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X and Y are independent exponential distributions with mean 1. Find E(X|Y<X)

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Consider the conditional density $$f_{X|A}$$ where A is the event that \(\{Y<X\}\). Find this by first finding the conditional CDF. We have

$$P(X<x|Y<X)= \frac{P(X<x,Y<X)}{P(Y<X)} = \frac{P(Y<X<x)}{P(Y<X)}= 2\int_{0}^{x} \int_{0}^{t}e^{-s}e^{-t}dsdt = \int_{0}^{x}2e^{-t}(1-e^{-t})dt$$ since \(P(Y<X)\) is 1/2 and the joint distribution is the product of marginals. We can then find the density

$$f_{X|A}(x)= \frac{\partial}{\partial x} \int_{0}^{x}2e^{-t}(1-e^{-t})dt = 2e^{-x}(1-e^{-x})$$ by the fundamental theorem of calculus. So we have $$E(X|Y<X)=\int_{0}^{\infty} 2xe^{-x}(1-e^{-x})dx =\frac{3}{2}$$