Past exam of the martingales and financial mathematics course of the University of Oxford / 2008 / A / 3

A version of $\mathbb{E}(X\mid \mathcal{G})$ is an integrable $\mathcal{G}$-measurable random variable $Y$ satisfyingThe measurability requirement restricts $Y$ to the available information; the integral identities preserve all averages visible to that information. This is the definition of conditional expectation.

Let $Y,Z$ be two versions. For every $A\in \mathcal{G}$, their defining identities imply ${\int }_A(Y-Z)d\mathbb{P}=0$. The set $A_m=\{Y\ge Z+1/m\}$ belongs to $\mathcal{G}$, and thereforeEvery $A_m$ is null. Their union is $\{Y>Z\}$, so it too is null. Reversing $Y,Z$ proves the same for $\{Z>Y\}$. Hence versions of conditional expectation are equal almost surely.

A martingale is an adapted process $X_n$ with $\mathbb{E}|X_n|<\infty$ for every $n$ andAdaptedness means that $X_n$ is ${\mathcal{F}}_n$-measurable. Iterating the tower property gives the analogous identity for any two deterministic times $m\le n$.

The supermartingale property and linearity of conditional expectation give, for $n\ge 1$,Each $G_n$ is integrable and ${\mathcal{F}}_{n-1}$-measurable, while $G_0=0$. Thus $Z_n$ is integrable andMoreover, all terms $G_j$, $j\le n+1$, are ${\mathcal{F}}_n$-measurable, so $Z_{n+1}\in m{\mathcal{F}}_n$. This is predictability of the decreasing component in the Doob decomposition theorem.

The process $Y_n=X_n-Z_n$ is adapted and integrable. Since $Z_{n+1}=Z_n+G_{n+1}$ is already ${\mathcal{F}}_n$-measurable,Therefore $Y$ is a martingale, and $X=Y+Z$ is the Doob decomposition in discrete time. Its normalization is $Z_0=0$, so the initial value is included in $Y_0=X_0$.

In any proposed decomposition, $Z_n^{\prime }=X_n-Y_n^{\prime }$ is integrable. Its increment is ${\mathcal{F}}_{n-1}$-measurable by predictability. Taking conditional expectation of the difference equation givesThe martingale increment has conditional mean zero, and a measurable integrable variable equals its own conditional expectation. Starting from $Z_0^{\prime }=0$, summation gives $Z_n^{\prime }={\sum }_{j=0}^n{G}_j=Z_n$. Hence $Y_n^{\prime }=X_n-Z_n^{\prime }=Y_n$. Thus the normalized predictable decomposition is unique, with a common probability-one event available for all integer times by countability.