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Let $X_n, n \in \mathbb N$ be a sequence of random variables with finite variances. As $n \to \infty$, are the following two equivalent:

  • $X_n \to N(0, \sigma^2)$ for some $\sigma^2 \in [0, \infty)$,

  • $\frac{X_n}{\sqrt{Var(X_n)}} \to N(0,1)$?

Motivation of my question:

The asymptotic normality of MLE is usually given with its asymptotic variance being inverse of Fisher information under some regularity conditions: $$ \sqrt{n}\big(\hat\theta_\mathrm{mle} - \theta_0\big)\ \xrightarrow{d}\ \mathcal{N}(0,\,I^{-1}(\theta_0)). $$

All of Statistics by Wasserman, however states enter image description here

I was wondering if the two results (or conclusion parts only) about asymptotic normality of MLE are equivalent?

Thanks and regards!

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  • $\begingroup$ One question you might like to ponder is how, if at all, $I_n(\hat{\theta}_n)$ might relate to $I(\theta_0)$ $\endgroup$ – Glen_b -Reinstate Monica May 29 '13 at 21:50
  • $\begingroup$ @Glen_b: Thanks! If I am correct, $I_n(\hat{θ}_n)$ is still an MLE of $I(θ_0)$, and therefore is consistent, i.e. converges to $I(θ_0)$ in probability. $\endgroup$ – Tim May 29 '13 at 22:02
  • $\begingroup$ @Glen_b: I think I was wrong, it is $I(\hat{\theta}_n)$ not $I_n(\hat{\theta}_n)$ which is still an MLE of $I(θ_0)$, and therefore is consistent, i.e. converges to $I(θ_0)$ in probability. $\endgroup$ – Tim May 29 '13 at 22:20
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The two conditions are in general not equivalent. Suppose that $$ X_n = Z_n + \frac{1}{n}T_n, $$ where $Z_n \sim N(0, 1)$ and $T_n$ is Student-$t$ with 2 degrees of freedom. Then, clearly, $$ X_n \Rightarrow N(0, 1), $$ meaning that the cdf of $X_n$ converges to $\Phi$. But, the Student-t has infinite variance, and so $$ Var[X_n] = \infty $$ for all $n$.

Granted, this is something of a pathological example. The "regularity conditions" that Wasserman alludes to are used to ensure that this kind of pathologies don't happen.

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  • $\begingroup$ You can still construct similar examples even if the variance of $X_n$ is finite. For example, suppose that $X_n = Z_n$ with probability $(n-1)/n$ and $\pm 1/\sqrt{n}$ with probability $1/n$. Then $X_n \Rightarrow N(0, 1)$, but $X_n/\sqrt{Var[X_n]} \Rightarrow N(0, 0.5^2)$. $\endgroup$ – Stefan Wager May 30 '13 at 1:11

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