# Find the conditional expectation $E(U\mid \min(U,1−U))$ where $U∼U[0,1]$

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!

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

• 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? Apr 22, 2021 at 16:14

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