Proving independent increments of $W_t:= t W(1/t)$ for a Brownian motion $(W_t)_{t \geq 0}$

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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?

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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.