Why Differences of Correlated Brownian Integrals Form a Multivariate Normal Vector

# Vector of differences of Brownian motion integrals is multivariate normal


# Vector of differences of Brownian motion integrals is multivariate normal












Given a 2-dimensional Wiener process $(W_{1},W_{2})$ with correlation $\rho$.

Let \begin{equation*} X(t):= F(t) + \int_{0}^{t} f(s) dW_{1}(s) + \int_{0}^{t} g(s) dW_{2}(s)\end{equation*} for some nice enough deterministic functions $F$, $f$ and $g$. Let now $0<t_{1}<t_{2}<\ldots < t_{n+1}$ and $k\in\{1,2,\ldots,n+1 \}$. We define $X_{i}:=X(t_{i})$ and \begin{equation*} Y_{k}:= (X_{k}-X_{i})_{i\neq k}=(X_{k}-X_{1},\ldots,\widehat{X_{k}-X_{k}},\ldots,X_{k}-X_{n+1})\in\mathbb{R}^{n}. \end{equation*} I know would like to understand the following claim:

$Y_{k}$ has a multivariate normal distribution.

Any help or references would be appreciated very much.

As a follow-up:

I figured that the definition of 2-dimensional Wiener process $(W_{1},W_{2})$ with correlation $\rho$ is not quite clear to me. I assume that $W_{1}$ and $W_{2}$ being one-dimensional Wiener processes and $corr(W_{1}(t),W_{2}(t))=\rho$ is not enough, isn't it!?

I would assume that we have to assume that $(W_{1}(t),W_{2}(t))$ has a two-dimensional normal distribution with mean 0.

Generally, is there a standard definition for a n-dimensional Wiener process with correlation? If so, I would be happy to get some references.

Else, my guess would be that a n-dimensional stochastic process $W=(W_{1},\ldots,W_{n})$ is a n-dimensional process with:

- $W(0)=0$ a.s.

- $W_{t}$ a.s.-continuous

- increments are independent

- $W_{t}-W_{s}\sim N(0,\Sigma)$, for $t>s$

where

$\Sigma$ is a positive-definite and symmetric matrix with diagonal elements equal to $t-s$

## Answer by Gordon (score 1, accepted)

https://quant.stackexchange.com/a/35973

As noted above, the random vector $Y_k$ is multi-normal if for any combinations \begin{align*} \sum_{i\ne k} a_i (X_k-X_i) \tag{1} \end{align*} is normal. WLOG, we assume that $1<k<n+1$. Note that \begin{align*} \sum_{i\ne k} a_i (X_k-X_i) &=-\sum_{i\ne k}a_i X_i +X_k \sum_{i\ne k} a_i\\ &=-a_{n+1}(X_{n+1}-X_n)\\ &\quad -(a_{n+1}+a_n)(X_n-X_{n-1})\\ &\quad - \cdots \\ &\quad -\sum_{i=k+1}^{n+1} a_i(X_{k+1}-X_k)\\ &\quad +\sum_{i=1}^{k-1}a_i(K_k-X_{k-1})\\ &\quad +\cdots\\ &\quad +a_1(X_2-X_1). \end{align*} Since $(W_1, W_2)$ is a 2-dimensional Brownian motion, by Cholesky decomposition, \begin{align*} W_1(t) &= B_1(t),\\ W_2(t) &= \rho B_1(t) + \sqrt{1-\rho^2}B_2(t), \end{align*} where $B_1$ and $B_2$ are two independent Brownian motions. Then, for $i=2,\ldots, n+1$, \begin{align*} X_i-X_{i-1} &= F(t_i)-F(t_{i-1}) + \int_{t_{i-1}}^{t_i}f(s)dW_1(s) + \int_{t_{i-1}}^{t_i}g(s)dW_2(s)\\ &= F(t_i)-F(t_{i-1}) + \int_{t_{i-1}}^{t_i}(f(s)+\rho g(s))dB_1(s) + \int_{t_{i-1}}^{t_i}\sqrt{1-\rho^2}g(s)dB_2(s) \end{align*} is normal. Moreover, since \begin{align*} (X_2-X_1),\, \ldots, \, (X_{n+1}-X_n) \end{align*} are independent, their combinations \begin{align*} \sum_{i\ne k} a_i (X_k-X_i) \end{align*} is also normal. That is, $Y_k$ is multi-normal.