5
$\begingroup$

The following is easy to prove and can't possibly be new. But I can't find it printed anywhere despite some effort. Can anyone tell me where it is published?

Let $X_1,X_2,\ldots$ be a sequence of real 1-D random variables with continuous densities $f_1,f_2,\ldots$. Let $\phi(x)$ be the standard normal density. We are given that $$\limsup_{u\in\mathbb R}\,\Bigl|\,\int_{-\infty}^u f_n(x)\,dx - \int_{-\infty}^u \phi(x)\,dx\,\Bigr| \to 0 \text{ as } n\to\infty.$$ It does not necessarily follow that $$\limsup_{u\in\mathbb R}\, \bigl|\, f_n(u)-\phi(u)\bigr|\to 0 \text{ as } n\to\infty,~~~~~~~~(1)$$ but the theorem is that (1) holds if each $f_n$ is log-concave.

An analogous result was published by Bender for integer-valued random variables, but I need the continuous case.

$\endgroup$
1
  • $\begingroup$ You're the Brendan McKay - from ANU? Welcome to CV. (I can't say I've seen this particular result, sorry.) $\endgroup$
    – Glen_b
    Jul 6, 2014 at 12:41

1 Answer 1

3
$\begingroup$

If I understand correctly, then Section 2 (and especially Proposition 2) of the paper "Theoretical properties of the log-concave maximum likelihood estimator of a multidimensional density" by Madeleine Cule and Richard Samworth (2010), may be of use to you.

The assumptions of proposition 2 are convergence in distribution of the corresponding (to the densities) measures, and log-concave densities. Then they prove that the sequence of densities converges to the corresponding density "$\mu$-almost everywhere". I have not checked the proof.

$\endgroup$
0

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.