How to solve $\mathrm dX(t)=B(t)X(t)\mathrm dt+B(t)X(t)\mathrm dB(t)$ with condition $X(0)=1$?

Question

I want to solve the stochastic differential equation $$\mathrm dX(t)=B(t)X(t)\mathrm dt+B(t)X(t)\mathrm dB(t)$$ with condition $X(0)=1$.

Answer 1

Solution I Dividing the SDE (formally) by $X_t$ yields

$$\frac{dX_t}{X_t} = B(t) \, dt+ B(t) \, dB_t.$$

This equation suggests using an approach similar to the separation-by-parts-approach for ordinary differential equations: We set $Z_t := \log X_t$ and apply Itô's formula

$$\begin{align*} Z_t - Z_0 &= \int_0^t \frac{1}{X_s} \, dX_s - \frac{1}{2} \int_0^t \frac{1}{X_s^2} \, d\langle X \rangle_s \tag{1} \end{align*}$$

where

$$dX_s = B(s) X(s) \, ds + B(s) X(s) \, dB(s) \qquad d\langle X \rangle_s = B(s)^2 X(s)^2 \, ds. \tag{2}$$

Combining $(1)$ and $(2)$, we obtain

$$Z_t - Z_0 = \int_0^t B(s) \, dB_s + \int_0^t \left( B(s) - \frac{1}{2} B_s^2 \right) \, ds.$$

By definition, $Z_t = \log X_t$, and therefore we find an explicit expression for $X_t$. Itô's formula shows that $(X_t)_t$ defines indeed a solution of the given SDE.

Solution II Consider the corresponding ordinary differential equation

$$dx(t) = B(t) \, x(t) \, dt \qquad x(0)=1.$$

Its unique solution is given by

$$x(t) = \exp \left( \int_0^t B(s) \, ds \right)$$

or, equivalently,

$$x_t \cdot \exp \left( - \int_0^t B(s) \, ds \right)=1$$

This indicates that we may try the following Ansatz:

$$Y_t := X_t \cdot \exp \left( -\int_0^t B(s) \, ds \right).$$

Applying Itô's formula to $f(x,y) := x \cdot e^{-y}$ and the (two-dimensional) Itô process $(X_t, \int_0^t B(s) \, ds)$ gives

$$\begin{align*} Y_t - Y_0 &= \int_0^t \exp \left(- \int_0^s B(r) \, dr \right) \, dX_s - \int_0^t Y_s B(s) \, ds = \int_0^t B(s) Y_s \, dB_s. \end{align*}$$

This SDE can be solved easily and therefore we obtain an expression for $X_t$. This approach looks more complicated and lengthy, but actually it works out in a much more general setting.