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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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Michael R. Chernick
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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
  • I wasn't asking for half normal. I just thought that it probably coul be done with half normals In a comment in the discussion to the last problem Dilip suggested a solution to this question that involved two half normals. I would like to see the details of that solution. – Michael R. Chernick May 20 '12 at 15:25
  • @cardinal I think if I understand your notation your example is too trivial and not what was intended by the question. A normal variable X is X+ when X is positive and -X- when X is negative by definition. – Michael R. Chernick May 20 '12 at 15:37
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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
  • @cardinal I was not saying that it was necessarily a bad answer. But it is based on the definition that decomposes random variables into their positive and negative parts. It does of course fit the criterion of the question. But if there is another solution involving half normals or any other pair of correlated random variables. – Michael R. Chernick May 20 '12 at 16:02
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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
  • The only reason Y needs to be a positive rv is that X+ + Y and X- + Y are both required to positive. Yes adding and substrating the same variable is a trivial addition to cardinal's solution. I would still like to see the half normal example. – Michael R. Chernick May 20 '12 at 19:30
  • Trivial as well: take two half-normal variables $Y$ and $Z$, then take $X=Y-(Y+Z)\delta$ where $\delta$ is $0$ with probability $1/2$ and $1$ with probability $1/2$... – Xi'an May 20 '12 at 20:21
  • Here I think it is a little less trivial because Y and Z can be indepedent and the dependence is between Y and (Y+Z)δ . – Michael R. Chernick May 20 '12 at 21:01
  • Especially since $Y+Z$ is also half-normal. – Xi'an May 21 '12 at 07:03
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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
  • I would still be interested in finding a way to do this with half normals. But even though it looked to me like to would work I haven't found a way and since Dilip said his method did not involve half normals maybe it can't be done. – Michael R. Chernick May 21 '12 at 14:37
  • @DilipSarwate: yes indeed, my mistake! – Xi'an May 21 '12 at 18:40
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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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Partially answered in comments:

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

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

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

kjetil b halvorsen
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