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Math Help - showing two distribution functions identical

  1. #1
    Senior Member Sambit's Avatar
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    showing two distribution functions identical

    Let X_1,X_2 be two i.i.d. Exp(\lambda) random variables and R is a Ber(\frac{1}{2}) variable. How to show that the distribution functions of X_1-X_2 and RX_1-(1-R)X_2 are identical?
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  2. #2
    Moo
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    Hello,

    Nice problem

    Let's call Z the rv 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)

    \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...

    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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