1

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

saz
  • 120,083
sharon
  • 13
  • One possibility is to show $\mathbb{E}(X_{t}^2) = t \mathbb{E}(X_1^2)$ for all $t \in \mathbb{Q}$ and then extend the identity using continuity. – saz May 15 '15 at 14:48
  • Thanks! I managed to show it for $t\in\mathbb{Q}$ but actually it's not clear to me how the homogeneity for $t\in\mathbb{R}$ follows... – sharon May 15 '15 at 18:47
  • Do you assume that $\mathbb{E}(X_t^2)<\infty$ for all $t$? – saz May 15 '15 at 19:02
  • I think I've to assume it. In the case $\mathbb{E}(X_t^2)=\infty$ for some $t>0$, then $\mathbb{E}(X_t^2)=\infty$, $\forall t\ge0$ by the additivity and the linearity follows trivially. While the other case is not trivial. Am I right? – sharon May 15 '15 at 21:07

1 Answers1

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 $$\mathbb{E}(X_t^2) = \lim_{n \to \infty} \mathbb{E}(X_{t_n}^2) \stackrel{t_n \in \mathbb{Q}}{=} \lim_{n \to \infty} t_n \mathbb{E}(X_1^2) = t \mathbb{E}(X_1^2).$$
saz
  • 120,083
  • Thank you very much! If I understood it properly, in your point 3 you used the right-continuity+ Doob $L^p$ inequality + dominated convergence in the first equality, the linearity $\forall t\in\mathbb{Q}$ in the second one, and the fact that $t_n\downarrow t$ in the third one. Am I right? – sharon May 18 '15 at 14:03
  • @sharon Yes, exactly. – saz May 18 '15 at 14:21