# Prove weak convergence (proof verification)

We are given a distribution, $$\mathbb{P}(Y_n=\frac k n)=2^{-k}$$ for $$k= 1, 2, 3,\dots$$. Check if this converges in distribution and if it does find the limit distribution.

I think it does, the CDF function for $$Y_n$$ looks as follows (?):

$$F_n=\begin{cases} 0 &nt<1 \\ \sum_{k=1}^{\lfloor nt \rfloor} 2^{-k} &nt\geq1 \end{cases}$$

in the limit we get $$F_n\to\begin{cases} 0 & t\leq0 \\ 1 &t>0 \end{cases}$$

This is not a distribution because it is not right continous however $$F(t)=\begin{cases} 0 & t<0 \\ 1 &t\geq0 \end{cases}$$

is a distribution and $$F_n$$ converges to $$F$$ in points of continuity of $$F$$ right?

So indeed $$F_n$$ converges.

• in the first $\sum_{k=1}^{\lfloor t \rfloor} 2^{-k}$ you should replace $t$ by $nt$. The rest is wrong because of this mistake. – justt Dec 5 '19 at 18:05
• @justt what about now? – Kran Dec 5 '19 at 18:51

## 1 Answer

It looks like you took a limit improperly. Try this hint instead:

Hint: Fix $$\epsilon > 0$$ and calculate $$\mathbb{P}(Y_n > \epsilon)$$. What happens as $$n \to \infty$$?

• I typed it wrong! I did not even considered your hint, but I think it is alright now. – Kran Dec 5 '19 at 18:13
• No in fact it was really wrong, however now after another editing it is right i think. – Kran Dec 5 '19 at 18:16