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Past exam of the mathematics course of the University of Oxford

2017

Prelims

Mathematics III

{c}

The products $TU$ and $UV$ share $U$, so they are not <independent>. Using the definition of <covariance> and the <independence> of $T,U,V$,
$$
\operatorname{Cov}(TU,UV)=\mathbb E(TU^2V)-\mathbb E(TU)\mathbb E(UV)
=(\mathbb E T)(\mathbb E U^2)(\mathbb E V)-1=2-1=1.
$$
By contrast, $TU$ and $VW$ depend on disjoint sets of <independent random variables>, so they are <independent> and have zero <covariance>. Hence
$$
\boxed{\operatorname{Cov}(TU,UV)=1,\qquad\operatorname{Cov}(TU,VW)=0.}
$$

Solution

Past exam of the mathematics course of the University of Oxford / 2017 / Prelims / Mathematics III / 6 / a / iii / Solution

Past exam of the mathematics course of the University of Oxford / 2017 / Prelims / Mathematics III / 6 / a / iii / Solution

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