How to solve the SDE $dX_t = - \frac{1}{1-t} X_t \, dt + dW_t$?

Question

Suppose we have the stochastic equation $dX_t=-\frac{1}{1-t}X_tdt+dW_t$ with $X_0=0$. I have to prove that exist some function $f=f(t)$ such that the following occurs: $$X_t=f(t)\int_0^t\frac{dW_s}{1-s}$$ and calculate $\text{Cov}(X_t,X_s)$,$\lim_{t\rightarrow 1}X_t$

Answer 1

Since you already know how they solution looks like, the easiest way is to apply Itô's formula and check that it's indeed a solution of the SDE.

Let $g(t,x) := f(t) \cdot x$ and $Y_t := \int_0^t \frac{1}{1-s} \, dW_s$. Then $X_t=g(t,Y_t)$ and by applying Itô's formula we obtain

$$\underbrace{g(t,Y_t)}_{X_t} - \underbrace{g(0,Y_0)}_{0} = \int_0^t f'(s) \cdot Y_s \, ds + \int_0^t \frac{1}{1-s} \cdot f(s) \, dW_s $$

Therefore $(X_t)_t$ is a solution of the given SDE if $$\int_0^t f'(s) \cdot Y_s \, ds + \int_0^t \frac{1}{1-s} \cdot f(s) \, dW_s = \int_0^t - \frac{1}{1-s} \cdot f(s) \cdot Y_s \, ds + W_t$$

Obviously this is equation is satisfied for $f(s):=1-s$. Hence

$$X_t = (1-t) \cdot \int_0^t \frac{1}{1-s} \, dW_s$$

(in particular $X_0 = 0$ and $\lim_{t \to 1} X_t = 0$). I leave it to you to calculate the covariation - that's straightforward:

$$\text{cov}(X_t,X_s) = \mathbb{E}(X_t \cdot X_s) =\mathbb{E}((X_t-X_s) \cdot X_s + X_s^2) = \ldots$$