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Let T and K be dependent continuous random variables, and note the Indicator function as I{.}:

Is it correct to say that $E[T|I\{T>t,K<k\}]=E[T|T>t,K<k]$? Is that a property of the Indicator function?

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No, because the left-most is a random variable, the right-most is a number. The first is the conditional expectation of $T$ conditioned on the sigma-field generated by $\{T>t,K<k\}$. To see their relation, see e.g. this answer – Stefan Hansen Jan 29 '13 at 11:39
@StefanHansen Oh, I missed your comment. Perhaps, it's better if I delete my answer (which happened to be almost identical) and you post your's as an answer. – Ilya Jan 29 '13 at 11:42
@Ilya: The answer is yours, I just gave the appropriate link. Go ahead and undelete :) – Stefan Hansen Jan 29 '13 at 11:49
Thanks! That means I have a problem with something else... Maybe I should open another question? I was asking this question because I know that $E[T \text{ x } I\{T>t,K<k\}]=E[T|T>t,K<k]Pr[T>a,k<b]$ I was trying to prove it by: $E[T \text{ x } I\{T>t,K<k\}]\\ =E[I\{T>t,K<k\}E[T|I\{T>t,K<k\}]]\\ =E[I\{T>t,K<k\}]E[T|I\{T>t,K<k\}]\\ =Pr[T>a,K<k]E[T|I\{T>t,K<k\}] =Pr[T>a,k<b]E[T|T>a,k<b] $.\\ But I see now the last step is not actually true... – EOO Jan 29 '13 at 12:11
Yes, you should post this as a new question :) – Stefan Hansen Jan 29 '13 at 12:25
up vote 2 down vote accepted

For any measurable $A$ you have $\mathsf E[\cdot|1_A] = \mathsf E[\cdot|\sigma(A)]$ where $\sigma(A)$ is given by $$ \sigma(A) = \{\emptyset,\Omega,A,A^c\}. $$ For the relation between this object and the conditional expectation you thought of, see this post.

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