# Is $t^{-\frac{1}{2}}B_{t^2}$ a Brownian Motion?

I think the title says it all. Let $X_t = t^{-\frac{1}{2}}B_{t^2}$, with $B_t$ being a brownian motion started at $0$.

I think I have proved continuity at $0$ by doing the following: $$X_t = t^{-\frac{1}{2}}B_{t^2} \stackrel{law}{=} t^{\frac{1}{2}}B_1$$

So now if $t = 0$ then $X_0 = 0$. However, I am not getting the desired results when proving stationarity of increments: \begin{align*} Var(X_t - X_s) &= \mathbb E\left(X_t - X_s \right)^2 \\ &= \mathbb EX_t^2 + \mathbb EX_s^2-2\mathbb EX_tX_s \\ &= t + s -2t^{-1/2}s^{-1/2}min(s^2,t^2) \\ & \ne t-s\\ \end{align*}

Similarly for the independence of increments. I guess I must be doing something wrong because by Levy's characterization theorem, the quadratic variation of $X_t$ is $<X,X>_t = t$ so this should be a Brownian Motion. Can somebody point out where the problem might be?.

• What's that "law" you are using to prove continuity at $0$? May 18, 2014 at 10:44
• Two points: (1) The continuity at time $0$ cannot be proved by the identity in law you mention. (2) Lévy's characterization theorem does not apply to your setting because an assumption of this theorem is that $X$ is a (local) martingale and for your process $E(X_t\mid\mathcal F^X_s)=X_s\sqrt{s/t}$, not $X_s$, for every $s\leqslant t$.
– Did
May 18, 2014 at 10:46
• @Did thanks for your answer, but could you explain me why can't I use the equality in law in this particular case? May 18, 2014 at 11:39
• The identity in distribution holds. In which ways do you use it to show the pathwise continuity? I know none.
– Did
May 18, 2014 at 11:41

We can use the following characterization:

The stochastic process $(X_t)$ is a Brownian motion if and only if:

• $(X_t)$ is almost-surely continuous;
• $(X_t)$ is a Gaussian process with mean function $0$ and covariance function $\Gamma(s,t)=s\wedge t$.

The process $(X_t)$ is clearly still a Gaussian process (if we define it to be $0$ at $t=0$), and its mean is $0$. The covariance function is given by $$\Gamma(s,t)=\mathbb E\left[X_tX_s\right]=(st)^{-1/2}t^2\wedge s^2\neq s\wedge t,$$ and therefore $(X_t)$ is not a Brownian motion.

As mentioned by Did in the comments, your proof of continuity does not work. Your mistake is that the equality $B_t\stackrel{law}{=}\sqrt tB_1$ holds for a fixed $t$, and not for the whole stochastic processes (and even if the result held as an equality in distribution of stochastic processes, the result would still not follow).

To show that the process is continuous at $0$, you may use the law of iterated logarithm. First, note that $(t^{-1/2}B_{t^2})_{t\ge0}$ has the same pathwise properties as $(t^{3/2}B_{1/t^2})_{t\ge0}$ since $(tB_{1/t})$ is a Brownian motion. Additionally, $$0\le\limsup_{t\rightarrow0}t^{3/2}\left|B_{1/t^2}\right|=\limsup_{t\rightarrow0}\sqrt{2t\log\log t^{-2}}\frac{\left|B_{1/t^2}\right|}{\sqrt{2t^{-2}\log\log t^{-2}}}=0,$$ almost surely. Therefore, $(X_t)$ is continuous at $t=0$ if you define $X_0=0$.

I should also mention that noting that $(X_t)$ is not a martingale is also sufficient to prove that it is not a Brownian motion.

• thanks, to prove continuity at 0 when $tB_{1/t}$ I can use the law of large numbers for Brownian Motion $(\frac{B_{1/t}}{1/t} \rightarrow 0$ when $t \rightarrow 0 )$, but I don't see a way of applying this in this example. Either way, your answer is more than enough, thanks again. May 19, 2014 at 9:59
• @Alfie, what I had written previously was not very rigorous, so I added a proof of the continuity of $(X_t)$ at $0$.
– Ian
May 19, 2014 at 13:55