# Prove the time inversion formula is brownian motion

Let $B=(B_t)_{t\geq 0}$ be a brownian motion. Show the time inversion formula $\hat{B}=(B_t)_t\geq0$ is a brownian motion, where for $t \geq 0$ we set $\hat{B}=0$ for $t=0$ and $\hat{B}=tB_{1/t}$ for $t>0$.

I cant figure out how to do this question , can someone help me ? Thanks

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What defines a Brownian motion? What do you have to check here? – Davide Giraudo Aug 13 '12 at 15:33

## 2 Answers

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 $e^{aW_t}$ 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_t \right|}{t} \geq \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)

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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 $Cov(\hat{B}_s,\hat{B}_t)=s\wedge t$. We want to use:

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

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.) $\mu$ should be 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

$$P[\lim_{t\downarrow 0}\hat{B}_t=0]=Q[x\in C(0,1]|\lim_{t\downarrow 0}x(t)=0]=\mu[x\in C(0,1]|\lim_{t\downarrow 0}x(t)=0]=P[\lim_{t\downarrow 0}B_t=0]=1$$

cheers

math

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this is exactly what I was looking for – mastro Jul 28 '15 at 9:40