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Let $X$ and $Y$ be independant exponential random variables with parameter $\beta$ = 1. Let $U = X+Y$ and $V = -\log(X/(X+Y))$ Are $U$ and $V$ independant? There's similar questions I found on here but $V$ is $X/(X+Y)$, I don't understand whether the $-\log(\cdot)$ can make them dependant.

For the joint PDF I got $$ f_{X,Y}(x,y) = \exp(-(x+y)) = \exp(-u)$$ and for the jacobian I found $$u\exp(-v), $$ which led to $$ f_{U,V}(u,v) = u\exp(-(u+v)). $$ But I don't know how to proceed.

Edit: I realise you can find $f_U(u)$ and $f_V(v)$ by integrating over $v$ and $u$ respectively, but is it sufficient to show that $U$ and $V$ are independant if the joint pdf can be written as a product of their pdfs?

Aaron Hendrickson
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tsasinc
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1 Answers1

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You are already finished because the joint PDF factorises:

$$f_{U,V}(u,v) = u\exp(-(u+v)) = [u\exp(-u)][\exp(-v)] \equiv f_U(u) f_V(v)$$

This means that $U$ and $V$ are independent.

Answer to your edit: Yes, that suffices. (See here for a more detailed proof.)

Andreas Tsevas
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