# Conditional Expectation of an Exponential and Uniform Random variable

Let's say X is an exponential random variable with $\theta$ and Y is an uniform random variable over $[0, T]$. How does one calculate $E[(X-Y)|(X>Y)]$?

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Let $X$ denote an exponential random variable and $Y$ any nonnegative random variable independent of $X$. Then, $\mathbb E(X-Y\mid X\gt Y)=\mathbb E(X)$ is independent of the distribution of $Y$.
To show this, call $\theta$ the parameter of the exponential distribution of $X$ and note that $\mathbb E(X-Y\mid X\gt Y)=\frac{N}D$ with $N=\mathbb E(X-Y;X\gt Y)$ and $D=\mathbb P(X\gt Y)$. The independence of $X$ and $Y$ yields $$D=\int_0^{+\infty}\int_y^{+\infty}\mathrm d\mathbb P_X(x)\mathrm d\mathbb P_Y(y)=\int_0^{+\infty}\mathbb P(X\gt y)\mathrm d\mathbb P_Y(y)=\int_0^{+\infty}\mathrm e^{-\theta y}\mathrm d\mathbb P_Y(y),$$ hence $$D=\mathbb P(X\gt Y)=\mathbb E(\mathrm e^{-\theta Y}).$$ Likewise, $$N=\int_0^{+\infty}\int_y^{+\infty}(x-y)\mathrm d\mathbb P_X(x)\mathrm d\mathbb P_Y(y)=\int_0^{+\infty}\mathbb E(X-y;X\gt y)\mathrm d\mathbb P_Y(y).$$ For every fixed $y\geqslant0$, $$\mathbb E(X-y;X\gt y)=\int_y^{+\infty}(x-y)\mathrm d\mathbb P_X(x)=\int_y^{+\infty}\int_y^x\mathrm dz\,\mathrm d\mathbb P_X(x),$$ hence $$\mathbb E(X-y;X\gt y)=\int_y^{+\infty}\int_z^{+\infty}\mathrm d\mathbb P_X(x)\,\mathrm dz=\int_y^{+\infty}\mathbb P(X\geqslant z)\,\mathrm dz,$$ that is, $$\mathbb E(X-y;X\gt y)=\int_y^{+\infty}\mathrm e^{-\theta z}\,\mathrm dz=\theta^{-1}\mathrm e^{-\theta y}.$$ Hence $$N=\mathbb E(X-Y;X\gt Y)=\theta^{-1}\mathbb E(\mathrm e^{-\theta Y}).$$ Finally, $$\mathbb E(X-Y\mid X\gt Y)=\frac{\theta^{-1}\mathbb E(\mathrm e^{-\theta Y})}{\mathbb E(\mathrm e^{-\theta Y})}=\theta^{-1}=\mathbb E(X).$$
Thanks! One would expect that the conditional expectation would be a function of $y$. But I guess the memory-less property of exponential distribution diminishes that? – jay-sun Oct 26 '12 at 23:38
Hint: I presume $X$ and $Y$ are supposed to be independent. Use the "lack of memory" property of the exponential distribution.
It's slightly subtle that the property works for a random variable $Y$ independent of $X$ as well as for a constant. You can prove that by approximating $Y$ by discrete random variables. – Robert Israel Oct 24 '12 at 1:50