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I'm looking for a specific kind of proof of the statement $$ \lim_{n\to\infty} \frac1{4^n}\binom{2n}{n} = 0 $$ I know how to show this using Stirling's formula; I have seen the very nice elementary estimates in this question which imply this limit; and I am aware of the proofs of this limit in this question. None of these are quite what I'm looking for right now.

What I want is a probabilistic proof which establishes a general upper bound on $P(X=EX)$, or perhaps $P(|X-EX|<t)$, in terms of some statistical parameters of the random variable $X$, and then applies that result for $X\sim B(2n,\frac12)$. For example, something like:

If $X$ is a random variable with $\operatorname{Var} X>0$ and $a\le X\le b$ a.s., then $P(X=EX)\le$ (something in terms of $\operatorname{Var} X$, $a$, and $b$).

This particular formulation cannot work, because the random variable $Y$ with $P(Y=0)=P(Y=2n)=\frac1{8n}$ and $P(Y=n)=1-\frac1{4n}$ has the same variance and a.s. bounds as a variable with distribution $B(2n,\frac12)$, but $P(Y=EY)\to 1$.

Is there something in the same spirit that does work?

Community
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3 Answers3

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Let $X_n, n\geq 1$ be i.i.d Bernoulli variable with parameter $\dfrac{1}{2}$, and $S_n = \sum_{k=1}^n X_k$. We have $P(S_{2n} = n) = \frac{1}{4^n}{2n \choose n}$

$P(S_{2n} = n) = P(\frac{S_{2n}}{2n} = \frac{1}{2}) = P(\sqrt{2n}(\frac{S_{2n}}{2n} - \frac{1}{2}) = 0) \to P(Y=0) = 0$

where $Y$ follows $\mathcal{N}(0,\frac{1}{4})$ by central limit theorem. The convergence $\to$ can be verified by equivalent conditions of convergence in distributions since the cumulative distribution function of normal variables is continuous.

Petite Etincelle
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  • Yes, this is okay, but it really uses more of the structure of the binomial distribution than I want to use. I'm looking for something that uses only, say, the first few moments, rather than the fact that it's a sum of iid summands. –  Nov 29 '14 at 23:41
  • @StevenTaschuk You cannot bound that on the moments alone: consider $Y_n$ having $P(Y_n=0)=1/2$ , $P(Y_n= \pm n)=1/4$ : it has zero mean, the variance tends to infinity, but $P(Y=E(Y))$ is constant. The CLT seems the way to go. – leonbloy Nov 29 '14 at 23:46
  • Then moments and a.s. bounds would be fine, for example, as in the attempt described in the question. CLT is a fine tool here, but simultaneously overpowered and over-assumptioned to be the kind of thing I'm looking for. –  Nov 29 '14 at 23:51
  • @StevenTaschuk more dangerously, proof of CLT is usually built on estimates of binomial coefficients – cactus314 Dec 04 '14 at 01:58
  • @johnmangual one can also avoid using binomial coefficient to get the proof. For example, write $x^n = e^{n\ln x}$ then use Taylor for $\ln(1+x)$ – Petite Etincelle Dec 04 '14 at 08:14
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Chebyshev's inequality gives an lower bound for the probability of lying sufficiently close to the mean.

Stirling's formula says $\displaystyle\frac{1}{4^n}\binom{2n}{n} \approx \frac{1}{\sqrt{\pi n}} \to 0 $. Another way to read is that for this particular distribution $X = \mathbb{E}[X]$ about $\frac{1}{\sqrt{n}}$ of the time. In fact, the CLT says the random walk at time $n$ is uniformly spread around an interval of size $\frac{1}{2\sqrt{n}}$.

cactus314
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I'm not so confident that what you are looking for is possible. I imagine that it is possible to have $2j-1$ possible outcomes $0,1,2,\cdots,j-2,n,2n-j+2,\cdots,2n$ and have the first $j$ moments correct and have the probability of getting $n$ go to $1$

The fact that $$\frac{\binom{2n}{n}}{4^n}=\frac{2n-1}{2n}\frac{\binom{2n-2}{n-1}}{4^{n-1}}$$ is sufficient to show $$\frac{1}{2\sqrt{n}} \lt \frac{\binom{2n}{n}}{4^n} \lt \frac{1}{\sqrt{3n}}$$ for $n \gt 1.$

One might use that there are many possible outcomes and that those near (enough) to the middle are roughly equal. $$\binom{2n}{n-k} =\binom{2n}{n}\prod_0^{k-1}\frac{n-j}{n+j+1} \gt \binom{2n}{n}\big(\frac{n-k+1}{n+k}\big)^k.$$ For $n$ not too small and $k \approx \sqrt{n},$ easy calculations (if I am not mistaken) give $\binom{2n}{n-k} \gt \frac{1}{e}\binom{2n}{n}.$ This gives $$\frac{\binom{2n}{n}}{4^n} \lt \frac{e}{2+2\sqrt{n}}.$$ One could do better than that by saying $\binom{2n}{n-j} \gt \binom{2n}{n}r^j$ for $r=\frac{n-\sqrt{n}}{n+\sqrt{n}+1}$ and $j \le \sqrt{n}.$

Aaron Meyerowitz
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