# Defining Brownian motion without normality

I read somewhere that Brownian motion can be defined to be a stochastic process with continuous paths, independent and stationary increments such that the mean of the process is zero at all times and the variance at $t$ is equal to $t$. So Gaussanity is not mentioned at all.

The claim is that any process that fits this definition is a Gaussian process. I am trying to understand why this is true. I thought about using Levy's characterization of BM but for that I need to show that a process with the definition above has a certain quadratic variation process. This requires me to make some statement on the fourth moments of the process but I don't see how that would follow from the definition above. Also, I want to use a more elementary method to show normality. Any hints/suggestions?

• Why do you need fourth moments to determine the quadratic variation?
– saz
Apr 8, 2018 at 11:15
• @saz The sum of quadratic increments until time $t$ converges to $t$ in mean but for quadratic variation I need convergence in probability, which is usually shown by proving that the convergence above is in fact in $L^2$. This brings in the fourth order terms. Apr 8, 2018 at 11:27
• In order to apply Lévy's characterization it suffices to show that $(X_t)_{t \geq 0}$ and $(X_t^2-t)_{t \geq 0}$ are martingales. You don't need 4th moments for this.
– saz
Apr 8, 2018 at 11:33
• @saz Thanks a lot. For a square integrable martingale, $M$, $M^2 - [M]$ is also a martingale and by continuity, $[M]=t$. Is there another way, preferably more elementary, to show that normality is implied? Apr 8, 2018 at 11:43

Recall that a stochastic process $(X_t)_{t \geq 0}$ with càdlàg sample paths is called a Lévy process if it has stationary and independent increments. The result which you mention can be formulated as follows:

Let $(X_t)_{t \geq 0}$ be a Lévy process such that $\mathbb{E}(X_t^2)=t$ and $\mathbb{E}(X_t)=0$ for all $t \geq 0$. If $(X_t)_{t \geq 0}$ has continuous sample paths, then $(X_t)_{t \geq 0}$ is Gaussian.

Let me remark that the assumptions on $(X_t)_{t \geq 0}$ can be weakened; in fact, the assumptions $\mathbb{E}(X_t^2)=t$ and $\mathbb{E}(X_t)=0$ can be droped. Any Lévy process with continuous sample paths is Gaussian.

There are several ways to prove the above result:

• Possibility 1: Apply Lévy's characterization of BM. For this you simply have to show that $(X_t)_{t \geq 0}$ and $(X_t^2-t)_{t \geq 0}$ are martingales which is straight-forward using the independence of the increments.
• Possibility 2: Calculate the characteristic function $\mathbb{E}e^{i \xi X_t}$. For this approach one uses the Taylor expansion of the mapping $x \mapsto e^{i \xi x}$. In order to make the approach work, one has to show that $(X_t)_{t \geq 0}$ has moments of a certain order; this can be done using elementary (but very lengthy) computations, see [1, Lemma 9.10], or using the strong Markov property of Lévy processes, see [2,Lemma 8.2]. For further details see [1, Theorem 9.12] or [2,2nd proof of Theorem 8.4].

References:

(1) Schilling & Partzsch: Brownian Motion - An Introduction to Stochastic Processes. De Gruyter. (2nd edition)

(2) Schilling: An Introduciton to Lévy and Feller Proceses. Online Version

• It may be worth emphasizing that using Possibility 1 one sees that the independence of the increments suffices -- the stationarity need not be assumed. Apr 8, 2018 at 16:44