Fall 2011 Homework 9: Question 7
Consider the following:
Consider a group of n roommate pairs at Harvard (so there are 2n students). Each of these 2n students independently decides randomly whether to take Stat 110, with probability p of "success" (where "success" is defined as taking Stat 110). Let N be the number of students among these 2n who take Stat 110, and let X be the number of roommate pairs where both roommates in the pair take Stat 110. Find E(X) and E(X|N).
Solution: Create an indicator r.v. Ij for the jth roommate pair, equal to 1 if both take Stat 110. Using indicators, symmetry, linearity, we have EX = np^2 and E(X|N) = N(N-1)/2(2n-1). Note that p no longer matters in the latter. Consult iTunes for detail. Historical note: an equivalent problem was first solved in the 1760s by Daniel Bernoulli, a nephew of Jacob Bernoulli, for whom the Bernoulli distribution is named.
"Mathematics is the logic of certainty, but statistics is the logic of uncertainty."
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