# Decomposing the normal distribution

Does there exist a positive-only distribution such that the difference of two independent samples from this distribution is normally distributed? If so, does it have a simple form?

• Interesting question! The normal distribution is infinitely decomposable, meaning you can always write it as the distribution of a sum $x_1+\ldots+x_n$ of an arbitrary number $n$ of random variables. But this is not the question. May 19, 2012 at 19:56
• If you get to the moment generating function, the question is whether or not $$e^{t\mu + \frac{1}{2}\sigma^2t^2}=\varphi(t)\varphi(-t)$$ allows for a solution (in $\varphi$) that is a moment generating function of a positive variable... May 19, 2012 at 19:57
• You are correct, @Dilip: a difference of half-normals does not have a normal distribution. The problem is not with the variance of the difference: the very shape of the distribution is not normal (its kurtosis is too great).
– whuber
May 19, 2012 at 22:17
• Although this is obvious, it may be worth noting that the statement is approximately correct. After all, the difference of an $N(\mu,\sigma^2/2)$ variable and an $N(\mu,\sigma^2/2)$ variable has a $N(0,\sigma^2)$ distribution and, by choosing $\mu$ sufficiently large, we can make the chance that either variable is negative as small as desired.
– whuber
May 21, 2012 at 14:45

The answer to the question is No, and it follows from a famous characterization of normal distributions.

Suppose that $X$ and $Y$ are independent random variables. Then so are $X$ and $-Y$ independent random variables, and of course we can write $X-Y$ as $X + (-Y)$, the sum of two independent random variables. Now, according to a theorem conjectured by P. Lévy and proved by H. Cramér (see Feller, Chapter XV.8, Theorem 1),

If $X$ and $Y$ are independent random variables and $X+Y$ is normally distributed, then both $X$ and $Y$ are normally distributed.

The OP asks whether there exist i.i.d. positive random variables $X$ and $Y$ such that $X-Y$ is normally distributed. But even if we dispense with positivity and identical distributions, and keep only the independence, normality of $X-Y = X + (-Y)$ requires that both $X$ and $-Y$ be normal random variables. As Feller says, "the normal distribution cannot be decomposed except in the trivial manner."

• I was somewhat hoping the answer would be yes, but thanks! I don't have easy access to a copy of Feller - is it possible to sketch a proof of the theorem? It seems quite counterintuitive. May 20, 2012 at 3:35
• Even Feller does not include the original proof claiming that it is based on analytic function theory and thus quite different from his approach to characteristic functions. May 20, 2012 at 3:42
• I thought that was the case but it opens the door for dependent variables. I was try to find a way to construct dependence between 2 positive half normals but couldn't quite get it to work. May 20, 2012 at 12:16
• well maybe someone should I was more interested in trying to solve it May 20, 2012 at 12:52
• I will make this a question and then you can spell out your answer. I am not quite following what this joint density looks like and are you taking Z=|X|-|Y|? May 20, 2012 at 13:59