i). suppose x has a binomial distribution with probability of success p over n trials. Show that E[1/(1+x)]=[1-(1-p)n+1]/[p(n+1)].
ii). suppose now x has a negative binomial distribution with density
P(X=k)=k-1Cn-1 pnqk-n with q=1-p and K>=n, prove that for any function f(x),
E[qf(x)]=E[(x-n)f(x-1)/(x-1)]
Hi,
I want the sum to begin at , in order to get the sum of probabilities of that binomial (n+1,p).
So I add the m=0 term by putting it in the sum, but I have to subtract it, otherwise there is no longer equality !
What is the value of when m=0 ? It's precisely the red stuff :
Is it clearer ?