Let $B(t)$ be standard Brownian motion on $[0.1]$

Define $W(t)$ as follows

$W(t) = B(t) - \int_0^t \frac{B(1)-B(s)}{1-s} \, ds$

Prove $W(t)$ is also Brownian motion

So I'm not sure how to deal with the integral here. In order to show it, too, is Brownian motion I think I would need to

  1. Make an argument that the transformation is linear and hence also Gaussian
  2. Show that $E[W(t_1) - W(t_2)] = 0$ for any $t_1>t_2>0$
  3. Show the variance of $W(t_1) - W(t_2) = t_1 - t_2$ or an equivalent covariance function
  4. Finally make an argument that separate increments are independent. Though this would follow trivially I think from the fact $B(t)$ is brownian motion with independent increments.

But yeah, how do I deal with the integral?

  • $\begingroup$ Hint: Integration is a linear functional. $\endgroup$ – A.S. Oct 31 '15 at 6:59
  • $\begingroup$ Is it valid to say $E[\int[B(1)-B(s)]/(1-s)ds$ $=$$ \int (EB(1) - EB(s))/(1-s)ds$ since the integral is Linear? As that would make arguing the expectation is 0 trivial I guess, and when calculating the variance it would just give me a usual integral in s. By the way I appreciate the help A.S. $\endgroup$ – Patrick Oct 31 '15 at 7:00
  • $\begingroup$ Yes. Expectation is just another integration and you can swap order of well-behaved integrals. $\endgroup$ – A.S. Oct 31 '15 at 7:04
  • $\begingroup$ When you say "well-behaved" would that just mean the expectation doesn't depend on the bounds of the integral, or would it be a variety of things? $\endgroup$ – Patrick Oct 31 '15 at 7:07
  • $\begingroup$ Look up Fubini's theorem $\endgroup$ – user223391 Oct 31 '15 at 7:10

Recall the following characterization of (one-dimensional) Brownian motion

A stochastic process $(W_t)_{t \geq 0}$ is a Brownian motion, if and only if,

  1. $(W_t)_t$ has continuous sample paths.
  2. $(W_t)_t$ is a Gaussian process with mean $0$ and covariance $\mathbb{E}(W_s W_t) = \min\{s,t\}$ for all $s,t \geq 0$.

As $(W_t)_t$ has obviously continuous sample paths, we just have to check the second property.

Since $(B_t)_{t \geq 0}$ is a Brownian motion, it is in particular a Gaussian process and so

$$B_t - \sum_{j=0}^{n-1} (B_1-B(t_j)) \frac{1}{1-t_j} (t_{j+1}-t_j)$$

is Gaussian for each $n \in \mathbb{N}$ where $t_j := \frac{t}{n} j$. If we let $n \to \infty$, then we get

$$W_t = \lim_{n \to \infty} \left( B_t - \sum_{j=0}^{n-1} (B_1-B(t_j)) \frac{1}{1-t_j} (t_{j+1}-t_j) \right)$$

is Gaussian as a limit of Gaussian random variables. Since this argumentation applies in exactly the same way to the joint distributions $(W_{s_1},\ldots,W_{s_m})$ where $s_j \geq 0$, we get that $(W_t)_{t \geq 0}$ is a Gaussian process. It remains to check mean and covariance.

By Fubini's theorem, we have

$$\begin{align*} \mathbb{E}(W_t) &= \underbrace{\mathbb{E}(B_t)}_{0} - \mathbb{E} \left( \int_0^t\frac{B_1-B_s}{1-s} \, ds \right) = - \int_0^t \underbrace{(\mathbb{E}(B_1-B_s)}_{0} \frac{1}{1-s} \, ds = 0. \end{align*}$$

Now fix $r \leq t$.

$$\begin{align*} \mathbb{E}(W_r W_t) &= \mathbb{E}(B_t B_r)- \mathbb{E} \left( B_t \int_0^r \frac{B_1-B_s}{1-s} \, ds \right) - \mathbb{E} \left( B_r \int_0^t \frac{B_1-B_s}{1-s} \, ds \right) \\ &\quad + \mathbb{E} \left( \int_0^t \int_0^r \frac{B_1-B_u}{1-u} \frac{B_1-B_v}{1-v} \, du \, dv \right) \\ &=: \mathbb{E}(B_r B_t) +I_2+I_3+I_4 \end{align*}$$

If we can show that $$I_2+I_3+I_4 = 0$$ we are done. Using $\mathbb{E}(B_u B_v) = \min\{u,v\}$ for any $u,v \in [0,1]$ and Fubini's theorem, we find

$$ \begin{align*} I_2 &= \int_0^r \frac{\mathbb{E}(B_1 B_t-B_tB_s)}{1-s} \, ds = \int_0^r \frac{t-s}{1-s} \, ds \\ &= - \log (1-r) t + r + \log(1-r) \end{align*}$$

as $r \leq t$. Similarly,

$$\begin{align*} I_3 &= \int_0^t \frac{r- \min\{r,s\}}{1-s} \, ds = \int_0^r \frac{r-s}{1-s} \, ds + \int_r^t \underbrace{\frac{r-r}{1-s}}_{0} \, ds = \int_0^r \frac{r-s}{1-s} \, ds \\ &= (1-\log(1-r)) r + \log(1-r) \end{align*}$$

and, finally,

$$\begin{align*} I_4 &= \int_0^t \int_0^r \frac{1-v-u+ \min\{u,v\}}{(1-u)(1-v)} \, du \, dv \\ &= \int_r^t \int_0^r \frac{1-v-u+ u}{(1-u)(1-v)} \, du \, dv + \int_0^r \int_0^r \frac{1-v-u+ \min\{u,v\}}{(1-u)(1-v)} \, du \, dv \\ &= (t-r) \int_0^r \frac{1}{1-u} \, du + 2 \int_0^r \int_v^r \frac{1}{1-v} \, du \, dv\\ &= -(t-r) \log(1-r) + 2 ((1-\log(1-r))r + \log(1-r)) \end{align*}$$

where we have used in the penultimate equation that

$$\begin{align*} \int_0^r \int_0^r \frac{1-v-u+ \min\{u,v\}}{(1-u)(1-v)} \, du \, dv &= \int_0^r \int_0^v \frac{1}{1-u} \, du \, dv + \int_0^r \int_v^r \frac{1}{1-v} \, du \, dv \\ &= \int_0^r \int_v^r \frac{1}{1-u} \, dv \, du + \int_0^r \int_v^r \frac{1}{1-u} \, dv \, du \\ &= 2 \int_0^r \int_v^r \frac{1}{1-v} \, du \, dv. \end{align*}$$

Adding all up, we get $I_2+I_3+I_4 = 0$ and this finishes the proof.

  • $\begingroup$ Thanks so much, your answer also helped me solve two similar proofs. I can't upvote because of my rep, but your help is very much appreciated. $\endgroup$ – Patrick Oct 31 '15 at 22:17
  • $\begingroup$ @Patrick You are welcome. Thanks for letting me know that the answer helped you. $\endgroup$ – saz Nov 1 '15 at 6:16

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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