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Let $X$ be a geometric random number, and $Y$ be an exponential random number. Then PDF of $X$ will be \begin{equation} f_X(x)=p(1-p)^x \end{equation} and \begin{equation} f_Y(y)=\frac{1}{t}\exp\left(-\frac{y}{t}\right). \end{equation} Let $Z=XY$, then what is the PDF of a new random number? Thank you.

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The best way to do this via MGF. Let $\phi$ be the MGF of $X$

$Ee^{uXY}=E[E[e^{uXY}|X]]=E[E[e^{uxY}]|_{x=X}]=E[(\frac{1/t}{1/t-u})^Y]=\phi(\log(\frac{1/t}{1/t-u}))=\frac{p/t}{p/t-u}$

that means the distribution of $XY$ is ? with parameter $p/t$

Fill in a few gaps.

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  • $\begingroup$ I may need to read about MGF, but the answer is similar to what I expected. Thank you very much. $\endgroup$ – user126830 Feb 10 '14 at 16:25

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