Consider the conditional entropy and mutual information for the binary symmetric channel. The input source has alphabet \(X = \{0, 1\}\) and associated probabilities \(\left\{\frac{1}{2}, \frac{1}{2}\right\}\). The channel matrix is \(\begin{pmatrix} 1-p & p \\ p & 1-p \end{pmatrix}\) where p is the transition probability. Then the conditional entropy is given by :
\(-p\log(p) - (1-p)\log(1-p)\)
Conditional entropy \(H(Y \mid X) = \sum_{x} P(x)\, H(Y \mid X = x)\).
Given either input, the output is correct with probability \(1-p\) and flipped with probability p, so \(H(Y \mid X = x) = -p\log p - (1-p)\log(1-p)\) for both inputs.
With weights \(\frac{1}{2}\) and \(\frac{1}{2}\): \(H(Y \mid X) = -p\log(p) - (1-p)\log(1-p)\).
Check: \(1 + p\log p + (1-p)\log(1-p)\) is the mutual information \(I(X;Y) = H(Y) - H(Y \mid X)\), not the conditional entropy.
Hence, \(-p\log(p) - (1-p)\log(1-p)\).
Suppose that from given statistics, it is known that meningitis causes stiff neck 50% of the time, that the proportion of persons having meningitis is \(\frac{1}{50000}\), and that the proportion of people having stiff neck is \(\frac{1}{20}\). Then the percentage of people who had meningitis and complain about stiff neck is :
A box contains 5 coins: 4 regular coins and 1 fake coin. When a regular coin is tossed, the probability 𝑃(ℎ𝑒𝑎𝑑) = 0.5 and for a fake coin, 𝑃(ℎ𝑒𝑎𝑑) = 1. You pick a coin at random and toss it twice, and get two heads. The probability that the coin you have chosen is the fake coin is _______. (rounded off to two decimal places)