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Find $\mathbb{E}(h(X) \mid U)$ where $h$ is measurable, $X, Y$ are independent and $U = \max(X,Y)$. $X$, $Y$ follow both an exponential distribution with parameter $\lambda = 1$.

I couldn't find a way to solve this problem as $(X,U)$ doesn't have a probability density function in $\mathbb{R^2}$.

Any other ways to proceed ?

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    $\begingroup$ See here for a related problem (although I am not sure there is a nice closed form for arbitrary measurable $h$): math.stackexchange.com/questions/1952493/… $\endgroup$ – Math1000 Feb 5 at 19:08
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    $\begingroup$ There are two equally likely possibilities; either $X>Y$, in which case $h(X)=h(U)$, or $X<Y$, in which case $X$ has a conditional density given by $f_{X|U=u}(x)=\frac1{P(X\le u)}e^{-x}$ for $0\le x\le u$. $\endgroup$ – Mike Earnest Feb 5 at 21:51
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Clearly, $$ E[ h(X) g(U)] = \underset{=(*)}{\underbrace{E[h(X)g(X),X>Y]}}+ \underset{=(**)}{\underbrace{E[h(X) g(Y),X<Y]}}.$$ Now \begin{align*} &(*) = \int_0^\infty g(u) h(u) e^{-u} (1-e^{-u})du\mbox{ and }\\ &(**) = \int_0^\infty g(u) e^{-u} H(u)du, \end{align*} where $$\boxed{H(u) = \int_0^u e^{-x} h(x) dx}$$

Thus,

$$ E[h(X) g(U) ] =\int_0^\infty g(u) e^{-u} [ h(u)(1-e^{-u}) + H(u)] du.$$

Pause and consider the particular case $h\equiv 1$, in which $H= 1-e^{-u}$. With this choice, the equation above shows that $U$ has density $f_U(u) = 2e^{-u} (1-e^{-u})$. Armed with this information, let's return to the general case:

\begin{align*} E [ h(X) g(U) ] &= \int_0^\infty g(u) f_U(u)\times \frac 12 \left (h (u) + \frac{H(u)}{1-e^{-u}}\right) du\\ & = E[ g(U) \frac 12 \left(h(U) + \frac{H(U)}{1-e^{-U}}\right)] \end{align*}

The answer is therefore

$$ \boxed{E[ h(X) | U ] = \frac 12 \left( h(U) + \frac{H(U)}{1-e^{-U}}\right)}$$

Note that the method can be used for more general cases.

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  • $\begingroup$ perfect ! that was very clear ! $\endgroup$ – rapidracim Feb 6 at 12:37
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As an alternative to the method mentioned in the @Math1000's comment (and a related post) you may compute this conditional expectation as follows (assuming that $h(X)$ is integrable): $$ \mathsf{E}(h(X)\mid U=u)=\lim_{\delta\downarrow 0}\frac{\mathsf{E}[h(X)1\{U\in [u,u+\delta)\}]}{\mathsf{P}(U\in [u,u+\delta))}. $$

For example, assuming that $h$ is continuous, $$ \mathsf{E}(h(X)\mid U=u)=\frac{1}{2}h(u)+\frac{1}{2}\int_0^u \frac{h(x)e^{-x}}{1-e^{-u}}dx $$ because \begin{align} \mathsf{E}[h(X)1\{U\in [u,u+\delta)\}]&=(e^{-u}-e^{-(u+\delta)})\int_0^u h(x)e^{-x}\,dx \\ &\quad+(1-e^{-(u+\delta)})\int_u^{u+\delta}h(x)e^{-x}\,dx, \end{align} and \begin{align} \mathsf{P}(U\in [u,u+\delta))&=(1-e^{-(u+\delta)})^2-(1-e^{-u})^2. \end{align}

The result follows from applying L'Hôpital's rule and taking the limit ($\delta\downarrow 0$).

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  • $\begingroup$ I’m guessing that means $\delta\to0^+$, the right-handed limit? $\endgroup$ – Chase Ryan Taylor Feb 6 at 5:06
  • $\begingroup$ @ChaseRyanTaylor Sure. $\delta\downarrow 0$ means the same. $\endgroup$ – d.k.o. Feb 6 at 5:07
  • $\begingroup$ and $1\{U\in[u,u+\delta)\}$ the indicator function of the interval $[u,u+\delta)$ at $U$? $\endgroup$ – Chase Ryan Taylor Feb 6 at 5:14
  • $\begingroup$ @ChaseRyanTaylor Yes. It's $1_{[u,u+\delta)}(U)$ $\endgroup$ – d.k.o. Feb 6 at 5:32

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