0
$\begingroup$

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 $U$?

  • 1
    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.2012-12-02

1 Answers 1

1

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.

  • 0
    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).2012-12-03