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$f(x,y) = \exp(-2(x+y))$ , $0<x<y$, and $0$ otherwise.

Find the joint distribution of $X$ and $X+Y$. My attempt consists of letting $ U = X + Y$ and then having $f_x,_u(x,u)=\exp(-2u)$ , $0 < x < u$. Is this correct?

TZakrevskiy
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Stephen
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1 Answers1

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Almost correct (in fact $x< u/2$), but just by a coincidence: the transform $x\mapsto x, y\mapsto x+y$ is a bijection with Jacobian equal to $1$. In general, you should find the inverse transform and divide the resulting density by (the absolute value of) its Jacobian.

zhoraster
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