I am working on an exercise stating as follows:
Let $(X_{t})_{t\in [0,1]}$ be an (uncountable) family of i.i.d random variables with non-degenerate distribution. Prove that no modification of this process can be continuous.
I have some attempt but cannot proceed:
Let $(\Omega,\mathcal{F},\mathbb{P})$ be the probability space over which $(X_{t})$ is defined. Suppose $X_{t}$ has a continuous modification $\tilde{X}_{t}$. That is, there exists a $\Omega_{0}\subset\Omega$ with $\mathbb{P}(\Omega_{0})=1$ such that we can define a random function $\tilde{X}:\Omega_{0}\times T\longrightarrow\mathbb{R}$ that satisfies $\tilde{X}_{t}(\omega)$ is continuous in $t$ for all $\omega\in\Omega_{0}$ and $X$ and $\tilde{X}$ only differ on a set $\Omega\setminus\Omega_{0}$ with probability $0$.
Then, note that, in the aspect of $X_{t}$, not all events are elements of $\mathcal{F}$. For instance $$A:=\{\omega\in\Omega:X(\omega,t)=0\ \text{for all}\ t\in [0,1]\}=\bigcap_{t\in[0,1]}\{X(\omega, t)=0\},$$ since this is an uncountable intersection of measurable events from $\mathcal{F}$.
However, since $\tilde{X}_{t}$ is continuous on, then we can depict $$B:=\{\omega\in\Omega_{0}:\tilde{X}(\omega,t)=0\ \text{for all}\ t\in[0,1]\}=\bigcap_{t\in D}\{\omega\in\Omega_{0}:\tilde{X}(\omega,t)=0\},$$ where $D$ is a dense countable subset of $[0,1]$, for instance, $D=\mathbb{Q}\cap [0,1]$. Therefore, $B\in\mathcal{F}$.
Then I got stuck, and for now I haven't even used the i.i.d and non-degenerate…. what should I do to get contradiction? Thank you!
Best Answer
Hint: let $t_n \in [0,1]$ be any sequence of distinct numbers converging to some $t$. Then the sequence of random variables $Y_n := \tilde{X}_{t_n}$ are iid with a non-degenerate distribution. If $\tilde{X}$ were really continuous, then this sequence would converge a.s. to $\tilde{X}_t$. However, show that a sequence of iid non-degenerate random variables converges with probability 0.
There are a number of ways to do the latter. One approach: use the non-degeneracy of the distribution to show that there exist two disjoint open sets $U,V \subset \mathbb{R}$ such that $P(Y_n \in U) > 0$ and $P(Y_n \in V) > 0$. Use the Borel zero-one law to show that, with probability 1, we have $Y_n \in U$ infinitely often and $Y_n \in V$ infinitely often. When this happens the sequence $Y_n$ is certainly not converging.