# Asymptotic normality implies consistency

I'm trying without success to solve the following exercise in my econometric textbook:

Show that $$\sqrt{N}\left(\widehat{\beta_1} - \beta_1 \right) \xrightarrow{d} \mathcal{N}(0,a^2)$$, where $$a^2$$ is constant, implies that $$\widehat{\beta}$$ is consistent. (Hint: use Slutsky's theorem).

My attempt is to consider $$Y_N \equiv \frac{1}{\sqrt{N}}$$, which converges in probability to the constant $$c=0$$, and $$X_N \equiv \sqrt{N}\left(\widehat{\beta_1} - \beta_1 \right)$$. Then by Slutsky's theorem I conclude that $$X_NY_N = \widehat{\beta_1} - \beta \xrightarrow{d} \mathcal{N}(0, \frac{a^2}{N})$$. Now intuitively since $$\frac{a^2}{N}$$ goes to $$0$$ as $$N \to \infty$$ and the mean is $$0$$, it seems clear that the estimator is consistent... However, I cannot figure out how to connect this result to the definition of convergence in probability.

Your choice of $$Y_N$$ and $$X_N$$ are good, but you applied the theorem incorrectly. It does not make sense to take a limit in $$N$$, and end up with some result that depends on $$N$$ like $$\mathcal{N}(0, a^2/N)$$.

Let $$X \sim \mathcal{N}(0, a^2)$$ so that $$X_N \overset{d}{\to}X$$. You have $$Y_N \overset{p}{\to} c = 0$$. Slutsky's theorem implies $$\hat{\beta}_N - \beta = X_N Y_N \overset{d}{\to} cX = 0.$$

Then use the fact that convergence in distribution to a constant implies convergence in probability.

• Thank you! I was so focused on figuring out the missing step that I didn’t even notice I was applying the theorem incorrectly Commented Feb 9 at 19:47

The accepted answer by the OP is wrong.

Assume that $$X_N = \sqrt{N}\left(\widehat{\beta_1} - \beta_1 \right) \xrightarrow{d} X \sim D$$, where $$D$$ is any proper statistical distribution. So we do not know nothing about whether this distribution has zero-mean, or anything else. Then for $$Y_N = 1/\sqrt{N}$$ we still get

$$X_N Y_N \to_d Xc = X\cdot 0 =0,$$

since $$X$$ is bounded.

This does not prove anything about the consistency of the estimator, obviously since we do not know anything specific about $$D$$, as already said.

Generally speaking, convergence in distribution does not imply convergence in probability. Some additional properties must hold, and these are listed and elaborated in the following thread

https://stats.stackexchange.com/a/379971/28746

In short, convergence in distribution to a zero-mean random variable implies consistency if

1) For the finite distribution of $$\sqrt{N}\left(\widehat{\beta_1} - \beta_1 \right)$$, the $$2+\delta,\; \delta >0$$ absolute moment exists and is finite.

2) The sequences of 1st and 2nd moments of $$\sqrt{N}\left(\widehat{\beta_1} - \beta_1 \right)$$ converge each to a constant.

The last step is intuitively appealing, but it is incorrect because it is imprecise. Think critically about $$\rightarrow_d$$: it is a limit on the distribution function $$F$$. So why then is $$N$$ appearing in the denominator? I think we have an intuitive notion that the normal distribution shrinks to a pointmass. You can make this explicit by applying the definition of convergence in probability, and use a delta-epsilon type argument to show that, for any given $$\epsilon$$ there is an $$N$$ for which the probabilistic statement is satisfied for each $$n\ge N$$.