Discussion
If \(p(x) = kp^{x}\), where \(0 \lt p \lt 1\) and \(x = 0, 2, 4, 6, \ldots\), then for what value of k p(x) becomes probability mass function of a r.v. ?
Answer: C
p(x) is a probability mass function when \(k = 1 - p^{2}\).
The probabilities must add to 1 over \(x = 0, 2, 4, \ldots\):
\[ \begin{aligned} k(1 + p^{2} + p^{4} + \ldots) &= 1 \\ \frac{k}{1 - p^{2}} &= 1 \Rightarrow k = 1 - p^{2} \end{aligned} \]
The geometric series converges because \(0 \lt p^{2} \lt 1\).
\(k = 1 - p\) is the constant for support 0, 1, 2, ..., which forgets that only even values are allowed. p and \(p^{2}\) do not make the total 1.
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