Past exam of the year 1 of the computer science course of the University of Oxford / 2022 / Discrete mathematics, continuous mathematics, and probability / Probability / 7 / b / iv

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- Solution iv

Solution

Let

q_{n} = (1 - 2/ n)^{n}

. The product of two distinct empty-box indicators in uniform allocation is the indicator random variable of both boxes being empty, so

E (I_{i} I_{j}) = q_{n}

. Thus

cov (I_{i}, I_{j}) = q_{n} - p_{n}^{2} = (1 - \frac{2}{n})^{n} - (1 - \frac{1}{n})^{2 n} < 0.

Indeed

0 \leq 1 - 2/ n < (1 - 1/ n)^{2}

for

n \geq 2

, and taking the positive integer power

n

preserves this strict inequality. This proves the negative covariance directly.

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