Suppose $\frac1{\sqrt{n}}\sum_{i=1}^nY_i\overset{d}\to N(0,V).$ What is the distribution of$\frac1{\sqrt{n}}\sum_{i=1}^nG(Y_i)$?

Question

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 $\frac1n\sum_{i=1}^nY_i^2\overset{p}\to V$?

Answer 1

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.

Answer 2

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 $$ \mathbb{E}Y^2=Var(Y)+\mathbb{E}^2Y=V+0=V. $$