Prove the time inversion formula is brownian motion Let $B=(B_t)_{t\geq 0}$ be a brownian motion. Show the time inversion formula $\hat{B}=(\hat B_t)_t\geq0$ is a brownian motion, where for $t \geq 0$ we set $\hat{B}_0=0$ and $\hat{B}_t=tB_{1/t}$ for $t>0$.
I cant figure out how to do this question , can someone help me ? Thanks 
 A: Using the characterisation of Brownian motion as a unique Gaussian process subject to certain properties, continuity at zero may be demonstrated differently, without the appeal to the Wiener measure. Indeed, we may write the event as:
\begin{align*}
\left\{ \lim_{t\ \to 0}\hat{B}_t=0 \right\} =  \bigcap_{n=1}^{\infty} \bigcup_{m=1}^{\infty} \bigcap_{q \in \mathbb{Q} \cap \left(0,\frac{1}{m}\right]} \left\{ |\hat{B}_q| \leq \frac{1}{n} \right\}.
\end{align*}
Under our probability measure for $t>0$,
\begin{align*}
\mathbb{P}\left[|\hat{B}_t| \leq \frac{1}{n}\right] = \mathbb{P}\left[|B_t| \leq \frac{1}{n}\right],
\end{align*}
so the almost sure continuity follows from that of the original Brownian motion.
A: The only difficulty is to prove the continuity at $\hat{B}_0$, i.e.
$$P\left(\lim _{t \rightarrow 0+} t B_{1/t}=0\right)=1.$$
In other words, we only need to prove
$$P\left(\lim _{t \rightarrow \infty} B_{t} / t=0\right)=1.$$

Lemma. Let $\xi_{1}, \xi_{2}, \cdots$ be i.i.d. random variables with
$E\left|\xi_{1}\right|<\infty$, then  $$  P\left(\lim _{n \rightarrow\infty} \frac{\xi_{n}}{n}=0\right)=1 .  $$

Proof. Notice that
\begin{equation}
   \sum_{n=1}^{\infty}P\left(\left|\xi_{n}\right|>n \varepsilon\right) = E\left\lfloor\frac{|\xi_1|}{\varepsilon}\right\rfloor.
  \end{equation}
Since $E\left|\xi_{1}\right|<\infty$, we have $\Sigma_{n} P\left(\left|\xi_{n}\right|>n \varepsilon\right)<\infty, \forall \varepsilon>0$. Thus we can derive $P\left(\left|\xi_{n}\right|>n \varepsilon\right.$ i.o. $)=0$ from  Borel-Cantelli Lemma, and the lemma is proved.

Corollary. $P\left(\lim _{t \rightarrow \infty} B_{t} / t=0\right)=1$.

Proof. For any integer $n \geqslant 0$, let
$$
  X_{n}=\sup _{n \leqslant t \leqslant n+1}\left(B_{t}-B_{n}\right),
  $$
and then $X_{0}, X_{1}, X_{2}, \cdots$ are i.i.d. random varibles with $E|X_1|<\infty$. The lemma above shows $X_{n} / n \rightarrow 0$ a.s..
Let
$$S_t = B_{\lfloor t\rfloor},$$
$$U_t = \sup _{\lfloor t\rfloor \leqslant r \leqslant \lfloor t\rfloor+1}\left(B_{r}-B_{\lfloor t\rfloor}\right),$$
$$V_t = \sup _{\lfloor t\rfloor \leqslant r \leqslant \lfloor t\rfloor+1}\left(B_{\lfloor t\rfloor} - B_{r}\right)$$
be 3 random processes (which are all piece-wise constant).
The SLLN shows that $B_{n} / n \rightarrow 0$ a.s., and thus $S_{t} / t \rightarrow 0$ a.s.. Using the above fact that $X_{n} / n \rightarrow 0$ a.s., we can derive that $U_{t} / t \rightarrow 0$ a.s., and $V_{t} / t \rightarrow 0$ a.s. for the same reason. Finally, we have $|B_{t} / t|\le |S_{t} / t| + U_{t} / t + V_{t} / t \rightarrow 0$ a.s..
A: We refer to the following theorem:

Theorem. [1] If $B$ is a process such that all the finite-dimensional distributions are jointly normal, $\mathbb{E}B_s = 0$ for all $s$, $\mathrm{Cov}(B_s,B_t) = s$ when $s \leq t$, and the paths of $B_t$ are continuous, then $B$ is a Brownian motion.

Let $\hat{B}_t = t B_{1/t}$ for $t > 0$ and $\hat{B}_0 = 0$. Then everything is clear except for $\mathrm{Cov}(\hat{B}_s,\hat{B}_t) = s$ when $s \leq t$ and the continuity of sample paths.
First, observe that $0 \leq s \leq t$ implies $1/t \leq 1/s$. Thus
$$ \mathrm{Cov}(\hat{B}_s,\hat{B}_t) = \mathrm{Cov}(s B_{1/s}, t B_{1/t}) = st \, \mathrm{Cov}(B_{1/s}, B_{1/t}) = st (1/t) = s.$$
Next, for the continuity of $\hat{B}$, it suffices to show the continuity at zero, i.e.,
$$\lim_{t\to  0} \hat{B}_t = \lim_{t\to  0} t B_{1/t} = \lim_{t\to\infty} \frac{B_t}{t} = 0 \quad \text{a.s.},$$
or equivalently, for any $\epsilon > 0$,
$$ \mathbb{P} \left( \limsup_{t\to\infty} \frac{\left| B_t \right|}{t} > \epsilon \right) = 0. $$
To prove this, we claim the following inequality: For $\lambda, t > 0$,
$$ \mathbb{P} \left( \sup_{s\leq t} \left| B_s \right| \geq \lambda \right) \leq 2e^{-\lambda^2 / 2t}. $$
Indeed, let $a > 0$. Then $e^{aB_t}$ is a submartingale, thus Doob's martingale inequality shows that
$$ \mathbb{P} \left( \sup_{s\leq t} B_s \geq \lambda \right) = \mathbb{P} \left( \sup_{s\leq t} e^{aB_s} \geq e^{a\lambda} \right) \leq \frac{\mathbb{E}\left[ e^{aB_t} \right]}{e^{a\lambda}} = e^{a^2t/2 - a\lambda}. $$
Putting $a = \lambda / t$ to this inequality yields
$$ \mathbb{P} \left( \sup_{s\leq t} B_s \geq \lambda \right) \leq e^{-\lambda^2/2t}. $$
Then the claim follows by the symmetry of the Brownian motion.
Note that
$$ \begin{align*}
\sup_{n \leq s \leq n+1} \frac{\left| B_s \right|}{s} > \epsilon 
& \quad \Longrightarrow \quad \exists s \in [n, n+1] \text{ such that } \left| B_s \right| > \epsilon s \\
& \quad \Longrightarrow \quad \exists s \in [n, n+1] \text{ such that } \left| B_s \right| > \epsilon n \geq \frac{\epsilon (n+1)}{2} \\
& \quad \Longrightarrow \quad \sup_{s \leq n+1} \left| B_s \right| \geq \frac{\epsilon (n+1)}{2}.
\end{align*}$$
Thus we have
$$ \begin{align*}
\mathbb{P} \left( \sup_{t \geq N} \frac{\left| B_t \right|}{t} > \epsilon \right)
&\leq \sum_{n=N}^{\infty} \mathbb{P} \left( \sup_{n \leq t \leq n+1} \frac{\left| B_t \right|}{t} > \epsilon \right)
\leq \sum_{n=N+1}^{\infty} \mathbb{P} \left( \sup_{s \leq n} \left| B_s \right| \geq \frac{\epsilon n}{2} \right) \\
&\leq 2 \sum_{n=N+1}^{\infty} e^{-\epsilon^2 n/8}
= O_{\epsilon}\left( e^{-\epsilon^2 N/8} \right).
\end{align*}$$
Therefore we have
$$\mathbb{P} \left( \limsup_{t\to\infty} \frac{\left| B_t \right|}{t} > \epsilon \right)
= \lim_{N\to\infty} \mathbb{P} \left( \sup_{t \geq N} \frac{\left| B_t \right|}{t} > \epsilon \right) = 0.$$

[1] Richard F. Bass (2011), Stochastic Processes, Cambridge University Press (Theorem 2.4)
A: There is another way to prove this. That the process is Gaussian and continuous for $t>0$ is clear. Furthermore we have $E[\hat{B}_t]=0$ and $\mathrm{Cov}(\hat{B}_s,\hat{B}_t)=s\wedge t$. We want to use:

Brownian Motion is the unique Gaussian process $X$ with continuous path such that $E[X_t]=0$ and $\mathrm{Cov}(X_t,X_s)=s\wedge t:=\min\{t,s\}$.

So all you have to check is the continuity at $0$, i.e. $$P[\lim_{t\downarrow 0}\hat{B}_t=0]=1.$$
We denote with $Q$ the distribution of $\hat{B}$ on $C(0,1]$. (I prove this for $t\in[0,1]$, but there is no problem to extend it to the general case.) Let $\mu$ denote the Wiener measure on $C[0,1]$. Now we have $\mu = Q$ on $C(0,1]$ since they have the same finite dimensional marginal distributions. Therefore
\begin{align*}
P[\lim_{t\downarrow 0}\hat{B}_t=0]&=Q[x\in C(0,1]:\lim_{t\downarrow 0}x(t)=0]\\
&=\mu[x\in C(0,1]:\lim_{t\downarrow 0}x(t)=0]\\
&=P[\lim_{t\downarrow 0}B_t=0]\\
&=1.
\end{align*}
cheers 
math
