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Thread: showing two distribution functions identical

  1. #1
    Senior Member Sambit's Avatar
    Oct 2010

    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?
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  2. #2
    Moo is offline
    A Cute Angle Moo's Avatar
    Mar 2008
    P(I'm here)=1/3, P(I'm there)=t+1/3

    Nice problem

    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.

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