I have the following system of stochastic differential equations:
$dX_{1}=(-\mu X_{1}-X_{2})dt+\sigma dW_{1}$
$dX_{2}=(-\mu X_{2}+X_{1})dt+\sigma dW_{2}$
Where $W_{1}$ and $W_{2}$ are independent standard one dimensional Brownian motions and $\sigma$ and $\mu$ are constants.
I am looking to show the following:
$\lim_{t\rightarrow \infty }\mathbb{E}[X_{1}^{2}+X_{2}^{2}]=\frac{\sigma^{2}}{\mu}$
So far I have applied the multi-dimensional form of the Ito equation with $U(t)=u(X(t),t)$ such that $u(x,t)=\underline{x} \cdot \underline{x}$. This gives that:
$dU = [-2\mu X_{1}^{2}-2\mu X_{2}^{2}+2\sigma ^{2}]dt+2X_{1}dW_{1}+2X_{2}dW_{2}$
Now seeing as I should solve for $\mathbb{E}[X_{1}^{2}+X_{2}^{2}]$ I know the last two terms will be zero since they are Ito integrals. I am however unsure about the $dt$ term. Should I solve for $X_{1}$ and $X_{2}$ and then plug that in or is there some other way to simplify this expression?
Best Answer
We can focus on the fact that $$d(X_1^2+X_2^2)+2\mu(X_1^2+X_2^2)dt=2\sigma^2dt+(...)$$ $$d(X_1^2+X_2^2)e^{2\mu t}+2\mu e^{2\mu t}(X_1^2+X_2^2)dt=2\sigma^2e^{2\mu t}dt+(...)$$ $$d((X_1^2+X_2^2)e^{2\mu t})=2\sigma^2e^{2\mu t}dt+(...)$$ Therefore by integrating $$(X_1^2+X_2^2)_t=(X_1^2+X_2^2)_0e^{-2\mu t}+2\sigma^2\int_{[0,t]} e^{-2\mu (t-s)}ds+(...)$$ $$(X_1^2+X_2^2)_t=(X_1^2+X_2^2)_0e^{-2\mu t}+\frac{\sigma^2}{\mu}(1-e^{-2\mu t})+(...)$$ The $(...)$ term has expectation zero, so $$E[(X_1^2+X_2^2)_t]\to \frac{\sigma^2}{\mu}$$