Central limit theorem replacing radical n with n Does $n(y_n-\mu) \overset{\mathbb d}{\to} \mathcal N(0,\sigma^2/ n^2),$ remembering that $n^{1/2}(y_n-\mu) \overset{\mathbb d}{\to} \mathcal N(0,\sigma^2)?$

Notice a few things.

*

*If we compute the variance of the LHS, $\text{Var}(n(y_n-y)) = n^2\text{Var}(y_n)$. However, $y_n = \frac{1}{n}(x_1+...+x_n)$ where the $(x_i)_{1\leq i\leq n}$ are i.i.d. (I am assuming based on your notation). Therefore, $\text{Var}(y_n) = \tfrac{1}{n^2}(n\cdot \sigma^2) = \tfrac{\sigma^2}{n}$. We conclude from here that the variance of the LHS is equal to $n\cdot \sigma^2$. As $n\to \infty$ this quantity blows up to infinity and there is no limit.

*As $n\to \infty$ the quantity $\frac{\sigma^2}{n^2}\to 0$, therefore you are suggesting that the distribution of $n(y_n-y)$ converges in distribution to $N(0,0)$ which looks very suspicious.

