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I am trying to compute the distribution of the following $$Z=\bigl(X+Y\bigl)^2$$ BUT I have that both $X$ and $Y$ are Nakagami with parameter $m$. (A Nakagami random variable is the square root of a Gamma random variable.) So the above is hard to derive in general ( I would have to take the convolution assuming the summands are independent..)

So my alternative is to solve $$W=X^2+Y^2$$ which is easier to compute as it will be the sum of two Gamma distributions with scale and shape parameter $m$ and is also Gamma distributed.

My question is, when can one argue that $W$ and $Z$ have the same distribution, is it when X and Y are uncorrelated then they are equivalent? or will they never be equivalent?

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Without further information about the Nakagami distribution or even the dimension of $X$ and $Y$, here is one case when $Z$ and $W$ have the same distribution:

If you expand $W=(X+Y)^2$, you get $$ W=(X+Y)^2 = X^2+2YX+Y^2=Z+2XY $$ therefore $W=Z$ when $XY=0$ with probability 1.

I suspect this is the only non-trivial case when $Z$ and $W$ share the same distribution.

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    $\begingroup$ And the simplest non-trivial example of such $X$ and $Y$ is a pair of discrete random variables $(X,Y)$ taking on values $(0,1),(1,0),(0,-1),(-2,0)$. (Replacing $(-2,0)$ with the more expected $(-1,0)$ makes $(X+Y)^2$ and $X^2+Y^2$ degenerate random variables that equal $1$ with probability $1$. $\endgroup$ – Dilip Sarwate Nov 7 '14 at 21:52
  • $\begingroup$ But what does it mean that we have random variables $XY=0$, does it mean they are independent, uncorrelated.. i,e can we say something special about the summands? @Xi'an $\endgroup$ – Tyrone Nov 8 '14 at 5:22
  • $\begingroup$ Thanks, I understand this example you provided, but for Nakagami random variables how can one argue about computing $X^2+Y^2$ and not $(X+Y)^2$, I assume one in general cannot say that $XY=0$ with probability one @Dilip Sarwate. $\endgroup$ – Tyrone Nov 8 '14 at 5:33
  • $\begingroup$ @Tyrone: yes, obviously, if $XY=0$ with probability 1, $X$ and $Y$ are dependent. Can you add an explanation of what a Nakagami random variable is in your question (and why this matters)? Thank you. $\endgroup$ – Xi'an Nov 8 '14 at 8:25
  • $\begingroup$ @Xi'an So the reason I want to do is is as I said finding the distribution of $|X+Y|^2$ doesnt have a nice closed for where as $|X|^2+|Y|^2$ does if $X,Y$ are $Nakagami(m,1)$, does that make sense ? $\endgroup$ – Tyrone Nov 8 '14 at 16:26

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