Given a sequence of probability measure $P_n$ on $\mathbb R$ with the Boreal sigma algebra, define $f_n = \int e^{ixt}dp_n$, and suppose that $f_n \to f$ for some $f$ bounded measurable, is it true that there exists some $p$ such that $f = \int e^{ixt}dp$?

My intuition tells me that this is false, but I cannot come up with an example. The example that I tried: Let $p_n$ be the Gaussian with mean $0$, variance $\frac{1}{n^2}$, then the characteristic function converges to the constant $1$. However, $1$ is the characteristic function of some dirac delta function.

up vote 1 down vote accepted

Instead of having Gaussian with decreasing variance, let $p_n$ be Gaussian with mean 0 and variance $n$. This gives the counterexample you are looking for. In general the continuity theorem states that if the limiting function $f$ is continuous, then there exists such a $p$, but not otherwise.

Your Answer

By clicking "Post Your Answer", you acknowledge that you have read our updated terms of service, privacy policy and cookie policy, and that your continued use of the website is subject to these policies.

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