Help to understand martingale example from Billingsley

Question

I am studying Billingsley's section 35 about martingales and I have some difficulties understanding one of the examples. This is the example 35.4.

Suppose we have a measurable space $(\Omega,\mathcal F)$ and let $Q$ and $P$ be probability measures on the latter. Let $\mathcal F_n=\sigma (Y_1,\cdots,Y_n)$ be the sigma algebra generated by the random variables $Y_1,\cdots,Y_n$.

Suppose that under $P$ the distribution of the random vector $(Y_1,\cdots,Y_n)$ has density $p_n(y_1\cdots,y_n)$ with respect to the n-dimensional Lebesgue measure, and under $Q$ it has density $q_n(y_1\cdots,y_n)$. (To avoid technicalities assume that $p_n,q_n$ are everywhere positive.

Then the Radon-Nikodym derivative of $Q$ with respect to $P$ on $\mathcal F_n$ is $$X_n=\frac{q_n(y_1\cdots,y_n)}{p_n(y_1\cdots,y_n)}$$

So far so good, but the author then concludes that $X_n$ is a martigale under $P$ w.r.t. the filtration $\mathcal F_n$.

How can I see this last point,

why is that the following holds?

$$\int_F X_n dP=\int_F X_{n+1} dP, \forall F\in\mathcal F_n$$

Thanks in advance.

EDIT: I may have found a possible solution, but I am not sure about the validity of it.

I can assume that there is a random variable $\Phi\in\mathcal M(\mathcal F)$ such that $$Q(F)=\int_F \Phi dP,\forall F\in\mathcal F_n$$ (One problem might be that then $Q$ is not a probability measure on $\mathcal F$).

Then I can interpret the Radon-Nikodym derivative $X_n$ as the condition expectation $E[\Phi|\mathcal F_n]$. Finally I have that $$E[X_{n+1}|\mathcal F_n]=E[E[\Phi|\mathcal F_{n+1}]|\mathcal F_n]=E[\Phi|\mathcal F_n]=X_n$$

$$E[X_{n+1}]=E[E[X_{n+1}|\mathcal F_n]]=E[X_n]$$

Answer 1

It's a little strange to say that $X_n=\frac{q_n(y_1\cdots,y_n)}{p_n(y_1\cdots,y_n)}$ is the "the Radon-Nikodym derivative of $Q$ with respect to $P$ on $\mathcal{F}_n$." In general, a Radon-Nikodym derivative $\frac{dQ}{dP}$ is a random variable defined on $(\Omega,\mathcal{F})$, whereas $X_n=\frac{q_n(y_1\cdots,y_n)}{p_n(y_1\cdots,y_n)}$ is defined on $\mathbb{R}^n$.

I will assume a specific $\Omega$ so that $X_n$ is actually $\frac{dQ}{dP}$. Take $\Omega = \mathbb{R}^{\infty}$, with the Borel $\sigma$-algebra given by the product topology, and $Y_n$ is the $n$-th coordinate map. So $\mathcal F_n=\sigma (Y_1,\cdots,Y_n)$ consists of cylinder sets given by the first $n$ coordinates. In this case, $X_n = \frac{q_n(y_1\cdots,y_n)}{p_n(y_1\cdots,y_n)} = \frac{dQ}{dP}$.

The answer to your question now simply follows from the factorization $$ q_{n+1}(y_1\cdots,y_{n+1}) = q_{n}(y_1\cdots,y_{n}) q_{n+1|n}(y_{n+1}|y_1\cdots,y_{n}) $$ where $q_{n+1|n}(y_{n+1}|y_1\cdots,y_{n})$ is the conditional density of $y_{n+1}$ given $y_1\cdots,y_{n}$.

To be more explicit, a general element of $\mathcal{F}_n$ takes the form $F_n = B_n \times \mathbb{R} \times \mathbb{R} \cdots$, where $B_n$ is a Borel set in $\mathbb{R}^n$. So

\begin{align*} \int_{F_n} X_{n+1} dP &= \int_{B_n \times \mathbb{R}} \frac{q_{n+1}}{p_{n+1}} p_{n+1} dy_1 \cdots dy_{n+1}\\ &= \int_{B_n \times \mathbb{R}} q_{n+1} dy_1 \cdots dy_{n+1}\\ &= \int_{B_n \times \mathbb{R}} q_{n} q_{n+1 | n} dy_1 \cdots dy_{n+1}\\ &= \int_{B_n} (\int_{\mathbb{R}} q_{n+1 | n} ) q_{n} dy_1 \cdots dy_{n}\\ &= \int_{B_n} q_{n} dy_1 \cdots dy_{n}\\ &= \int_{B_n} \frac{q_{n}}{p_n} p_n dy_1 \cdots dy_{n}\\ &= \int_{F_n} X_{n} dP.\\ \end{align*}

(This really just a martingale of the form $E[X|\mathcal{F}_n]$ where $X$ is an $L^1$ random variable and $\{ \mathcal{F}_n \}$ is a filtration.)