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Chapter 1: Problem 33

Suppose that the conditional entropy \(\mathrm{H}[y \mid x]\) between two discrete random variables \(x\) and \(y\) is zero. Show that, for all values of \(x\) such that \(p(x)>0\), the variable \(y\) must be a function of \(x\), in other words for each \(x\) there is only one value of \(y\) such that \(p(y \mid x) \neq 0\).

Short Answer

Expert verified
Since the conditional entropy \( \mathrm{H}[y \mid x] = 0 \), for each \( x \), there is only one \( y \) with non-zero probability, making \( y \) a function of \( x \).

Step by step solution

01

Understand Conditional Entropy

Conditional entropy \( \mathrm{H}[y \mid x] \) measures the uncertainty remaining about \( y \) when \( x \) is known. The formula is given by \( \mathrm{H}[y \mid x] = \sum_{x} p(x) \mathrm{H}[y \mid x = x] \), where \( \mathrm{H}[y \mid x = x] \) is the entropy of \( y \) given a specific value of \( x \).
02

Evaluate Given Condition

We know that \( \mathrm{H}[y \mid x] = 0 \). This implies that for each specific value of \( x \), \( \mathrm{H}[y \mid x = x] = 0 \). Since entropy is zero, there must be no uncertainty about \( y \) given \( x \).
03

Determine Implications of Zero Entropy

If \( \mathrm{H}[y \mid x = x] = 0 \), the probability distribution of \( y \) given \( x \) must have all mass concentrated on a single point. This means that for each \( x \), there is exactly one \( y \) such that \( p(y \mid x) = 1 \) and for all other values of \( y \), \( p(y \mid x) = 0 \).
04

Conclude the Function Relationship

Since for each value of \( x \), there is exactly one \( y \) for which \( p(y \mid x) = 1 \), \( y \) must be a deterministic function of \( x \). This means \( y \) can be described as \( y = f(x) \), where \( f \) is a function mapping each \( x \) to a specific value of \( y \).

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Key Concepts

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

Discrete Random Variables
Discrete random variables are a fundamental concept in probability and statistics. These are variables that take on a set of distinct and separate values. Think of them like a die that can only land on one of six faces鈥�1 through 6. These possible outcomes are discrete because the die doesn't land on any value between these numbers.
The beauty of discrete random variables is that they allow us to list all possible outcomes, and this makes it easier to compute probabilities. For example, if you were throwing a die, you could easily calculate the probability of landing on any number between 1 and 6, which should be \(\frac{1}{6}\). It's straightforward because there are a limited number of possible outcomes.
The concept of randomness here is key. Despite knowing all possible outcomes, we don't know which outcome will occur in a single trial. In summation, discrete random variables are basic yet powerful tools for understanding probabilities in a finite set of separate possible outcomes.
Function Relationship
A function relationship in mathematics and statistics establishes a connection between two variables such that one variable depends on the other. In simpler terms, if you know the value of one variable, you can determine the other using a function.
Consider the exercise you are working on: if the conditional entropy \(\mathrm{H}[y \mid x]\) is zero, it implies a very clear function relationship between the variables \(x\) and \(y\). This means for every value of \(x\), there is precisely one outcome for \(y\). Mathematically, this relationship is expressed as \(y = f(x)\), where \(f\) is the function mapping each \(x\) to its unique \(y\).
This kind of relationship is vital because it eliminates unpredictability for \(y\) as long as \(x\) is known. It's like having a locked-in GPS route where selecting a starting point \(x\) always dictates the destination \(y\). Every input has only one output, and that's the essence of a deterministic function relationship.
Probability Distribution
A probability distribution is a statistical function that describes all the possible values and likelihoods that a random variable can take within a given range. It's like a recipe that tells us how probable different outcomes are.
For discrete random variables, the probability distribution is relatively simple to understand because it can be listed explicitly. For every possible outcome \(x_i\) of a discrete random variable \(X\), there is an associated probability \(p(x_i)\). These probabilities must sum to 1, as this represents the certainty that one of the possible outcomes will occur.
When dealing with conditional situations, like when examining \(p(y \mid x)\), the probability distribution adapts to show probabilities given the condition. If this distribution shows all probabilities centered on one outcome, as in the case of zero conditional entropy, it becomes deterministic for that condition, meaning there's no randomness left鈥攐nly certainty about the result when \(x\) is known.
Deterministic Function
A deterministic function is a type of statistical function that ensures a unique outcome for a given input. Unlike random processes where multiple outcomes can occur, deterministic functions leave no room for uncertainty.
In the exercise you worked on, a deterministic function between \(x\) and \(y\) is established by having zero conditional entropy. This means that for every value of \(x\), there's only one possible \(y\), effectively linking them through a function that operates predictably and without ambiguity鈥擻(y = f(x)\).
Think of it like a vending machine: you choose a specific button (the input \(x\)), and it always dispenses the same snack (the output \(y\)). This kind of predictability is crucial in systems where the exact relationship between variables is needed, without any noise or uncertainty that could potentially disrupt the output.

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Most popular questions from this chapter

Show that an arbitrary square matrix with elements \(w_{i j}\) can be written in the form \(w_{i j}=w_{i j}^{\mathrm{S}}+w_{i j}^{\mathrm{A}}\) where \(w_{i j}^{\mathrm{S}}\) and \(w_{i j}^{\mathrm{A}}\) are symmetric and anti-symmetric matrices, respectively, satisfying \(w_{i j}^{\mathrm{S}}=w_{j i}^{\mathrm{S}}\) and \(w_{i j}^{\mathrm{A}}=-w_{j i}^{\mathrm{A}}\) for all \(i\) and \(j\). Now consider the second order term in a higher order polynomial in \(D\) dimensions, given by $$ \sum_{i=1}^{D} \sum_{j=1}^{D} w_{i j} x_{i} x_{j} $$ Show that $$ \sum_{i=1}^{D} \sum_{j=1}^{D} w_{i j} x_{i} x_{j}=\sum_{i=1}^{D} \sum_{j=1}^{D} w_{i j}^{S} x_{i} x_{j} $$ so that the contribution from the anti-symmetric matrix vanishes. We therefore see that, without loss of generality, the matrix of coefficients \(w_{i j}\) can be chosen to be symmetric, and so not all of the \(D^{2}\) elements of this matrix can be chosen independently. Show that the number of independent parameters in the matrix \(w_{i j}^{\mathrm{S}}\) is given by \(D(D+1) / 2\).

In this exercise, we prove the normalization condition (1.48) for the univariate Gaussian. To do this consider, the integral $$ I=\int_{-\infty}^{\infty} \exp \left(-\frac{1}{2 \sigma^{2}} x^{2}\right) \mathrm{d} x $$ which we can evaluate by first writing its square in the form $$ I^{2}=\int_{-\infty}^{\infty} \int_{-\infty}^{\infty} \exp \left(-\frac{1}{2 \sigma^{2}} x^{2}-\frac{1}{2 \sigma^{2}} y^{2}\right) \mathrm{d} x \mathrm{~d} y $$ Now make the transformation from Cartesian coordinates \((x, y)\) to polar coordinates \((r, \theta)\) and then substitute \(u=r^{2}\). Show that, by performing the integrals over \(\theta\) and \(u\), and then taking the square root of both sides, we obtain $$ I=\left(2 \pi \sigma^{2}\right)^{1 / 2} $$ Finally, use this result to show that the Gaussian distribution \(\mathcal{N}\left(x \mid \mu, \sigma^{2}\right)\) is normalized.

In Section 1.6, we introduced the idea of entropy \(h(x)\) as the information gained on observing the value of a random variable \(x\) having distribution \(p(x)\). We saw that, for independent variables \(x\) and \(y\) for which \(p(x, y)=p(x) p(y)\), the entropy functions are additive, so that \(h(x, y)=h(x)+h(y)\). In this exercise, we derive the relation between \(h\) and \(p\) in the form of a function \(h(p)\). First show that \(h\left(p^{2}\right)=\) \(2 h(p)\), and hence by induction that \(h\left(p^{n}\right)=n h(p)\) where \(n\) is a positive integer. Hence show that \(h\left(p^{n / m}\right)=(n / m) h(p)\) where \(m\) is also a positive integer. This implies that \(h\left(p^{x}\right)=x h(p)\) where \(x\) is a positive rational number, and hence by continuity when it is a positive real number. Finally, show that this implies \(h(p)\) must take the form \(h(p) \propto \ln p\).

The gamma function is defined by $$ \Gamma(x) \equiv \int_{0}^{\infty} u^{x-1} e^{-u} \mathrm{~d} u $$ Using integration by parts, prove the relation \(\Gamma(x+1)=x \Gamma(x)\). Show also that \(\Gamma(1)=1\) and hence that \(\Gamma(x+1)=x !\) when \(x\) is an integer.

By applying Jensen's inequality (1.115) with \(f(x)=\ln x\), show that the arithmetic mean of a set of real numbers is never less than their geometrical mean.
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