# Does finding this conditional expectation boil down to finding a conditional probability?

For a random variable $T \geq 0$ with distribution function $F$, a real number $t > 0$ and $\mathcal{G}= \{ \emptyset, \{ T > t \}, \, \{ T \leq t \}, \, \Omega \}$, I need to evaluate the conditional expectation of $E|T-t||\mathcal{G}$, where $\mathcal{G}$ is my sub-$\sigma$-algebra.

Now, I asked somebody how to do this, and they told me that because the $\sigma$-algebra is so small, if I want to find this conditional expectation, all I have to do is find $P(T\leq x| T>t)$, which seems very strange to me. What is the reasoning behind this.

This person then said to apply this formula and then express everything in terms of $F$, but I don't know what the distribution is in this case, do I? So then, how would I go about doing this?

I am extremely confused...

I thank you in advance for your time and patience in helping me to become un-confused!

• Can you give what $F$ is? Or the answer should be given in terms of $F$ without specifying what $F$ is? – Shashi Dec 21 '17 at 18:19
• @Shashi the answer should be given in terms of $F$ wihtout specifying what $F$ is. .Can you do that? – ALannister Dec 21 '17 at 18:24
• One more question is $T$ a continuous random variable? I don't think it affects the answer much, but it is nice to know – Shashi Dec 21 '17 at 18:29
• @Shashi again, that is unknown. I gave all the information I had in my question above. – ALannister Dec 21 '17 at 18:31

## 1 Answer


It is given that $\mathcal G=\{\emptyset, A,A^c,\Omega\}$, where $A=\{T>t\}$ and to make it a little bit interesting let us assume $\PM(A)\in(0,1)$ otherwise this exercise is boring $(\star)$. One knows that $Y:=\E[ |T-t|\mid \mathcal G]$ is the only (up to a null set) $\mathcal G$-measurable function that satisfies: \begin{align}\tag{1} \int_B |T-t|\,d\PM = \int_B Y\,d\PM ,\hspace{15pt} \forall_{ B\in\mathcal G} \end{align} I claim the following: \begin{align} Y(\omega)= \begin{cases} \E[\mathbf{1}_A|T-t|](\PM(A))^{-1} & \text{ if } & \omega\in A\\ \E[\mathbf{1}_{A^c}|T-t|](\PM(A^c))^{-1} & \text{ if } & \omega\in A^c\\ \end{cases} \end{align} You may verify that this indeed satisfies $(1)$.

We have a problem now, because we did not put everything in terms of $F$. Let's do it now. We have: \begin{align} \E[\mathbf{1}_A |T-t|](\PM(A))^{-1}&\stackrel{?}{=}\E[\mathbf{1}_AT](\PM(A))^{-1}-\E[\mathbf{1}_At](\PM(A))^{-1}\\ &=\E[\mathbf{1}_AT](\PM(A))^{-1}-t\\ &=\frac{\E[\mathbf{1}_AT]}{1-F(t)}-t \end{align} And you think about the question mark. Similarly we get: \begin{align} \E[\mathbf{1}_{A^c}| T-t|](\PM(A^c))^{-1}=t-\frac{\E[\mathbf{1}_{A^c}T]}{F(t)} \end{align} Putting these things together yields: \begin{align} \E[|T-t|\mid \mathcal G] = \mathbf{1}_{A^c}\left(t-\frac{\E[\mathbf{1}_{A^c}T]}{F(t)}\right)+\mathbf{1}_{A}\left(\frac{\E[\mathbf{1}_AT]}{1-F(t)}-t \right) \end{align} Do you now understand why we assumed $\PM(A)\in(0,1)$ at ($\star$)? What would change?