# Showing that $X_t=\int_0^t e^{A(t-s)}KX_s\,dB_s+\int_0^te^{A(t-s)}M\,dB_s$ satisfies $dX_t$

Consider $$dX_t=AX_t\,dt+KX_t\,dB_t+M\,dB_t,$$ where $$A=\begin{pmatrix} 0 & 1 \\ -w^2 & -a_0\end{pmatrix}$$, $$K=a_o\eta\begin{pmatrix} 0 & 0 \\ 0 & -1\end{pmatrix}$$ and $$M=T_0\eta\begin{pmatrix} 0 \\ 1\end{pmatrix}$$ for constants $$a_0,w,T_0,a_0$$ and $$\eta$$. Here $$B_t$$ denotes a Brownian motion at time $$t$$ ($$B_0=0$$).

I am trying to show that $$X_t$$ satisfies the integral equation $$X_t=\int_0^t e^{A(t-s)}KX_s\,dB_s+\int_0^te^{A(t-s)}M\,dB_s,$$ if $$X_0=0$$.

Clearly $$X_t$$ satisfies the initial condition. To see if $$X_t$$ satisfies $$dX_t$$, I attempt to write $$X_t$$ in the form of $$dX_t$$. \begin{align} X_t&=X^1_t+X^2_t \\ \implies dX_t&=dX^1_t+dX^2_t \\ &=\left(KX_t\,dB_t+AX^1_t\right)+\left(M\,dB_t+AX^2_t\right) \\ &=AX_t+KX_t\,dB_t+M\,dB_t. \end{align} For some reason the first term in the final line is missing a $$dt$$. How does this appear?

This question is taken from Oksendal's Stochastic Differential Equations Vol. $$5$$, question $$5.13$$

Update

With thank to TheBridge, I have tried to use the multidimensional Ito Lemma. Let $$X_t=(X^1_t,X^2_t)$$. Then

\begin{align} X_t&=\int_0^t e^{A(t-s)}KX_s\,dB_s+\int_0^te^{A(t-s)}M\,dB_s \\ e^{-At} X_t&=\int_0^t e^{-As}KX_s\,dB_s+\int_0^te^{-As}M\,dB_s \\ e^{-At}X^1_t+e^{-At}X^2_t&=\int_0^t e^{-As}KX_s\,dB_s+\int_0^te^{-As}M\,dB_s \end{align}

I let $$g(t,x_1,x_2)=e^{-At}x_1+e^{-At}x_2$$. Then by Ito's multidimensional lemma, \begin{align} d\left(e^{-At}X_t\right)&=\frac{\partial g}{\partial t} (t,X^1_t,X^2_t)\,dt + \left(\frac{\partial g}{\partial x_1}(t,X^1_t,X^2_t)\,dX^1_t+\frac{\partial g}{\partial x_2}(t,X^1_t,X^2_t)\,dX^2_t\right)+\frac{1}{2}\left(\frac{\partial^2 g}{\partial x_1^2}(t,X^1_t,X^2_t)\,\left(dX^1_t\right)^2+\frac{\partial^2 g}{\partial x_2^2}(t,X^1_t,X^2_t)\,\left(dX^2_t\right)^2\right) \\ &=-Ae^{-At}X(t)\,dt+e^{-At}\,dX_1(t)+e^{-At}\,dX_2(t) \end{align}

• You just forgot to write it. The term $AX_t^1$ appears when you took a partial derivative with respect to $t$, so it should really be $AX_t^1dt$. Similarly the $AX_t^2$ should be $AX_t^2dt$. Commented May 4, 2021 at 2:44
• I rather think that you have missed that $X_t$ is a two dimensional process, so $X_t\not=X^1_t+X^2_t$ but $X_t=(X^1_t+X^2_t)$, you should rather use the multidimensional version of Itô 's lemma on $e^{-A.t}.X_t$). Commented May 4, 2021 at 3:36
• @TheBridge How do I use the multidimensional version of Ito's lemma on $e^{-At}$? I'm unsure how this is done.
– Bell
Commented May 4, 2021 at 3:42
• not on $e^{-A.t}$, on $e^{-A.t}.X_t=\int_0^t e^{-s.A}KX_s\,dB_s+\int_0^te^{-s.A}M\,dB_s$. Commented May 4, 2021 at 3:56
• @TheBridge I will post an update. I have made some progress but am unable to finish the solution
– Bell
Commented May 4, 2021 at 4:02

Well Let's write $$g(t,x_1,x_2) = e^{-A.t}(x_1,x_2)$$ then by multidimensional Itô's lemma : $$d\left(e^{-At}X_t\right)=$$(1) $$\frac{\partial g}{\partial t} (t,X^1_t,X^2_t)\,dt$$ (2)

$$+\left( \nabla g(t,X^1_t,X^2_t)\,dX_t\right)$$(3)

$$+\frac{1}{2}\left(dX_t\right)^T\Delta g(t,X^1_t,X^2_t)\,\left(dX_t\right)$$

Where $$\nabla g$$ is the gradient of $$g$$ with respect to vector $$X$$ and $$\Delta$$ is the Laplacian.

So $$(1) =-Ae^{-At}X(t)\,dt$$

$$(3)= (0,0)$$ because the Laplacian with respect to $$(x=(x_1,x_2)$$ of the function $$(x_1,x_2)$$ is the $$0$$ 2 by 2 matrix.

We are left with :

$$(2)=e^{-At}.Id.dX_t$$ where Id is the 2 dim identity matrix.

And putting it all together we have :$$d\left(e^{-At}X_t\right)= -Ae^{-At}X(t)\,dt +e^{-At}.dX_t$$ $$=-Ae^{-At}X(t)\,dt+ e^{-At}.(AX_t\,dt+KX_t\,dB_t+M\,dB_t)$$ $$=e^{-At}.(KX_t\,dB_t+M\,dB_t)$$

Now integrate (knowing $$X_0=0$$ at hypothesis in Oksendal's exercise you forgot to mention) : $$e^{-At}X_t-1.X_0=e^{-At}X_t=\int_0^t e^{-As}KX_s\,dB_s+\int_0^t e^{-As}M\,dB_s)$$ So $$X_t=\int_0^t e^{A(t-s)}KX_s\,dB_s+\int_0^t e^{A(t-s)}M\,dB_s)$$

N.B. : $$e^{A}$$ needs to be invertible. QED