I am trying to solve this exercise which my professor has "solved" (he says what the result but not how he gets it). This is in a problem sheet which is about the Euler-Maruyama scheme. What I get wrong is part B. Thanks in advance!
QUESTION: Consider $X^{(i)}(t)>0$, $i=1,2$ of the SDE $dX^{(i)}=\mu X^{(i)}dt+\sigma X^{(i)}dW$, where $\mu$ and $\sigma$ are real constants.
(A) Write down the SDE satisfied by the separation $Y(t)=X^{(1)}(t)-X^{(2)}(t)$ between these solutions.
(B) By solving this equation or otherwise, find the region of the $(\mu,\sigma)$-plane where the equation is stable in the sense that $\lim_{t\to\infty} \mathbb{E}Y^2(t)=0$.
ATTEMPT (mix of my calculations/assumptions and my lecturer's few hints and steps)
(A) Subtracting 2 linear solutions for $X^{(1)}(t)$ and $X^{(2)}(t)$ we get $Y(t)=X^{(1)}(t)-X^{(2)}(t)$. Therefore, we have $dY(t)=\mu Y(t)dt+\sigma Y(t)dW$
(B)
The unique solution of $dY(t)=\mu Y(t)dt+\sigma Y(t)dW$ is $$Y(t)=Y(0)e^{\left(\mu-\frac{\sigma^2}{2}\right)t+\sigma W(t)}.$$ So we have $$Y^2(t)=Y^2(0)e^{2\left[\left(\mu-\frac{\sigma^2}{2}\right)t+\sigma W(t)\right]}=Y^2(0)e^{\left[\left(2\mu-\sigma^2\right)t+2\sigma W(t)\right]}=Y^2(0)e^{\left(2\mu-\sigma^2\right)t}e^{2\sigma W(t)}$$ And taking expectation of it gives: $$\mathbb{E}Y^2(t)=\mathbb{E}Y^2(0)e^{\left(2\mu-\sigma^2\right)t}e^{2\sigma W(t)}=Y^2(0)e^{\left(2\mu-\sigma^2\right)t}\mathbb{E}e^{2\sigma W(t)}=Y^2(0)e^{\left(2\mu-\sigma^2\right)t}e^{2\sigma^2 t}=Y^2(0)e^{\left(2\mu+\sigma^2\right)t}$$
But according to my professor, $\lim_{t\to\infty} \mathbb{E}Y^2(t)=0$ for $\mu<\sigma^2$, however, I must have done something wrong because the exponent is never negative in my solution.