Conditioning on an union of sigma algebras

Question

Let $\Omega = [0,1]$, $\mathcal{F} = \mathcal{B}(0,1)$, P=Lebesgue measure.

Let $X(w)= \begin{cases} 1 \quad w \in [0,1/2] \\ 0 \quad w \not\in [0,1/2] \end{cases}$

Let $Y(w)= \begin{cases} 1 \quad w \in [0,3/4] \\ 0 \quad w \not\in [0,3/4] \end{cases}$

Let $Z(w)= \begin{cases} 1 \quad w \in [1/4,3/4] \\ 0 \quad w \not\in [1/4,3/4] \end{cases}$

I succeeded in proving that X is independent from Z and E(X|Y=1)=2/3 conditioning on the sub-sigma algebras generated by Z and Y, respectively.

Now my aim is discovering the value of E{X|(Y,Z)} but I don't know how to condition X on an union of sigma algebras.

Answer 1

• $\mathbb E[X\mid Y=1,Z=1]=P(w\in[0,\frac12]\mid w\in[\frac14,\frac34])=P(w\in[\frac14,\frac12]/P(w\in[\frac14,\frac34])=\frac12$
• $\mathbb E[X\mid Y=1,Z=0]=P(w\in[0,\frac12]\mid w\in[0,\frac14))=P(w\in[0,\frac14)/P(w\in[0,\frac14))=1$
• $\mathbb E[X\mid Y=0,Z=0]=P(w\in[0,\frac12]\mid w\in(\frac34,1])=P(w\in\varnothing)/P(w\in(\frac34,1])=0$

Observe that $\{Y=0,Z=1\}=\varnothing$ so conditioning on that event can be left out.

This shows that: $$\mathbb E[X\mid (Y,Z)]=Y-\frac12Z$$

Answer 2

Hint: Note that $$ Y^{-1}(\{1\})=[0,\frac{3}{4}], $$ $$ Z^{-1}(\{1\})=[\frac{1}{4},\frac{3}{4}]. $$ It says that the $\sigma$-algebra generated by $(Y,Z)$ is also generated by $ [0,\frac{1}{4}]=Y^{-1}(\{1\})\setminus Z^{-1}(\{1\}) $ and $[\frac{1}{4},\frac{3}{4}]$.

Answer 3

For whatever it is worth, let me describe a general procedure for answering such questions. It is easy to write $\sigma(Y,Z)$ as the sigma algebra generated by a partition $\{A_1,A_2,\cdots ,A_n\}$. Once you identify the sets $A_i$ you can compute $E(X|Y,Z)$ by the formula $E(X|Y,Z)=\sum c_iI_{A_i}$ where $c_i=\frac {\int_{A_i} XdP} {P(A_i)}$.