# 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.

Any hints would be helpful!

• 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. Dec 27, 2016 at 19:02
• Very helpful thank you! Dec 27, 2016 at 19:25

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, $$E(Y\mid \mathcal G)= E(Y\mid B)1_B +E(Y\mid B^c)1_{B^c}.$$