7
$\begingroup$

I found this exercise in a book and I don't know how to start:

With the help of probabilistic methods, prove that $$ \lim_{n\rightarrow\infty} \frac{1}{4^n} \sum_{k=0}^n \binom{2n}{k} = 1/2 $$

One cannot use the binomial theorem because the sum just goes to $n$ and not to $2n$ and besides that, am I supposed to use one of the limit theorems in stochastics? If so, which one and how can I approach this exercise.

$\endgroup$
1
  • 5
    $\begingroup$ Yes, you can use the binomial theorem! Hint: $\binom{2n}{k}=\binom{2n}{2n-k}$ $\endgroup$
    – Robert Z
    Feb 2, 2023 at 14:20

2 Answers 2

7
$\begingroup$

Let $(\xi_j)_{j\geqslant 1}$ be an i.i.d. sequence of random variables such that $\mathbb P(\xi_1=1)=\mathbb P(\xi_1=0)=1/2$. Then $\sum_{j=1}^{2n}\xi_j$ has a binomial distribution with parameters $(2n,1/2)$ hence $$ \frac{1}{4^n} \sum_{k=0}^n \binom{2n}{k}=\mathbb P\left(\sum_{j=1}^{2n}\xi_j\leqslant n\right). $$ The central limit theorem will give the answer.

$\endgroup$
1
$\begingroup$

We obtain \begin{align*} \color{blue}{\sum_{k=0}^n\binom{2n}{k}}&=\frac{1}{2}\sum_{k=0}^{n-1}\binom{2n}{k}+\binom{2n}{n}+\frac{1}{2}\sum_{k=n+1}^{2n}\binom{2n}{k}\tag{1}\\ &=\frac{1}{2}\sum_{k=0}^{2n}\binom{2n}{k}+\frac{1}{2}\binom{2n}{n}\\ &\,\,\color{blue}{=\frac{1}{2}2^{2n}+\frac{1}{2}\binom{2n}{n}}\tag{2} \end{align*} In (1) we use the symmetry $\binom{2n}{k}=\binom{2n}{2n-k}$. We use Stirling's approximation formula \begin{align*} n!\sim \left(\frac{n}{e}\right)^n\sqrt{2\pi n} \end{align*} and get \begin{align*} \color{blue}{\binom{2n}{n}}=\frac{(2n)!}{n!n!} &\sim\left(\frac{2n}{e}\right)^{2n}\sqrt{4\pi n}\left(\frac{e}{n}\right)^{2n}\frac{1}{2\pi n}\\ &\,\,\color{blue}{\sim 4^n\frac{1}{\sqrt{\pi n}}}\tag{3} \end{align*}

We conclude from (2) and (3) \begin{align*} \color{blue}{\frac{1}{4^n}}\color{blue}{\sum_{k=0}^n\binom{2n}{k}} &=\frac{1}{4^n}\left(\frac{1}{2}2^{2n}+\frac{1}{2}\binom{2n}{n}\right)\\ &=\frac{1}{2}+\frac{1}{2}\,\frac{1}{4^n}\binom{2n}{n}\\ &\sim\frac{1}{2}+\frac{1}{2}\frac{1}{\sqrt{\pi n}}\color{blue}{\underset{n\to\infty }{\longrightarrow} \frac{1}{2}} \end{align*} and the claim follows.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .