Hitting time or normalized Brownian motion (divided by square root of t)

What I mean by a normalized Brownian motion is the following: $$U(t) = \frac{W(t)}{\sqrt{t}},$$ where W(t) is a standard Brownian motion (with $\mu = 0$ and $\sigma = 1$). Is there a name for such a stochastic process $U(t)$?

My question is that, for a given two-sided boundary $a > 0$, what is the distribution of the hitting time $T$? $$T(a) = \inf_{t:t>0} \{|U(t)| > a\}$$ Or equivalently, for a given $t>0$, what is the distribution of its supremum before $T$? $$M(t) = \sup_{s:s<t}\{U(s) \}$$

For my particular interest, it would be more helpful to consider all the above with respect to a Brownian motion in discrete time field $W(n), n = 1,2,\ldots$, but only considering $W(t)$ is good enough. Thank you!

• If $W_t$ be a standard Brownian motion then $\mathbb{Var}(W_t)=t$. Indeed $U_t$ is called time scaled Brownian motion such that $\mathbb{E}(U_t)=0$ and $\mathbb{Var}(U_t)=1$ – Behrouz Maleki Jul 11 '16 at 22:31

1 Answer

By a variant of the Law of the Iterated Logarithm, we have $$\limsup\limits_{t\rightarrow 0^{+}}{\frac{|W(t)|}{\sqrt{2t\log\log(1/t)}}} = 1$$ with probability one. It follows that $$\limsup\limits_{t\rightarrow 0^{+}}{|U(t)|} = \limsup\limits_{t\rightarrow 0^{+}}{\frac{|W(t)|}{\sqrt{2t\log\log(1/t)}}\sqrt{2\log\log(1/t)}} = \infty$$ with probability one, since $\lim\limits_{t\rightarrow 0^{+}}{\sqrt{2\log\log(1/t)}} = \infty$. Hence $$T(a) = \inf\{t>0: |U(t)|> a\} = 0$$ with probability one for all $a$.

• Thanks! I see where the problem of my argument is. Then what if we consider a Gaussian random walk instead, in which case we only consider discrete time points so we cannot take t to be infinitesimal? – sjasonma Jul 12 '16 at 1:21