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I am a bit rusty on this. So let us consider the following two dimensional standard Brownian motion issued from zero defined on the probability space $(\Omega, \mathcal{F},\mathbb{P})$ (note that, in particular, $W^0$ and $W^1$ are stochastically independent). We assume that $\mathcal{F}= \sigma(W^0,W^1)$. For every $r \in (-1,1)$, we define the stochastic process $$W^r_t = rW_t^1 + \sqrt{1-r^{2}}W_t^0,\;\;t \geq 0.$$ We write $\mathcal{F}= \sigma(W_u^0, W_u^1: u \leq t)\vee\mathcal{N}$ where $\mathcal{N}$ are the $\mathbb{P}-$null sets of $\mathcal{F}$.

Now I'd like to describe the law of the process $W^r = \lbrace W_t^r: t \geq 0 \rbrace?$ What is the law of the two dimensional process $$\lbrace W_t^r, W^1_t : t\geq 0 \rbrace ?$$ And finally I want to compute explicitly the following quantities: $$\mathbb{E}\left[(W^1_t W^r_t)^2\right],\;\;\mathbb{E}\left[(W^1_t W^r_t)^3\right],\;\;\mathbb{E}\left[(W^1_t W^r_t)^4\right]$$

Now I think that the odd moments all vanish due to the properties of independent brownian motions. Any suggestions?

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  • $\begingroup$ What dont you just make a substitution for W^r and calculate the expectation after expanding everything and use independence? $\endgroup$
    – Lost1
    Jan 21, 2014 at 12:26
  • $\begingroup$ W^r and W^1 are correlated brownian motions with correlation r, unsurprisingly... $\endgroup$
    – Lost1
    Jan 21, 2014 at 12:27
  • $\begingroup$ Ok I'll write up and answer to my own question and someone can correct it, or not. $\endgroup$ Jan 21, 2014 at 12:35

1 Answer 1

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It's enough to determine the finite-dimensional distributions of the process $\{(W_t^r,W_t^1)\}_{t\geqslant 0}$. Using mutual independence and independence of the increments, it's enough to deduce the distribution of $(W_t^r,W_t^1)$ for a fixed $t$. We have in distribution $(W_t^r,W_t^1)=t(W_1^r,W_1^1)$, so the problem reduced to find the distribution of $(rX+\sqrt{1-r^2}Y, X)$ for $X$ and $Y$ independent standard Gaussian random variable.

Since we now the distribution of $(X,Y)$, we write for $\phi\colon\mathbb R^2\to\mathbb R$ measurable and bounded $$\mathbb E[\phi(rX+\sqrt{1-r^2}Y, X)]=\frac 1{\pi}\iint_{\mathbb R^2}\phi(rx+\sqrt{1-r^2}y,x)\exp(-(x^2+y^2)/2)\mathrm dx\mathrm dy.$$ For a fixed $x$, use the substitution $t:=rx+\sqrt{1-r^2}y$ in order to get a density.

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  • $\begingroup$ I will handle the moments question - thanks for the answer Davide. Let us calculate explicitly the following quantities. $\mathbb{E}[(W^1_t W^r_t)^k]$ for k={2,3,4}. For k = 3 we know from results about Brownian motion that it becomes zero. For k=2 we expand and get $\mathbb{E}[(rW_t^1 W_t^1 + \sqrt{1-r^{2}}W_t^0 W_t^1)^2$ we notice that the terms with $W_t^0.W_t^1$ cancel by independence. We get $r\sigma^{2}$. For k=4 we get by another result from moments of Brownian motion $3r\sigma^{4}$. $\endgroup$ Jan 22, 2014 at 9:24

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