Calculate $\mathbb E(Y\mid\min(Y,t))$ 
How to calculate $\mathbb E(Y\mid\min(Y,t))$, if $P(Y>t)=e^{-t}$ for $t>0$ ?

Do I have to use Bayes ? I mean the expectation must give a function of $\min(Y,t)$ and $\mathbb E(\min(Y,t)\mid Y)=Y\mathbf 1_{Y\le t}+t\mathbf 1_{t\le Y}$ or is it completely wrong ?
 A: So, call $\min(Y,t) = y$.
If $y < t$, this implies that $Y = y$, so in this case $E[Y \mid \min(Y,t) = y] = y$
If instead $y \ge t$, you can only conclude that $Y \ge t$; so you get 
$$E[Y \mid \min(Y,t) = t] = E_Q[Y]$$
where $Q$ is defined as $\displaystyle Q(A) = P(A \mid Y \ge t) = \frac{P(A \cap \{Y \ge t\})}{P(Y \ge t)} $. 
Therefore 
$$E_Q[Y] = \frac 1{\int_t^\infty e^{-s}ds} \int_t^\infty se^{-s}ds = \frac 1{e^{-t}}\int_t^\infty se^{-s}ds = t + 1$$
To sum up you have 
$$E[Y \mid \min(Y,t)] = \begin{cases} \min(Y,t) & \text{if $\min(Y,t)< t$} \\ t + 1 & \text{otherwise}\end{cases}$$
or $$E[Y \mid \min(Y,t)] = \min(Y,t)1_{\min(Y,t) < t} + (t+1)1_{\min(Y,t) \ge t}$$
if you want you could also say that since $\min(Y,t) < t \implies \min(Y,t) = Y$, then 
$$E[Y \mid \min(Y,t)] = Y1_{Y < t} + (t+1)1_{Y \ge t}$$
which is a little weird though because you are given only $\min(Y,t)$, not $Y$. Clearly it's the same thing but you should note that! 
Another way of writing it, probably more elegant, is
$$E[Y \mid \min(Y,t) ] = \min(Y, t) + 1_{\min(Y,t) \ge t} = \min(Y, t) + 1_{\min(Y,t) = t}$$
It really depends on what you're after in any case 
