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Thread: Independent exponential random variables

  1. #1
    Senior Member I-Think's Avatar
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    Independent exponential random variables

    If $\displaystyle X_1$ and $\displaystyle X_2$ are independent exponential random variables with parameters $\displaystyle \lambda_{1}$ and $\displaystyle \lambda_{2}$, find the distribution of $\displaystyle Z=\frac{X_1}{X_2}$

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  2. #2
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    Re: Independent exponential random variables

    Quote Originally Posted by I-Think View Post
    If $\displaystyle X_1$ and $\displaystyle X_2$ are independent exponential random variables with parameters $\displaystyle \lambda_{1}$ and $\displaystyle \lambda_{2}$, find the distribution of $\displaystyle Z=\frac{X_1}{X_2}$

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    Quotient of two random variables

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    MHF Contributor chisigma's Avatar
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    Re: Independent exponential random variables

    As in the reference reported by Mr Fantastic if X and Y are independent random variables with p.d.f. $\displaystyle p_{x}(*)$ and $\displaystyle p_{y}(*)$, then the random variable $\displaystyle Z=\frac{X}{Y}$ has p.d.f...

    $\displaystyle p_{z}(z)= \int_{- \infty}^{+ \infty} |y|\ p_{x}(z y)\ p_{y}(y)\ dy$ (1)

    If X and Y are exponential random variables then...

    $\displaystyle p_{x}(x)= \lambda_{1}\ e^{- \lambda_{1}\ x}\ u(x)$ (2)

    $\displaystyle p_{y}(y)= \lambda_{2}\ e^{- \lambda_{2}\ y}\ u(y)$ (3)

    ... where u(*) is the Haeviside Step Function and (1) becomes after some steps...

    $\displaystyle p_{z}(z)= \frac{\lambda_{1}}{\lambda_{2}}\ \frac{1}{(1+\frac{\lambda_{1}}{\lambda_{2}}\ z)^{2}}$ (4)

    Surprisingly simple!...



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