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I have two independent random variables say $X$, $Y$. Both of them follow exponential distribution with parameter $λ$ i.e $X\sim λe^{−λx}$ and $Y\sim λe^{−λy}$. I want to find the pdf of $Z=XY$ given $X>c$ i.e $p_{Z/X>c}(z/x>c)$ where $c$ is a positive fixed number.

Can I write it like this: $$p_{Z/X>c}(z/x>c)=\frac{p_{Z}(z)}{P_{X}(X>c)},$$ or is the following formula right $$\int_{c}^{\infty}p_{Y}(z/x)p_{X}(x)dx\ ?$$

Did
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Frank Moses
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    $$p_{Z\mid X>c}(z)=P(X>c)^{-1}\int_{c}^{\infty}p_{Y}(x^{-1}z)p_{X}(x)x^{-1}dx$$ – Did Dec 31 '15 at 09:47
  • Can you please provide a reference? – Frank Moses Dec 31 '15 at 09:51
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    A method, rather: the task is to compute $E(u(Z)\mathbf 1_{X>c})$ for every function $u$, and to reach the formula $$E(u(Z)\mathbf 1_{X>c})=P(X>c)\int_\mathbb Ru(z)g(z)dz$$ valid for every measurable bouned function $u$ since then, one can be sure that $g=p_{Z\mid X>c}$. But the LHS is simply $$E(v(X,Y))=\iint v(x,y)p_X(x)p_Y(y)dxdy$$ for some specific function $v$ depending on $u$, hence one can apply the general approach explained there on a specific case. – Did Dec 31 '15 at 09:55
  • Thanks again Did for your time and great help. I will try to understand it. – Frank Moses Dec 31 '15 at 10:10

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The conditional pdf is: $$\begin{align} f_{XY\mid X>c}(z) & = \frac{\mathrm d \;}{\mathrm d z}\mathsf P(XY\leq z\mid X>c) \\[1ex] & = \frac{\mathrm d \;}{\mathrm d z}\frac{\mathsf P(XY\leq z, X>c)}{\mathsf P(X>c)} \\[1ex] & = \frac{\frac{\mathrm d \;}{\mathrm d z}\int_c^\infty\mathsf P(Y\leq z/x)f_X(x)\operatorname d x}{\mathsf P(X>c)} \\[1ex] & = \frac{\int_c^\infty f_X(x)\frac{\mathrm d \;}{\mathrm d z}\mathsf P(Y\leq z/x)\operatorname d x}{\mathsf P(X>c)} \\[1ex] & = \frac{\int_c^\infty f_X(x)x^{-1}f_Y(z/x)\operatorname d x}{\mathsf P(X>c)} \\[1ex] & = \frac{\int_c^\infty x^{-1}\lambda^2 e^{-\lambda x}e^{-\lambda z/x}\operatorname d x}{e^{-\lambda c}} \\[1ex] & = e^{\lambda c}\lambda^2 \int_c^\infty x^{-1} e^{-\lambda x}e^{-\lambda z/x}\operatorname d x \\[2ex] & \vdots \end{align}$$

Graham Kemp
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