Problem statement:
Let $\mu _n = N(0, 1/n)$ be a normal distribution with mean 0 and variance $1/n$. Does the sequence $\{\mu _n \}$ converge weakly to some probability measure? If yes, to what measure?
Relevant theorems & definitions:
- If $\{ X_n \} \to X$ in probability then $\{X_n \} \to X$ in distribution.
- Let $Y$ be an arbitrary random variable with finite mean $m_Y$. Then for all $\alpha > 0$ we have $P(|Y - m_Y| \geq \alpha) \leq Var(Y) / \alpha ^2$. (Chebychev)
- A sequence of random variables $X_1, X_2, \ldots$, converges in probability to a random variable $X$ if for every $\varepsilon> 0$ $\lim _{n\to \infty} P(|X_n - X| \geq \varepsilon) = 0$.
Attempt at solution:
Take $\mu _n$ with mean 0. From Chebychev's inequality we get $$ \begin{align} P(|\mu _n - 0| \geq \varepsilon) &= P(\mu _n ^2 \geq \varepsilon ^2) \\ &\leq \frac{Var(\mu _n)}{\varepsilon ^2} \\ &= \frac{1}{\varepsilon ^2 n}. \end{align} $$ Then $\lim _{n\to \infty} P(|\mu_n - 0| \geq \varepsilon) = 0$ so $\mu _n \to 0$ in probability and therefore $\mu _n \to 0$ in distribution.
If a sequence of random variables $\mu _n$ converges to a constant $c$ then the distribution is $$ \delta _c (A) = 1_A (c) = \begin{cases} 0 & \mbox{if } c \notin A \\ 1 & \mbox{if } c \in A \end{cases} $$
Let $F$ be the distribution for a point mass at 0.
The problem:
I'm having some trouble finishing the problem. I know the $A$ should be the interval $[x, \infty)$ but I don't know how to tie things together. Any help will be appreciated.