Convergence in distribution and calculation of variance I encountered the following argument, and I don't quite understand them. 


  The sequence of random variables $(\sqrt n X_n)$ converges in distribution to the random variable $Y \sim \frac{1}{2}\delta_0+\frac{1}{2}N(0, \sigma^2)$. The variance of $Y$ is $\frac{1}{2}\sigma^2$. Half of the time, the sequence converges to $0$ faster than $1/\sqrt n$.


My questions are the followings:

(i) How can I calculate the expectation and variance of $Y$?

(ii) What does it mean by "converges faster"?

Thanks for your help ;)
(i) $E(Y)=0$ and $Var(Y)=\frac{1}{2}\sigma^2$
(ii) $Y =O_{p}(1)$.
(i) $E(Y)=0$ and $Var(Y)=\frac{1}{2}\sigma^2$
(ii) $Y =O_{p}(1)$.