I'm wondering if there is an easy way to compute the following expectation :
$E[X(T)max(S(T)-K,0)]$
if $X(T)$ and $S(T)$ are two lognormals with parameters ($\mu_1,\sigma_1)$ and ($\mu_2,\sigma_2) $ with correlation parameter $\rho$.
Thanks a lot in advance !
For such type of question, you can compute by a brute force. However, measure change is always the most natural approach. We assume the following system of SDEs under the probability measure $P$: \begin{align*} dX(t) &= X(t)\left[\mu_1dt + \sigma_1 dW_t^1 \right],\\ dS(t) &= S(t) \left[\mu_2dt + \sigma_2 \left(\rho dW_t^1 + \sqrt{1-\rho^2} dW_t^2 \right) \right], \end{align*} where $\{W_t^1, t\ge 0\}$ and $\{W_t^2, t\ge 0\}$ are two independent standard Brownian motions.
We define the new probability measure $Q$ such that \begin{align*} \frac{dQ}{dP}\big|_t = e^{-\frac{1}{2}\sigma_1^2 t +\sigma_1 W_t^1}. \end{align*} Then, under $Q$, $\{\widehat{W}_t^1, t\ge 0\}$ and $\{\widehat{W}_t^2, t\ge 0\}$ are two independent standard Brownian motions, where \begin{align*} \widehat{W}_t^1 &= W_t^1 - \sigma_1 t,\\ \widehat{W}_t^2 &= W_t^2. \end{align*} Moreover, \begin{align*} X(T) &= X(0) e^{\left(\mu_1 +\frac{1}{2}\sigma_1^2 \right)T + \sigma_1 \widehat{W}_T^1}, \\ S(T) &= S(0) e^{\left(\mu_2+\rho \sigma_1\sigma_2-\frac{1}{2}\sigma_2^2 \right)T + \sigma_2\left(\rho \widehat{W}_T^1 + \sqrt{1-\rho^2} \widehat{W}_T^2\right)},\\ \frac{dQ}{dP}\big|_T &= e^{\frac{1}{2}\sigma_1^2 T +\sigma_1 \widehat{W}_T^1}. \end{align*} Then \begin{align*} E\big(X(T) \max(S(T)-K, 0) \big) &= E_Q\left(\left(\frac{dQ}{dP}\big|_T\right)^{-1}X(T) \max(S(T)-K, 0) \right)\\ &=X(0)e^{\mu_1T} E_Q\left(\max(S(T)-K, 0) \right). \end{align*} The remaining can now be based on the standard Black-Scholes formula.