Uniform integrability of $\sqrt{n} Y_n$ for $Y_n \sim \mathcal{N}(0, \frac{\sigma^2}{n})$

Question

Let $Y_n \sim \mathcal{N}(0, \frac{\sigma^2}{n})$.

Then the set $\{\sqrt{n} Y_n\}_{n \ge 1}$ is uniformly integrable since

$$ \sqrt{n} Y_n \sim \sqrt{n}\mathcal{N}\bigg(0, \frac{\sigma^2}{n}\bigg) = \mathcal{N}(0, \sigma^2), $$ is a normal distribution that is not dependent on $n$.

Now suppose instead $Y_n$ is not normally distributed, but it is asymptotically normally distributed, i.e., $Y_n \stackrel{a}{\sim} \mathcal{N}(0, \frac{\sigma^2}{n})$. For example $Y_n$ could be the sample mean for $n$ iid observations.

Once again, consider the set $\{\sqrt{n} Y_n\}_{n \ge 1}$. Is it still uniformly integrable now that $Y_n$ is only asymptotically distributed as $\mathcal{N}(0, \frac{\sigma^2}{n})$?

If it isn't, are there additional assumptions that can be placed on $Y_n$ so that $\{\sqrt{n} Y_n\}_{n \ge 1}$ becomes uniformly integrable?

Answer 1

Here we interpret the notation $Y_n \stackrel{a}{\sim} \mathcal{N}(0, \frac{\sigma^2}{n})$ as there exists sequences of random variables $(\varepsilon_n)$ and $(Z_n)$ such that $\varepsilon_n\to 0$ in probability, $Z_n\sim \mathcal{N}(0, \frac{\sigma^2}{n})$ and $Y_n=Z_n+\varepsilon_n$. Then uniform integrability of $\{\sqrt{n} Y_n\}_{n \ge 1}$ is equivalent to the uniform integrability of $\{\sqrt{n} \varepsilon_n\}_{n \ge 1}$, but it does not need to hold. For example, $\varepsilon_n$ could take the value $n$ with probability $1/n$ and $0$ with probability $1-1/n$.

Answer 2

Any sequence of random variables with bounded second moments is uniformly integrable.