1
$\begingroup$

Suppose $(\mu_n)_{n \in \mathbb{N}}$ is a sequence of probability measures (on $\mathcal{B}(\mathbb{R}$)) that I wish to show converges weakly to $\mu$. One way of doing is of course to show that $\lim\limits_{n \to \infty} F_{\mu_n}(x) = F_\mu(x)$, where $F$ denotes the distribution function) for every continuity point $x \in \mathbb{R}$.

Say I have already proven that this is true for $x$ with either $F_\mu(x) \geq F_{\mu_n}(x)$ for all $n \in \mathbb{N}$ or $F_\mu(x) \leq F_{\mu_n}(x)$ for all $n \in \mathbb{N}$. Can I use this to show this for an arbitray continuity point $x$? I was thinking of taking an arbitrary subsequence and showing that this sequence has a convergent subsequence - but I am not sure how to utilize what I have shown.

$\endgroup$
1
  • $\begingroup$ The answer seems to be no. For example, fix $m\in\mathbb{N}$ an conducer $\mu_n=\delta_{\frac{1}{m}+\frac{1}{n}}$ and $\mu=\delta_0$. We have that $\mu_n\Longrightarrow\nu:=\delta_{\frac{1}{m}}$, and $\lim_n F_{\mu_n}(x)=F_\nu(x)\leq F_\mu(x)$ $\endgroup$ Jun 6 '21 at 11:03
1
$\begingroup$

It might not be sufficient. Take a standard normal random variable $Z$, and for $n\geq 1$ define a sequence of random variables $Z_n:=Z+\frac{(-1)^n}{n}$, Clearly $Z_n \to Z$ in distribution. Now pick an $x\in \mathbb{R}$, note that for even $n$, $F_{Z_n}(x)=\mathbb{P}(Z+\frac{1}{n}\leq x)$ is strictly smaller than $F_{Z}(x)=\mathbb{P}(Z\leq x)$, and for odd $n$,$F_{Z_n}(x)=\mathbb{P}(Z-\frac{1}{n}\leq x)$ is strictly larger than $F_{Z}(x)=\mathbb{P}(Z\leq x)$. Therefore there is no $x$ satisfying your assumptions. Hope this clarifies your doubts.

$\endgroup$

Your Answer

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

Not the answer you're looking for? Browse other questions tagged or ask your own question.