Conditional Second Moment of a Diffusion with Correlated Noise and Drift Input

Summary

The document addresses the conditional second moment of a process with linear drift, a deterministic time-varying additive drift term, and two correlated Brownian shocks. The additive term prevents the logarithm of the process from having dynamics independent of the process itself, so the usual direct lognormal argument does not apply.

The proposed method constructs an integrating factor that removes the linear drift and stochastic terms when applied to the product of the factor and the process. The product then has a simpler finite-variation differential, yielding an integral representation for the process at the terminal time in terms of its value at the conditioning time and the deterministic input. This representation is a starting point for taking conditional moments. The excerpt stops before carrying out the conditional expectation or giving an explicit second-moment formula, so further calculation is needed; the result depends on the Brownian correlation and the time-varying coefficients.

Key ideas

• An additive drift term prevents the logarithm of the process from having state-independent dynamics.
• An integrating factor can simplify the product of the process and the factor by canceling stochastic terms.
• The transformed process has a terminal representation involving its initial conditional value and an integral of the deterministic input.
• The excerpt gives a route toward the conditional second moment but does not complete its calculation.

Tags

# Expected payoff at future time


# 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 by Gordon (score 4, accepted)

https://quant.stackexchange.com/a/43960

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.

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