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?

  • 2
    $\begingroup$ $\phi$ is more commonly used for the standard normal density; $\Phi$ is more often used for the cdf. $\endgroup$
    – Glen_b
    Commented Mar 13, 2016 at 9:11

1 Answer 1


The binomial distribution of size $n$ and probability $p$ has probability mass function $P(Y_n=k)=\binom{n}{k}p^k(1-p)^{n-k}$. Setting $p=2/3$ gives:


It follows that the left-hand-side of your equation sneakily represents the probability:


$Y_n$ has mean $\frac{2}{3}n$. Do you now see how to apply the CLT (after a few manipulations)?


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.