Discussion
Suppose that the conditional distribution of X given that \(N = n\) is binomial with parameters \(n\) and 0.25 and that the marginal distribution of N is geometric with mean 2 and variance 2. Then, variance (X) is equal to :
Answer: C
\(\text{Var}(X) = \frac{8}{16}\).
Use the law of total variance, \(\text{Var}(X) = E[\text{Var}(X \mid N)] + \text{Var}(E[X \mid N])\). Given \(N\), X is binomial \((N, 0.25)\), so \(E[X \mid N] = N/4\) and \(\text{Var}(X \mid N) = N \cdot \frac{1}{4} \cdot \frac{3}{4} = \frac{3N}{16}\). With \(E(N) = 2\) and \(\text{Var}(N) = 2\):
\[ \begin{aligned} \text{Var}(X) &= \frac{3}{16}E(N) + \frac{1}{16}\text{Var}(N) \\ &= \frac{6}{16} + \frac{2}{16} = \frac{8}{16} \end{aligned} \]
So the variance can be determined: only the mean and variance of N are needed, not its full distribution. The value 10/16 adds the two terms with the wrong weights, and 1 ignores the thinning by 0.25.
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