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

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 $\text{Cov}(X_t,X_s)$,$\lim_{t\rightarrow 1}X_t$

-
It seems that OP is serial posting, which arouses some suspicion about the nature of his posts. –  Jeel Shah Feb 17 '13 at 15:26
What do you mean by $\lim_{x \to 1} X_t$? There is no $x$ at all. And what have you tried? –  saz Feb 17 '13 at 18:00
I'm sorry.. I mistaked, it was $t\rightarrow 1$ –  user62662 Feb 17 '13 at 18:19

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:

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

-

Apply Itô formula for $(X_t)_{t<1}$ and $g(x,t)=x e^{\int_0^t \frac{1}{1-s}ds}$ and let $Y_t:=g(X_t,t)$. Then : $$dY_t = \frac{1}{1-t}Y_tdt + e^{\int_0^t \frac{1}{1-s}ds}dX_t$$ $$dY_t = \frac{1}{1-t}Y_tdt -\frac{1}{1-t}Y_tdt + e^{\int_0^t \frac{1}{1-s}ds} dW_t = e^{\int_0^t \frac{1}{1-s}ds} dW_t$$

Thus $$X_t = e^{-\int_0^t \frac{1}{1-s}ds} \int_0^t e^{\int_0^s \frac{1}{1-u}du} dW_s$$

As $$e^{\int_0^t \frac{1}{1-s}ds} = \frac{1}{1-t}$$ We get $f(t)=1-t$ and $$X_t = f(t)\int_0^t \frac{1}{f(s)}dW_s$$

It converges in probability to $0$ (series of gaussians whose variance tend to $0$ so convergence in law to the constant $0$ thus in probability). As for the covariance function : $$\textrm{cov}(X_t,X_s) = \mathbb{E}(X_t X_s) = f(t)f(s)\int_0^{t\wedge s} \frac{1}{f(u)^2}du = 1- t \vee s$$

Bonus : To show that $X$ converges almost surely to zero you must show that the convergence in probability is so fast that : $$\forall \epsilon>0, \ \sum_{n=1}^{\infty} P(|X_{1-1/n}| >\epsilon) < \infty$$ (see Borel-Cantelli lemma). I believe this is the case here but this needs more work.

-