Doob's decomposition of a brownian motion.

Question

Let $B_n$ be a discrete Brownian motion. I need to find the Doob decomposition for ($B_n^2$). Can someone help me please.

Thank you in advance.

Answer 1

By definition, the Doob decomposition of a (discrete) $\mathcal{F}_n$-measurable process $(X_n)_n$ is of the form

$$X_n = X_0 + M_n+A_n \tag{1}$$

where $(M_n)_n$ is a martingale and $(A_n)_n$ an increasing (predictable) process. If we take the conditional expectation with respect to $\mathcal{F}_{n-1}$ we get

$$\mathbb{E}(X_n \mid \mathcal{F}_{n-1}) = X_0 + M_{n-1} + \mathbb{E}(A_n \mid \mathcal{F}_{n-1}) = X_0 + M_{n-1} + A_n$$

where we used in the first step that $(M_n)_n$ is a martingale and in the second step that $(A_n)_n$ is predictable. Now if we substract $X_{n-1}$ on both sides, we find from $(1)$ that

$$\mathbb{E}(X_n - X_{n-1} \mid \mathcal{F}_{n-1}) = A_n-A_{n-1}$$

or, equivalently,

$$A_n = \sum_{j=1}^n \mathbb{E}(X_j-X_{j-1} \mid \mathcal{F}_{j-1}). \tag{2}$$

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In the given example we consider $X_n := B_n^2$ where $B$ is a Brownian motion. In order to find the Doob decomposition, we have to calculate $(2)$, i.e.

$$A_n = \sum_{j=1}^n \mathbb{E}(B_j^2-B_{j-1}^2 \mid \mathcal{F}_{j-1}).$$

To this end, write

$$B_j^2 - B_{j-1}^2 = \big( (B_j-B_{j-1})+B_{j-1} \big)^2 - B_{j-1}^2$$

and use that $B$ has independent increments as well as $B_n \sim N(0,n)$.

Answer 2

@saz last question :) Can you confirm this \begin{align*} A_n &= \sum_{j=1}^n \mathbb{E}(B_j^2-B_{j-1}^2 \mid \mathcal{F}_{j-1})\\ &= \sum_{j=1}^n \mathbb{E}\big[\big( (B_j-B_{j-1})+B_{j-1} \big)^2 - B_{j-1}^2 \mid \mathcal{F}_{j-1}\big]\\ &= \sum_{j=1}^n \bigg[\mathbb{E}\big[\big( (B_j-B_{j-1})+B_{j-1} \big)^2\mid \mathcal{F}_{j-1}\big]-\mathbb{E}[ B_{j-1}^2 \mid \mathcal{F}_{j-1}\big]\bigg] \\ &= \sum_{j=1}^n \mathbb{E}\big[ (B_j-B_{j-1})^2\mid \mathcal{F}_{j-1}\big] \\ &= \sum_{j=1}^n \mathbb{E}\big[ (B_j-B_{j-1})^2\big] \\ &= \sum_{j=1}^n \mathbb{E}[ (B_j)^2] -\mathbb{E}[ (B_{j-1})^2] \\ &= \sum_{j=1}^n n-n \\ &= 0 \end{align*}