Lévy process + scaling property $\implies$ Brownian motion

Question

How can I show that if $\xi_t$ is a Lévy process distributed as $\xi_{t+s}- \xi_s$ for all $t,s \in [0,\infty)$ and has independence of increments, and also is distributed as $\lambda\xi_{\lambda^{-2}t}$ for all $\lambda\in (0,\infty)$, then $\xi_t$ is a Brownian motion of some diffusivity $k\in[0,\infty)$?

Answer 1

Since $(\xi_t)_{t \geq 0}$ has, by assumption, stationary and independent increments, it suffices to show that $\xi_t \sim N(0,t)$ for any $t>0$. Denote by $\psi$ the characteristic exponent of $(\xi_t)_{t \geq 0}$, i.e. $$\mathbb{E}\exp(i \eta \xi_t) = \exp(-t \psi(\eta)), \qquad t \geq 0, \eta \in \mathbb{R}. \tag{1}$$ Fix $t>0$. As

$$\xi_t \sim \lambda \xi_{\lambda^{-2}t}, \qquad \lambda>0 \tag{2}$$

we have

$$\mathbb{E}\exp(i \eta \xi_t) = \mathbb{E}\exp(i \eta \lambda \xi_{t \lambda^{-2}})$$

i.e.

$$\exp(-t \psi(\eta)) = \exp(- t \lambda^{-2} \psi(\lambda \eta)).$$

Because of the uniqueness of the characteristic exponent $\psi$ this implies

$$\psi(\eta) = \lambda^{-2} \psi(\lambda \eta) \quad \text{for all $\lambda>0$, $\eta \in \mathbb{R}$}. \tag{3}$$

If we choose $\eta:=1$ it follows that the limit $$d := \lim_{\lambda \to 0} \frac{\psi(\lambda)}{\lambda^2}$$ exists and $$d = \psi(1).$$

As

$$\frac{\psi(\eta)}{\eta^2} \stackrel{(3)}{=} \frac{\psi(\lambda \eta)}{(\lambda \eta)^2} \xrightarrow[]{ \lambda \to 0} d = \psi(1)$$

we conclude that

$$\psi(\eta) = \psi(1) \eta^2, \qquad \eta \in \mathbb{R},$$

which means that $(\xi_t)_{t \geq 0}$ is a Brownian motion with scaling parameter $\sigma \geq 0$.

Remark: In the multidimensional framework, i.e. if $(\xi_t)_{t \geq 0}$ is a $d$-dimensional Lévy process, we can apply the above reasoning to each of the coordinates.