Answer to Question #277892 in Statistics and Probability for Johnnie

Question

If X is a random variable such that P (X=x)=Pn,Gp(x) is the pgf f X:n=0,1,2...

Define P[X≥n]=qn.Obtain the pgf of qn in terms of GP(x)

Answer 1

The probability generating function (PGF) of X is G_X(s) = E(s^X), for all s ∈ R for which the sum converges:

$G_{xn}(s)=\displaystyle\sum_{k=1}^{n} s^kP(x=n)=\displaystyle\sum_{k=1}^{n} s^kP_n$

$G_{qn}(s)=1-\displaystyle\sum_{k=1}^{n} s^kP(x\le n)$

$G_{qn}(s)=1-( sP(x=1)+\displaystyle\sum_{k=1}^{2} s^2P(x=2)+...+\displaystyle\sum_{k=1}^{n} s^nP(x=n))$

$G_{qn}(s)=1-\displaystyle\sum_{k=1}^{n}G_{xn}(s)$