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Use the central limit theorem to show that for x>0, $$\lim_{n \rightarrow \infty} \frac{1}{3^n} \sum_{k:|3k-2n| \leq \sqrt{2n}x} \binom{n}{k} 2^k = \int^{x}_{-x} \frac{1}{\sqrt{2\pi}}e^{-\frac{u^2}{2}}du.$$

Ok so I know $\int^{x}_{-x} \frac{1}{\sqrt{2\pi}}e^{-\frac{u^2}{2}}du = \phi(x)-\phi(-x)$. I also know the CLT is, if we let $X_1, X_2,...,X_n$ denote the observations of a random sample from a distribution that has mean $\mu$ and variance $\sigma^2$. Then the random variable $Y_n = (\sum^{n}_{i=1}X_i-n\mu)/\sqrt{n}\sigma$ converges in distribution to a random variable which has a normal distribution with mean zero and variance 1. I expanded the RHS for a small n, n=4 for example. The number of k terms that we can accept depends on how big we make x. I am confused on how I am supposed to manipulating the RHS to put in a form for which we could use the CLT. Can any one help me?

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Toss a coin $n$ times, where the probability of head on any toss is $\frac{2}{3}$. Make the natural independence assumption.

If $X$ is the number of heads obtained, then $X$ has binomial distribution, mean $\frac{2}{3}n$, and variance $\frac{2}{9}n$. If $Y=\frac{X}{n}$ then $Y$ has mean $\frac{2}{3}$ and variance $\frac{2}{9n}$, so standard deviation $\frac{\sqrt{2}}{3\sqrt{n}}$.

Rewrite $|3k-2n|\le \sqrt{2n}x$ as $$\left|\frac{k}{n}-\frac{2}{3}\right|\le x\cdot \frac{\sqrt{2}}{3\sqrt{n}}, $$ and then as $$\frac{\left|\frac{k}{n}-\frac{2}{3}\right|}{(\sqrt{2})/(3\sqrt{n})}\le x.$$ The sum of the problem, divided by $3^n$, is the probability that $$\frac{Y}{(\sqrt{2})/(3\sqrt{n})}$$ is $\le x$.

This is because $\frac{1}{3^n}\sum_A \binom{n}{k}2^k$ can be rewritten as $$\sum_A \binom{n}{k}\left(\frac{2}{3}\right)^k\left(\frac{1}{3}\right)^{n-k}.$$

By the Central Limit Theorem, this probability has limit $\Pr(Z\le x)$, where $Z$ is standard normal.

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