I have the following two-dimensional SDE: $dX_1=(-\mu X_1 - X_2)dt +\sigma dW_1$ and $dX_2=(-\mu X_2 + X_1)dt +\sigma dW_2$
I then have to show that $E(X_1^2 + X_2^2) = \frac{\sigma^2}{\mu}$.
I know I have to use the multidimensional Ito formula, however I am not too sure how to set everything up. Any help is greatly appreciated.
Edit: The initial conditions are $X_1 = 1$ and $X_2=0$
On
If you apply Ito to $Y=X_1^2+X_2^2$, then $$\begin{align} dY &= 2(X_1dX_1+X_2dX_2)+(dX_1)^2+(dX_2)^2 \\ &= -2\mu(X_1^2+X_2^2)\,dt+2\sigma(X_1dW_1+X_2dW_2)+2\sigma^2\,dt, \end{align}$$ using the informal notation for the increment of the quadratic variation.
Going over to the expectation removes the terms with Brownian increments, resulting in $$ dE(Y)=-2\mu\,E(Y)\,dt+2\sigma^2\,dt. $$ This now is a standard first-order linear ODE, converging to an equilibrium that is indeed located at $\frac{\sigma^2}{\mu}$
The one dimensional SDE $$ dX_t=AX_t\,dt+\sigma W_t $$ has the solution $$\tag{1} X_t=e^{At}\textstyle(X_0+\sigma\int_0^te^{-As}\,dW_s)\,. $$ When you replace the constant $A$ by the matrix $$ A=\left(\begin{matrix}-\mu&-1\\1&-\mu\end{matrix}\right) $$ and $X$ resp. $W$ by their two dimensional sisters then you will see that (1) is the solution to your system.