assume I have two stocks with known volatilities and a known correlation coefficient of returns - does anyone know how to determine the correlation between the prices and NOT THE RETURNS

  • $\begingroup$ so, you're asking how to determine co-movement of two series' drift terms rather than residuals? cointegration is a related technique that might be useful $\endgroup$
    – Chris
    Commented May 30, 2019 at 22:28

1 Answer 1


We can obtain a closed-form expression for price correlation given (log) return correlation when the two stocks follow geometric Brownian motion:

$$S_1(t) = S_1(0)e^{(\mu_1- \frac{1}{2} \sigma_1^2)t}e^{\sigma_1Z_1(t)},\\ S_2(t) = S_2(0)e^{(\mu_2- \frac{1}{2} \sigma_2^2)t}e^{\sigma_2Z_2(t)},$$

where $\text{corr}(Z_1(t),Z_2(t)) = E[Z_1(t)Z_2(t)]=\rho t$. The correlation of log returns over an interval of length $\delta t$ is

$$\text{corr}\left(\log \frac{S_1(t+\delta t)}{S_1(t)} , \log \frac{S_2(t + \delta t)}{S_2(t)} \right) = \rho \delta t$$

The price correlation is

$$\tag{*}\rho_{S_1S_2}=\frac{E[(S_1(t) - E(S_1(t))(S_2(t) - E(S_2(t))]}{\sqrt{\text{var}(S_1(t))}\sqrt{\text{var}(S_2(t))}}$$

Recalling that $E(e^{\sigma_1 Z_1(t)}) = e^{\frac{1}{2} \sigma_1^2 t}$, we obtain $$E(S_1(t)) = S_1(0)e^{\mu_1t}, \quad E(S_2(t)) = S_2(0)e^{\mu_2t} \\\text{var}(S_1(t)) = S_1(0)^2e^{2 \mu_1 t}( e^{\sigma_1^2t}-1), \quad \text{var}(S_2(t)) = S_2(0)^2e^{2 \mu_2 t}( e^{\sigma_2^2t}-1) $$

Note that

$$E[(S_1(t) - E(S_1(t))(S_2(t) - E(S_2(t))] = E[S_1(t)S_2(t)] - E(S_1(t)) E(S_2(t)) \\ = S_1(0)S_2(0)e^{\mu_1t}e^{\mu_2t}\left(e^{-\frac{1}{2}\sigma_1^2t}e^{-\frac{1}{2}\sigma_2^2t}E[e^{\sigma_1Z_1(t) + \sigma_2Z_2(t)}] - 1\right)$$

Substituting into (*) we obtain

$$\tag{**}\rho_{S_1S_2} = \frac{e^{-\frac{1}{2}\sigma_1^2t}e^{-\frac{1}{2}\sigma_2^2t}E[e^{\sigma_1Z_1(t) + \sigma_2Z_2(t)}] - 1}{\sqrt{ e^{\sigma_1^2t}-1}\sqrt{ e^{\sigma_2^2t}-1}}$$

Since $Z_1(t)$ and $Z_2(t)$ are both normally distributed with mean $0$ and variance $t$, it follows that $\sigma_1Z_1(t) + \sigma_2 Z_2(t)$ is normally distributed with mean $0$ and variance

$$\text{var}(\sigma_1Z_1(t)+\sigma_2Z_2(t)) = E[(\sigma_1Z_1(t)+\sigma_2Z_2(t))^2 \\ = (\sigma_1^2 + \sigma_2^2 + 2\rho \sigma_1\sigma_2)t$$

We then have

$$E[e^{\sigma_1Z_1(t) + \sigma_2Z_2(t)}] = e^{\frac{1}{2}\sigma_1^2t}e^{\frac{1}{2}\sigma_2^2t}e^{\rho\sigma_1\sigma_2t},$$

and after substituting into (**)

$$\rho_{S_1S_2} = \frac{e^{\rho\sigma_1\sigma_2t} - 1}{\sqrt{ e^{\sigma_1^2t}-1}\sqrt{ e^{\sigma_2^2t}-1}}$$

  • $\begingroup$ thanks, very much appreciated !! $\endgroup$
    – ZRH
    Commented Jun 3, 2019 at 12:18
  • $\begingroup$ I think you could add some assumptions about dividends in order to generalize from price returns to total returns correlation. $\endgroup$ Commented May 8, 2020 at 20:19

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.