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If $X,Y,Z$ are independent random variables having identical density functions $f(x) = e^{-x} $ , $0 < x < \infty$ , derive the joint probability density of the vector $(U,V,W)$ , where

\begin{align} U &= X + Y \\ V &= X + Z \\ W &= Y + Z \end{align}

I'm not sure on how to do this but, if were to find the joint probability density... I would do moment generate function for each $U,$ $V$ and $W$ and then multiple those.

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  • $\begingroup$ If you knew that $U,$ $V$ and $W$ were already independent, then you would be able to use the method you want, but they don't seem independent at first glance (and they aren't, look at my solution below). $\endgroup$ – Will M. Dec 18 '16 at 0:36
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One idea is to use the change of variables theorem. Consider the random vector $(X,Y,Z)$ and notice that: $$f_{(X,Y,Z)}(x,y,z) = e^{-(x+y+z)} \mathbf{1}_{\mathbb{R}_+^3}(x,y,z).$$ Define the transformation $$T:(x,y,z) \mapsto (x + y, x + z, y + z)$$ which is linear and possesses the inverse $$S:(u,v,w) \mapsto \left(\dfrac{u+v-w}{2},\dfrac{u - v + w}{2}, \dfrac{-u+v+w}{2} \right).$$ By the change of variables theorem, the random vector $(U,V,W) = T(X,Y,Z)$ has density $$f_{(U,V,W)}(u,v,w) = f_{(X,Y,Z)}(S(u,v,w)) |\det \mathbf{D}S(u,v,w)|.$$ Now, $\det \mathbf{D}S(u,v,w) = \det S = -\dfrac{1}{2},$ so $\begin{align} f_{(U,V,W)}(u,v,w) &=\dfrac{1}{2} f_{(X,Y,Z)}\left(\dfrac{u+v-w}{2},\dfrac{u - v + w}{2}, \dfrac{-u+v+w}{2} \right) \\ &= \dfrac{1}{2} e^{-\frac{u+v+w}{2}} \mathbf{1}_{\mathbb{R}_+^3} \left(\dfrac{u+v-w}{2},\dfrac{u - v + w}{2}, \dfrac{-u+v+w}{2} \right) \\ &= \dfrac{1}{2} e^{-\frac{u+v+w}{2}} \mathbf{1}_{\mathrm{G}}(u,v,w), \end{align}$

where $\mathrm{G}$ is the set of $(u,v,w)$ such that the three relations $u + v \geq w,$ $u + w \geq v$ and $v + w \geq u$ hold. (Naturally, these three relations imply $u,v,w \geq 0$).

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