Suppose $X$ is a random variable with exponential PDF $f_X(x)=e^{-x}$ with $\lambda=1$.
Also suppose $$U_1=\min(X,t) \quad \text{ and } \quad U_2 = \max(X,t) $$ for $t>0$. How do I compute $E[X|U_i]$ for $i=1,2$?
Here is how I am attempting this problem: $E[X|U_1] = E[X|\min(X,t)]$. Also, from conditional expectations, I know
$$E[X|U_1] = \int_{-\infty}^\infty xf_{X|U_1}(x|u_1)\,dx$$
I also know from Law of Iterated Expectation $$E[X] = E[ E[ X|U]]$$ Finally, I also know $$E[X] = \int_{-\infty}^\infty xf_{X}(x)\,dx = \int_{0}^\infty x e^{-x}dx = 1$$
Therefore $E[ E[ X|U]] = 1$.
After, after this point I am lost. I'm not sure where I am going with this.
Best Answer
Firstly observe that $$U_1=\begin{cases} X, & \text{ if } X<t, \\\\ t& \text{ if } X\ge t\end{cases}$$ or equivalently $U_1=X\cdot1_{\{X<t\}}+t\cdot1_{\{X\ge t\}}$. Now it is s traightforward that $$E[X|U_1=x<t]=x$$ and that $$E[X|U_1=t]=E[X|X>t]$$ But since $$f_{X|X>t}(x|X>t)=\frac{f_X(x)}{P(X>t)}=\frac{e^{-x}}{e^{-t}}=e^{t-x}$$ for all $x>t$, you have that $$E[X|U_1=t]=E[X|X>t]=\int_{t}^{\infty}xe^{t-x}dx=\int_{0}^{\infty}(t+u)e^{-u}du=t+1$$ Hence, $$E[X|U_1]=U_1\cdot 1_{\{U_1<t\}}+(t+1)\cdot 1_{\{U_1\ge t\}}$$ and similarly for $E[X|U_2]$.
Note that for any given $t>0$ you have that $$P(U_1=t)=P(X\ge t)=e^{-t}$$ and therefore $$f_{U_1}(u)=\begin{cases}f_X(u), &u<t\\\\e^{-t}, &u = t\end{cases}$$ Hence, by the law of total expectation $$\begin{align*}E[E[X|U_1]]&=\int_{0}^{t}E[X|U_1=x]f_{U_1}(x)dx+E[X|U_1=t]P(U_1=t)\\\\&=\int_{0}^{t}xe^{-x}dx+(t+1)e^{-t}=1\\\\&=E[X]\end{align*}$$