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!


2 Answers 2


Here is a short proof which makes use of symmetry.

Let $A\in \sigma(Y)$ and write $A=Y^{-1}(B)$ for some Borel set $B$, so that $1_A=1_{Y\in B}$.

Note that $1-U\stackrel{d}{=}U$ and let $g:u\mapsto 1_{\min(1-u,u)\in B}\cdot u$, so that $g(1-U)\stackrel{d}{=}g(U)$, which rewrites $$1_{Y\in B}(1-U) \stackrel{d}{=} 1_{Y\in B}U.$$

Consequently, $$\begin{align}E[1_A U] &= E[1_{Y\in B}U]\\ &= \frac 12 E[1_{Y\in B}(1-U)] + \frac 12 E[1_{Y\in B}U] \\ &= \frac 12 E[1_{Y\in B}(1-U+U)] \\ &= E[1_{Y\in B}\frac 12]\\ &= E[1_A \frac 12]. \end{align} $$

Hence $E[U|Y] = \frac 12.$

  • $\begingroup$ Thanks for this! This is a very nice proof. I’m just wondering whether my proof works in general. Or there are some arguments that need to be finessed? $\endgroup$ Apr 22, 2021 at 16:14

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


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