normal distribution and weak convergence

626 Views Asked by At

is true that if I have the normal distribution $\mu_n \sim N(0,\frac{1}{n^2})$, then it converges weakly to $\mu \sim N(0,0)$?

1

There are 1 best solutions below

1
On

If $X_n$ has $N(0,\frac 1{n^{2}})$ distribution then $EX_n^{2}=\frac 1{n^{2}} \to 0$. This implies that $X_n \to 0$ in probability, hence weakly. (You can interpret $N(0,0)$ as the distribution of the zero random variable though this notation is rarely used).