Question

15. Consider the mode m of the binomial (n, p) distribution. Use the formula m - int (np+ p) to show the following a) If np happens to be an integer, then m np. b) If np is not an integer, then the most likely number of successes m is one of the two integers to either side of np. c) Show by examples that m is not necessarily the closest integer to np. Neither is m always the integer above np, nor the integer below it.

Answer 1

$The~pmf~of~X,~where~X\sim Bin(n,p):\\ f(x)=\binom{n}{x}p^x(1-p)^{n-x};~x=0,1,2,...,n\\ =0~otherwise\\ \frac{f(x)}{f(x-1)}=\frac{n-x&plus;1}{x}\frac{p}{1-p}\\ Now,~f(x)>f(x-1)~according~as~\frac{n-x&plus;1}{x}\frac{p}{1-p}>1~or,~x<np&plus;p\\ f(x)<f(x-1)~according~as~\frac{n-x&plus;1}{x}\frac{p}{1-p}<1~or,~x>np&plus;p\\ f(x)=f(x-1)~according~as~\frac{n-x&plus;1}{x}\frac{p}{1-p}=1~or,~x=np&plus;p\\ (a)~If~np~is~integer~then~np&plus;p(<np&plus;1)~is~a~fraction~and\\ f(0)<f(1)<...<f(np)>f(np&plus;1)>...>f(n)\\ Hence~mode=np=m\\ (b)~np~is~not~an~integer~then~np&plus;p~is~also~not~an~integer.\\ Take~m=[np&plus;p]~then~m<np<np&plus;p<m&plus;1~or~np<m<np&plus;1<m&plus;1\\ f(0)<f(1)<...<f(m)>f(m&plus;1)>...>f(n)\\ Hence~m~is~one~of~the~two~integers~to~either~side~of~np.\\ (c)~Take~n=10,~p=0.728\\ m=[10*0.728&plus;0.728]=8\\ np=7.28, ~hence~closest~integer~of~np~is~7.~However~the~mode=8.\\$