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.

  • 1
    $\begingroup$ 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. $\endgroup$ – justt Dec 5 '19 at 18:05
  • $\begingroup$ @justt what about now? $\endgroup$ – Kran Dec 5 '19 at 18:51

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 $n \to \infty$?

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

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