Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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 $U$?

share|cite|improve this question
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
up vote 1 down vote accepted

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.

share|cite|improve this answer
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

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.