Every zero-mean Lévy process has linear variance (wrt $t$)

Question

I'd like to show that every Lévy process with $\mathbb{E}X_t=0, \:\forall t\ge0$ has linear variance, namely $t\mapsto\mathbb{E}X^2_t$ is linear. I showed that indeed the additivity holds, i.e. $t+s\mapsto\mathbb{E}X^2_s+\mathbb{E}X^2_t$, but I still have some problem in showing the homogeneity, i.e. $at\mapsto a\mathbb{E}X_t^2$.

That's my attempt

$t+s\mapsto E[X_t^2]+E[X_s^2]$, in fact

\begin{align} E[X_{t+s}^2]=&E[(X_{t+s}-X_s+X_s)^2]\\ =&E[(X_t-X_s)^2+X_s^2+2X_s(X_{t+s}-X_s)]\\ =&E[(X_t+X_s)^2]+E[X_s^2]+2E[X_{s}(X_{t+s}-X_s)]\\ =&E[X_t^2]+E[X_s^2]+2E[X_s]E[X_{t+s}-X_s]\\ =&E[X_t^2]+E[X_s^2]+2E[X_s]E[X_{t}]\\ =&E[X_t^2]+E[X_s^2] \end{align}

Regarding the homogeneity, it is equivalent to show that for every $t>0$ one has $t\mapsto tE[X^2_1]$, but doing the computation iteratively I cannot manage to obtain the desired result.

Any help is appreciated. Tnx

Answer 1

Hints:

1. It suffices to consider the case that $\mathbb{E}(X_t^2)<\infty$ for all $t \geq 0$.
2. Show that $\mathbb{E}(X_t^2) = t \mathbb{E}(X_1^2)$ for any $t \in \mathbb{Q} \cap [0,\infty)$ using the additivity you have already shown. (Hint: write $t= \frac{m}{n}$ for $m,n \in \mathbb{N}$.)
3. Fix arbitrary $t \in [0,\infty)$ and a sequence $(t_n)_{n \in \mathbb{N}} \subseteq \mathbb{Q} \cap [0,T]$ such that $t_n \downarrow t$. As $(X_t)_{t \geq 0}$ is a martingale, it follows from Doob's inequality that $\sup_{s \in [0,T]} X_s^2 \in L^1$. Using the right-continuity of $(X_t)_{t \geq 0}$ and the dominated convergence theorem, we get $$\mathbb{E}(X_t^2) = \lim_{n \to \infty} \mathbb{E}(X_{t_n}^2) \stackrel{t_n \in \mathbb{Q}}{=} \lim_{n \to \infty} t_n \mathbb{E}(X_1^2) = t \mathbb{E}(X_1^2).$$