# Explicit conditional expectation with respect to a $\sigma$-algebra

I have a feeling that this question is supposed to be easy, but I'm having a hard time with it (probably because I'm not familiar with conditional expectations with respect to $\sigma$-algebras):

Let $\Omega = \{(x, y)\in [0, 1]^2 \mid x\geq y\}$ and $P$ the uniform probability in $\Omega$. Define the $\mathcal{B}(\Omega)$-measurable random variables $X_1, X_2:\Omega\to\mathbb{R}$ with

$$X_1(x, y)=x$$ $$X_2(x, y)=y$$

for every $(x, y)\in \Omega$. Find the explicit formula for the conditional expectation $E[X_2\mid \sigma(X_1)]$ in terms of $X_1$ and $X_2$ (where $\sigma(X_1)=\{X_1^{-1}(A)\mid A \in \mathcal{B}(\mathbb{R})\}$).

I only know the formal definition of conditional expectation, but this problem made me realize that I have no clue how to find the explicit formula in concrete examples.

• You can get proper spacing by using \mid instead of |. – joriki Jun 2 '16 at 16:21
• The formal definition works like a charm here and yields $$E(X_2\mid X_1)=\frac{X_1}2.$$ Alternatively, one can use the general formula, probably in your notes, valid when the distribution of $(X_1,X_2)$ has PDF $f$, which states that $$E(u(X_2)\mid X_1)=v(X_1),$$ where $$v(x)=\frac{\int_\mathbb R u(y)f(x,y)dy}{\int_\mathbb R f(x,y)dy}.$$ – Did Jun 2 '16 at 16:22

It's a general fact that if $X$ and $Y$ are random variables (with $X$ having finite expectation) and $\mathcal{G}=\sigma(Y)$, then there is a Borel function $f:\mathbb{R}\to\mathbb{R}$ such that $$\mathbb{E}[X\mid\mathcal{G}]=f(Y)$$
So it's enough to determine $f$. In this case $X_1$ and $X_2$ have a joint pdf $$f_{X_1,X_2}(x,y)=2\cdot1_{0\leq y\leq x\leq 1}$$ hence the marginal pdf of $X_1$ is $$f_{X_1}(x)=\int f_{X_1,X_2}(x,y)\;dy=\int_0^x2\;dy=2x\cdot 1_{0\leq x\leq 1}$$ so the conditional pdf is $$f_{X_2|X_1}(y|x)=\frac{1}{x}\cdot 1_{0\leq y\leq x}$$ for $0\leq x\leq 1$. From this we can conclude that $X_2$ is uniformly distributed on $[0,X_1]$, hence $$\mathbb{E}[X_2\mid X_1]=\frac{X_1}{2}$$
• carmichael561, do we still have to prove by formal definition that $$\mathbb{E}[X_2\mid X_1]=\frac{X_1}{2}$$ ? If not, is there a way to generalise this problem that allows us to bypass formal definition for a certain class of problems? – BCLC Apr 15 '18 at 21:07
• check 3 things: 1. integrable 2. $X_1$-measurable 3. $E[(X_1/2) 1_A] = E[X_2 1_A]$ for A in $\sigma(X_1)$ – BCLC Apr 16 '18 at 1:17
• Well, then yes the formal definition can be "bypassed", using the general method for computing $\mathbb{E}[X\mid Y]$ when $X$ and $Y$ have a joint pdf. This is precisely the approach taken in the answer above. – carmichael561 Apr 16 '18 at 1:46