User:Ajoyt/sandbox

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 probabiity 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 sd from the mean in the unscaled curve.

The normal distribution is defined by the differential equation (DE)


 * $$f'\!(x)\!=\!-\!\,\frac{x-\mu}{\sigma^2}f(x)$$ with 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 $$\textstyle n\!\to\!\infty$$, the discrete derivative becomes the continuous derivative. Hence the proof need show only that, for the unscaled binomial distribution,


 * $$\frac{f'\!(x)}{f\!(x)}\!\cdot\!-\!\,\frac{\sigma^2}{x-\mu} \!\to\! 1$$ as $$ n\!\to\!\infty$$.

The required result can be shown directly:



\begin{align} \frac{f'\!(x)}{f\!(x)}\frac{npq}{np\!\,-\!\,k}\!&=\frac{p\left(n, k + 1\right) - p\left(n, k\right)}{p\left(n, k\right)}\frac{\sqrt{npq}}{c} \\ &= \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.