1
$\begingroup$

Suppose $\{P_n\}$ and P are probability measures on the real line with corresponding distribution functions $\{F_n\}$ and $F$, respectively.

Prove that $P_n$ converges weakly to P if and only $$\lim_{n \rightarrow \infty} F_n(x) = F(x)$$ at every point x where F is continuous.

We define weak convergence as: Let S be a metric space with its Borel $\sigma$-algebra $\Sigma$. We say that a bounded sequence of positive probability measures $P_n$ on $(S, \Sigma)$, $n = 1, 2, ...,$ converges weakly to the finite positive measure P, and write: $P_n \Rightarrow P$

Could we somehow use the fact that the probability measure for an interval $(a,b)$ = $F(b)-F(a)$ to answer this question?

$\endgroup$
  • $\begingroup$ What is your definition of weak convergence again? I get the double arrow notation, but not your actual working definition. $\endgroup$ – kimchi lover Jan 18 '18 at 3:28
  • $\begingroup$ Indeed, you are missing a condition that defines the convergence. $\endgroup$ – cgrudz Jan 18 '18 at 3:29
  • 1
    $\begingroup$ Weak convergence of measures. Essentially what is given to us by Portmanteau theorem. Here is a link: en.wikipedia.org/wiki/… $\endgroup$ – Francois Wassert Jan 18 '18 at 3:50
  • $\begingroup$ Your question can only make sense as a request to prove part of the Portmanteau theorem, which (in effect) asserts the equivalence of many alternative definitions of weak convergence. So you have to be more precise about exactly what we are assuming "weak convergence" means. Under one definition, what you want to prove is the definition of weak convergence. $\endgroup$ – kimchi lover Jan 18 '18 at 13:16

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.