Covariance of Overlapping Brownian Motion Increments

Summary

The document explains how to calculate the expected product of two Brownian motion increments defined over unit-length intervals. By independent increments, intervals that do not overlap have zero covariance. When the intervals overlap, the shared time segment contributes to the expected product, and its duration determines the result.

The worked derivation splits the increments into nonoverlapping pieces around their shared interval. Independence removes cross terms, leaving the second moment of the shared Brownian increment. This reasoning also applies when finding lagged covariance for a process formed from scaled Brownian motion plus an independent noise process: the Brownian contribution depends on interval overlap, while independent components contribute according to their own covariance structure. The result assumes standard Brownian motion and the specified increment intervals; the document does not develop the noise process’s covariance model.

Key ideas

• Brownian increments over disjoint time intervals are independent and have zero covariance.
• Overlapping increments share a Brownian segment, so their covariance equals the variance of that shared increment.
• For unit-length increments, the overlap length determines the expected product when the intervals overlap.
• A scaled Brownian component inherits this covariance pattern, while independent noise must be analyzed separately.

Tags

# Differenced Brownian Motion covariance


# Differenced Brownian Motion covariance












I am having some difficult showing what the following equals, where $x$ and $y$, $x>y$, distinct times:

$\mathbb{E}[\Delta W_x \Delta W_y]$

where each $\Delta W_t = W_t - W_{t-1}$.

I have decomposed it into its four terms, which allows taking expectations of the product of some W term and another, which I think is 0 by independence, but that makes this entire expectation 0, which makes the overall covariance I am solving for to be 0, which seems off to me.

The full problem I am trying to solve is:

$Cov(\Delta Z_t + \Delta\epsilon_t, \Delta Z_{t-i} + \Delta\epsilon_{t-i})$,

where $Z_t = \kappa W_t$ and $i = 1,2,3,...$, $W$ and $\epsilon $ independent, so guidance with how to would be even more appreciated, really.

## Answer by pbr142 (score 3, accepted)

https://quant.stackexchange.com/a/10919

A key property of Brownian motion is independent increments. So if $x-1 > y$, then $$ \mathbb{E}[\Delta W_x \Delta W_y] = 0 $$ because the time intervals [x-1,x] and [y-1,y] do not overlap. If they do overlap, i.e. $x-1 \leq y < x$, then \begin{align} \mathbb{E}[\Delta W_x \Delta W_y] =&\ \mathbb{E}[(W_x - W_{x-1}) (W_y-W_{y-1})] \\ =&\ \mathbb{E}[(W_x - W_y + W_y - W_{x-1}) (W_y - W_{x-1} + W_{x-1} -W_{y-1})] \\ =&\ \mathbb{E}[(W_y - W_{x-1})^2] = y-x+1 \end{align} where the second to last step uses the independent increments property and the last step uses the fact that the second moment of Brownian motion is equal to the time step.

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.