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

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.