Aufgaben:Exercise 3.3: Entropy of Ternary Quantities

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Given entropy functions

On the right you see the entropy functions  $H_{\rm R}(p)$,  $H_{\rm B}(p)$  and  $H_{\rm G}(p)$, where  $\rm R$  stands for "red",  $\rm B$  for "blue" and  $\rm G$  for "green".  The probability functions are for all random variables:

$$P_X(X) = [\hspace{0.05cm}p_1\hspace{0.05cm},\ p_2\hspace{0.05cm},\ p_3\hspace{0.05cm}]\hspace{0.3cm}\hspace{0.3cm} \Rightarrow \hspace{0.3cm} |X| = 3\hspace{0.05cm}.$$

For the questionnaire, the relationship  $p_1 = p$  and  $p_2 = 1 - p_3- p$.

The probability function of a random variable

$$X = \big \{\hspace{0.05cm}x_1\hspace{0.05cm}, \hspace{0.15cm} x_2\hspace{0.05cm},\hspace{0.15cm} \text{...}\hspace{0.1cm} ,\hspace{0.15cm} x_{\mu}\hspace{0.05cm}, \hspace{0.05cm}\text{...}\hspace{0.1cm} , \hspace{0.15cm} x_{M}\hspace{0.05cm}\big \}$$

with the symbolic range  $|X| = M$  is generally:

$$P_X(X) = [\hspace{0.05cm}p_1\hspace{0.05cm}, \hspace{0.15cm} p_2\hspace{0.05cm},\hspace{0.05cm} ...\hspace{0.1cm} ,\hspace{0.15cm} p_{\mu}\hspace{0.05cm}, \hspace{0.05cm}...\hspace{0.1cm} , \hspace{0.15cm} p_{M}\hspace{0.05cm}]\hspace{0.05cm}.$$

The entropy (uncertainty) of this random variable is calculated according to the equation

$$H(X) = {\rm E} \big [\log_2 \hspace{0.05cm} {1}/{P_X(X)} \big ]\hspace{0.05cm},$$

and always lies in the range  $0 \le H(X) \le \log_2 \hspace{0.05cm} |X|$.

• The lower bound  $H(X) = 0$  results when any probability  $p_\mu = 1$  and all others are zero.

• The upper bound is to be derived here as in the lecture "Information Theory" by  Gerhard Kramer  at the TU Munich:

Upper bound estimate for the natural logarithm

• By extending the above equation by  $|X|$  in the numerator and denominator, using  $\log_2 \hspace{0.05cm}x= \ln(x)/\ln(2)$, we obtain:

$$H(X) = \frac{1}{{\rm ln}(2)}\cdot {\rm E} \left [{\rm ln} \hspace{0.1cm} \frac{1}{|X| \cdot P_X(X)} \right ] + {\rm log}_2 \hspace{0.1cm}|X| \hspace{0.05cm}.$$

• As can be seen from the graph opposite, the estimation  $\ln(x) \le x-1$  holds with the identity for  $x=1$.  Thus, it can be written:

$$H(X) \le \frac{1}{{\rm ln}(2)}\cdot {\rm E} \left [\frac{1}{|X| \cdot P_X(X)} -1 \right ] + {\rm log}_2 \hspace{0.1cm}|X| \hspace{0.05cm}.$$

• In   Exercise 3.2  the expected value  ${\rm E} \big [\log_2 \hspace{0.05cm} {1}/{P_X(X)} \big ] =|X|$  was calculated for the case  $p_\mu \ne 0$  for all  $\mu$ .  Thus, the first term disappears and the known result is obtained:

$$H(X) \le {\rm log}_2 \hspace{0.1cm}|X| \hspace{0.05cm}.$$

Hints:

• The exercise belongs to the chapter  Some preliminary remarks on 2D random variables.
• In particular, reference is made to the page  Probability function and entropy.
• The same constellation is assumed here as in  Exercise 3.2.

• The equation of the binary entropy function is:

$$H_{\rm bin}(p) = p \cdot {\rm log}_2 \hspace{0.1cm} \frac{1}{p} +(1-p) \cdot {\rm log}_2 \hspace{0.1cm} \frac{1}{1-p} \hspace{0.05cm}.$$

Questions

1 Which statements are true for the red entropy function  $H_{\rm R}(p)$?

	$H_{\rm R}(p)$  results, for example, with  $p_1 = p$,  $p_2 = 1- p$  and  $p_3 = 0$.
	$H_{\rm R}(p)$  is identical to the binary entropy function  $H_{\rm bin}(p)$.

2 What are the properties of the binary entropy function  $H_{\rm bin}(p)$ ?

	$H_{\rm bin}(p)$  is concave with respect to the parameter  $p$.
	$\text {Max } [H_{\rm bin}(p)] = 2$  bit applies.

3 Which statements are true for the blue entropy function  $H_{\rm B}(p)$?

	$H_{\rm B}(p)$  ergibt sich beispielsweise mit  $p_1 = p$,  $p_2 = 1/2- p$  und  $p_3 = 1/2$.
	Es gilt  $H_{\rm B}(p = 0)= 1$  bit.
	Es gilt  $\text {Max } [H_{\rm B}(p)] = \log_2 \hspace{0.1cm} (3)$  bit.

4 Which statements are true for the green entropy function  $H_{\rm G}(p)$?

	$H_{\rm G}(p)$  results, for example, with  $p_1 = p$,  $p_2 = 2/3- p$  and  $p_3 = 1/3$.
	$H_{\rm G}(p = 0)= 1$  bit is valid.
	$\text {Max } [H_{\rm G}(p)] = \log_2 \hspace{0.1cm} (3)$ bit applies.

Solution

Solution

(1)  Both statements are correct:

• If we set  $p_3 = 0$ and formally  $p_1 = p$   ⇒    $p_2 = 1- p$, we get the binary entropy function

$$H_{\rm bin}(p) = p \cdot {\rm log}_2 \hspace{0.1cm} \frac{1}{p} +(1-p) \cdot {\rm log}_2 \hspace{0.1cm} \frac{1}{1-p} \hspace{0.05cm}.$$

(2)  Only the proposed solution 1 is correct:

• One can also put the binary entropy function into the following form because of  $\log(x) = \ln(x)/\ln(2)$ :

$$H_{\rm bin}(p) = \frac{-1}{{\rm ln}(2)} \cdot \big [ p \cdot {\rm ln}(p) +(1-p) \cdot {\rm ln}(1-p) \big ] \hspace{0.05cm}.$$

• The first derivation leads to the result

$$\frac {{\rm d}H_{\rm bin}(p)}{{\rm d}p} = \frac{-1}{{\rm ln}(2)} \cdot \big [ {\rm ln}(p) + p \cdot \frac{1}{p} -{\rm ln}(1-p) - (1-p) \cdot \frac{1}{1-p} \big ] =\frac{1}{{\rm ln}(2)} \cdot \big [ {\rm ln}(1-p) - {\rm ln}(p) \big ] = {\rm log}_2 \hspace{0.1cm} \frac{1-p}{p} \hspace{0.05cm}.$$

• By setting this derivative to zero, we obtain the abscissa value  $p = 0.5$, which leads to the maximum of the entropy function:   $H_{\rm bin}(p =0.5) = 1$ bit
  ⇒   the proposed solution 2 is wrong.
• By differentiating again, one obtains for the second derivative

$$\frac {{\rm d}^2H_{\rm bin}(p)}{{\rm d}p^2} = \frac{1}{{\rm ln}(2)} \cdot \left[ \frac{-1}{1-p} - \frac{1}{p} \right ] =\frac{-1}{{\rm ln}(2) \cdot p \cdot (1-p)} \hspace{0.05cm}.$$

• This function is negative in the entire definition domain  $0 ≤ p ≤ 1$    ⇒   $H_{\rm bin}(p)$ is concave   ⇒   the proposed solution 1 is correct.

Three entropy functions with  $M = 3$

(3)  Propositions 1 and 2 are correct here:

• For  $p = 0$  one obtains the probability function  $P_X(X) = \big [\hspace{0.05cm}0\hspace{0.05cm}, \hspace{0.15cm} 1/2\hspace{0.05cm},\hspace{0.15cm} 1/2 \hspace{0.05cm} \big ]$   ⇒   $H(X) = 1$  bit.
• The maximum under the condition  $p_3 = 1/2$  is obtained for  $p_1 = p_2 = 1/4$:

$$P_X(X) = \big [\hspace{0.05cm}1/4\hspace{0.05cm}, \hspace{0.05cm} 1/4\hspace{0.05cm},\hspace{0.05cm} 1/2 \hspace{0.05cm} \big ]\hspace{0.3cm} \Rightarrow \hspace{0.3cm}{\rm Max} \ [H_{\rm B}(p)] = 1.5 \ \rm bit\hspace{0.05cm}.$$

• In compact form,  $H_{\rm B}(p)$  with the constraint  $0 ≤ p ≤ 1/2$  can be represented as follows:

$$H_{\rm B}(p) = 1.0\,{\rm bit} + {1}/{2} \cdot H_{\rm bin}(2p)\hspace{0.05cm}.$$

(4)  The first and last statements are correct here::

• The green curve with  $p = 1/3$  also contains the equal distribution of all probabilities

$$ {\rm Max} \ [H_{\rm G}(p)] = \log_2 (3)\ \text{bit}.$$

• In general, the entire curve in the range  $0 ≤ p ≤ 2/3$  can be expressed as follows:

$$H_{\rm G}(p) = H_{\rm G}(p= 0) + {2}/{3} \cdot H_{\rm bin}(3p/2)\hspace{0.05cm}.$$

• From the graph on the information page, one can also see that the following condition must be fulfilled:

$$H_{\rm G}(p = 0) + {2}/{3}= {\rm log}_2 \hspace{0.01cm} (3)\hspace{0.3cm} \Rightarrow \hspace{0.3cm}H_{\rm G}(p= 0) = 1.585 - 0.667 = 0.918 \,{\rm bit}\hspace{0.05cm}.$$

• The second suggested solution is therefore wrong.  The same result can be obtained with the equation

$$H_{\rm G}(p = 0) = {1}/{3} \cdot {\rm log}_2 \hspace{0.01cm} (3)+{2}/{3} \cdot {\rm log}_2 \hspace{0.01cm} (3/2) = {\rm log}_2 \hspace{0.01cm} (3) -2/3 \cdot{\rm log}_2 \hspace{0.01cm} (2)\hspace{0.05cm}.$$