Assume we have two geometric Brownian motions $$ dX_t = \mu X_t dt + \sigma X_t dW^1_t, \qquad \qquad dY_t = \mu Y_t dt + \sigma Y_t dW^2_t $$ where the Wiener processes are correlated with $E[dW^1_t dW_t^2] =\rho dt$. I want to show that $$ \text{Cov}(X_t, Y_t) = X_0Y_0e^{2\mu t}(e^{\rho\sigma^2t}-1). $$ (This holds accoring to wikipedia )

My try:

If one takes the Ito solutions of the differentials, we have $$ X_t = X_0e^{(\mu-\sigma^2/2)t+\sigma W^1_t}\\ Y_t = Y_0e^{(\mu-\sigma^2/2)t+\sigma W^2_t} $$ Now caluclating $\text{Cov} = E[X_t Y_t]-E[X_t]E[Y_t]$, I start with $E[X_t Y_t]$: $$ E[X_t Y_t] = X_0 Y_0 e^{(2\mu - \sigma^2)t} E[e^{\sigma(W_t^1 + W_t^2)}] $$ Here is where I am not sure about how to calculate the last expectation. I can think of first deriving the joint distribution of the Wiener processes and then caluclate the integral but it seems like a huge detour. I would hope there is some more direct way, involving properties of the Brownian motion?


Basically, you can simply apply Itô's formula: $$\begin{align*} f(W_t^1, W_t^2)-f(0,0) &= \int_0^t f_x(W_s^1,W_s^2) \, dW_s^1 + \int_0^t f_y(W_s^1,W_s^2) \, dW_s^2 \\ &\quad +\frac{1}{2} \bigg( \int_0^t f_{xx} (W_s^1,W_s^2) \, \underbrace{dW_s^1 dW_s^1}_{ds} + 2 \int_0^t f_{xy}(W_s^1,W_s^2) \, \underbrace{dW_s^1 dW_s^2}_{\varrho \, ds} \\ &\quad + \int_0^t f_{yy}(W_s^1,W_s^2) \, \underbrace{dW_s^2 dW_s^2}_{ds} \bigg). \end{align*}$$

If we choose $f(x,y) := \exp(\sigma(x+y))$ and take the expectation on both sides, it follows that

$$\mathbb{E}e^{\sigma (W_t^1+W_t^2)}-1 = \sigma^2(\varrho+1) \int_0^t \mathbb{E}e^{\sigma (W_s^1+W_s^2)} \, ds.$$

Setting $\varphi(t) := \mathbb{E}e^{\sigma (W_t^1+W_t^2)}$ we find that $\varphi$ is a solution to the ordinary differential equation

$$\varphi'(t) = \sigma^2 (\varrho+1) \varphi(t) \qquad \varphi(0)=1.$$

This ODE can be solved explicitly,

$$\varphi(t) = \exp(\sigma^2(\varrho+1)t).$$

Combining the results, we conclude

$$\mathbb{E}(X_t\cdot Y_t) = X_0 Y_0 e^{2\mu t} e^{\sigma^2 \varrho t}.$$

  • $\begingroup$ Beautiful. Could you quickly explain the justification for interchanging the expectation and integral? $\endgroup$
    – Slug Pue
    Dec 5 '14 at 22:09
  • $\begingroup$ @Slungpue This follows from Fubini's (or Tonelli's) theorem. $\endgroup$
    – saz
    Dec 6 '14 at 6:47
  • $\begingroup$ Shouldn't the second term on the RHS be $\int_0^t f_y(W_s^1,W_s^2) \, dW_s^2$? $\endgroup$
    – Sandu Ursu
    Feb 13 '20 at 17:26
  • $\begingroup$ @SanduUrsu Yes, thanks. $\endgroup$
    – saz
    Feb 13 '20 at 17:59

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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