# 5.3 Entropy

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Entropy

The self information gives the information in a single outcome. In most cases, e.g in data compression, it is much moreinteresting to know the average information content of a source. This average is given by the expected value of the self information with respect to the source's probabilitydistribution. This average of self information is called the source entropy.

## Definition of entropy

Entropy
If symbol has zero probability, which means it never occurs, it should not affect the entropy. Letting $0\lg 0=0$ , we have dealt with that.

In texts you will find that the argument to the entropy function may vary. The two most common are $H(X)$ and $H(p)$ . We calculate the entropy of a source X, but the entropy is,strictly speaking, a function of the source's probabilty function p. So both notations are justified.

## Calculating the binary logarithm

Most calculators does not allow you to directly calculate the logarithm with base 2, so we have to use a logarithm base that mostcalculators support. Fortunately it is easy to convert between different bases.

Assume you want to calculate $\log_{2}x$ , where $x> 0$ . Then $\log_{2}x=y$ implies that $2^{y}=x$ . Taking the natural logarithm on both sides we obtain

$\log_{2}x=\frac{\ln x}{\ln 2}$

## Examples

When throwing a dice, one may ask for the average information conveyed in a single throw. Using the formula for entropy we get $H(X)=-\sum_{i=1}^{6} {p}_{X}({x}_{i})\lg {p}_{X}({x}_{i})=\lg 6\mathrm{bits/symbol}$

If a soure produces binary information $\{0, 1\}$ with probabilities $p$ and $1-p$ . The entropy of the source is

$H(X)=-(p\log_{2}p)-(1-p)\log_{2}(1-p)$
If $p=0$ then $H(X)=0$ , if $p=1$ then $H(X)=0$ , if $p=1/2$ then $H(X)=1$ . The source has its largest entropy if $p=1/2$ and the source provides no new information if $p=0$ or $p=1$ .

An analog source is modeled as a continuous-time random process with power spectral density bandlimited to the bandbetween 0 and 4000 Hz. The signal is sampled at the Nyquist rate. The sequence of random variables, as a result ofsampling, are assumed to be independent. The samples are quantized to 5 levels $\{-2, -1, 0, 1, 2\}$ . The probability of the samples taking the quantized values are $\{\frac{1}{2}, \frac{1}{4}, \frac{1}{8}, \frac{1}{16}, \frac{1}{16}\}$ , respectively. The entropy of the random variables are

$H(X)=-\sum_{i=1}^{5} {p}_{X}({x}_{i})\lg {p}_{X}({x}_{i})=\frac{1}{2}+\frac{1}{2}+\frac{3}{8}+\frac{1}{4}+\frac{1}{4}=\frac{15}{8}\mathrm{bits/sample}$
There are 8000 samples per second. Therefore, the source produces $8000\frac{15}{8}=15000$ bits/sec of information.

Entropy is closely tied to source coding. The extent to which a source can be compressed is related to its entropy.There are many interpretations possible for the entropy of a random variable, including

• (Average)Self information in a random variable
• Minimum number of bits per source symbol required to describe the random variable without loss
• Description complexity
• Measure of uncertainty in a random variable

## References

• ien, G.E. and Lundheim,L. (2003) Information Theory, Coding and Compression , Trondheim: Tapir Akademisk forlag.

#### Questions & Answers

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Source:  OpenStax, Information and signal theory. OpenStax CNX. Aug 03, 2006 Download for free at http://legacy.cnx.org/content/col10211/1.19
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