Expectation and variance of correlated exponential brownian motions 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.
$$F = A\exp\left[\sigma\left(\rho\mathrm dW_1 + \sqrt{1-\rho^2}\mathrm dW_2\right)\right].$$
 A: 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.
