# 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 $$0$$. 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 $$t$$.

• There is a small typo in the matrix multiplication. The $h$ should be $1$. Commented Oct 20, 2017 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 $t$.