Discussion
Suppose that X is a non-negative integer valued random variable having \(\varphi\) as its probability generating function. Then, variance (X) = 1, when :
Answer: D
\(\text{Var}(X) = 1\) for \(\varphi(s) = (s + s^{3})/2\).
That pgf puts probability 1/2 on each of the values 1 and 3, so \(E(X) = 2\) and \(\text{Var}(X) = \frac{1}{2}(1 - 2)^{2} + \frac{1}{2}(3 - 2)^{2} = 1\). Equivalently \(\varphi'(1) = 2\), \(\varphi''(1) = 3\), and \(\text{Var} = \varphi''(1) + \varphi'(1) - \varphi'(1)^{2} = 3 + 2 - 4 = 1\).
The others give different variances:
- \((1 + s)/2\) is Bernoulli(1/2), variance 1/4.
- \(1/(2 - s) = \frac{1/2}{1 - s/2}\) is geometric on {0, 1, 2, ...} with \(p = 1/2\), variance \(q/p^{2} = 2\).
- \(s/(3 - 2s) = \frac{s/3}{1 - 2s/3}\) is geometric on {1, 2, ...} with \(p = 1/3\), variance \(q/p^{2} = 6\).
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