# Durrett Probability: Theory and Examples, Example 1.1 in Section 5.1: $P(X_{n+1}\in B|\mathcal{F}_n)=\mu_{n+1}(B-X_n)$.

Update before the body of the question: The problem is about the text of 3rd edition of the book, and Durrett gave a more detailed argument in the 4th edition of the book.

This question comes from a step in Durrett's probability textbook, example 1.1 in section 5.1. I summarized the hypothesis rather than typing verbatim:

Let $$(\Omega, \mathcal{F},P)$$ be a probability space and $$X_0,\xi_1,\xi_2,\cdots,\xi_n,\cdots:\Omega\to\mathbb{R}^d$$ be independent random variables. $$X_n=X_0+\xi_1+\cdots+\xi_n$$. Set the $$\sigma$$-field $$\mathcal{F}_n$$ to be $$\mathcal{F}_n=\sigma(X_0,\cdots,X_n)$$ (the smallest $$\sigma-$$field subject to $$X_i:(\Omega,\mathcal{F}_n)\to(\mathbb{R}^d,\mathcal{B})$$ ($$0\le i\le n$$) are measurable where $$\mathcal{B}$$ is the Borel sets). By $$P(A|\mathcal{F})$$ where $$A$$ is an event and $$\mathcal{F}$$ is a $$\sigma-$$field, one means the conditional expectation $$E(1_A| \mathcal{F})$$ where $$1_A(\omega)=1$$ if $$\omega\in A$$ and $$1_A(\omega)=0$$ otherwise.

The author then wrote the following equation: $$\begin{equation*} P(X_{n+1}\in B|\mathcal{F}_n)=\mu_{n+1}(B-X_n)=P(X_{n+1}\in B|\sigma(X_n)), \end{equation*}$$ where $$\mu_{n+1}(A)=P(X_{n+1}\in A)$$ for $$A\subseteq\mathbb{R}^d$$ and the meaning of $$\sigma(X_n)$$ is a $$\sigma-$$field defined similarly above.

I can prove the second equality, but got stuck on the first.

To include my work, the following is the proof of the second equality.

The author said the following two facts might be needed:

1, If $$\mathcal{F} \subseteq \mathcal{G}$$ and $$E(X|\mathcal{G})$$ is $$\mathcal{F}$$-measurable then $$E(X|\mathcal{F}) = E(X|\mathcal{G})$$.

2, Suppose $$X$$ and $$Y$$ are independent. Let $$\phi$$ be a function with $$E|\phi(X, Y)| < \infty$$ and let $$g(x) = E(\phi(x, Y))$$. Then $$E(\phi(X, Y)|\sigma(X)) = g(X) .$$

I think by $$\mu_{n+1}(B-X_n)$$ the author means $$P(\xi_{n+1}\in \{x-X_n(\omega): x\in B\})$$ where $$\omega$$ is the point which the conditional expectation is evaluated at. The second equality follows from the second fact with $$\phi(X_n,\xi_{n+1})=1_{(X_n+\xi_{n+1}\in B)}$$.

$$\begin{gather*} P(X_{n+1}\in B|\mathcal{F}_n)(\omega)=P(X_n+\xi_{n+1}\in B|\mathcal{F}_n)(\omega)=P(X_n(\omega)+\xi_{n+1}\in B|\mathcal{F}_n)(\omega)=\\=P(X_n(\omega)+\xi_{n+1}\in B)=P(\xi_{n+1}\in B-X_n(\omega)) \end{gather*}$$ Same for $$P(X_{n+1}\in B|\sigma(X_n))$$
• Thank you. Could you please explain why $P(X_n(\omega)+\xi_{n+1}\in B|\mathcal{F}_n)(\omega)=P(X_n(\omega)+\xi_{n+1}\in B)$? I think this is the key step of the argument. Commented Aug 27 at 3:07
• $X_n(\omega)+\xi_{n+1}$ is independent of $\mathcal{F}_n$. Commented Aug 27 at 7:04
• Thanks! (This is from Durrett Exercise 1.4.1 and Example 4.1.2). Would you please also elaborate on $P(X_n+\xi_{n+1}\in B|\mathcal{F}_n)(\omega)=P(X_n(\omega)+\xi_{n+1}\in B|\mathcal{F}_n)(\omega)$? Thanks! Commented Aug 27 at 20:12