operatorname{Poisson}(\lambda_2)$, and form the random variable $X+Y$. Its MGF is $$M_{X+Y}(t) = M_X(t) M_Y(t) = e^{\lambda_1(e^t - 1)} e^{\lambda_2(e^t - 1)} = e^{(\lambda_1 + \lambda_2)(e^t - 1)},$$ and clearly we see that this is the MGF of a Poisson distribution with rate parameter $\mu = \lambda_1 + \lambda_2$.