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Suppose X and Y are independent and follow uniform distribution (-a,a). How do we find the distribution of X-Y? I tried finding the area with the help of a diagram for cases when x-y>0 and x-y<0. I want to know what is the right way to do it.

user584534
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    Please add the [tag:self-study] tag & read its [wiki](https://stats.stackexchange.com/tags/self-study/info). Then tell us what you understand thus far, what you've tried & where you're stuck. We'll provide hints to help you get unstuck. Please make these changes as just posting your homework & hoping someone will do it for you is grounds for closing. – kjetil b halvorsen May 12 '21 at 08:20
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    Consider that $-Y$ and $Y$ has the same distributios, and have a look at https://stats.stackexchange.com/questions/41467/consider-the-sum-of-n-uniform-distributions-on-0-1-or-z-n-why-does-the – kjetil b halvorsen May 12 '21 at 08:23
  • Who knows what the "right" way might be -- but many different ways are given (in great detail) at https://stats.stackexchange.com/questions/41467. – whuber May 12 '21 at 15:54

1 Answers1

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So the individual means of X and Y are both 0, and so the mean of (X - Y) will also be 0.

The covariance of the two is 0, so we can just add up the variances. The variance of one variable is $(2a)^2 / 12=a^2/3$, so the variance of (X - Y) will be $2a^2 / 3$.

Dirk Nachbar
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