An event has true probability 1/50. A sampler makes it happen with probability 1/5, and each draw scores the weight (1/50)/(1/5) if the event happens and 0 if not. What is the variance of one score under the sampler? Answer as a fraction in lowest terms.

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Variance of a weighted indicator

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An event has true probability $\frac{1}{50}$ . A sampler makes it happen with probability $\frac{1}{5}$ , and each draw scores the weight $\frac{\frac{1}{50}}{\frac{1}{5}}$ if the event happens and $0$ if not. What is the variance of one score under the sampler? Answer as a fraction in lowest terms.

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Hint

Find the first and second moments of the score under the sampler.

Solution

The weight is $\frac{\frac{1}{50}}{\frac{1}{5}} = \frac{1}{10}$ , so the mean score is $\frac{1}{5} \times \frac{1}{10} = \frac{1}{50}$ .
The second moment is $\frac{1}{5} \times \frac{1}{100} = \frac{1}{500}$ .
The variance is $\frac{1}{500} - \frac{1}{2500} = \frac{1}{625}$ , so the answer is $\frac{1}{625}$ .

Answer: $\frac{1}{625}$

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