Determine $\Bbb E[{\langle S \rangle}]$ where $S$ is simple symmetric random walk

Question

Let $(Z_k)_{k \in \Bbb N}$ be i.i.d. with $\Bbb P[Z_1 = 1] = \Bbb P[Z_1 = -1] = {1 \over 2}$. Consider the process $S = (S_n)_{n \in \Bbb N_0}$ with $S_n = \sum_{k=1}^n Z_k$. I want to determine $\Bbb E[{\langle S \rangle}]$ for $n \ge 0$, where ${\langle S \rangle} := (A_n)_{n \in \Bbb N}$ which is given by the Doob decomposition $X_n = M_n + A_n$ with $M_n$ a martingal, $A_n$ is predictable and $X_n := {S_n}²$ (quadratic variation process).

Answer 1

Elaborating on @saz's hint, we can write \begin{align} S_n^2 &= (S_n-S_{n-1})^2 + 2S_{n-1}(S_n-S_{n-1}) + S_{n-1}^2\\ &= Z_n^2 + 2S_{n-1}Z_n + S_{n-1}^2. \end{align} Letting $\{\mathcal F_n\}$ be the filtration generated by $\{Z_n\}$, it is clear that $Z_n^2$ is independent of $\mathcal F_{n-1}$ and $S_{n-1}$, $S_{n-1}^2$ are $\mathcal F_{n-1}$ measurable, hence \begin{align} \mathbb E[X_n\mid\mathcal F_{n-1}] &= \mathbb E[S_n^2\mid\mathcal F_{n-1}]\\ &= \mathbb E[Z_n^2 + 2S_{n-1}Z_n + S_{n-1}^2\mid \mathcal F_{n-1}]\\ &= \mathbb E[Z_n^2] + 2S_{n-1}\mathbb E[Z_n] + S_{n-1}^2\\ &= 1 + S_{n-1}^2. \end{align} It follows that the (predictable) quadratic variation of $X_n$ is \begin{align} A_n &= \sum_{k=1}^n (\mathbb E[X_k\mid\mathcal F_{k-1}] - X_{k-1})\\ &= \sum_{k=1}^n (1 + S_{n-1}^2-S_{n-1}^2)\\ &= n. \end{align}

It follows that $M_n=S_n^2-n$ as expected (since $S_n$ is a martingale).