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Is there any special technique to deal with the distribution of sum of two random variables where they are not independent?

For example I have concluded that if $X =_p W$ and $Y=_pZ$ ($=_p$ means having same distribution) then these two sum must be equal

$$\ \int_{t}P({X+Y< t<W+Z}) = \int_t P({W+Z< t< X+Y}) . $$

But I don't know how to do it technically. It seems to be true by intuition!

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What do you mean by $=_p$? – martini Nov 2 '12 at 10:23
means having same distributions – peanut Nov 3 '12 at 4:34
It would be interesting to know how you have concluded that .../... these two probabilities must be equal, since they are not always equal. – Did Nov 3 '12 at 10:38
Actually my conclusion was $$\ \int_t P({X+Y< t<W+Z}) = \int_t P({W+Z< t< X+Y}) . $$ and I thought that maybe the integrand are equal ( as a stronger guess!) – peanut Nov 3 '12 at 10:38
Briefly put, the pointwise version fails (as demonstrated by @martini) but the integrated version holds (see my answer). – Did Nov 3 '12 at 10:59

On $\Omega = [0,1]$ with the Lebesgue measure, let $X(\omega) = W(\omega) = Z(\omega)=\omega$ and $Y(\omega) = 1- \omega$. Then all four variables are uniformly $[0,1]$-distributed. We have, that $X+Y$ is constant, and $W+Z$ is uniformly $[0,2]$-distributed. No, for say $t= \frac 12$ we have \[ P\left(X+Y < \frac 12 < W+Z \right) = P\left(1 < \frac 12 < W+Z \right) = 0 \] and \[ P\left(W+Z < \frac 12 < X+Y \right) = P\left(W+Z < \frac 12 < 1 \right) = P\left[0,\frac 14\right) = \frac 14 \]

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Thanks but my fault! The main phrase was the integral of both sides on $t$ from $-/inf$ to $+/inf$. I have updated the right one – peanut Nov 3 '12 at 10:42

Considering $U=X+Y$ and $V=W+Z$, the LHS is $$ \int_{-\infty}^{+\infty}\mathbb P(U\lt t\lt V)\,\mathrm dt=\mathbb E((V-U)^+), $$ and the RHS is $$ \int_{-\infty}^{+\infty}\mathbb P(V\lt t\lt U)\,\mathrm dt=\mathbb E((U-V)^+). $$ The LHS and the RHS coincide as soon as $\mathbb E(U)=\mathbb E(V)$ since, for every integrable random variable $R$, $\mathbb E(R)=\mathbb E(R^+)-\mathbb E((-R)^+)$. Finally, $X\stackrel{(d)}{=}W$ and $Y\stackrel{(d)}{=}Z$ hence, if all these random variables are integrable, then $\mathbb E(X)=\mathbb E(W)$ and $\mathbb E(Y)=\mathbb E(Z)$, which implies $\mathbb E(U)=\mathbb E(V)$, hence we are done.

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I actually wanted to use this to conclude that E[X+Y]=E[Z+W] ( I meant that I was trying to use the main question in order to prove something else using your note about their expectation!) – peanut Nov 3 '12 at 10:57
What if they are not integrable but the sums are? This is actually my end – peanut Nov 3 '12 at 11:01
Please stop this. You should state once and for all the hypotheses you are interested in (which are not very clear, even now), and not modify them afterwards. If you are interested in a different question, post it as a separate, carefully written, post. – Did Nov 3 '12 at 11:06

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