I'm trying to prove the following implication:
Every sequence of random variables which converges weakly is also stochastically bounded.
I know that this is an implication of Prohorov's Theorem but I would prefer a direct approach. The overall aim is to deduce tightness from weak convergence but I know already that tightness and stochastic boundedness are equivalent.
Could you provide any hint? I've tried a lot but I don't have any idea how to use the weak convergence.
Best Answer
Let $\epsilon >0$. There exists $M$ such that $P\{|X| >M\} <\epsilon$. [ Because the events $\{|X| >M\}$ decrease to the empty set]. There exists $M_1 >M$ such that $M$ and $-M$ are continuity points for the distribution of $X$. [ Because there are at most countably many points where the distribution function of $X$ is not continuous]. Then $P\{|X_n| >M_1\}\to P\{|X| >M_1\}<\epsilon $. Hence there exists $m$ such that $P\{|X_n| >M_1\}<\epsilon $ for $n>m$. Now consider $P\{|X_n| >K\}$ for $n=1,2,...m$. By the argument in the beginning of the proof we can find $K_1,K_2,...,K_m$ such that $P\{|X_n| >K_n\}<\epsilon $ for $n=1,2...,m$. Let $K_0=\max \{K_1,K_2,...,K_m,M_1\}$. Then $P\{|X_n| >K_0\}<\epsilon $ for all $n$.