# Distribution of solution to SDE

Let $$X_0$$ be a standard normal random variable and suppose that $$dX_t=-\frac{1}{2}X_tdt+dB_t.$$ $$X_0$$ is independent of the Brownian motion. Find the distribution of $$X_t$$ for $$t\geq0$$ and find $$\text{cov}(X_t,X_s)$$ for all $$t,s$$.

So far, I have attempted to solve the SDE using Ito's lemma, but this has seemed to imply that there exists no solution of the form $$X_t=f(t,B_t)$$. So I am beginning to think that maybe it is possible to describe the distribution without explicitly solving the SDE. But I'm not sure how to do this - any advice would be greatly appreciated!

• This SDE has a well known explicit solution. May 17 at 10:26

$$X_t=e^{-\frac{1}{2}t}\left(X_0+\int_0^t e^{\frac{1}{2}s}dB_t\right)$$
So what do we have here ? A deterministic term $$e^{-\frac{1}{2}t}X_0$$ and a martingale term (why ?) with deterministic integrand. So such stochastic integral are known as Wiener integrals and are gaussian with expectation $$0$$, and variance equals to the expectation of quadratic variation of the integral (please look up in this forum multiple proofs of this fact) . Using Ito's isometry we get by aggregating all of this : $$X_t \sim \mathcal{N} (e^{-\frac{1}{2}t}.X_0, 2\sinh (t))$$