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!
Best Answer
From the general solution to linear SDEs, you have :
$$X_t=e^{-\frac{1}{2}t}\left(X_0+\int_0^t e^{\frac{1}{2}s}dB_t\right)$$
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 : $$X_t \sim \mathcal{N} (e^{-\frac{1}{2}t}.X_0, 2\sinh (t)) $$
unless mistaken (I let you check the algebra) here