Past exam of the year 1 of the computer science course of the University of Oxford / 2019 / Discrete mathematics, continuous mathematics, and probability / 9 / c

Write the card values as

a_{1}, \dots, a_{1000}

. Uniform selection gives

E [X] = \frac{1}{1000} \sum_{i} a_{i} = 0, E [X^{2}] = \frac{1}{1000} \sum_{i} a_{i}^{2} = 1.

Thus $E [X] = 0$ and $Var (X) = 1$ , by the variance identity

Var (X) = E [X^{2}] - E [X]^{2}

ii

Every particular card is second with probability

\sum_{999 other first cards} \frac{1}{1000} \frac{1}{999} = \frac{1}{1000} .

Hence $X$ and $Y$ have the same distribution, with means zero and variances one. They are not independent, since a card cannot be drawn twice.

Every ordered pair of distinct cards has probability

1/ (1000 \cdot 999)

. Therefore

Cov (X, Y) = E [X Y] = \frac{\sum _{i \neq = j} a _{i} a _{j}}{1000 \cdot 999} = \frac{( \sum _{i} a _{i} ) ^{2} - \sum _{i} a _{i}^{2}}{1000 \cdot 999} = - \frac{1}{999} .

This negative covariance is the effect of sampling without replacement, not a contradiction of their identical marginals.

Represent the removed half as the first

m = 500

cards in a uniformly random ordering. Every unordered half is then equally likely. Let

X_{1}, \dots, X_{500}

be their values. Each has mean zero and variance one, and every distinct pair has covariance

- 1/999

by the same ordered-pair argument as in (ii).

The general covariance expansion from (b), applied without imposing independence, gives

E [Z] = 0, Var (Z) = 500 - \frac{500 \cdot 499}{999} = \frac{250000}{999} .

This is the finite population correction for a half-sample; treating the draws as independent would give the wrong variance.

Finally

{Z \geq 50} \subseteq {∣ Z ∣ \geq 50}

P (Z \geq 50) \leq \frac{Var ( Z )}{5 0 ^{2}} = \frac{250000}{999 \cdot 2500} = \frac{100}{999} .

The cards returned before this sampling step restore the original population; no conditioning on the earlier two observed values is imposed.

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