# Proof of truncated expectation

I want to prove that $E[T \times I\{T>t,K<k\}]=E[T|T>t,K<k]\Pr[T>t,K<k]$, where $T$ and $K$ are continuous random variables and $I{}$ is the indicator function.

I was trying to prove this result, as follows: $$E[T \times I\{T>t,K<k\}]\\[8pt] =E[I\{T>t,K<k\}E[T|I\{T>t,K<k\}]]\\[8pt] =E[I\{T>t,K<k\}]E[T|I\{T>t,K<k\}]\\[8pt] =\Pr[T>t,K<k]E[T|I\{T>t,K<k\}]\\[8pt] =\Pr[T>t,K<k]E[T|T>t,K<k]$$

I know my proof is wrong (on a previous post -here- it was noted that the last step does not follow).

-
I changed $T\text{ x }I$ to $T\times I$ in two places in this question. –  Michael Hardy Jan 29 '13 at 15:31

## 1 Answer

Let $A=\{T>t,K<k\}$ and $I_A$ denote the indicator function of $A$. Note that on the set $A$ we have that $E[T\mid I_A]$ and $E[T\mid A]$ are identical (according to Ilya's answer), i.e. $$E[T\mid I_A]\cdot I_A=E[T\mid A]\cdot I_A$$ and hence your second line becomes $$E\big[I_A\cdot E[T\mid I_A]\big]=E\big[I_A\cdot E[T\mid A]\big].$$ But $E[T\mid A]$ is just a number and so $$E\big[I_A\cdot E[T\mid A]\big]=E[T\mid A]E[I_A]=E[T\mid A]P(A).$$

-
Thanks for spelling it out! –  EOO Jan 29 '13 at 13:39
You're welcome. –  Stefan Hansen Jan 29 '13 at 13:46