Proving Brownian Motion has Stationary Increments

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 $t$. 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: $$\sigma^2=\left(\begin{array}{cc}-1 & 1\end{array}\right)\left(\begin{array}{cc}t & t \\ t & t+h \end{array}\right)\left(\begin{array}{c}-1 \\ h\end{array}\right)=h$$

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

• There is a small typo in the matrix multiplication. The $h$ should be $1$. – Harto Saarinen Oct 20 '17 at 7:56

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 $B_{t+h}-B_t \sim N(0,h)$.
I used another simpler route by looking at the characteristic function, not sure if it is as tight though: $B_{t+h}-B_{t}$ is normally distributed with mean $0$ and variance $t+h-t=h$. Let's call that random variable $Z$
$\hat\phi(u)=E[e^{ui(B_{t+h}-B_{t})}]=E[e^{uiZ}]=e^{-\frac{1}{2}hu^2}$
As you can see it does not depend on $t$, now since the characteristic function defines the distribution of $Z$, the distribution of $Z$ does not depend on $t$.