Stochastic Processes – Proving Brownian Motion Has Stationary Increments

brownian motionprobability distributionsstochastic-processes

In Oksendal's 'Stochastic Differential Equations', we define Brownian Motion as follows:
Fix $x\in\mathbb{R}^n$ and define for $y\in\mathbb{R}^n$:
$$p(t,x,y)=(2\pi t)^{-n/2}\exp\left(-\frac{|x-y|^2}{2t}\right)$$
If $0\le t_1\le t_2\le\cdots\le t_k$, define a measure $\mu_{t_1,\ldots,t_k}$ on $\mathbb{R}^{nk}$ by:
$$\mu_{t_1,\ldots,t_k}(F_1\times\cdots\times F_k)$$
$$=\int_{F_1\times \cdots\times F_k}p(t_1,x,x_1)p(t_2-t_1,x_1,x_2)\cdots p(t_k-t_{k-1},x_{k-1},x_k)dx_1\cdots dx_k$$
Extend this definition of all finite sequences of $t_i$s by first sorting them in increasing order. Then by Kolmogorov's extension theorem there exists a probability space and a stochastic process $\{B_t\}$ such that the finite distributions of $B_t$ are given by the above measure. We call this Browninan motion starting at $x$.

As an exercise we are asked to show that $\{B_{t+h}-B_t\}_{h\ge 0}$ has the same distribution for all $t$. I however am having trouble showing this. I am not sure how to go from the distributions for the individual $B_t$ to the linear combination $B_{t+h}-B_t$.

EDIT: Following saz's hints:

Note that $B_t$, $B_{t+h}$ are Gaussian random variables with mean $0$. We shall assume they represent Browninan motion in 1 dimension originating at the origin. We have the identities (Oksendal eq 2.2.9):

$$E[(B_t-0)^2]=t$$
$$E[(B_t-0)(B_s-0)]=\min(s,t)$$

Thus the Covariance matrix is given by:

$$
\left(\begin{array}{cc}
t & t \\
t & t+h \\
\end{array}\right)
$$

As $B_t$, $B_{t+h}$ are Gaussian with mean $0$, their sum $B_{t+h}-B_t$ is also Gaussian with mean $0$ and variance:
$$
\sigma^2=\left(\begin{array}{cc}-1 & 1\end{array}\right)\left(\begin{array}{cc}t & t \\ t & t+h \end{array}\right)\left(\begin{array}{c}-1 \\ 1\end{array}\right)=h
$$

Hence the sum is normally distributed with mean $0$ and variance $h$, which is independent of $t$.

Best Answer

Hints:

  1. Set $t_1 = t$, $t_2=t+h$. Conclude from the definition of $\mu_{t_1,t_2}$ that $(B_t,B_{t+h})$ is (jointly) Gaussian with mean vector $m=(0,0)$ and covariance matrix $$C = \begin{pmatrix} t & t \\ t & t+h \end{pmatrix}.$$
  2. Recall that if $(X,Y)$ is (jointly) Gaussian with mean vector $m$ and covariance matrix $C$, then $a \cdot X + b \cdot Y$ is Gaussian with mean $m^T \cdot (a,b)$ and variance $(a,b) \cdot C \cdot (a,b)^T$. Apply this, in order to conclude that $B_{t+h}-B_t \sim N(0,h)$.
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