Chapter 15.8, Problem 8P

Carry through the following details of a derivation of (8.3). Start with (7.3); we want an approximation to (7.3) for large n. First approximate the factorials in $C(n,x)$ by Stirling’s formula (Chapter 11, Section 11 ) and simplify to get $f(x)~{\left(\frac{np}{x}\right)}^x{\left(\frac{nq}{n-x}\right)}^{n-x}\sqrt{\frac{n}{2\pi x(n-x)}}$.

Show that if $\delta =x-np,$ then $x=np+\delta$ and $n-x=nq-\delta .$ Make these substitutions for $x$ and $n-x$ in the approximate $f(x).$ To evaluate the first two factors in $f(x)$ (ignore the square root for now): Take the logarithm of the first two factors; show that $\ln \frac{np}{x}=-\ln \left(1+\frac{\delta }{np}\right)$ and a similar formula for $\ln [nq/(n-x)];$ expand the logarithms in a series of powers of $\delta /(np),$ collect terms and simplify to get $\ln {\left(\frac{np}{x}\right)}^x{\left(\frac{nq}{n-x}\right)}^{n-x}~-\frac{{\delta }^2}{2npq}\left(1+\text{ powers of }\frac{\delta }{n}\right)$.

Hence ${\left(\frac{np}{x}\right)}^x{\left(\frac{nq}{n-x}\right)}^{n-x}~e^{-{\delta }^2/(2npq)}$ for large n. [We really want $\delta /n$ small, that is, $x$ near enough to its average value np so that $\delta /n=(x-np)/n$ is small. This means that our approximation is valid for the central part of the graph (see Figures 7.1 to 7.3) around $x=np$ where $f(x)$ is large. Since $f(x)$ is negligibly small anyway for $x$ far from np, we ignore the fact that our approximation may not be good there. For more detail on this point, see Feller, p. 192]. Returning to the square root factor in $f(x),$ approximate $x$ by np and $n-x$ by nq (assuming $\delta \ll np$ or nq) and obtain (8.3).