I have a sample of $n$ i.i.d. "observations" of a random process $X(t)$ where $t\in [a,b]=T$.
I want to show that there exists a constant $0<D<\infty$ such that $$\lim_{n\to \infty} P(\inf_{t\in T} \frac{1}{n} \sum_{i=1}^n X_i(t)^2 \geq D a_n) = 1,$$ where $a_n>0$ is a sequence with $a_n \to 0$, as $n \to \infty.$
what i know:
- I know that for all $t\in [a,b]$: \begin{equation} 0< D_1 a_n \leq E(X(t)^2) \leq D_2 a_n <\infty \end{equation} for some constants $D_1$ and $D_2$ (i.e. $\inf$ and $\sup$ of $E(X(t)^2)$ can be bounded.)
- Moreover I know that there exist a constant $0<D_3<\infty$ such that for some sequence $\tilde{a}_n$ with $\tilde{a}_n/a_n \to 0$ as $n \to \infty$: $$\lim_{n\to \infty}P(\sup_{t\in T}|\frac{1}{n}\sum_{i=1}^n X_i(t)^2 - E(X_i(t)^2)|\leq D_3 \tilde{a_n}) = 1.$$
How can I use these results to show the assertion?
Usually I would apply the triangle inequality to get
$$|\frac{1}{n}\sum_{i=1}^n X_i(t)^2 | = |\frac{1}{n}\sum_{i=1}^n X_i(t)^2 - E(X_i(t)^2) + E(X_i(t)^2)| \leq |\frac{1}{n}\sum_{i=1}^n X_i(t)^2 - E(X_i(t)^2)| + E(X_i(t)^2).$$
and hence $$\inf_t \frac{1}{n}\sum_{i=1}^n X_i(t)^2 \leq \sup_{t} |\frac{1}{n}\sum_{i=1}^n X_i(t)^2 - E(X_i(t)^2)| + D_2 a_n.$$ Obivously this attemp then goes in the wrong direction.
edit: the attemp was obviously very wrong. we can't conclude $$\inf_t \frac{1}{n}\sum_{i=1}^n X_i(t)^2 \leq \sup_{t} |\frac{1}{n}\sum_{i=1}^n X_i(t)^2 - E(X_i(t)^2)| + D_2 a_n$$ from the triangle inequality.
However: I think I found a solution (see the answer below ): It turns out (unexpected) that Assertion (2.) was too weak: I updated it.