# showing two distribution functions identical

• Jun 11th 2011, 10:35 PM
Sambit
showing two distribution functions identical
Let $\displaystyle X_1,X_2$ be two i.i.d. $\displaystyle Exp(\lambda)$ random variables and $\displaystyle R$ is a $\displaystyle Ber(\frac{1}{2})$ variable. How to show that the distribution functions of $\displaystyle X_1-X_2$ and $\displaystyle RX_1-(1-R)X_2$ are identical?
• Jun 12th 2011, 01:27 AM
Moo
Hello,

Nice problem :p

Let's call Z the rv $\displaystyle RX_1-(1-R)X_2$.

We'll use mgf's and conditional expectations :

(expectations with the rv's in subscript mean that it's the expectation with respect to the distributions of the rv's in the subscript - to avoid any confusion)

\displaystyle \begin{aligned} M_Z(t)&=E_{R,X_1,X_2} \left[e^{tRX_1} e^{-t(1-R)X_2}\right] \\ &=E_R \left[E_{X_1}\left[e^{tRX_1}|R\right] E_{X_2}\left[e^{-t(1-R)X_2}|R\right]\right] \\ &=E_R\left[\frac{1}{1-\frac{tR}{\lambda}} \cdot \frac{1}{1+\frac{t(1-R)}{\lambda}}\right] \\ &=\frac 12 \left(1\cdot \frac{1}{1+\frac{t}{\lambda}}\right)+\frac 12 \left(\frac{1}{1-\frac{t}{\lambda}}\cdot 1\right) \\ &=\frac{1}{\left(1+\frac{t}{\lambda}\right) \left(1-\frac{t}{\lambda}\right)}\end{aligned}

And then comparing to the other one...

$\displaystyle M_{X_1-X_2}(t)=E\left[e^{tX_1}\right]E\left[e^{-tX_2}\right]=\frac{1}{1-\frac{t}{\lambda}}\cdot\frac{1}{1+\frac{t}{\lambda }}$

Same mgf implies same distribution. (Happy)

P.S. : being of the lazy type, some steps have been put in silent mode :D