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?

  • $\begingroup$ What dont you just make a substitution for W^r and calculate the expectation after expanding everything and use independence? $\endgroup$ – Lost1 Jan 21 '14 at 12:26
  • $\begingroup$ W^r and W^1 are correlated brownian motions with correlation r, unsurprisingly... $\endgroup$ – Lost1 Jan 21 '14 at 12:27
  • $\begingroup$ Ok I'll write up and answer to my own question and someone can correct it, or not. $\endgroup$ – Peadar Coyle Jan 21 '14 at 12:35

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.

  • $\begingroup$ 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 $3r\sigma^{4}$. $\endgroup$ – Peadar Coyle Jan 22 '14 at 9:24

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.