Past exam of the mathematics course of the University of Oxford / Undated / Prelims / Mathematics III resit / 6 / a / i

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- Solution i

Solution

A joint probability density function integrates to one. Here

1 = C \int_{0}^{1} \int_{0}^{1} u (1 + 2 v) d v d u = C (\frac{1}{2}) (2),

so

C = 1.

This value also makes the probability density function nonnegative everywhere.

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