# Weak convergence to probability measure

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:

1. If $\{ X_n \} \to X$ in probability then $\{X_n \} \to X$ in distribution.
2. 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)
3. 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.

• You have to mind the difference between random variables and their distributions. – Michael Greinecker Mar 7 '18 at 22:45
• You can use Skorohod, but it is probably easier in your case to do things directly. The answer will also depend on which of the several equivalent definitions of weak convergence of random variables you use. – Michael Greinecker Mar 7 '18 at 23:00
• So I can just adjust my solution by saying, for example: Let $X_n$ be a random variable with distribution $\mu _n$. Then $X_n \to 0$ in probability so $X_n \to 0$ in distribution? – docjay Mar 7 '18 at 23:17
• Yes, pretty much. But the probability measures do not converge to "0", they converge to $\delta_0$, the point mass at $0$. – Michael Greinecker Mar 7 '18 at 23:20
• Ok thank you so much for your help. I'm going to try to finish this by myself. – docjay Mar 7 '18 at 23:22