# probability theory, conditional probability

This is exercise 13.4.1 from "A First Look at Rigorous Probability". The question statement is below.

Let $A$ and $B$ be events, with $0 < P(B) < 1$. Let $\mathcal{G} = \sigma(B)$ be the $\sigma$-algebra generated by $B$.

(a) Describe $\mathcal{G}$ explicitly.

(b) Compute $P(A \mid \mathcal{G})$ explicitly.

(c) Relate $P(A \mid \mathcal{G})$ to the earlier notion of $P(A \mid B) = \frac{P(A \cap B)}{P(B)}$.

(d) Similarly, for a random variable $Y$ with finite mean, compute $E(Y \mid \mathcal{G})$, and relate it to the earlier notion of $E(Y \mid B) = \frac{E(Y 1_B)}{P(B)}$.

For part (a), the sigma algebra generated by $B$ is just $\mathcal{G} = \{\varnothing, \Omega, B, B^c\}$.

For part (b), we are looking for a $\mathcal{G}$ measurable function $P(A \mid \mathcal{G})$ with the property that $E[P(A\mid \mathcal{G}) 1_S] = P(A \cap S)$ for $S \in \mathcal{G}$. I'm having trouble guessing this function. The only functions I could think of which are $\mathcal{G}$ measurable are $1_B$ or $P(A)1_B$ but neither of these satisfy the expectation property. The problem with the function $1_A$ is even though it satisfies the expectation property, it is not $\mathcal{G}$ measurable.

Earlier in the reading, Rosenthal defines this $\mathcal{G}$ measurable function to be the Radon Nikodym derivative $\frac{d\nu}{dP_0}$ of $\nu(E) \doteq P(A \cap E)$ with respect to $P_0 = P$ restricted to $\mathcal{G}$. But again I'm not sure how to compute this derivative.

• Hint: A function $h$ is $\mathcal G$-measurable if and only if $H=c_11_B+c_21_{B^c}$, where $c_1$ and $c_2$ are constants. – John Dawkins Dec 27 '16 at 19:02
it is easy to check that $$P(A\mid \mathcal G)= P(A\mid B)1_B +P(A\mid B^c)1_{B^c},$$ which answers b, and c.
Similarly, $$E(Y\mid \mathcal G)= E(Y\mid B)1_B +E(Y\mid B^c)1_{B^c}.$$