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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.

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  • $\begingroup$ 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 $\lambda$ $\endgroup$ – Azlif Nov 1 '19 at 3:54
  • $\begingroup$ @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. $\endgroup$ – The Pointer Nov 1 '19 at 3:59
  • $\begingroup$ 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. $\endgroup$ – Azlif Nov 1 '19 at 4:03
  • $\begingroup$ @Azlif Ok, understood. Thanks. $\endgroup$ – The Pointer Nov 1 '19 at 4:10
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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 $\mu = \lambda_1 + \lambda_2$.

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  • $\begingroup$ 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? $\endgroup$ – The Pointer Nov 1 '19 at 4:32
  • $\begingroup$ @ThePointer Because if $3X+Y$ were in fact Poisson, $\mu$ would be its rate parameter, and it clearly cannot be zero or negative. $\endgroup$ – heropup Nov 1 '19 at 4:34
  • $\begingroup$ Ahh, you're right. I was mixing up the support of the distribution with the rate parameter en.wikipedia.org/wiki/Poisson_distribution . This is where my confusion was stemming from. Thank you for the clarification. $\endgroup$ – The Pointer Nov 1 '19 at 4:36
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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.

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