# Correlation between stock prices given correlation between returns

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

• 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 – Chris May 30 '19 at 22:28

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

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

Substituting into (*) we obtain

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

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 (**)

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

• thanks, very much appreciated !! – ZRH Jun 3 '19 at 12:18