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Past exam of the year 1 of the computer science course of the University of Oxford

2022

Discrete mathematics, continuous mathematics and probability resit

Equality can occur only at the two pairs $(0,0)$ and $(1,1)$. Their combined <probability> is
$$
e^{-p}-p(1-e^{-p})+pe^{-p}=(1+2p)e^{-p}-p.
$$
Taking the complementary <probability> gives
$$
\boxed{\mathbb P(X\ne Y)=1+p-(1+2p)e^{-p}.}
$$
Since $1+2p>0$, the supplied exponential bound implies
$$
\mathbb P(X\ne Y)\leqslant1+p-(1+2p)(1-p)=\boxed{2p^2}.
$$
This is a bound on the mismatch <probability> for this particular <coupling> of the two <marginal distributions>; the variables are not presumed independent.

Solution

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

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

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

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