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A question was asked whether or not two independent variables $X$ and $Y$ that take on only positive values can have $X-Y$ be a normal distribution. I was shown that the answer is no. But I think that this can be done with two half normals that are dependent. But I could not quite figure out how to structure the dependence.

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Of course. Just take $X = X^+ - X^-$. (They aren't half normal, but are nonnegative in the spirit of the title and the original question.) – cardinal May 20 '12 at 14:13
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What do you mean "too trivial"? They are (dependent) identically distributed random variables such that the difference is a normal. It exactly answers the question. The construction has the benefit of drawing out exactly how and why one might expect it to be true. :) – cardinal May 20 '12 at 15:46
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You will call this trivial as well, but any decomposition of the form $X=(X^++Y)-(X^-+Y)$ where $Y$ is a positive rv works. Including a half-normal rv. – Xi'an May 20 '12 at 19:09
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@Xi'an Why is $Y+Z$ half-normal? Is the sum of two half-normal random variables half normal? I very much doubt it. $Y+Z$ is, of course, nonnegative as Michael wants. – Dilip Sarwate May 21 '12 at 10:59
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Despite the "too trivial" remark, I find this to be an interesting question and upvoted it long ago. My comments and the extension provided by @Xi'an answer the first part. I have spent a little time thinking about the second part. There is one fairly obvious attack on this problem that could prove the second part to be false and also potentially prove an interesting property of the normal. I have not been able to resolve it either way at the present time. – cardinal Jul 23 '12 at 17:51
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