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Let the conditional pdf of $X$,given $Y=y$ be given by $f(x|y)=e^{y-x} , x>y$

and let $Y$ have the pdf $g(y)=\lambda{e^{-\lambda y}},y>0,\lambda>0,\lambda \neq 1$ We need to find the marginal pdf of $X$.

Of course $f_{X}(x)=\int_{0}^{x}f(x|y)g(y) dy= \int_{0}^{x} \lambda e^{(1-\lambda)y-x} dy= \frac{\lambda}{1-\lambda} e^{-x}(e^{x-\lambda x}-1)$. Am I correct?

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Yes, your calculations are correct.

Observe that in general for two random variables $X$ and $Y$ \begin{equation} p(X) = \int p(X,Y) dY = \int p(Y|X)p(X)dY = \int p(X|Y) p(Y) dY \end{equation}

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