As
n grows large, for
k in the
neighborhood of
np we can approximate : {n \choose k}\, p^k q^{n-k} \simeq \frac{1}{\sqrt{2 \pi npq}}\,e^{-\frac{(k-np)^2}{2npq}}, \qquad p+q=1,\ p, q > 0 in the sense that the ratio of the left-hand side to the right-hand side converges to 1 as
n → ∞.
Proof The theorem can be more rigorously stated as follows: \left(X\!\,-\!\, np\right)\!/\!\sqrt{npq}, with \textstyle X a binomially distributed random variable, approaches the standard normal as n\!\to\!\infty, with the ratio of the probability mass of X to the limiting normal density being 1. This can be shown for an arbitrary nonzero and finite point c. On the unscaled curve for X, this would be a point k given by :k=np+c\sqrt{npq} For example, with c at 3, k stays 3 standard deviations from the mean in the unscaled curve. The normal distribution with mean \mu and standard deviation \sigma is defined by the
differential equation (DE) :f'\!(x)\!=\!-\!\,\frac{x-\mu}{\sigma^2}f(x) with an
initial condition set by the probability axiom \int_{-\infty}^{\infty}\!f(x)\,dx\!=\!1. The binomial distribution limit approaches the normal if the binomial satisfies this DE. As the binomial is discrete the equation starts as a
difference equation whose limit morphs to a DE. Difference equations use the discrete derivative, \textstyle p(k\!+\!1)\!-\!p(k), the change for step size 1. As \textstyle k\!\to\!\infty, the discrete derivative becomes the
continuous derivative. Hence the proof need show only that, for the unscaled binomial distribution, :\frac{p(n, k+1)-p(n,k)}{p(n,k)}\!\cdot\! \left(-\frac{\sigma^2}{k-\mu}\right) \!\to\! 1 as n\!\to\!\infty, where p(n,k) = {n \choose k}\, p^k q^{n-k}, \mu = np, and \sigma = \sqrt{npq}. The required result can be shown directly: : \begin{align} \frac{p\left(n, k + 1\right) - p\left(n, k\right)}{p\left(n, k\right)}\frac{npq}{np\!\,-\!\,k}\!&= \frac{np - k -q}{kq+q}\frac{\sqrt{npq}}{-c} \\ &= \frac{-c\sqrt{npq} -q}{npq + cq\sqrt{npq}+q}\frac{\sqrt{npq}}{-c} \\ & \to 1 \end{align} The last holds because the term -cnpq dominates both the denominator and the numerator as n\!\to\!\infty. As \textstyle k takes just integral values, the constant \textstyle c is subject to a rounding error. However, the maximum of this error, \textstyle {0.5}/\!\sqrt{npq}, is a vanishing value.
Alternative proof The proof consists of transforming the left-hand side (in the statement of the theorem) to the right-hand side by three approximations. First, according to
Stirling's formula, the factorial of a large number
n can be replaced with the approximation : n! \simeq n^n e^{-n}\sqrt{2 \pi n}\qquad \text{as } n \to \infty. Thus : \begin{align}{n \choose k} p^k q^{n-k} & = \frac{n!}{k!(n-k)!} p^k q^{n-k} \\& \simeq \frac{n^n e^{-n}\sqrt{2\pi n} }{k^ke^{-k}\sqrt{2\pi k} (n-k)^{n-k}e^{-(n-k)}\sqrt{2\pi (n-k)}} p^k q^{n-k}\\&=\sqrt{\frac{n}{2\pi k\left(n-k\right)}}\frac{n^n}{k^k\left(n-k\right)^{n-k}}p^kq^{n-k}\\&=\sqrt{\frac{n}{2\pi k\left(n-k\right)}}\left(\frac{np}{k}\right)^k\left(\frac{nq}{n-k}\right)^{n-k}\end{align} Next, the approximation \tfrac{k}{n} \to p is used to match the
square root above to the desired square root on the right-hand side. : \begin{align}{n \choose k} p^k q^{n-k} & \simeq \sqrt{\frac{1}{2\pi n\frac{k}{n}\left(1-\frac{k}{n}\right)}}\left(\frac{np}{k}\right)^{k} \left(\frac{nq}{n-k}\right)^{n-k}\\&\simeq\frac{1}{\sqrt {2\pi npq}}\left(\frac{np}{k}\right)^{k} \left(\frac{nq}{n-k}\right)^{n-k} \qquad p+q=1\\ \end{align} Finally, the expression is rewritten as an exponential, x = \frac{k-np}{\sqrt{npq}} (the standardized value for
k) is introduced, and the
Taylor Series approximation for ln(1+w) is used: : \ln\left(1+w\right)\simeq w-\frac{w^2}{2}+\frac{w^3}{3}-\cdots Then : \begin{align}{n \choose k} p^k q^{n-k} &\simeq \frac{1}{\sqrt {2\pi npq}}\exp\left\{\ln \left( \left(\frac{np}{k}\right)^{k} \right)+\ln \left( \left(\frac{nq}{n-k}\right)^{n-k}\right)\right\}\\ &=\frac{1}{\sqrt {2\pi npq}}\exp\left\{-k\ln\left(\frac{k}{np}\right)+(k-n)\ln \left(\frac{n-k}{nq}\right)\right\}\\&=\frac{1}{\sqrt {2\pi npq}}\exp\left\{-k\ln\left(\frac{np+x\sqrt{npq}}{np}\right)+(k-n)\ln \left(\frac{n-np-x\sqrt{npq}}{nq}\right)\right\}\\&=\frac{1}{\sqrt {2\pi npq}}\exp\left\{-k\ln\left({1+x\sqrt\frac{q}{np}}\right)+(k-n)\ln \left({1-x\sqrt\frac{p}{nq}}\right)\right\}\qquad p+q=1\\&=\frac{1}{\sqrt {2\pi npq}}\exp\left\{-k\left({x\sqrt\frac{q}{np}}-\frac{x^2q}{2np}+\cdots\right)+(k-n) \left({-x\sqrt\frac{p}{nq}-\frac{x^2p}{2nq}}-\cdots\right)\right\}\\&=\frac{1}{\sqrt {2\pi npq}}\exp\left\{\left(-np-x\sqrt{npq}\right)\left({x\sqrt\frac{q}{np}}-\frac{x^2q}{2np}+\cdots\right)+\left(np+x\sqrt{npq}-n\right) \left(-x\sqrt\frac{p}{nq}-\frac{x^2p}{2nq}-\cdots\right)\right\}\\&=\frac{1}{\sqrt {2\pi npq}}\exp\left\{\left(-np-x\sqrt{npq}\right)\left(x\sqrt\frac{q}{np}-\frac{x^2q}{2np}+\cdots\right)-\left(nq-x\sqrt{npq}\right) \left(-x\sqrt\frac{p}{nq}-\frac{x^2p}{2nq}-\cdots\right)\right\}\\&=\frac{1}{\sqrt {2\pi npq}}\exp\left\{\left(-x\sqrt{npq}+\frac{1}{2}x^2q-x^2q+\cdots\right)+\left(x\sqrt{npq}+\frac{1}{2}x^2p-x^2p-\cdots\right) \right\}\\&=\frac{1}{\sqrt {2\pi npq}}\exp\left\{-\frac{1}{2}x^2q-\frac{1}{2}x^2p-\cdots\right\}\\&=\frac{1}{\sqrt {2\pi npq}}\exp\left\{-\frac{1}{2}x^2(p+q)-\cdots\right\}\\&\simeq\frac{1}{\sqrt {2\pi npq}}\exp\left\{-\frac{1}{2}x^2\right\}\\&=\frac{1}{\sqrt {2\pi npq}}e^\frac{-(k-np)^2}{2npq}\\ \end{align} Each "\simeq" in the above argument is a statement that two quantities are asymptotically equivalent as
n increases, in the same sense as in the original statement of the theorem—i.e., that the ratio of each pair of quantities approaches 1 as
n → ∞. == See also ==