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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?

kjetil b halvorsen
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Martin O'Leary
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  • 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. – Xi'an May 19 '12 at 19:56
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    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... – Xi'an May 19 '12 at 19:57
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    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 '12 at 22:17
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    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 '12 at 14:45

1 Answers1

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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."

Dilip Sarwate
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  • 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. – Martin O'Leary May 20 '12 at 03:35
  • 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. – Dilip Sarwate May 20 '12 at 03:42
  • 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. – Michael R. Chernick May 20 '12 at 12:16
  • well maybe someone should I was more interested in trying to solve it – Michael R. Chernick May 20 '12 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|? – Michael R. Chernick May 20 '12 at 13:59
  • @DilipSarwate I didn't have a clear picture as to what your example demonstrated. I had said that I was attempting to construct a normal as the difference of two dependent half normals. I was not sure it could be done but thought it was possible. I thought you were present an example of that. – Michael R. Chernick May 20 '12 at 22:10