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

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

  • 3
    $\begingroup$ A continuous version of this question (with time index "t" over the nonnegative reals rather than the discrete "n" of the present question) is answered on the Maths Exchange sister site here: math.stackexchange.com/questions/2273549/… Note that in the definition of the martingale as a quotient of densities, the lowercase "y" variables appearing in the numerator and demoninator ought to be replaced by capitals "Y", since these are random variables. $\endgroup$ Nov 24, 2019 at 14:52

1 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.)


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