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

  • $\begingroup$ Where did you get the expression for $\bar{X}$? It doesen't seems right to me. $\endgroup$
    – FKaria
    May 14, 2013 at 8:46
  • $\begingroup$ From the same paper, it is pretty much a generic way to write the geometric average, sometimes the integrand is written as $\log(S_X(t))$, sometimes instead of writing it $\bar{X}$, it is $I$... $\endgroup$
    – zebullon
    May 15, 2013 at 1:43
  • $\begingroup$ It looks like you're assuming that if you have $X$ and $Y$ which are equal in distribution, then for any $Z$, $\text{cov}(X,Z) = \text{cov}(Y,Z)$. This isn't true in general. $\endgroup$
    – quasi
    May 16, 2013 at 0:16

1 Answer 1


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 $$ \text{corr}(\overline{X},W_t) = \frac{\frac{t}{2}}{\sqrt{t}\cdot\frac{\sqrt{t}}{\sqrt{3}}} = \frac{\sqrt{3}}{2}. $$


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