TITLE: Find the conditional expectation $E(U\mid \min(U,1−U))$ where $U∼U[0,1]$
QUESTION [7 upvotes]: Assume that $U$ is a random variable on $(Ω,\mathcal{F},P)$ with $U∼U[0,1]$. Let $Y= \min(U,1−U)$. I am asked to find $E(U\mid Y)$.
I have figured out a proof for this and I want some sanity check.
My train of thought for tackling this:
First, by heuristic: Suppose $Y = y \in [0, 1/2]$. What this mean is that $U = y$ or $U = 1-y$ with equal probability $1/2$. Therefore, the conditional expectation of $U$ given $Y = y$ is then
$$
E(U\mid Y=y) = y/2 + (1-y)/2 = 1/2.
$$
Since $y$ here is arbitrary, it should imply that
$$
E(U\mid Y) = 1/2 \ \text{ a.s}.
$$
Now to prove this by definition, first note that $1/2$ is a constant and measurability w.r.t. $\sigma(Y)$ is trivial.
Secondly, choose $A \in \sigma(Y)$, we then have
$$
\int_A 1/2\,dP  = P(A)/2.
$$
On the other hand, we have
$$
\int_{A}U(\omega)\,dP(\omega) = \int_\Omega U(\omega)\boldsymbol{1}(\omega \in A)\,dP(\omega).
$$
Since $U$ is uniform $(0,1)$ R.V., it has a density of $1$ w.r.t. the Lebesgue measure. Furthermore, there is a $B \subset [0,1]$ s.t. $U^{-1}(B) = A$. Hence, the above integral can be written as
$$
\int_\Omega U(\omega)\boldsymbol{1}(\omega \in A)\,dP(\omega) = \int_{B}u\,du,
$$
where the integration on the RHS is w.r.t. the Lebesgue measure. Now consider the set $B$, it is easy to see that it is an element of the sigma-algebra generated by the following sets:
$$
[0,t]\cup[1-t,1], \ t\in[0,1/2],
$$
since $\sigma(Y)$ is generated by $\sigma(\{Y \leq t, \ t\in[0,1/2]\})$ and $Y \leq t$ is equivalent to $U \leq t$ or $U \geq 1-t$. Now since $[0,t]\cup[1-t,1]$ is symmetric w.r.t. $1/2$, then $B$ should also be a set that is symmetric w.r.t. $1/2$. This implies that
$$
\int_{B}u\,du = \lambda(B)/2 = P(A)/2,
$$
which concludes the proof.
Does this proof work? Thanks a lot!

REPLY [2 votes]: "Single-line" way of computing this conditional expectation. Let $Y=U\wedge(1-U)$. For $y\in(0,1/2)$,
$$
\mathsf{E}[U\mid Y=y]=\lim_{\epsilon\downarrow 0}\frac{\mathsf{E}[U1\{Y\in [y-\epsilon,y+\epsilon]\}]}{\mathsf{P}(Y\in [y-\epsilon,y+\epsilon])}=\lim_{\epsilon\downarrow 0}\frac{2\epsilon}{4\epsilon}=\frac{1}{2}.
$$