What is the expectation and variance of correlated exponential Brownian motions for the random variable $F$, where $A$ is real constant, $\sigma$ is a real constant and $\rho$ is the correlation.

$$F = A\exp\left[\sigma\left(\rho\mathrm dW_1 + \sqrt{1-\rho^2}\mathrm dW_2\right)\right].$$

  • $\begingroup$ What do the differentials inside the exponential mean? $\endgroup$ – Chris Janjigian May 1 '14 at 21:14
  • $\begingroup$ $W_1$ and $W_2$ are weiner processes or brownian motions $\endgroup$ – user146654 May 1 '14 at 21:22
  • $\begingroup$ As stated the question doesn't make sense at all; expressions of the form $f(dW)$ (where $W$ is a Brownian motion) are simply not well-defined. I guess you mean $$F = A \exp \bigg( \sigma (\varrho W_t^1 + \sqrt{1-\varrho^2} W_t^2) \bigg)$$ where $(W_t^1)$ and $(W_t^2)$ are Brownian motions. And what about $\varrho$? Is $\varrho$ the correlation of the Brownian motions $W^1$ and $W^2$ or is it simply a given constant (and the Brownian motions are independent)? $\endgroup$ – saz May 2 '14 at 14:29
  • $\begingroup$ yes, sorry - Brownians are independent and a given constant. $\endgroup$ – user146654 May 2 '14 at 17:55
  • $\begingroup$ Duplicate of a recent question (maybe by the same OP) where the differentials inside the exponential had already been signalled in comments as being problematic, with no visible reaction from the OP. $\endgroup$ – Did May 3 '14 at 9:33

If $(W_t^1)_{t \geq 0}$ and $(W_t^2)_{t \geq 0}$ are independent Brownian motions, then $W_t^1$ and $W_t^2$ are independent for any $t \geq 0$. Hence,

$$\mathbb{E}(F) = A \cdot \mathbb{E} \exp \bigg[ \sigma (\varrho W_t^1+\sqrt{1-\varrho^2} W_t^2) \bigg] = A \cdot \mathbb{E}\exp(\sigma \varrho W_t^1) \cdot \mathbb{E}\exp(\sigma \sqrt{1-\varrho^2} W_t^2).$$

Now as $W_t^1$ and $W_t^2$ are Gaussian random variables with mean $0$ and variance $t$, we find

$$\begin{align*} \mathbb{E}(F) &= A \cdot \exp \left(\frac{1}{2} \sigma^2 \varrho^2 t \right) \cdot \exp \left(\frac{1}{2} \sigma^2 (1-\varrho^2) t \right) = A \exp \left(\frac{\sigma^2}{2} t \right). \end{align*}$$

Remark An alternative argumentation goes as follows: If $(W_t^1)_t$ and $(W_t^2)_t$ are independent Brownian motions, it is not difficult to see that $$B_t := \varrho W_t^1+ \sqrt{1-\varrho^2} W_t^2$$ also defines a Brownian motion. In particular, $B_t \sim N(0,t)$. Using again that the exponential moments are known, the claim follows.

  • $\begingroup$ Thanks for your answer which I agree with, have you got an answer for the variance please? $\endgroup$ – user146654 May 2 '14 at 20:16
  • $\begingroup$ @user146654 Well, $$\text{var} F = \mathbb{E}(F^2)-(\mathbb{E}F)^2$$,i.e. it suffices to calculate $\mathbb{E}(F^2)$. In order to do so, note that $\exp(x) \cdot \exp(x) = \exp(2x)$. $\endgroup$ – saz May 2 '14 at 20:27
  • $\begingroup$ I think the line before the remark should be $E(F)=A \exp\left(\frac{1}{2}\sigma^2\rho^2t\right)\exp\left(\frac{1}{2}\sigma^2 \left(1-\rho^2\right)t\right)$ just a minor typo? $\endgroup$ – user146654 May 3 '14 at 9:24
  • $\begingroup$ @user146654 Yes, you are right. $\endgroup$ – saz May 3 '14 at 9:30
  • $\begingroup$ Is $Var(F) = A^2\exp(\sigma^2t)\left(\exp(\sigma^2 t) - 1\right)$ ? $\endgroup$ – user146654 May 3 '14 at 12:46

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