1
$\begingroup$

Problem

Let $\Omega,F,P$ be a probability space with $\Omega = \{a,b,c,d\} $ and

$ P(a) = 1/6 , P(b) = 1/3, P(c) = 1/4, P(d) = 1/4 $, with the probability of every other set in $F$ being the sum of probabilities of the elements in the set.

Let $X,Y$ be random variables:

$ X(a) = 1, X(b) = 1, X(c) = -1, X(d) = -1 $

$ Y(a) = 1, Y(b) = -1, Y(c) = 1, Y(d) = -1 $

Determine $E[Y|X]$ (i.e. specify the values of this random variable for $a,b,c,d$). Verify the partial averaging property is satisfied.

Attempt

The partial averaging property is $$\int_{A} E[X|G](w)dP(w) = \int_{A} X(w)dP(w)$$.

So for this problem, $\sigma(X) = \{\emptyset,\Omega,\{a,b\},\{c,d\}\}$ takes the place of $G$. And $E[X|Y]$ is given by:

$$\sum_{ \alpha\in\{a,b,c,d\} } E[Y|X=\alpha] P(X=\alpha) = \int_{A} X(w)dP(w) $$

Which equals:

$$\int_{A} E[X|G](w)dP(w) = \sum_{ \alpha\in\{\{a,b\},\{c,d\} \}} E[Y|X=\alpha] P(X=\alpha) $$.

Is my reasoning correct, and if not how would I do this computation? I still am having trouble understanding what the right hand side of the integral represents and if either of these integrals actually gives $E[Y|X]$.

1 Answers 1

0

$X$ can take two values $\pm 1$. So, you need to calculate $E[Y|X=1]$ and $E[Y|X=-1]$.

When $X$ takes 1, what is the average of $Y$? Note that $X=1$ only if $a$ or $b$ is realized. So $$E[Y|X=1]= \frac{Y(a)P(a)+Y(b)P(b)}{P(a)+P(b)}$$

  • 0
    So by writing it out in this manner we show that the two sides of the partial averaging property are the same computation, makes sense. Thanks for the clarification! – 2017-02-07