# Are the random variables $T$ and $U$ independent?

Let $X$ and $Y$ both be independent random variables distributed $Exponential(\lambda)$. Let $T = X + Y$ and $U = \frac{X}{X+Y}$. Are $T$ and $U$ independent?

Using Jacobians, I got that the joint probability density function (PDF) of $T$ and $U$ to be $f_{T, U}(t, u) = \lambda^2 e^{-\lambda t}t$. I did this to check whether the PDFs of $T$ and $U$ factors into the joint PDF.

I am not sure how to proceed from here though. How do I find the PDFs of $T$ and $U$?

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Hint: If a joint pdf $f_{X,Y}(x,y)$ factors into $g(x)h(y)$ where $g(x)$ and $h(y)$ both are nonnegative functions (or both are nonpositive functions!) then there is a constant $c \neq 0$ such that $f_X(x) = cg(x)$ and $f_Y(y)=c^{-1}h(y)$. Try and prove this first (it is not hard). – Dilip Sarwate Dec 2 '12 at 22:08
You should also be careful to state the domain on which that formula for the PDF is valid. It certainly is not true when $t$ or $u$ is negative. It's not a problem in this case, but the factorization result @DilipSarwate stated would not be valid if the domain was not a product of intervals. – Robert Israel Dec 2 '12 at 22:12

The PDF is not what you wrote but $f_{U,T}:(u,t)\mapsto\lambda^2t\mathrm e^{-\lambda t}\,\mathbf 1_{t\geqslant0}\,\mathbf 1_{0\leqslant u\leqslant1}$. (Yes, the only difference is the indicator functions, and they make all the difference in the world.) From here, your task is to write $f_{U,T}$ as $f_{U,T}(u,t)=f_U(u)f_T(t)$ for some PDFs $f_U$ and $f_T$... which I am sure you can do.
Thanks! By indicator functions, you mean boundaries? Also, I'm a bit confused about writing the PDFs as 2 products because (1) $u$ isn't even in the joinf PDF and (2) how do I find the PDF of $T$? – David Faux Dec 3 '12 at 0:11
For every property $P$, $\mathbf 1_P$ is $1$ if $P$ holds and $0$ otherwise. // The variable $u$ is definitely in the joint PDF since $u$ enters the last indicator function. // As always, the PDF of $T$ at $t$ is the integral of $f_{U,T}(\ ,t)$. – Did Dec 3 '12 at 7:12
Thank you! So would the PDF of $T$ at $t$ be $\lambda^2 e^{-\lambda t}t(1) - \lambda^2 e^{-\lambda t}t(0)$ after integrating the PDF over $u$? – David Faux Dec 3 '12 at 7:23
It is. And the PDF of $U$ is $\mathbf 1_{[0,1]}$. In fact, since $f_{U,T}(u,t)=g(u)h(t)$ for some functions $g$ and $h$, one knows that $f_U=cg$ and $f_T=h/c$ for some nonzero constant $c$ (as @Dilip already explained). – Did Dec 3 '12 at 7:29