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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

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  • $\begingroup$ 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 May 30 at 22:28
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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 (**)

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

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  • $\begingroup$ thanks, very much appreciated !! $\endgroup$ – ZRH Jun 3 at 12:18

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