I'm reading a paper on noise and had a question about the stochastic integral. In the paper, they consider the SDE:
$$dX = \lambda Xdt + \epsilon dW$$
which has the solution $$ X(t) = \epsilon e^{\lambda t} \int_0^t e^{-\lambda s} dW(s) $$
I understand up to there. The next part is a little confusing to me. They state:
The integral in the r.h.s. converges a.s. to a centered Gaussian r.v. $N$ so that for large t, $$|X(t)| \approx \epsilon e^{\lambda t}N $$
I thought the stochastic integral already had a normal distribution (from a time change theorem so it has the same distribution as a Brownian Motion). So why is it that it converges to a Gaussian r.v. and not that it already is a Gaussian r.v?
You are correct that for each finite $t$, the stochastic integral $\int_0^t e^{-\lambda s}\,dW(s)$ has a normal distribution. What is being asserted is that the limit $$N = \lim_{t \to \infty} \int_0^t e^{-\lambda s} \,dW(s)$$ exists almost surely and also has a normal distribution.