2
$\begingroup$

A stochastic process $(X_t)_{t\in\mathbb R}$ is said to be Gaussian with continuous paths if its paths are almost surely continuous and all its finite dimensional distributions are, possibly non centered, Gaussian. Is it possible to apply dominated convergence to argue that the mean function $$m:\mathbb R\to\mathbb R,\ t\mapsto m_t:=E(X_t)$$ or even the covariance function $$c:\mathbb R^2\to\mathbb R,\ (s,t)\mapsto c(s,t):=E(X_sX_t)-E(X_s)E(X_t)$$ of such a process are continuous?

$\endgroup$
4
$\begingroup$

Consider without loss of generality the case of the almost sure limit $X_t\to X_0$, This convergence implies the almost sure convergence $e^{isX_t}\to e^{isX_0}$ for every fixed real number $s$, hence, by Lebesgue dominated convergence, the convergence of the expectations $E(e^{isX_t})\to E(e^{isX_0})$.

Recall that each random variable $X_t$ is normal with mean $m_t$ and variance $\sigma_t^2$ hence $$E(e^{isX_t})=e^{ism_t-s^2\sigma_t^2/2}$$ Now, for $s=1$, $$e^{-\sigma_t^2/2}=|E(e^{iX_t})|\to |E(e^{iX_0})|=e^{-\sigma_0^2/2}$$ hence $\sigma_t^2\to\sigma_0^2$. Using this, one sees that, for every $s$, $$e^{s^2\sigma_t^2/2}E(e^{isX_t})\to e^{s^2\sigma_0^2/2}E(e^{isX_0})$$ that is, $e^{ism_t}\to e^{ism_0}$. This convergence holds for every $s$ hence $m_t\to m_0$.

$\endgroup$

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.