# Ballistic Growth

Consider the SDE
$dX(t)=b(X(t))dt +\sigma(X(t))dB(t)$
with $$b(x)\to b_0 >0$$ as $$x\to\infty$$ and with $$\sigma$$ bounded and positive. Suppose that $$b$$ and $$\sigma$$ are such that
$\lim_{t\to\infty}X(t)=\infty$, with probability one for any starting point. Show that
$P_x\Big\{\lim_{t\to\infty}\frac{X(t)}{b_0 t}=1\Big\}=1 \ .$
From
$X(t)=x+\int_0^{t}b(X(s))ds +\int_0^{t}\sigma(X(s))dB(s)$
and the hypotheses, note that the result follows from showing that
\begin{align*}
\mathbf P_x\Big\{\lim_{t\to\infty}\frac{1}{t}\int_0^{t}\sigma(X(s))dB(s)=0\Big\}=1 \ .
\end{align*}

There are a number of ways of thinking about this. In the end they all come down to essentially the same calculations. One way is to show that for some fixed $$\delta \in(0,1)$$ the following statement holds with probability one:

There exist a constants $$C(\omega)$$ so that
\begin{align*}
\int_0^{t}\sigma(X(s))dB(s) \leq Ct^\delta
\end{align*}
for all $$t >0$$.

To show this partition $$[0,\infty]$$ into blocks and use the Doob-Kolmogorov inequality to estimate the probability that the max of $$\int_0^{t}\sigma(X(s))ds$$ on each block excess $$t^\delta$$ on that block. Then use the Borel-Cantelli to show that this happens only a finite number of times.

A different way to organize the same calculation is to estimate
$\mathbf P_x\Big\{\sup_{t>a}\frac{1}{t}|\int_0^t \sigma(X(s))dB(s)|>\epsilon\Big\}$
by breaking the interval $$t>a$$ into the union of intervals of the form $$a2^k <t\leq a2^{k+1}$$ for $$k=0,1,\dots$$ and using Doob-Kolmogorov Martingale inequality. Then let $$a\to\infty$$.