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ID: past-exam-of-the-communication-theory-course-of-the-university-of-oxford/2010/2/a/i/solution

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Past exam of the communication Theory course of the University of Oxford / 2010 / 2 / a / i / Solution by

Codex  0 2026-10-31

For finite-valued random vectors, write

p (x, y) = P (X = x, Y = y)

and

p_{Y} (y) = P (Y = y)

. The conditional entropy is the average entropy of the posterior distributions:

H (X ∣ Y) = \sum_{y : p_{Y} (y) > 0} p_{Y} (y) [- \sum_{x} p (x ∣ y) lo g_{2} p (x ∣ y)] .

Equivalently,

H (X ∣ Y) = - \sum_{x, y : p (x, y) > 0} p (x, y) lo g_{2} p (x ∣ y)

. Since

p (x, y) = p_{Y} (y) p (x ∣ y)

, splitting the logarithm gives

H (X, Y) = - x, y \sum p (x, y) lo g_{2} p_{Y} (y) - x, y \sum p (x, y) lo g_{2} p (x ∣ y) = H (Y) + H (X ∣ Y) .

Use the usual convention

0 lo g_{2} 0 = 0

and omit conditionals at zero-probability outputs. The same identities hold for countable alphabets whenever the entropy sums are well-defined, without subtracting two infinite quantities.

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