/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 17 For any discrete random variable... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 9: Problem 17

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

Short Answer

Expert verified
To prove that \(H(f(X)) \leq H(X)\), we use the definitions of entropy and conditional entropy, the property that entropy is non-negative, and the chain rule for entropy. Consider \(Y = f(X)\), a discrete random variable obtained by applying function f to X. Since \(H(Y|X) \ge 0\), we have \(H(X, Y) \ge H(X)\). As the randomness in Y is entirely determined by the entropy of X, we have \(H(X, Y) = H(Y)\). Thus, we get \(H(Y) \geq H(X)\), which implies \(H(f(X)) \leq H(X)\).

Step by step solution

01

Definitions of entropy and conditional entropy

Entropy of a discrete random variable is defined as follows: \(H(X) = -\sum_{x \in X} p(x) \log p(x)\) Where p(x) represents the probability mass function of the random variable X. Conditional entropy is defined as: \(H(Y|X) = -\sum_{x \in X, y \in Y} p(x, y) \log \frac{p(x, y)}{p(x)}\) Where p(x, y) denotes the joint probability mass function of random variables X and Y. Now, let's say that \(Y = f(X)\) is a discrete random variable obtained by applying function f to X.
02

Property of entropy

One of the properties of entropy is that it is non-negative, i.e., \(H(X) \geq 0\) for any random variable X. This means that the entropy of the conditional random variable Y, given X, is also non-negative, i.e., \(H(Y|X) \geq 0\).
03

Chain rule for entropy

Using the chain rule for entropy, we can rewrite the joint entropy of X and Y (i.e. H(X, Y)) as follows: \(H(X, Y) = H(X) + H(Y|X)\) Since \(H(Y|X) \ge 0\), we can write the following inequality: \(H(X, Y) \ge H(X)\)
04

Relation between H(X, Y) and H(Y)

We know that random variable Y is a function of X, i.e., \(Y = f(X)\). Therefore, the joint entropy H(X, Y) is equal to the entropy of Y, i.e., \(H(X, Y) = H(Y)\), because the randomness or uncertainty in Y is entirely determined by the entropy of X.
05

Proving the inequality

Now, let's substitute the result from Step 4 into the inequality from Step 3: \( H(Y) \geq H(X) \) Which is equivalent to: \( H(f(X)) \leq H(X) \) So, we have proved that the entropy of a function of a discrete random variable is less than or equal to the entropy of the random variable itself.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!

Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Entropy
Entropy is a fundamental concept in information theory that measures the uncertainty or disorder in a system. For a discrete random variable \(X\), its entropy \(H(X)\) is defined as:
\[H(X) = -\sum_{x \in X} p(x) \log p(x)\]Here, \(p(x)\) is the probability mass function of \(X\). This formula effectively captures how unpredictable \(X\) is. The more spread out the probabilities, the higher the entropy, indicating more uncertainty.
  • When all outcomes are equally likely, the entropy is at its maximum.
  • If one outcome is certain (probability of 1), the entropy is zero, indicating no uncertainty.
  • The base of the logarithm in the entropy formula can vary, often being 2, 10, or \(e\), which results in measuring entropy in bits, digits, or nats, respectively.
Understanding entropy helps us quantify the amount of information contained within a random variable.
Conditional Entropy
Conditional entropy quantifies the amount of information needed to describe the outcome of a random variable \(Y\) given that another random variable \(X\) is known. It is defined as:
\[H(Y|X) = -\sum_{x \in X, y \in Y} p(x, y) \log \frac{p(x, y)}{p(x)}\]Here, \(p(x, y)\) is the joint probability mass function of \(X\) and \(Y\), while \(p(x)\) is the marginal probability of \(X\).
  • Conditional entropy measures the remaining uncertainty in \(Y\) after knowing \(X\).
  • If \(Y\) is completely determined by \(X\), \(H(Y|X)\) is zero.
  • When \(Y\) is independent of \(X\), \(H(Y|X)\) is simply \(H(Y)\).
Conditional entropy provides insights into the dependency between random variables and can be crucial when estimating the efficiency of data encoding systems.
Chain Rule for Entropy
The Chain Rule for Entropy is a powerful tool in information theory. It relates joint entropy to individual and conditional entropies of random variables. The rule is expressed as:
\[H(X, Y) = H(X) + H(Y|X)\]This formula shows that the total entropy of the system made up of \(X\) and \(Y\) is the sum of the entropy of \(X\) and the conditional entropy of \(Y\) given that \(X\) is known.
  • The chain rule helps in understanding how knowing some variables reduces the overall uncertainty.
  • It is used to derive important results in coding theory and probabilistic modeling.
  • The inequality derived using chain rule, \(H(X, Y) \ge H(X)\), highlights that knowing \(Y\) can only reduce or leave unchanged the uncertainty about \(X\).
This rule plays a vital role in comprehensive systems like communication networks and information processing.
Joint Probability Mass Function
The Joint Probability Mass Function (Joint PMF) provides a detailed picture of the probability distribution over multiple discrete random variables. For two random variables \(X\) and \(Y\), the joint PMF \(p(x, y)\) is defined as the probability that \(X = x\) and \(Y = y\) simultaneously:
\[p(x, y) = P(X = x, Y = y)\]The joint PMF is a cornerstone in studying interactions between random variables, as it encapsulates how the probabilities of various outcomes are distributed.
  • The sum of the joint PMF over all possible \(x\) and \(y\) must equal 1, due to the axioms of probability.
  • Using the joint PMF, one can derive marginal and conditional probabilities by summing over appropriate variables.
  • The joint PMF is used to construct other probabilistic measures, such as joint and marginal entropies.
Understanding joint PMFs enriches the analysis of real-world phenomena where multiple variables are involved, like in statistical inference or machine learning.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

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

Four out of every five trucks on the road are followed by a car, while one out of every six cars is followed by a truck. What proportion of vehicles on the road are trucks?

Events occur according to a Poisson process with rate \(\lambda=3\) per hour. (a) What is the probability that no events occur between times 8 and 10 in the morning? (b) What is the expected value of the number of events that occur between times 8 and 10 in the morning? (c) What is the expected occurrence time of the fifth event after 2 P.M?

Cars cross a certain point in the highway in accordance with a Poisson process with rate \(\lambda=3\) per minute. If \(A l\) blindly runs across the highway, then 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 it for \(s=2,5,10,20\).

A certain person goes for a run each morning. When he leaves his house for his run he is equally likely to go out either the front or the back door; and similarly when he returns he is equally likely to go to either the front or back door. The runner owns 5 pairs of running shoes which he takes off after the run at which猫ver door he happens to be. If there are no shoes at the door from which he leaves to go running he runs barefooted. We are interested in determining the proportion of time that he runs barefooted. (a) Set this up as a Markov chain. Give the states and the transition probabilities. (b) Determine the proportion of days that he runs barefooted.
See all solutions

Recommended explanations on Math Textbooks

Logic and Functions

Read Explanation

Applied Mathematics

Read Explanation

Pure Maths

Read Explanation

Calculus

Read Explanation

Discrete Mathematics

Read Explanation

Decision Maths

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.