Find $\mathbb E(X\mid Y)$ where $X$ is a random variable such that $P(X > t) =e^{−t}$ with $t >0$ and $Y=\min(X,t)$.
I have not done a problem before where we condition on the minimum or maximum value. I thought about breaking it up into cases as follows:
$$\mathbb E(X\mid X>t)\mathbb P(X>t)+\mathbb E(X\mid X<t)\mathbb P(X<t)$$
but this gives $\mathbb E(X)$ rather than $\mathbb E(X\mid Y)$. I tried thinking of this conceptually but what makes this problem tricky is that (if I'm understanding the problem correctly) we are not told whether $X<t$ or $X>t$ but rather we're just given the smaller of the two values. If we are (unknowingly) given the information that $Y=\min\{X,t\}=t$ with probability $e^{-t}$ then by the memoryless property, $\mathbb E(S\mid t)=t+1$ and if we are (unknowingly) given the information that $Y=\min\{X,t\}=X$ with probability $1-e^{-t}$ then $\mathbb E(X\mid X=x)=x$ so that
$$\mathbb E(X\mid Y)=(t+1)e^{-t}+x\left(1-e^{-t}\right)$$
but here I'm basically doing the same thing as above so I'm not sure how to correctly think about this type of problem. Any help would be appreciated!
First observe that $Y=\min\{X,t\}$ is supported on $(0,t]$ and that $E(X|Y=y)$ is necessarily a function of $y$. On one hand, if $y\in (0,t)$, we have $\{Y=y\}=\{X=y\}$ giving us $$E(X|Y=y)=E(X|X=y)=y$$ On the other hand, if $y=t$, then $\{Y=t\}=\{X\geq t\}$ so in this case $$E(X|Y=y)=E(X|X \geq t)$$ The RHS evaluates to $$\int_{t}^{\infty}\frac{xf_X(x)}{P(X\geq t)}dx=t+1$$ Here we're using the fact that $X\sim \text{Exponential}(1)$.