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.

  • $\begingroup$ 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

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 $\mu = \lambda_1 + \lambda_2$.

  • $\begingroup$ 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

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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.