I'm working on the following exercise:
Show that G(s)=1−α(1−s)β is the probability generating function of a nonnegative integer valued random variable when α,β∈(0,1).
I tried the following:
The probability generating function of a discrete random variable X is defined by GX(s)=E(sX)=∑∞k=0sk⋅P(X=k), and thus G(k)X(0)=k!⋅P(X=k), where G(k)X(s) denotes the k'th derivative with respect to s. Thus P(X=k)=G(k)X(0)k!. Working this out I find:
P(X=0)=G(0)X(0)0!=G(0)X(0)=GX(0)=1−αP(X=1)=G(1)X(0)1!=αβP(X=2)=G(2)X(0)2!=αβ(1−β)2! ⋮P(X=n)=G(n)X(0)n!=αβ(1−β)(2−β)⋯(n−β)n!
If α,β∈(0,1) it follows that all probabilities P(X=k), for k∈N∪{0} are in (0,1). For this nonnegative integer valued random variable I will assume that P(X=x)=0 for x∉N∪{0}.
To show that this is indeed the probability generating function of some nonnegative integer valued random variable I have to show that ∞∑k=0P(X=k)=1−α+∞∑k=1αβ(1−β)(2−β)⋯(k−β)k!=1−α+α∞∑k=1β(1−β)(2−β)⋯(k−β)k! equals 1. I didn't succeed to show this, but I have the feeling I have to use the Binomial Theorem somehow. Any ideas? Thanks in advance!
Answer
For α,β∈(0,1), your G satisfies
- 0<G(s)<1 for all s∈[0,1).
- G is infinitely differentiable on [0,1) with G(n)≥0.
- lim.
So G is a probability generating function.
Edit: You are trying to use that G(s)=\sum_{n=0}^{\infty}\frac{G^{(n)}(0)}{n!}s^n, but you do not have to calculate the power series on the RHS and plug in s=1. You already have its closed form (the LHS) and you can do this directly.
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