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Chapter 9: Problem 13

Prove that if \(X\) can take on any of \(n\) possible values with respective probabilities \(P_{1}, \ldots, P_{n},\) then \(H(X)\) is maximized when \(P_{i}=1 / n, i=1, \ldots, n .\) What is \(H(X)\) equal to in this case?

Short Answer

Expert verified
In summary, the entropy \(H(X)\) of a discrete random variable \(X\) is maximized when all probabilities are equal to \(1/n\), i.e., \(P_i = 1/n\) for all \(i\). In this case, the maximum entropy is \(-\log\frac{1}{n}\). We proved this using the method of Lagrange multipliers for constrained optimization.

Step by step solution

01

Write down the formula for entropy

The entropy \(H(X)\) of a discrete random variable \(X\) is given by: \[H(X) = -\sum_{i=1}^n P_i \log P_i\] Our goal is to maximize this function subject to the constraint that the probabilities sum to 1, i.e., \[\sum_{i=1}^n P_i = 1\]
02

Set up the Lagrangian function

In order to solve this constrained optimization problem, we will use the method of Lagrange multipliers. Let's introduce a Lagrange multiplier \(\lambda\) and form the Lagrangian function: \[L(P_1,\ldots,P_n,\lambda) = -\sum_{i=1}^n P_i \log P_i - \lambda \left(\sum_{i=1}^n P_i - 1\right)\] Now to find the optimal probability values, we will take partial derivatives with respect to each \(P_i\) and \(\lambda\), and set them to zero.
03

Find partial derivatives and set them to zero

Let's find the partial derivative with respect to \(P_i\) and set it to zero: \[\frac{\partial L}{\partial P_i} = -\log P_i - 1 - \lambda = 0\] \[\log P_i + 1 + \lambda = 0\] \[\log P_i = -1 - \lambda\] Now take the partial derivative with respect to \(\lambda\) and set it to zero: \[\frac{\partial L}{\partial \lambda} = -\left(\sum_{i=1}^n P_i - 1\right) = 0\] \[\sum_{i=1}^n P_i = 1\]
04

Solve the system of equations

We have obtained a system of equations: one equation for each \(P_i\): \[\log P_i = -1 - \lambda, \qquad i = 1,\ldots, n\] and one equation for the constraint: \[\sum_{i=1}^n P_i = 1\] Let's solve these equations: From the equation \(\log P_i = -1 - \lambda\), we can get: \[P_i = e^{-1 - \lambda}\] By plugging this into the constraint equation, we get: \[\sum_{i=1}^n e^{-1 - \lambda} = 1\] \[ne^{-1 - \lambda} = 1\] \[\lambda = -1 - \log \frac {1} {n}\] Plug this back into the equation for \(P_i\): \[P_i = e^{-1 - \lambda} = \frac{1}{n}\] This shows that the entropy is maximized when all probabilities are equal to \(1/n\).
05

Compute the entropy for uniform probabilities

Now let's find the value of \(H(X)\) when all probabilities are equal to \(1/n\). Plugging the values back into the entropy formula: \[H(X) = -\sum_{i=1}^n P_i \log P_i = -\sum_{i=1}^n \frac{1}{n} \log \frac{1}{n} = -\log\frac{1}{n}\] Thus, the maximum entropy \(H(X)\) is \(-\log\frac{1}{n}\) when all probabilities are equal to \(1/n\).

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Most popular questions from this chapter

A pair of fair dice is rolled. Let $$ X=\left\\{\begin{array}{ll} 1 & \text { if the sum of the dice is } 6 \\ 0 & \text { otherwise } \end{array}\right. $$ and let \(Y\) equal the value of the first die. Compute (a) \(H(Y),(b) H_{Y}(X),\) and \((c) H(X, Y)\).

A coin having probability \(p=\frac{2}{3}\) of coming up heads is flipped 6 times. Compute the entropy of the outcome of this experiment.

Cars cross a certain point in the highway in accordance with a Poisson process with rate \(\lambda=3\) per minute. If Al runs blindly across the highway, what is the probability that he will be uninjured if the amount of time that it takes him to cross the road is \(s\) seconds? (Assume that if he is on the highway when a car passes by, then he will be injured.) Do this exercise for \(s=2,5,10,20\).

Show that, for any discrete random variable \(X\) and function \(f\) $$ H(f(X)) \leq H(X) $$

A transition probability matrix is said to be doubly stochastic if $$ \sum_{i=0}^{M} P_{i j}=1 $$ for all states \(j=0,1, \ldots, M .\) Show that such a Markov chain is ergodic, then \(\prod_{j}=1 /(M+1), j=\) \(0,1, \ldots, M\).
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