Conditional expectation of $X\sim U[0,2]$ given $\min(X,t)$, where $t\in [0,2]$. I want to determine the conditional expectation of $X\sim U[0,2]$ given $\min(X,t)$, where $t\in [0,2]$. Using the definition $$\mathbb E[X|\min(X,t)] = \frac{\mathbb E[X\chi_{\min(X,t)}]}{\mathbb P(\min(X,t))}$$
does not look that inviting to me. I am unsure if I should consider the probability for a specific value $\mathbb P(\min(X,t) = x)$ for $x\in [t,2]$ if the minimum is larger than $t$ or the probability that $\mathbb P(\min(X,t) \ne t)$
Intuitively I would say that since we know $t$ if $\min(X,t)$ is not $t$, it has to be $X$, and thus the conditional expectation has to be $\min(X,t) = X$ which seems tautological. If $\min(X,t)$ we know that $X>t$ and because $X\sim U[0,2]$ it follows that $X$ is uniformly distributed between $t$ and $2$. The conditional expectation, in that case, should be $\frac{2+t}{2}.$
Is this correct? If so, how does this follow from the definition given above?
 A: Note that there is no such thing as $P[\min[X,t]]$ since $\min[X,t]$ is a random variable, not an event.

If we fix $t \in [0,2]$ and define $Y=\min[X,t]$ then $E[X|Y]$ is required to be a Borel measurable function of $Y$. One version of the conditional expectation is $$E[X|Y] = Y1_{\{Y<t\}} + \left(\frac{2+t}{2}\right)1_{\{Y\geq t\}} \quad \mbox{[Equation *]}$$ and any other version is the same as the above almost surely. Here I am using indicator functions of the type
$$ 1_{\{Y<t\}} = \left\{\begin{array}{cc}
1 & \mbox{if $Y<t$} \\
0 & \mbox{else}\end{array}\right.  $$
It is not formally correct to say
$$E[X|Y]=X1_{\{X\leq t\}} + \left(\frac{2+t}{2}\right)1_{\{X>t\}} \quad [\mbox{Incorrect equation}]$$
since, even though the right-hand-side is a random variable that is the same as the right-hand-side of equation (*) almost surely, it is not a function of $Y$ alone: What would be its value if $Y=t$?

Formal definitions of $E[X|Y]$ can be found in many places. The formal definition requires $E[X|Y]$ to be a Borel function of $Y$, and also some other requirements.
Informally, you can compute $E[X|Y]$ by using $E[X|Y]=g(Y)$ where $g(y)$ is a Borel measurable function that satisfies $g(y)=E[X|Y=y]$. This can usually be computed by considering either discrete or continuous cases for each relevant $y$.  In the above example, for $y=t$ we have $P[Y=t]>0$ so we compute $E[X|Y=t]$ using a standard conditional expectation given an event of positive probability:
$$ E[X|Y=t] = \frac{E[X1_{\{Y=t\}}]}{P[Y=t]}$$
which is consistent with the alternative computation $E[X|Y=t]=\int_0^2x f_{X|Y=t}(x)dx$.
If $y<t$ then $P[Y=y]=0$ and we use $f_{X|Y}(x|y)=\delta(x-y)$ to obtain $E[X|Y=y]=y$. You can verify the result (given in Equation (*)) is indeed a Borel measurable function of $Y$ and satisfies the other requirements for a conditional expectation given $Y$.
A: The conditional expectation is a random variable: essentially, $\mathbb{ E}[X|Y]$ is the random variable that takes the value "expectation of $X$ when $Y=y$" with probability $\mathbb{P}(Y=y)$. With probability $\frac{2-t}{2}$, $X$ is greater than $t$ so with this probability the random variable takes the value $\frac{2+t}{2}$ (your argument for the expectation in this stuation is correct). Then, from the case where $X$ is less than $t$, the random variable is equally likely to take every value in $[0, t]$ (is uniformly distributed on this interval), but with total probability $\frac{t}{2}$ (so with density $\frac{1}{2}$).
