Exercice about conditional expectation

Question

Let $X$ and $Y$ independemt continuous uniform random variables on $[0,1]$. If $0<t<1$ and $A=\{\min\{X,Y\}\leq t\}$, evaluate:

\begin{equation*} \int_{A}{\mathbb{E}(\max\{X,Y\}|\min\{X,Y\})\,d\mathbb{P}}. \end{equation*}

Firstly, as $X$ and $Y$ are random variables over a measurable space $(\Omega,\Sigma)$ and, $\max$ and $\min$ are continuous functions, then $\max\{X,Y\}$ and $\min\{X,Y\}$ are random variables. Let's see if the random variable $\max\{X,Y\}\in\mathcal{L}^1_{\mathbb{P}}(\Omega,\mathbb{R})$:

$$\int_{\Omega}{|\max\{X,Y\}|\,d\mathbb{P}}\leq\int_{\Omega}{(|X|+|Y|)\,d\mathbb{P}}\leq \int_{\Omega}{(1+1)\,d\mathbb{P}}=2\cdot\mathbb{P}(\Omega)=2\cdot 1=2<+\infty.$$

By definition of conditional expectation, we have the following:

$$\int_{A}{\mathbb{E}(\max\{X,Y\}|\min\{X,Y\})\,d\mathbb{P}}=\int_{A}{\max\{X,Y\}\,d\mathbb{P}}=\int_{A\cap\{X\leq Y\}}{Y\,d\mathbb{P}}+\int_{A\cap\{X>Y\}}{X\,d\mathbb{P}}=\int_{\{X\leq\min\{Y,t\}\}}{Y\,d\mathbb{P}}+\int_{\{Y\leq\min\{X,t\}\}}{X\,d\mathbb{P}}.$$

I don't know what to do with those integrals or even this is the right way to do this exercice. Does anyone have any advice?

Thanks in advance.

Answer 1

The set $A$ is measurable with respect the $\sigma$-algebra generated by $\min\{X,Y\}$ hence by definition of conditional expectation, \begin{equation*} \int_{A}{\mathbb{E}(\max\{X,Y\}|\min\{X,Y\})\,d\mathbb{P}}=\int_{A}{ \max\{X,Y\} \,d\mathbb{P}}. \end{equation*} In order to compute this expectation, we can show and use the fact that the vector $\left(\min(X,Y),\max(X,Y)\right)$ has density $f(u,v)=2\mathbf{1}_{0<u<v<1}$. We thus derive that \begin{equation*} \int_{A}{\mathbb{E}(\max\{X,Y\}|\min\{X,Y\})\,d\mathbb{P}}=2\int\int v\mathbf{1}_{0<u<v<1}\mathbf{1}_{u\leqslant t}dudv. \end{equation*} The integral over $u$ is simply $\min\{v,t\}$ hence \begin{align} \int_{A}{\mathbb{E}(\max\{X,Y\}|\min\{X,Y\})\,d\mathbb{P}}&=2\int_0^1v\min\{v,t\}dv\\ &=2\int_0^tv^2dv+2\int_t^1vtdv\\ &=\frac{2}3t^3+t(1-t^2)\\ &=t-t^3/3. \end{align}