Sign up ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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

share|cite|improve this question
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

2 Answers 2

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:

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

share|cite|improve this answer

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.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.