Convergence of Covariance Matrix and SDE

Question

I am learning Stochastic Calculus and there is a question which stuck me for a long while, so I'd like ask for help. Here is the question description:

Consider the linear SDE $dX_{t}=AX_{t}dt+BdW_{t}$, where W_{t} is a multi-variate standard Brownian motion. Define $X_{t} = S(t)X_{0}+\int_{0}^{t}S(t-t')BdW_{t'}$, where the $S(t)$ is fundamental solution with property $\frac{d}{dt}S(t)=AS(t)=S(t)A,S(0)=I.$

(a)calculate the $cov(X_{t})$, which involves an integral involving $S(t)$.

(b)Assume that $\int_{0}^{\infty}||S(t)||dt < \infty$, show that the limit of $cov(X_{t})$ exists and write an integral formula for it.

So the way how I started is that, I first verified the $X_{t}$ defined is in the question is indeed a solution to the linear SDE, then I proceed to try to calculate the covariance matrix.

I use $cov(X_{t})=\mathbf{E}[X_{t}X_{t}^T]-\mathbf{E}X_{t}\mathbf{E}X_{t}^T$, but I don't know how to properly deal with the Expectation of the $It\hat{o}$ integral part, after searching the Internet for a while, I think I might need to apply It$\hat{o}$ isometry, but I am not sure here.

Then I go to part(b), which confused me that, how does the integral of the norm of $S_{t}$ less than $\infty$ affect the problem and how to make use of the condition. Now I think the norm's convergence might have something to do with the "steady state" of the matrix in the long term.

So, I sincerely ask for some help here. Particularly, I would appreciate if there is a theorem or lemma involved, could someone please tell me, because I am not sure if I have learned that yet or not, so I need to pick it up.

Thank you for your time

Answer 1

Hi guys: I am still working on it. So I tried again and I found I could cancel out a few terms, but still stuck by the "Expectation of product of Ito integral".The way I did it Could someone proceed here to help me a bit?

Answer 2

It is indeed a direct result of the Ito Isometry, https://en.wikipedia.org/wiki/It%C3%B4_isometry. If we work in $\mathbb{R}$, we can directly compute \begin{align*} E[X_tX_t]=E[\int_0^tS(t-s)BdW_s\int_0^tS(t-s)BdW_s]=\int_0^tS(t-s)BB^TS(t-s)^Tds. \end{align*} The last integral expression is the one you are looking for. This expression also holds in any dimensions (also in infinite dimensions for SPDEs). However, the computation only makes sense in 1d. If you would want to show this in any Hilbert space H, you would have to do the following computation: Take two elements $v,w\in H$ and compute \begin{align*} E[\langle X_t,v\rangle_{H}\langle X_t,w\rangle_H]=\langle Q_tv,w\rangle_H. \end{align*} If you did everything correctly, you will find that \begin{align*} Q_t=\int_0^tS(t-s)BB^TS(t-s)^Tds \end{align*} as expected.