Conditional expectation with minimum and exponential r.v. Let Y be an exponential r.v. such that $P(Y>t)=e^{-t}$ for $ t > 0$.
Compute $E[Y|Y \land t]$.
I tried with the following but is my first time that I meet a minimum on the conditional expectation, so probably will be wrong.
I did $E[Y]\mathbf 1_{\{Y \ge t\}} + Y\mathbf 1_{\{Y<t\}}$. I'm quite sure that I can't split in this way. 
How should I solve it?
 A: Well, you're certainly right that $Y=1_{Y\geq t} Y+1_{Y<t} Y.$ Let's argue that the second variable is $\sigma(Y\wedge t)$ measurable. To see this, note that for any $a<b$, we have $$
(1_{Y<t} Y\in[a,b))=(Y\in [a,b\wedge t))=(Y\wedge t\in [a,b\wedge t))\in \sigma(Y\wedge t),$$
which implies the desired, since the half-open intervals generate the Borel $\sigma$-algebra.
Hence, $\mathbb{E}(Y|Y\wedge t)=1_{Y<t} Y+\mathbb{E}(1_{Y\geq t} Y|Y\wedge t)$
Now, for any borel set $B\subseteq (-\infty,t),$ we note that $(Y\wedge t\in B)=(Y\in B)\subseteq (1_{Y\geq t} Y=0),$ so and if $B\subseteq [t,\infty)$, the $(Y\wedge t\in B)=(Y\geq t)=(1_{Y\geq t} Y\neq 0).$ Hence, we get for a general Borel $B$ set that
$$
\mathbb{E}(1_{Y\wedge t\in B}1_{Y_t\geq t} Y)=1_{t\in B} \int_t^{\infty} ye^{-y}\textrm{d}y=\mathbb{E} \left(1_{Y\wedge t\in B} 1_{Y\wedge t=t} \int_t^{\infty} y e^{-y}\textrm{d}y\right),
$$
implying that $\mathbb{E}(1_{Y\geq t}Y|Y\wedge t)=1_{Y\wedge t=t}\int_t^{\infty} y e^{-y}\textrm{d}y=1_{Y\wedge t=t}\mathbb{E}1_{Y\geq t} Y$, with the indicator function written in a way to emphasise that this variable is $\sigma(Y\wedge t)$-measurable.
So yes, your split is correct.
