1
$\begingroup$

This theorem is stated as a footnote in a lecture:

Let $X,Y$ be independent random variables

  1. If $X+Y$ follows a normal distribution, then $X$ and $Y$ follow a normal distribution.

  2. If $X+Y$ follows a Poisson distribution, then $X$ and $Y$ follow a Poisson distribution.

Note that $X$ and $Y$ are not supposed to be identically distributed. $X$ may follow Poisson(1) while $Y$ follows Poisson(2).

I've been trying to prove these statements without success. Sums of independent random variables hint at something involving characteristic functions (inverse Fourier transform or some such), but I've been unsuccessful so far.

$\endgroup$
  • $\begingroup$ Can you provide a link to the mentioned lecture? My original belief was that 1. does not hold, but I am having a hard time in contructing a counter-example which meets Bochner's constraints (en.wikipedia.org/wiki/Bochner%27s_theorem), so I am starting to think that such constraints can be met only if both $X$ and $Y$ are normally distributed. $\endgroup$ – Jack D'Aurizio Nov 1 '17 at 22:07
  • $\begingroup$ @JackD'Aurizio it seems to be a famous theorem by Cramer, see here $\endgroup$ – Gabriel Romon Nov 1 '17 at 22:13
  • $\begingroup$ Indeed, a proof of 1. is hidden at page 443 here $\endgroup$ – Jack D'Aurizio Nov 1 '17 at 22:22
  • $\begingroup$ And a proof by entropy has been proposed by Per Vognsen on MO $\endgroup$ – Jack D'Aurizio Nov 1 '17 at 22:29
  • $\begingroup$ The book by Stoyanov called "Counterexamples in probability" or a similar title, has these and tons of other similar results. $\endgroup$ – Did Nov 1 '17 at 23:17
0
$\begingroup$

Different proofs of 1. (Cramer's theorem) have been given in the comments: the most common approaches are to employ characteristic functions, Bochner's theorem and complex analysis, or to exploit inequalities about entropy and pressure (see this short proof by Cert).

So, long story short, the normal distribution is an indecomposable distribution.

Let us try to outline a similar approach in the discrete case of the Poisson distribution.

To be clear, the following paragraph is just an attempt.

The moment generating function of a Poisson distribution with parameter $\lambda$ is given by $\exp\left(\lambda(e^t-1)\right)$. In general, if $f(t)$ is the MGF of a distribution supported on $\mathbb{R}_{\geq 0}$, the sequence given by $\{f^{(n)}(0)\}_{n\geq 0}$ is non-negative and log-convex by the Cauchy-Schwarz/Holder inequality. The Poisson distribution has the remarkable property of having constant cumulants. If we consider the cumulant generating functions $K_X(t)$ and $K_Y(t)$ and assume that $\kappa_n=K_X^{(n)}(0)$ is non-constant, we should reach a contradiction by exploiting log-convexity and the inverse Laplace transform.

$\endgroup$

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.