Correlation Between Brownian Motion and Its Time Integral

Summary

The note calculates the correlation between Brownian motion at a fixed time and the time integral of that same process. It corrects the initial conjecture that the variables are uncorrelated: both have zero mean, but their covariance is positive because each earlier Brownian value shares increments with the value at the endpoint. Using the covariance identity for Brownian motion, the covariance is obtained by integrating the earlier time over the interval.

The excerpt also gives the variances of the endpoint value and the integral, then combines covariance and standard deviations to obtain a correlation of √3/2. This is a direct calculation for standard Brownian motion over a common interval starting at zero. It does not extend the result to other stochastic processes, shifted intervals, or differently weighted integrals; those require their own covariance calculations.

Key ideas

• Brownian motion at time t and its integral from zero to t are positively correlated.
• The covariance follows by integrating the shared-time covariance of Brownian motion over the interval.
• The variance of the integral is obtained from a double integral of Brownian cross-moments.
• For the variables in the note, the resulting correlation is √3/2.

Tags

# Correlation coeffitiont between two stochastic processes


# Correlation coeffitiont between two stochastic processes












I want to find correlation coeffitiont between $W_t$ and $\int_{0}^{t}W_s ds$.

I think that these are uncorrelated. But Why?

So thanks

## Answer by M. Jeunesse (score 12, accepted)

https://quant.stackexchange.com/a/29495

if you talk about correlation then:

- compute expectation: $$\mathbb{E}(W_t)=0\text{ and }\mathbb{E}(\int_0^tW_d ds)=0$$

- variance: $$\text{Var}(W_t)=t\text{ and }\text{Var}(\int_0^tW_s ds)=\frac{t^3}{3}$$

- covariance: $$\mathbb{E}(W_t\int_{0}^tW_sds)=\int_{0}^t\mathbb{E}(W_tW_s)ds=\int_0^tsds=\frac{t^2}{2}$$

then you get: $$\text{Corr}(W_t,\int_0^tW_s ds)= \frac{\sqrt{3}}{2}$$

#### hint

$$\mathbb{E}(W_uW_s)=\min(u,s)$$

$$\text{Var}(\int_0^tW_sds)=\mathbb{E}(\int_0^t\int_0^tW_sW_u duds)$$

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.