Let $X_1, X_2, …$ be i.i.d. random variables such that $\mathbb{E}[X_n^2] < \infty$ and $\mathbb{E}[X_n] = 0$ for all $n$. Let $M = \sum_{k = 1}^n X_k$. We consider everything relative to the filtration $\mathcal{F}$ generated by the $X_i$ and want to calculate predictable process $\langle M \rangle$ such that $M^2 – \langle M \rangle$ is Martingale.
So the requirement $\mathbb{E}[X_n^2] < \infty$ allows us to use Doob's decomposition theorem, which means that $\Delta \langle M \rangle_n = \mathbb{E}[\Delta(M_n^2) | \mathcal{F}_{n – 1}] = \mathbb{E}[X_n^2 | \mathcal{F}_{n – 1}] – 2M_{n – 1} \mathbb{E}[X_n | \mathcal{F}_{n – 1}]$. If we can evaluate this and determine $\langle M \rangle_1$ we know $\langle M \rangle_n$ for all $n$, but I struggle with both of these. We have not used the requirement $\mathbb{E}[X_n] = 0$ yet. The only way I can think of using this, is that this implies that $M$ is martingale. I don't know why that is relevant though. Can we even determine $\langle M \rangle_1$ instead of leaving that arbitrary? The definition of martingales does not require anything of the first element of the stochastic process.
Best Answer
Suppose $Y=(Y_n)$ is an adapted sequence of integrable random variables, and you seek the Doob decomposition $Y_n = K_n+V_n$ in which $K$ is a martingale and $V$ is predictable. Observe that you would then have $\Bbb E[Y_n\mid\mathcal F_{n-1}] = K_{n-1}+V_n$ for $n= 1,2,\ldots$. Taking differences: $\Bbb E[Y_n\mid\mathcal F_{n-1}]-Y_{n-1} = V_n-V_{n-1}$. Therefore (assuming $V_0=0$) $$ V_n=\sum_{k=1}^n \left(\Bbb E[Y_k\mid\mathcal F_{k-1}]-Y_{k-1}\right) $$ for $n=1,2,\ldots$. In the same way, $$ K_n=Y_0+\sum_{k=1}^n \left(Y_k-\Bbb E[Y_k\mid\mathcal F_{k-1}]\right). $$
Now you can apply the discussion above to $Y_n=M_n^2$, and your $\langle M\rangle$ would be my $V$.