Weak convergence of probability measures with fixed second moments

Question

Consider Borel probability measures $\mu_n, \mu$ ($n \in \mathbb{N}$) on $\mathbb{R}$ such that

1. $\mu_n \to \mu$ weakly (test functions are continuous and bounded);
2. $E_{X \sim \mu_n}[X] = 0$ for each $n \in \mathbb{N}$;
3. $E_{X \sim \mu_n}[X^2] = 1$ for each $n \in \mathbb{N}$.

Can we conclude that (i) $E_{X \sim \mu}[X] = 0$ or (ii) $E_{X \sim \mu}[X^2] = 1$?

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Famously, $\mu_n \to \mu$ weakly does not imply convergence in first moments. The typical example is $\mu_n := (1-n^{-1})\delta_0 + n^{-1} \delta_n$, so that $\mu_n \to \delta_0$ while $E_{X \sim \mu_n}[X]=1$ and $E_{X \sim \delta_0}[X] = 0$. This isn't a counterexample for our purposes, however, because $E_{X \sim \mu_n}[X^2] = n$.

Answer 1

As I was writing the question out, I ended up working out the answer.

For question (i), the answer is yes. This is a consequence of uniform integrability. Since

$$ \sup_n E_{X \sim \mu_n}[X^2] = 1 < +\infty $$

we obtain $\sup_n E_{X \sim \mu_n}[|X|] < +\infty$ and consequently $E_{X \sim \mu}[X] = 0$; see, e.g., Theorem 3.5 and equation (3.18) in Billingsley, Convergence of Probability Measures (1999) for the details.

For question (ii), the answer is no. This answer gives an example which can be reworked to fit my question. Taking

$$ X_n = \begin{cases} 0 & \text{w.p. } 1-n^{-1} \\\\ \sqrt{n} & \text{w.p. } n^{-1} \end{cases} $$

we see that $E[X_n] = n^{-1/2}$ and $E[X_n^2] = 1$, while $X_n \to 0$ in distribution. Therefore, the centered sequence $Y_n := X_n - n^{-1/2}$ has $E[Y_n] = 0$ and $\mathrm{Var} (Y_n) = 1$, but converges to $0$ almost surely. Formally, $Y_n \sim \mu_n$, where

$$ \mu_n := (1-n^{-1}) \delta_{-n^{-1/2}} + n^{-1} \delta_{n^{1/2} - n^{-1/2}} $$

has $\mu_n \to \delta_0$ weakly but fails to converge in second moments.

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I had a hard time understanding this sequence, so I made it on Desmos.