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Correlation Between Brownian Motion and Its Time Average

Article Quant Q&A · Author: zebullon
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Summary

The document explains why the correlation between a Brownian-motion process at time t and its time average is not one, even though the average has a distribution that can be written using a scaled Brownian variable. Equality in distribution preserves marginal behavior but does not identify the average with that particular scaled variable on the same sample path; using the substitute when calculating covariance therefore gives the wrong dependence.

Instead, compute covariance from the original time average by integrating the covariance between the process at each intermediate time and its terminal value. The response evaluates this integral, then combines it with the variances of the terminal process and its average to obtain the stated correlation of √3/2. It sketches justification by approximating the integral with Riemann sums and passing to the limit using dominated convergence. The derivation assumes the stated Brownian model and does not extend the result to other processes or averaging schemes.

Key ideas

  • The time average and a scaled Brownian variable can share a distribution without being the same random variable.
  • Covariance must be calculated using the original pathwise time average.
  • The covariance follows by integrating the covariance between intermediate Brownian values and the terminal value.
  • Combining that covariance with the two variances gives a correlation of √3/2 under the stated model.
  • A Riemann-sum approximation and a limit argument can justify exchanging covariance and integration.

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# Covariance of brownian motion and its time average


# Covariance of brownian motion and its time average












It's a question pertaining to the correlation of a log asset process (following BM) and its time average, to put it into form, if

$$X(t)=\mu t+\sigma W(t)$$ then $$ \bar{X}(t):=\frac{1}{t}\int_0^tX(\tau)d\tau\,{\buildrel d \over =}\,\mu\frac{t}{2}+\frac{\sigma}{\sqrt3}W(t) $$ where we see that the mean and variance are $\mu t/2$ and $\sigma^2t/3$. Now I have a paper here (Path integral approach to Asian options in the Black–Scholes mode) that says that the correlation coefficient of $X$ and $\bar{X}$ is equal to $\sqrt3/2$...

Now if I do it from definition $$ \rho_{X(t),\bar{X}(t)}=\frac{Cov(\sigma W(t),\frac{\sigma}{\sqrt3}W(t))}{\sigma\sqrt t\frac{\sigma\sqrt t}{\sqrt3}}=\frac{\frac{\sigma^2}{\sqrt3} Var(W(t))}{\frac{\sigma^2 t}{\sqrt3}}=1$$

So I guess I missed something somewhere, if anyone could give me his 2 cents... Thanks

## Answer by quasi (score 2, accepted)

https://quant.stackexchange.com/a/7995

You want to work directly with $\overline{X}$, and not some other r.v. with the same distribution, since equivalence in distribution doesn't imply that correlation remains the same. For ease of notation, I'll assume that $\mu = 0$ and $\sigma = 1$. I claim that $$ \text{cov}\left(\overline{X},X \right) = \frac{1}{t} \int_0^t s \ ds. $$

Note that this is what you get if you calculated $\frac{1}{t} \int_0^t \text{cov}(W_s, W_t) \ ds$. For now, I will leave it to you to justify this step, but the approach you should take is 1) discretize the Riemann integral, and show that things work there, and 2) Use the Dominated Convergence Theorem to pass to the limit.

In any case, what we get is $$ \text{cov}\left(\overline{X},W_t \right) = \frac{1}{t} \cdot \frac{t^2}{2} = \frac{t}{2}. $$

Using your formula for the variance of $\overline{X}$, this leads to $$ \text{corr}(\overline{X},W_t) = \frac{\frac{t}{2}}{\sqrt{t}\cdot\frac{\sqrt{t}}{\sqrt{3}}} = \frac{\sqrt{3}}{2}. $$

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.