Matrix representation for a solution

Question

Suppose $W_t$ is a 1-dimensional brownian motion . I have this equation : $dx_t=a.x_t.dt+b.x_t.dW_t$ so I do like this : $$\frac{dx_t}{x_t}=a.dt+b.dW_t\\y_t=\ln x_t \to dy_t=0dt+\frac{1}{x}dx+\frac{1}{2}.\frac{-1}{x^2}.(dx)^2\\ dy_t=0dt+\frac{1}{x}(a.x_t.dt+b.x_t.dW_t)+\frac{1}{2}.\frac{-1}{x^2}.(a.x_t.dt+b.x_t.dW_t)^2\\dy_t=adt+bdW_t-\frac{1}{2}(a^2dt^2+2abdt.dW_t+b^2dW_t^2)\\ dy_t=adt+bdW_t-\frac{1}{2}(0+b^2dW_t^2)\\ dy_t=adt+bdW_t-\frac{1}{2}b^2dt\\ dy_t=(a-\frac{1}{2}b^2)dt+b.dW_t\\ \to x_t=x_0.e^{\large (a-\frac{1}{2}b^2)t+b.W_t}$$ this is clear , and now my question is : When a,b are matrces (for example $2 \times2) $ how can I rewrite that operation ? Or how can I represent above solution ,when A,B are constant matrices ?($W_t$ is 1D brownian motion) $$X=\left(\begin{array}{c}x_1(t)\\ x_2(t)\end{array}\right)\\dX=A.X.dt+B.X.dW_t\\ dX=\begin{bmatrix}a & b \\c & d \end{bmatrix}.X.dt+\begin{bmatrix}a' & b' \\c' & d' \end{bmatrix}.X.dW_t\\$$ It seems simple ,but I can't go on ... because we can't write $\frac{dX}{X}$ or $X^{-1}.dX$ when $2\times1 $ matix Thanks in advanced ?

Answer 1

If you have $dX_t=AX_t\,dt+BX_t\,dW_t$, where $A$ and $B$ are $2\times 2$ real matrices, $X_t$ is a $2$-dimensional column vector and $W_t$ is a $1$-dimensional Brownian motion, then the solution is $$X_t=e^{(A-\frac12 B^2)t+BW_t}X_0,$$ where $X_0$ is the initial condition (a $2$-dimensional column vector) and the exponential of a matrix is defined here.

Apply Itô's formula to $f(t,W_t)=e^{(A-\frac12 B^2)t+BW_t}X_0$, using the fact that $(e^{Ct})'=Ce^{Ct}$ when $C$ is a matrix:$$f_t(t,x)=(A-\frac12 B^2)f(t,x),\quad f_x(t,x)=Bf(t,x),\quad f_{xx}(t,x)=B^2f(t,x).$$ Then $df(t,W_t)=(A\,dt+B\,dW_t)f(t,W_t)$.