Conditional Second Moment of Brownian Motion Given Its Filtration

Summary

The document works through the conditional expectation of the squared value of a Brownian motion at a later time, given its filtration at an earlier time. It expands the later value into the known earlier value plus an independent increment, then asks why the cross term should equal the time difference. The accepted explanation corrects that interpretation: the cross term has conditional expectation zero, not the time difference.

Because the earlier Brownian value is measurable with respect to the earlier filtration, it can be taken outside the conditional expectation. The later increment is independent of that filtration and has mean zero, so the cross term vanishes. The squared increment’s conditional expectation equals its unconditional second moment, which is the elapsed time. Together these facts yield the conditional second moment as the square of the earlier value plus elapsed time. The result relies on standard Brownian motion properties and the stated filtration; the note does not discuss extensions to other processes or filtrations.

Key ideas

• A Brownian value at a later time can be decomposed into its earlier value and an independent increment.
• The earlier value is measurable with respect to the filtration at the earlier time.
• The increment has conditional mean zero, so the cross term vanishes.
• The conditional second moment of the increment equals the elapsed time.

Tags

# Properties of Brownian motion and filtration, Exercise 6.22, Joshi Concepts and applications to mathematical finance


# Properties of Brownian motion and filtration, Exercise 6.22, Joshi Concepts and applications to mathematical finance












Let $W_t$ be a Brownian motion, and let $F_t$ be its filtration then for $t > s$ we are asked to compute

$$\mathbb{E}\left[W_t^2|F_s\right]$$

We have $$W_t = W_s + (W_t - W_s)$$

and

$$W_t^{2} = W_s^{2} + 2W_s(W_t - W_s) + (W_t - W_s)^2$$

So

$$\mathbb{E}\left[W_t^{2}|F_s\right] = W_s^{2} + t - s$$

I don't see how

$$2\mathbb{E}\left[W_s(W_t - W_s)|F_s\right] = t - s$$

## Answer by LocalVolatility (score 2, accepted)

https://quant.stackexchange.com/a/37849

\begin{equation} \mathbb{E} \left[ \left. W_s \left( W_t - W_s \right) \right| \mathfrak{F}_s \right] = W_s \mathbb{E} \left[ W_t - W_s \right] = 0 \end{equation}

The first step uses that $W_s$ is $\mathfrak{F}_s$-measureable and that the increment $W_t - W_s$ is independent of $\mathfrak{F}_s$. Next,

\begin{equation} \mathbb{E} \left[ \left. \left( W_t - W_s \right)^2 \right| \mathfrak{F}_s \right] = \mathbb{E} \left[ \left( W_t - W_s \right)^2 \right] = \mathbb{E} \left[ W_{t - s}^2 \right] = t - s. \end{equation}

Here we used again independence in the first step. In the second one we use that the unconditional distribution of $W_t - W_s$ is the same as that of $W_{t - s}$.

Shown in full with attribution under the source's licence. Licence: CC BY-SA 4.0 (Stack Exchange)

This summary was written by Stratmill's research agent from the original; it is not a copy of the source.