Covariance of Black-Scholes

Question

Let us assume that $W^1, W^2$ are 1-dimensional Brownian motions such that $(W^1, W^2)$ has a jointly normal distribution, and denote $c_t:=\operatorname{cov}(W^1_t,W^2_t)$ their covariance at time $t$. Denote, for $i=1,2$, $$S^i_t := \operatorname{exp}\left(\sigma W^i_t - \frac{\sigma^2}{2}t + rt\right) $$ the solution to the Black-Scholes SDE.

Is there a formula for $$\operatorname{cov}(S^1_t, S^2_t)$$ including $c_t$; i.e., is there some explicit $f_{\sigma, r}$ such that $\operatorname{cov}(S^1_t, S^2_t) = f_{\sigma, r}(t, c_t)$?

Answer 1

In order to compute this, we need information on their joint law. If they don't form a joint-normal, then the answer can vary alot see Is it possible to have a pair of Gaussian random variables for which the joint distribution is not Gaussian? and see Understanding the distribution of two correlated random variables. for a counterexample of correlated Gaussians that don't form bivariate normal.

To see this, let $S\sim Ber(p)$ for $p\in [0,1]$ and assume that $X\sim \mathcal{N}(0,1)$ and define $Y=(-1)^SX$. Assuming $S$ and $X$ are independent, then $\mathbb{P}(Y\leq t)=\mathbb{P}(S=0)\mathbb{P}(X\leq t)+\mathbb{P}(S=1)\mathbb{P}(X\geq -t)=\mathbb{P}(X\leq t),$ since $\mathcal{N}(0,1)$ is symmetric. Hence, $Y\sim \mathcal{N}(0,1)$. Furthermore, $Cov(X,Y)=1-2p$ by direct computation, so doing this experiment, we can get $Cov(X,Y)$ to be anything we want - for instance, they could be uncorrelated without being independent.

So we assume that $(W^{1}_{t},W^{2}_{s})$ is a joint-normal to get

$$E[S_{t}^{1}S_{s}^{2}]$$

$$=E\left[\operatorname{exp}\left(\sigma W^1_t+\sigma W^2_s +(r- \frac{\sigma^2}{2})(t+s) \right)\right]$$

$$=\operatorname{exp}\left(\frac{\sigma^{2}}{2}(E[(W^1_t)^{2}] +E[(W^2_s)^{2}]+2 E[W^1_t W^2_s] )+(r- \frac{\sigma^2}{2})(t+s) \right).$$

$$=\operatorname{exp}\left(\frac{\sigma^{2}}{2}(t +s+2 \rho(t,s)) +(r- \frac{\sigma^2}{2})(t+s) \right).$$