$g(z)=z^2~$ What will $~g(y ̅_n )~$ converge to in probability?

Question

Another homework question here.

$~Y_1, Y_2, \cdots, Y_n~$ is i.i.d.

$$g(z)=z^2$$

What will $~g(y ̅_n )~$ converge to in probability?

I'm not sure if Slutsky Theorem has something to do with it.

I'm assuming its simply $~\mu~$?

Answer 1

I assume your $\overline y_n$ is given by $\overline y_n=(Y_1+\cdots Y_n)/n$.

By the continuous mapping theorem (which can be considered Slutsky's theorem's big sister) if $\overline y_n$ converges in probability to a constant $c$, then $g(\overline y_n)$ converges to $c^2$. But without further info about the distribution of the $Y_i$ one does not know if such a $c$ exists. If $E|Y_i|<\infty$, for instance, $c$ exists and is given by $c=EY_1$. But if the $Y_i$ are Cauchy distributed (so $E|Y_i|=\infty$) no such $c$ exists.