 Solution

ID: 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/ii/solution

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.

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