Show that $X(t)=t W(1/t)$ is a Brownian motion if $W(t)$ is a Brownian motion.

Question

I am trying to solve a past exam question for which I have its answers. I've got to the end, but the very last and simplest line has confused me. I've spotted some errors and corrected them, but I think that this line is correct and I just don't know some theory. Thanks!

QUESTION: Show that $X(t)=t W(1/t)$ is a Brownian motion if $W(t)$ is a Brownian motion. As a hint, we are told that we need to check that $\lim_{t\to 0}X(t)=0$ a.s.

ANSWER:

Like $W(t)$, $X(t)$ is also a Gaussian process.

We need to check that $\mathbb{E}(X(t)X(s))=\min(t,s)$.

I checked it and it is correct.

Now, I have that $$\mathbb{E}\left[X(t)-X(s)\right]^2=t-s$$

for all $s<t$, and

$$\mathbb{E}\left[(X(t)-X(s))(X(s)-X(u))\right]=s-u-s-u=0$$ for $u<s<t$. From this we have that the increments are independent $N(0,t-s)$.

PART THAT CONFUSES ME

According to the answer, by the law of large numbers we have

$$\lim_{t\to 0}X(t)=\lim_{s\to \infty}\frac{W(s)}{s}=0$$

But how does $\lim_{t\to 0}X(t)=\lim_{s\to \infty}\frac{W(s)}{s}$ happen if $X(t)=tW(1/t)$?

Answer 1

It's a basic substitution $s=1/t$.

$$t \rightarrow 0\implies s\rightarrow\infty$$

Answer 2

Here we see that the new process is continuous and 0 at origin. We have $E[Y_t]=tE[W_{1/t}]= 0$ and $ Y_t$ is a Gaussian process also we have the covariance function $ E[Y_sY_t]= st(\dfrac{1}{s} \wedge \dfrac{1}{t}) = s \wedge t.$

Answer 3

I am just a student and i jsut solved the same type of question and i think i found an other elegant way to proove the continuity at zero. First we make a substitution of the variable $t=1/s$ so we get: $ lim_{t\rightarrow 0 }tX_{\frac{1}{t}}=lim_{s\rightarrow \infty }\frac{1}{s}X_{s} $

Now we can writte:
$|\frac{1}{s}X_{s}| = |\frac{1}{s}X_{s} \frac{\sqrt{2slog(log(s))}}{\sqrt{2slog(log(s))}}| = |\frac{X_s}{\sqrt{2slog(log(s))}}||\frac{\sqrt{2slog(log(s))}}{s}|$
Now on one hand by the law of the iterated logarithm . More preciselly this paricular caset of the theorem on the Brownian motion:

We can writte: $lim_{s\rightarrow \infty } |\frac{X_s}{\sqrt{2slog(log(s))}}|=1 \ a.s.$
On the other hand we have: $lim_{s\rightarrow \infty } |\frac{\sqrt{2slog(log(s))}}{\sqrt{s}}|=0 \ a.s.$
So easily we can conclude: $lim_{s\rightarrow \infty } |\frac{1}{s}X_{s}|=1\cdot 0=0 \ a.s.$

Q.E.D.