1
$\begingroup$

Consider a sample-continuous stochastic process $\left\{ X_t \right\}_{t \in T}$ s.t. each $X_t$ is real-valued and $$\int_\Omega | X_t(\omega) | ^p \, \mathrm{d} P(\omega)< \infty$$ for all $1 \leq p < \infty$. We denote $\mathbb{E}[X_t]=\mu_t$.

Assume that $$\mathbb{E}\left[\, \left|X_t - \mu_t \right|^2 \, \middle| \, \,|X_0-\mu_0|^2 > \delta \right] > \delta t,$$ i.e. that the paths with an offset in the beginning are expected to "spread out". Is there a nice lower bound for $\text{Var}[X_t]$ in terms of $\text{Var}[X_0]$?

$\endgroup$
  • 1
    $\begingroup$ Your condition doesn't really make sense - the $E$ on the right side is really an integral, and it looks like $\omega$ is the variable of integration, so $\omega$ is not a free variable on the right side. $\endgroup$ – Nate Eldredge Mar 31 '15 at 21:36
  • $\begingroup$ Not writing the $\omega$ doesn't make it better. The left side of that implication is inherently a pointwise (pathwise) statement, and the only way it makes sense without a free variable is to assume it holds for every (or almost every) $\omega$. The right side is integrated over all paths. $\endgroup$ – Nate Eldredge Apr 1 '15 at 3:02
  • $\begingroup$ A conditional statement would make sense: $E\left[|X_t - \mu_t|^2 \left| |X_0 - \mu_0|^2 > \delta\right.\right] > \delta t$. However, I think you need to spend more time understanding how the mechanisms of probability work, and what a random variable really is. $\endgroup$ – Nate Eldredge Apr 1 '15 at 3:04
  • $\begingroup$ Actually I did think about using conditional expectation. Thank you, at least I know what I'm actually asking now! $\endgroup$ – Matias Heikkilä Apr 1 '15 at 5:19

Your Answer

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

Browse other questions tagged or ask your own question.