Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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

share|improve this question
1  
What defines a Brownian motion? What do you have to check here? –  Davide Giraudo Aug 13 '12 at 15:33
add comment

2 Answers

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

share|improve this answer
add comment

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)

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.