Suppose $X$ is a random exponential variable with $\lambda=1$ such that the PDF of $X$ is $f_X(x)=e^{-x}$.
Suppose $Y_1=min(X,t)$
What is the $E[X|Y_1=t]$?
Apparently the answer is $\frac{\int_{t}^{\infty}xe^{-x}dx}{ \int_{t}^{\infty} e^{-x}dx }$ but why ?
The definition of conditional expectation as I understand it is as follows:
$E[X|Y_1=t] = \int_{-\infty}^{\infty}xf_{X|Y1}(x|y_1) = \int_{-\infty}^{\infty}x\frac{f_{X,Y1}(x,y_1)}{f_{Y_1}(y1)}dx$
How do I apply the textbook formula for conditional expectation to get the answer above?
I think first you've got to think about $$ P(Y_1=t). $$
$$ P(Y_1= t)=P(min(X,t)=t)=P(X>t)=\int_t^\infty{e^{-x}}dx $$
And by following your idea, you'll get the answer