# Thread: [SOLVED] &quot;Easy&quot; Convergence Problems

1. ## [SOLVED] &quot;Easy&quot; Convergence Problems

Hello, this is supposed to be an easy problem, but I am lost.

Let $\displaystyle X_{n}$ be a sequence of random variables with

$\displaystyle p_{n}(x) = \left\{ %always put [\] { \begin{array}{cl} 1,& x=2+\frac{1}{n}\\ 0,& elsewhere \end{array} \right.$

Why $\displaystyle \lim_{n\rightarrow \infty} p_{n}=0$ for all values of $\displaystyle x$? is it because x just approaches 2? how would you state that more formally?

Also, I don't get this.
The cdf of $\displaystyle X_n$ is
$\displaystyle F_{n}(x) = \left\{ \begin{array}{cl} 0,& x < 2+\frac{1}{n}\\ 1,& x \geq 2+\frac{1}{n} \end{array} \right.$

Why $\displaystyle \lim_{n \rightarrow \infty} F_{n}(x) = \left\{ \begin{array}{cl} 0,& x \leq 2\\ 1,& x > 2 \end{array} \right.$ ?? why the inequalities change?

2. For the first part:

pick an x. The sequence $\displaystyle p_n(x)$ is non-zero at most 1 time. So there exists some m such that $\displaystyle p_n(x)=0$ for $\displaystyle n>m$. So the tail of the sequence converges to 0, since it is identically 0, and thus so does the whole sequence.

For the second part, look at the limit in 3 cases: $\displaystyle x<2,\;x=2,\;x>2$

For $\displaystyle x<2:\,F_n(x)\rightarrow 0$ since $\displaystyle F_n(x)=0$ for all n.

For $\displaystyle x>2:\,F_n(x)\rightarrow 1$ since again for a given x there is some m at which $\displaystyle F_n(x)=1$ for all n>m.

For $\displaystyle x=2:\,F_n(x)\rightarrow 0$ since $\displaystyle F_n(x)=0$ for all n.