If $X$ is a Lévy process, then $\operatorname{Var}[X_n]=n\operatorname{Var}[X_1]$

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Let $(\Omega,\mathcal A,\operatorname P)$ be a probability space, $H$ be a $\mathbb R$-Hilbert space, $(X_t)_{t\ge0}$ be an $H$-valued Lévy process on $(\Omega,\mathcal A,\operatorname P)$ and $\mu_t:=\mathcal L(X_t)$ for $t\ge0$.

How can we show that $$\operatorname{Var}[X_n]=n\operatorname{Var}[X_1]\tag1$$ for all $n\in\mathbb N$.

Since $\mu_n=\mu_1^{\ast n}$, we easily see that $$\operatorname E[X_n]=\int x\:\mu_1^{\ast n}({\rm d}x)=n\mu(H)^{n-1}\int x\:\mu({\rm d}x)=n\operatorname E\left[X_1\right]\tag1,$$ but I'm not able to derive $(1)$.

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$$X_n=(X_n-X_{n-1})+(X_{n-1}-X_{n-2})+\cdots (X_2-X_1)+X_1$$ so $$Var (X_n)$$ $$= Var (X_n-X_{n-1})+Var (X_{n-1}-X_{n-2})+\cdots Var (X_2-X_1)+ Var (X_1)$$ $$=n Var(X_1).$$