# Using moment generating functions to determine whether $3X + Y$ is Poisson if $X$ and $Y$ are i.i.d. Pois($\lambda$)

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.

• What you conclude from there is $3X + Y$ has poisson distribution if and only if $\lambda = 0$. But from what I know, $\lambda > 0$ for poisson distribution. Thus, $3X + Y$ never has poisson distribution for all $\lambda$ – Azlif Nov 1 '19 at 3:54
• @Azlif Yes, that's my point: I think we require $e^{\lambda(e^{3t} + e^t - 2)} = e^{\mu(e^t - 1)}$ to be true for all $\mu$, rather than saying that "there must exist some $\mu$" such that it is true. The author's solution is what confused me here. – The Pointer Nov 1 '19 at 3:59
• The author is correct though. If $3X + Y$ were poisson distributed with parameter $\mu$, then the MGF is of the form $$e^{\mu (e^t - 1)}.$$ But then such $\mu$ cannot exist from what you did. – Azlif Nov 1 '19 at 4:03
• @Azlif Ok, understood. Thanks. – The Pointer Nov 1 '19 at 4:10

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 $$\mu = \lambda_1 + \lambda_2$$.
• Thanks for the answer. Why $\mu > 0$? $0$ is in the support of the Poisson distribution, so it seems to me that it should be allowed? – The Pointer Nov 1 '19 at 4:32
• @ThePointer Because if $3X+Y$ were in fact Poisson, $\mu$ would be its rate parameter, and it clearly cannot be zero or negative. – heropup Nov 1 '19 at 4:34
You need a choice of $$\mu$$ here. Certainly, all $$\mu$$ cannot work because different $$\mu$$ give different generating functions. What we must show is that no $$\mu$$ works.
Indeed, the equality $$e^{\lambda (e^{3t} + e^t -2)} = e^{\mu(e^t - 1)}$$ for all $$\mathit{t \in \mathbb R}$$ can be contradicted as follows : suppose for some $$\mu$$ it held true, then take logarithms on both sides and transpose to conclude that $$\lambda e^{3t} + (\lambda - \mu)e^t + (\mu - 2 \lambda) = 0$$ for all $$t$$. However, the polynomial $$\lambda x^3 + (\lambda - \mu)x + (\mu - 2 \lambda) = 0$$ has only at most three real roots, but we are saying that $$e^t$$ is a real root for all $$t$$. This is an infinite set : thus we have a contradiction, showing that all the coefficients are zero : we must have $$\lambda = \mu = 0$$ then, otherwise we get that $$3X+Y$$ isn't Poisson.