# Doob's decomposition of $X_n^2$, where $X_n$ symmetric random walk.

I need to find Doob's decomposition of $$X_n^2$$, where $$X_n$$ is symmetric random walk. I have tried induction expressions.

Let $$X_n^2 = M_n+A_n$$, where $$M_n$$ is martingale and $$A_n$$ is integrable predictable process. Consider $$X_{n+1}^2 = (X_n+\Delta X_n)^2 = X_n^2+2X_n\Delta X_n+\Delta X_n^2$$. Using $$\Delta X_n^2 = 1$$ and $$X_n^2 = M_n+A_n$$ we get $$X_{n+1}^2 = M_n+2X_n\Delta X_n + A_n +1$$.

Hence $$M_{n+1}=M_n+2X_n\Delta X_n$$ and $$A_{n+1}=A_n+1$$.

Am I correct? Do I need to prove something else?

• Nov 15, 2022 at 22:10

Suppose $$\{ \xi_{i} : i \geq 1 \}$$ be i.i.d. random variables with $$\mathbb{E} ( \xi_i ) = 0$$ and $$\text{Var} (\xi) = \sigma^2$$. Let $$S_0 = 0$$ and for $$n \geq 1$$, $$S_n = \sum_{ i = 1}^n \xi_i$$.
Let $$Y_n = S_n^2 - n \sigma^2$$ for $$n \geq 0$$. Then $$\{ Y_n : n \geq 1 \}$$ is a $${\cal F}_n := \sigma ( \xi_k : 1 \leq k \leq n )$$ martingale. Therefore, we will have $$X_n^2 = Y_n + n \sigma^2$$ so that $$A_n = n \sigma^2$$ is the predictable process.
To show this, just note that $$S_{n+1} = S_n + \xi_{n+1}$$ so that $$S_{n+1}^2 = S_n^2 + 2 S_n \xi_{n+1} + \xi_{n+1}^2$$. Thus, using the fact that $$S_n$$ is $${\cal F}_n$$-measurable and $$\xi_{n+1}$$ is independent of $${\cal F}_n$$, we have \begin{align*} \mathbb{E} ( S_{n+1}^2 \mid {\cal F}_n ) & = \mathbb{E} ( S_n^2 + 2 S_n \xi_{n+1} + \xi_{n+1}^2 \mid {\cal F}_n ) \\ & = S_n^2 + 2 S_n \mathbb{E} ( \xi_{n+1} \mid {\cal F}_n ) + \mathbb{E} ( \xi_{n+1}^2 \mid {\cal F}_n ) \\ & = S_n^2 + 2 S_n \mathbb{E} ( \xi_{n+1} ) + \mathbb{E} ( \xi_{n+1}^2 ) \\ & = S_n^2 + \sigma^2. \end{align*} This proves that $$\mathbb{E} \bigl( S_{n+1}^2 - (n+1) \sigma^2 \mid {\cal F}_n \bigr) = S_n^2 + \sigma^2 - (n+1) \sigma^2 = S_n^2 - n \sigma^2.$$ This proves the result.
In the case of SSRW, we have $$\xi_{n+1} = \pm{1}$$ with probability $$1/2$$ so that $$\sigma^2 = 1$$. Thus, the decomposition is $$A_n = n$$ for $$n \geq 1$$.