# 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 $X_{n}$ be a sequence of random variables with

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

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

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

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

2. For the first part:

pick an x. The sequence $p_n(x)$ is non-zero at most 1 time. So there exists some m such that $p_n(x)=0$ for $n>m$. 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: $x<2,\;x=2,\;x>2$

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

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

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