options - Expected payoff at future time

Let $a$, $b$, $c$, and $e$ be constants, $W_1$ and $W_2$ be Brownian motions with correlation $\rho$, and $f(t)$ and $g(t)$ be deterministic functions of time. Let $X$ satisfy $$d(X(t))=(aX(t)+ef(t)g(t))dt+f(t)X(t)dW_1(t)+g(t)X(t)dW_2(t).$$ Compute the expected value of $X(T)^2$ given $X(t)$ for some $0\le t\le T$.

If $e=0$, we can use Ito's rule to write $d(\log X)$ as an expression independent of $X$. Integrating gives that $X(T)|X(t)$ is log-normal. If $e\neq 0$, $d(\log X)$ is no longer independent of $X$. I can't think of a way around this issue.

Answer

Based on ideas from this question, let \begin{align*} M_t = e^{-at+\frac{1}{2}\int_0^t (f^2+g^2+2\rho fg)ds -\int_0^t(f dW_1(s)+gdW_2(s))}. \end{align*} Then \begin{align*} dM_t = M_t\Big[\big(-a + f^2+g^2 + 2\rho fg \big)dt - f dW_1(t)- gdW_2(t)\Big]. \end{align*} Moreover, \begin{align*} d(M_tX_t) &= M_t dX_t + X_t dM_t + d\langle M, X\rangle_t\\ &=e M_t f g dt. \end{align*} Then, \begin{align*} X_T = \frac{M_t}{M_T}X_t + e\int_t^T\frac{M_s}{M_T} f(s)g(s)ds. \end{align*} Now, you should be able to compute the conditional expectation.