3
$\begingroup$

Suppose that we have a continuous Gaussian process $(X_t)_{t \ge 0}$ with independent increments and $X_0=0$. If the increments are also identically distributed, meaning that $X_b-X_a \stackrel{D}{=} X_t-X_s$ whenever $b-a=t-s$, then $\mathbb{E}[X_t]=ct$ and $Cov(X_t,X_s)=\sigma (s \wedge t)$, where $c$ and $\sigma$ are constants.


Let $a(t)=\mathbb{E}[X_t]$. Since $X_0=0$, we have that the function $a$ satisfies $a(t+s)=a(t)+a(s)$, for all $s,t \ge 0$. Also, notice that if $s<t$, then $Cov(X_t,X_s)=Cov((X_t-X_s)+(X_s-X_0),X_s-X_0)=Var(X_s)$. Putting $b(t)=Var(X_t)$ implies by independence that $b(x+y)=b(x)+b(y)$, for all $x,y \ge 0$. So, if we proof that $a(t)$ and $b(t)$ are continuous functions, we are done, since every continuous function $f$ which satisfies the identity $f(u+v)=f(v)+f(u)$ for all $u, v \in \mathbb{R}$ is linear.

So, how can I proof that $\mathbb{E}[X_t]$ and $Var(X_t)$ are continuous functions on $t$?

$\endgroup$
  • $\begingroup$ do you have any information on the regularity of the sample paths? $\endgroup$ – saz Feb 27 '16 at 12:15
  • $\begingroup$ sorry, the trajectories are also continuous $\endgroup$ – leticia Feb 27 '16 at 12:23
  • $\begingroup$ If $X_b-X_a \stackrel{d}{=} X_t-X_s$ whenever $b-a=t-s$, it is called stationary increments, not independent increments. $\endgroup$ – Shalop Feb 28 '16 at 3:26
  • $\begingroup$ I didn't say this $\endgroup$ – leticia Feb 28 '16 at 12:50
0
$\begingroup$

Let $t_0 \in \mathbb{R}$. Since the trajectories are continuous, we have $\lim \limits_{t \to t_0} X_t(\omega)=X_{t_0}(\omega)$. But, we know that the pointwise convergence implies convergence in distribution. By Paul Lévy theorem, for all sequence $(t_k)_{k \ge 1}$ such that $\lim \limits_{k \to \infty}t_k=t_0$, one has

$$\varphi_{X_{t_k}} \to \varphi_{X_{t_0}} \iff X_{t_k} \stackrel{D}{\to} X_{t_0}$$

Then, since $X_t$ is normal distributed for all $t \ge 0$, it follows that $\mathbb{E}[X_{t_k}]\to \mathbb{E}[X_{t_0}] $ and $Var[X_{t_k}]\to Var[X_{t_0}] $. Since this is valid for all sequences $(t_k)_{k \ge 1}$ such that $\lim \limits_{k \to \infty}t_k=t_0$, one can conclude that the functions given by $a(t)=\mathbb{E}[X_t]$ and $b(t)=Var(X_t)$ are continuous in $t$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.