Is $\left(\overline{X}\right)^2$ a consistent estimator of $(E[X_1])^2$

Question

There are iids, $X_1,X_2,\dots$ and $\mathbb E\left[X^4_1\right]$ has an finite value.

$\theta = \mathbb E[X_1]$, and $\overline{X}$ is a sample mean.

When $n\rightarrow \infty$, is $\left(\overline{X}\right)^2$ a consistent estimator of $\theta^2$?

Answer 1

Because $E(X_1^4)$ is finite the conditions for both the weak and strong law of large numbers apply. Applying the weak law $\bar{X} \overset{p}{\rightarrow} E(X_1)$ as $n \rightarrow \infty$. Because $x^2$ is a continuous function:

$$(\bar{X})^2 \overset{p}{\rightarrow} E(X_1)^2=\theta^2$$

so yes it is a consistent estimator.