Completely noisy binary symmetric channel Created 2026-09-24 Updated 2026-09-24
At crossover probability , the two rows of the transition matrix of a binary symmetric channel coincide. Its output is then independent of its input and its capacity is zero.
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The ideal observer performs maximum a posteriori decoding: for received word , it maximizes , equivalently . Maximum-likelihood decoding maximizes the latter conditional likelihood alone. Minimum-distance decoding minimizes the Hamming distance . Ties may be resolved arbitrarily among maximizers.
For independent errors in a binary symmetric channel,For , its ratio on increasing the distance by one is . Thus maximum-likelihood decoding and minimum-distance decoding select exactly the same codewords. At the allowed endpoint , they agree on a possible observation ; impossible observations have zero likelihood for every codeword, so an arbitrary maximum-likelihood tie rule need not agree with minimum distance. This qualification is needed for the literal assumption .
For the specified observation, the Hamming distances from 000 and 111 are two and one. Their likelihoods are and , respectively. After multiplying by the priors, their joint probabilities are and , so their posterior probabilities are and . Therefore
Past exam of the mathematics course of the University of Cambridge 2018 ii Paper 2 3H Solution Created 2026-09-24 Updated 2026-10-03
The binary symmetric channel with crossover probability has channel matrixMaximum-likelihood decoding chooses a codeword maximizing , while minimum-distance decoding chooses one minimizing the Hamming distance . For a block of length sent through a binary symmetric channel,When , the final factor decreases strictly with the distance. Thus maximum-likelihood and minimum-distance decoding agree.
For the length-three repetition code , majority decoding is wrong exactly when at least two bits flip. Therefore
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At , the output of the binary symmetric channel is independent of its input. It conveys no mutual information, and therefore
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For a chosen coordinate and symbol , the shortened code retains the words having that symbol in the chosen coordinate and then deletes that coordinate. For the last coordinate,At least one of the two coordinate fibres contains at least words. Choose that fibre and, if necessary, discard surplus words. Because all retained words agree in the deleted coordinate, their mutual distances do not change. Hence one can always obtainWithout discarding words, its size is that of the chosen fibre and its minimum distance is at least .
For the final calculation, the given code is the whole space , so its parity extension is the length-four even-weight binary code. In a binary symmetric channel, parity fails to notice a nonzero error exactly when an even number of bits flips. The possible error weights are therefore two and four. Their total probability isThis excludes the weight-zero event, since it is not an undetected error.
Past exam of the mathematics course of the University of Cambridge 2021 ii Paper 2 3K Solution Created 2026-09-23 Updated 2026-09-25
For a binary symmetric channel with crossover probability , the capacity iswhere is the binary entropy. Shannon second coding theorem says that for every transmission rate and every , sufficiently long block codes exist with rate at least and decoding-error probability below ; conversely, a sequence of codes whose error tends to zero cannot have limiting rate above .
For the general channel let and be its input and output. Every conditional output distribution is a permutation of , hencefor every input distribution. Since ,The uniform input makes every output probability equal to : the column-permutation assumption and the total sum imply that all column sums are equal to one. It therefore attains , proving
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A binary symmetric channel independently flips each input bit with probability , so, with rows indexed by the input and columns by the output, its channel matrix isIf , complementing every received bit converts the channel into one with crossover probability without changing its information-carrying ability. The case has identical rows and capacity zero, so it suffices to study .
The Shannon second coding theorem states that for every rate below the channel capacity there are arbitrarily long codes whose decoding error tends to zero, whereas no sequence of codes with rate above capacity can have vanishing error. It also identifiesFor the present channel,The output entropy is at most one bit, with equality for a uniform input. Hence the binary symmetric channel capacity isbits per channel use.
Past exam of the mathematics course of the University of Cambridge 2024 ii Paper 1 11K a Solution Created 2026-09-24 Updated 2026-09-25
For a received word , the ideal-observer rule chooses a message maximizing the posterior probabilityMaximum-likelihood decoding chooses maximizingand minimum-distance decoding chooses a codeword minimizing its Hamming distance from .
Bayes' formula givesEqual message priors therefore make ideal-observer and maximum-likelihood decoding identical. On a binary symmetric channel, if , thenFor , this strictly decreases with , so maximum likelihood and minimum distance agree.
Past exam of the mathematics course of the University of Cambridge 2025 ii Paper 2 3K Solution Created 2026-09-23 Updated 2026-09-25
The received word has Hamming distances from and from , soAfter multiplying by the priors, the unnormalized posterior probabilities are and . Thus the ideal observer decodes as , while maximum-likelihood and minimum-distance decoding both choose .
The ideal observer minimizes average error but needs priors and channel statistics. Maximum likelihood needs the channel law but not priors, and can be suboptimal for unequal codeword probabilities. Minimum distance is simple and agrees with maximum likelihood for a binary symmetric channel with crossover probability below , but ignores unequal priors and general channel asymmetry.