Suppose $\frac1{\sqrt{n}}\sum_{i=1}^nY_i\overset{d}\to N(0,V).$ What is the distribution of$\frac1{\sqrt{n}}\sum_{i=1}^nG(Y_i)$? Suppose 
$$\frac1{\sqrt{n}}\sum_{i=1}^nY_i\overset{d}\to N(0,V).$$ 
Let $G(x)=\int_{-\infty}^xk(u)du$ be a kernel distribution function. 
Can we obtain the asymptotic distribution of $\frac1{\sqrt{n}}\sum_{i=1}^nG(Y_i)$?
Another question is if 
$$\frac1{\sqrt{n}}\sum_{i=1}^nY_i\overset{d}\to N(0,V).$$ 
Can we obtain that
$\frac1n\sum_{i=1}^nY_i^2\overset{p}\to V$?
 A: For the second question - assuming i.i.d and finite fourth moment, using the WLLN you have 
$$
\frac{1}{n}\sum_i Y_i^2\xrightarrow{p}\mathbb{E}Y^2,
$$
where 
$$
\mathbb{E}Y^2=Var(Y)+\mathbb{E}^2Y=V+0=V.
$$
A: For the second question, the answer is no in general. Let $\left(\varepsilon_i\right)_{i\geqslant 1}$ be i.i.d. random variables which take the valued $-1$ and $1$ with probability $1/2$. Define 
$$
Y_i:=\varepsilon_i+Z_i-Z_{i-1},
$$
where $\left(Z_i\right)_{i\geqslant 0}$ is an i.i.d. sequence independent of $\left(\varepsilon_i\right)_{i\geqslant 1}$. Then 
$$
\frac 1{\sqrt n}\sum_{i=1}^nY_i=\frac 1{\sqrt n}\sum_{i=1}^nY_i+\frac 1{\sqrt n}\left(Z_n-Z_0\right)
$$
and since the second term goes to $0$ in probability, the central limit theorem shows that $\left(\frac 1{\sqrt n}\sum_{i=1}^nY_i\right)_{n\geqslant 1}$ converges in distribution to a centered normal law with variance $1$. Moreover, 
$$
\frac 1n\sum_{i=1}^nY_i^2=\frac 1n\sum_{i=1}^n\varepsilon_i^2
+\frac 2n\sum_{i=1}^n\varepsilon_i\left(Z_i-Z_{i-1}\right)+\frac 1n\sum_{i=1}^n\left(Z_i-Z_{i-1}\right)^2.
$$ 
The first term is equal to $1$, the second converges to $0$ and the third to 
$\mathbb E\left[\left(Z_1-Z_{0}\right)^2\right]$ which may be not zero.
