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I need to prove that if $W$ is a Brownian motion then $W'(t)=tW(1/t), t >0, W'(0)=0$ is a Brownian motion.

It is continuous and by using the law of large numbers for Brownian motion it is continuous in 0 aswell. I am able to prove stationary increments, and that the increments are normally distributed with variance of the length of the increment. My difficulty is independent increments.

Let $0<s_1<s_2<\ldots s_n$.

I need to prove that $W'(s_n)-W'(s_{n-1}),W'(s_{n-1})-W'(s_{n-2}),\ldots, W'(s_1)$, are mutually independent. This means I need to prove that:

$s_nW(1/s_n)-s_{n-1}W(1/s_{n-1}), s_{n-1}W(1/s_{n-1})-s_{n-2}W(1/s_{n-2}), \ldots,s_1W(1/s_1)$ are mutually independent.

Since we have that $1/s_n<1/s_{n-1}<\ldots <1/s_1$. I have from the independence of Brownian increments that: $W(1/s_1)-W(1/s_2), W(1/s_2)-W(1/s_3),\ldots, W(1/s_{n-1})-W(1/s_n), W(1/s_n)$ are mutually independent.

I have tried calculating the characteristic functions of the increments, but I don't get it to work. I was only able to do this for two points. Is there any other way to do it? Do you have any idea on how to show the independent increments?

saz
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user119615
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  • Have a look at this question. – saz Oct 01 '16 at 19:46
  • "It is continuous and by using the law of large numbers for brownian motion it is continuous in 0". The whole point of this exercise is actually proving the law of large numbers for Brownian motion. Are you sure you got that part right? – Calculon Oct 01 '16 at 19:55
  • @saz I looked at that question, but in the answers they do not talk about the independent increments, they are proving continuity at 0, that I have seen how to do. They say that everything else is clear. I can't see how the independent increments part is clear. – user119615 Oct 01 '16 at 19:55
  • @Calculon No I have already seen the law of large numbers for Brownian motion, the problem is the independent increments part of the new process. – user119615 Oct 01 '16 at 19:56
  • Rather than independent increments I would suggest that you show the finite dimensional distributions of the time inverted BM. You only need to look at pairs of time instants for that. – Calculon Oct 01 '16 at 19:56
  • @user119615 Do you know the following characterization of Brownian motion? Let $(B_t){t \geq 0}$ be a stochastic process with continuous sample paths. If $(B_t){t \geq 0}$ is Gaussian (i.e. all finite-dimensional distributions are Gaussian), $\mathbb{E}(B_t) = 0$ and $\mathbb{E}(B_s B_t) = \min{s,t}$ for all $s,t \geq 0$, then $(B_t)_{t \geq 0}$ is a (one-dimensional) Brownian motion. – saz Oct 01 '16 at 19:59
  • @Calculon If I only look at two points, I am able using the characteristic function that we have binormal distribution. But how does that solve it? – user119615 Oct 01 '16 at 20:15
  • @saz No I wasn't, what is the theorem called? But if that is to work, do we have to find the finite dimensional distribution for an arbitrary number of points, and show that it is multinormal? Is that any easier than showing that every increment is independent? – user119615 Oct 01 '16 at 20:16
  • @user119615 Not sure if I get your question but you indeed need to show that $W'$ is a Gaussian process. For that take an arbitrary number of distinct time points and show that the joint distribution of $W'$ at those time instants is multivariate normal. Then you show that the mean and the covariance structure are that of BM. Since you also have continuity, you are done. One final thing, since $B_0 = 0$ a.s., you can and should allow degenerate multivariate normality, which makes things easier. – Calculon Oct 01 '16 at 20:21
  • @user119615 As far as I know, this theorem doesn't have a specific name. To show that $(W_t'){t \geq 0}$ is Gaussian, show that you can write any vector of the form $Y := (W{t_1}',\ldots,W_{t_n}')$ as $$Y = A \cdot X$$ for some suitable (deterministic) matrix $A$ and $X := (W_{1/t_1},\ldots,W_{1/t_n})$. Since $X$ is Gaussian (as $(W_t)_t$ is Gaussian), this implies that $Y$ is Gaussian. – saz Oct 01 '16 at 20:24
  • @saz Thank you, but is it so then, that the proof of the independent increments is in the proof of the result you are referring to? – user119615 Oct 01 '16 at 20:28
  • @user119615 Well, yes... see my answer below. – saz Oct 01 '16 at 20:37

1 Answers1

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Hints:

  1. Show that $(W_t')_{t \geq 0}$ is a Gaussian process. To this end, show that any vector $Y := (W_{t_1}',\ldots,W_{t_n}')$ can be written in the form $$Y = A \cdot X$$ for some (deterministic) matrix $A$ and $X := (W_{1/t_1},\ldots,W_{1/t_n})$.

  2. Recall the following statement:

    Let $Y = (Y_1,\ldots,Y_n)$ be a Gaussian random vector. Then $Y_1,\ldots,Y_n$ are (mutually) independent if, and only if, $\text{cov}(Y_i,Y_j) = 0$ for all $i \neq j$.

  3. Since $(W_t')_{t \geq 0}$ is Gaussian, we know that $Y:=(W_{t_1}',W_{t_2}'-W_{t_1}',\ldots,W_{t_n}'-W_{t_{n-1}}')$ is a Gaussian random vector. Check that $\mathbb{E}(W_{t_j}'-W_{t_{j-1}}')=0$ and $$\mathbb{E}\bigg((W_{t_j}'-W_{t_{j-1}}') \cdot (W_{t_k}'-W_{t_{k-1}}')\bigg)=0$$ for all $j \neq k$.
  4. Conclude from step 2 that $(W_t')_{t \geq 0}$ has independent increments.
saz
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