Almost sure convergence of stochastic process

Suppose that we have a (almost surely) continuous stochastic process $\{ X_{t} \}_{t \geq 0}$ on $[0,1]$ with non-stochastic initial value $X_{0} = x_{0} \in [0,1]$ and exponentially decreasing expectation $E(X_{t}) = x_{0} e^{-\gamma t}$, where $\gamma > 0$.

For the corresponding discrete-time process $\{ X_{n} \}_{n = 0}^{\infty}$, an application of the Markov inequality and the Borel–Cantelli lemma shows that $\lim_{n \rightarrow \infty} X_{n} = 0$ almost surely. Is the same true for the continuous-time process $\{ X_{t} \}$, i.e. do we have $\lim_{t \rightarrow \infty} X_{t} = 0$ almost surely?

Originally, the process $\{ X_{t} \}$ is the (almost surely) unique, continuous, and Markovian solution of the Ito stochastic differential equation \begin{equation*} dX_{t} = -\gamma X_{t} dt + k \sqrt{\gamma} \sqrt{X_{t}^{3}(1-X_{t})}dW_{t}, \ X_{0} = x_{0} \in [0,1] \end{equation*} where $k > 0$ and $W_{t}$ is a Brownian motion. Does this ensure the almost sure convergence towards $0$?

• @Ilya Yes, I did. Using Ito's lemma, I get $dE(X_{t}^{2}) = - 2\gamma E(X_{t}^{2}) dt + 2 \gamma E( X_{t}^{3}(1-X_{t}) ) dt$ and thus $dVar(X_{t}) = 2\gamma E(X_{t}^{3}(1-X_{t})) dt$. But I don't know where that gets me? – tot Jul 12 '13 at 9:25
• @Ilya What do you mean? The Gronwall's inequality gives me a lower bound on $E(X_{t}^{2})$, but no upper bound, as far as I can see. In any event, a bounded/decreasing mean and variance will not imply almost sure convergence. – tot Jul 12 '13 at 10:12
• You are right, nevermind these comments – Ilya Jul 12 '13 at 19:09

By applying Ito's formulq we get $e^{\gamma t}X_t$ is a local martingale, so it is a supermartingale for it is positive, and so it is for $X_t$. Then by some classical argument(for example an exercise in Chapter 1 of Brownian motion and stochastic calculus) we know any positive supermartingale converge almost surely. Denote its limit by $X_{\infty}\geq 0$, Fatou's lemma gives

$$E[X_{\infty}]\leq\liminf_{t\rightarrow\infty}E[X_t]=0$$

which allows to show $X_{\infty}=0$ a.s.

• @Ilya Thanks for your time anyway. – tot Jul 14 '13 at 12:40

@Higgs88 Thank you very much for taking the time to answer this question. I have decided to write out the details myself to see if I understand it right.

We want to prove that if $\{ X_{t} \}_{t \geq 0}$ is a continuous, supermartingale with asymptotically vanishing expectation $\lim_{t \rightarrow \infty} E(X_{t}) = 0$, then $\lim_{t \rightarrow \infty } X_{t} = 0$ almost surely. Following @Higgs88, by Doob's first martingale theorem, $\{ X_{t} \}_{t \geq 0}$ has a limit $X_{\infty} \in [0, \infty[$ almost surely. In particular, we have almost sure convergence in discrete time, i.e. $\lim_{n \rightarrow \infty} X_{n} = X_{\infty}$ almost surely, and thus, by Fatou's lemma applied to the non-negative sequence $\{ X_{n} \}$, we get $$E(X_{\infty}) = E( \lim_{n \rightarrow \infty} X_{n} ) = E( \liminf_{n \rightarrow \infty} X_{n} ) \leq \liminf_{n \rightarrow \infty} E(X_{n}) = \lim_{n \rightarrow \infty} E(X_{n})= 0$$ But by the Markov inequality, $P(X_{\infty} \geq \epsilon ) \leq \frac{1}{\epsilon} E(X_{\infty}) = 0$ for all $\epsilon > 0$, and $X_{\infty} = 0$ almost surely.

To see how to apply the above result to the Ito solution $\{ X_{t} \}$, we need to show that it is a supermartingale. However, put $Y_{t} = e^{\gamma t} X_{t}$ and notice that $dY_{t} = k \sqrt{\gamma} \sqrt{e^{-3\gamma t} Y_{t}^{3} (1-e^{-\gamma t} Y_{t})}dW_{t}$ by Ito's lemma, and $\{ Y_{t} \}$ is a non-negative local martingale and thus a supermartingale. Consequently, $\{ X_{t}\}$ is a supermartingale.

However, I wonder if some of the assumptions might be weakened or removed, in particular the supermartingale assumption.