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Let $W$ and $B$ be correlated one dimensional Brownian motions with constant correlation coefficient $r \in (-1, 1)$, that is, we have $d\langle W, B \rangle_t = r \, dt.$ We assume we have $B_0 = v$ for some arbitrary initial condition $v \in \mathbb R$.

Consider the SDE

$$dX_t = B_t X_t \, dW_t$$

with initial condition $X_0 = x_0$ a.s.

Question: Is there an explicit solution to the above SDE, modulo stochastic/deterministic integrals without explicit $X$ terms?

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    $\begingroup$ Have you actually tried? Hint: rewrite the equation in Stratonovich form… $\endgroup$ Commented Oct 8, 2023 at 14:12
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    $\begingroup$ @MartinHairer I should have written things out, maybe taking the logs would have popped up as a natural idea... $\endgroup$
    – Nate River
    Commented Oct 9, 2023 at 9:32

1 Answer 1

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First, we move to independent BMs

$$W_t^2 = \rho W_t^1 + \sqrt{1-\rho^2} Z_t=c_{1}W_{t}^{1}+c_{2}Z_{t},$$

where $Z_t$ is a BM independent of $W_t^1$. So we condition on $Z_{t}$ to make it deterministic

$$dX_{t}=(c_{1}W_{t}+c_{2}Z_{t})X_{t}dW_{t}.$$

By applying Itô for $Y=\ln(X)$ we have

$$dY=\frac{1}{X}dX+\frac{-1}{2X^{2}}((c_{1}W_{t}+c_{2}Z_{t})X_{t})^{2}dt$$

$$=c_{1}W_{t}+c_{2}Z_{t}dW_{t}-\frac{1}{2}(c_{1}W_{t}+c_{2}Z_{t})^{2}dt$$

and so

$$X=\exp\left(\int c_{1}W_{t}+c_{2}Z_{t}dW_{t}-\frac{1}{2}\int (c_{1}W_{t}+c_{2}Z_{t})^{2}dt\right).$$

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  • $\begingroup$ Thank you so much! I should've thought to take logs. $\endgroup$
    – Nate River
    Commented Oct 9, 2023 at 9:31

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