# Exercise 1.1: Entropy of the Weather

(Redirected from Aufgabe 1.1: Wetterentropie)

A weather station queries different regions every day and receives a message  $x$  back as a response in each case, namely

• $x = \rm B$:   The weather is rather bad.
• $x = \rm G$:   The weather is rather good.

The data were stored in files over many years for different regions, so that the entropies of the  $\rm B/G$–sequences can be determined:

$$H = p_{\rm B} \cdot {\rm log}_2\hspace{0.1cm}\frac{1}{p_{\rm B}} + p_{\rm G} \cdot {\rm log}_2\hspace{0.1cm}\frac{1}{p_{\rm G}}$$

with the  base-2 logarithm

$${\rm log}_2\hspace{0.1cm}p=\frac{{\rm lg}\hspace{0.1cm}p}{{\rm lg}\hspace{0.1cm}2}.$$

Here,  "lg"  denotes the logarithm to the base  $10$.  It should also be mentioned that the pseudo-unit  $\text{bit/enquiry}$  must be added in each case.

The graph shows these binary sequences for  $60$  days and the following regions:

• Region  "Mixed":              $p_{\rm B} = p_{\rm G} =0.5$,
• Region  "Rainy":              $p_{\rm B} = 0.8, \; p_{\rm G} =0.2$,
• Region  "Enjoyable":     $p_{\rm B} = 0.2, \; p_{\rm G} =0.8$,
• Region  "Paradise":        $p_{\rm B} = 1/30, \; p_{\rm G} =29/30$.

Finally, the file  "Unknown"  is also given, whose statistical properties are to be estimated.

Hinss:

• For the first four files it is assumed that the events  $\rm B$  and  $\rm G$  are statistically independent, a rather unrealistic assumption for weather practice.
• The task was designed at a time when  Greta  was just starting school.  We leave it to you to rename  "Paradise"  to  "Hell".

### Questions

1

What is the entropy  $H_{\rm M}$  of the file  "Mixed"?

 $H_{\rm M}\ = \$ $\ \rm bit/enquiry$

2

What is the entropy  $H_{\rm R}$  of the file  "Rainy"?

 $H_{\rm R}\ = \$ $\ \rm bit/enquiry$

3

What is the entropy  $H_{\rm E}$  of the file  "Enjoyable"?

 $H_{\rm E}\ = \$ $\ \rm bit/enquiry$

4

How large are the information contents of events  $\rm B$  and  $\rm G$  in relation to the file  "Paradise"?

 $I_{\rm B}\ = \$ $\ \rm bit/enquiry$ $I_{\rm G}\ = \$ $\ \rm bit/enquiry$

5

What is the entropy  (that is:  the average information content)  $H_{\rm P}$  of the file  "paradise"?  Interpret the result.

 $H_{\rm P}\ = \$ $\ \rm bit/enquiry$

6

Which statements could be true for the file  "Unknown"?

 Events  $\rm B$  and  $\rm G$  are approximately equally probable. The sequence elements are statistically independent of each other. The entropy of this file is  $H_\text{U} \approx 0.7 \; \rm bit/enquiry$. The entropy of this file is  $H_\text{U} = 1.5 \; \rm bit/enquiry$.

### Solution

#### Solution

(1)  For the file  "Mixed"  the two probabilities are the same:   $p_{\rm B} = p_{\rm G} =0.5$.  This gives us for the entropy:

$$H_{\rm M} = 0.5 \cdot {\rm log}_2\hspace{0.1cm}\frac{1}{0.5} + 0.5 \cdot {\rm log}_2\hspace{0.1cm}\frac{1}{0.5} \hspace{0.15cm}\underline {= 1\,{\rm bit/enquiry}}\hspace{0.05cm}.$$

(2)  With  $p_{\rm B} = 0.8$  and  $p_{\rm G} =0.2$,  a smaller entropy value is obtained:

$$H_{\rm R} \hspace{-0.05cm}= \hspace{-0.05cm}0.8 \cdot {\rm log}_2\hspace{0.05cm}\frac{5}{4} \hspace{-0.05cm}+ \hspace{-0.05cm}0.2 \cdot {\rm log}_2\hspace{0.05cm}\frac{5}{1}\hspace{-0.05cm}=\hspace{-0.05cm} 0.8 \cdot{\rm log}_2\hspace{0.05cm}5\hspace{-0.05cm} - \hspace{-0.05cm}0.8 \cdot {\rm log}_2\hspace{0.05cm}4 \hspace{-0.05cm}+ \hspace{-0.05cm}0.2 \cdot {\rm log}_2 \hspace{0.05cm} 5 \hspace{-0.05cm}=\hspace{-0.05cm} {\rm log}_2\hspace{0.05cm}5\hspace{-0.05cm} -\hspace{-0.05cm} 0.8 \cdot {\rm log}_2\hspace{0.1cm}4\hspace{-0.05cm} = \hspace{-0.05cm} \frac{{\rm lg} \hspace{0.1cm}5}{{\rm lg}\hspace{0.1cm}2} \hspace{-0.05cm}-\hspace{-0.05cm} 1.6 \hspace{0.15cm} \underline {= 0.722\,{\rm bit/enquiry}}\hspace{0.05cm}.$$

(3)  In the file  "Enjoyable"  the probabilities are exactly swapped compared to the file  "Rainy" .  However, this swap does not change the entropy:

$$H_{\rm E} = H_{\rm R} \hspace{0.15cm} \underline {= 0.722\,{\rm bit/enquiry}}\hspace{0.05cm}.$$

(4)  With  $p_{\rm B} = 1/30$  and  $p_{\rm G} =29/30$,  the information contents are as follows:

$$I_{\rm B} \hspace{0.1cm} = \hspace{0.1cm} {\rm log}_2\hspace{0.1cm}30 = \frac{{\rm lg}\hspace{0.1cm}30}{{\rm lg}\hspace{0.1cm}2} = \frac{1.477}{0.301} \hspace{0.15cm} \underline {= 4.907\,{\rm bit/enquiry}}\hspace{0.05cm},$$
$$I_{\rm G} \hspace{0.1cm} = \hspace{0.1cm} {\rm log}_2\hspace{0.1cm}\frac{30}{29} = \frac{{\rm lg}\hspace{0.1cm}1.034}{{\rm lg}\hspace{0.1cm}2} = \frac{1.477}{0.301} \hspace{0.15cm} \underline {= 0.049\,{\rm bit/enquiry}}\hspace{0.05cm}.$$

(5)  The entropy  $H_{\rm P}$  is the average information content of the two events  $\rm B$  and  $\rm G$:

$$H_{\rm P} = \frac{1}{30} \cdot 4.907 + \frac{29}{30} \cdot 0.049 = 0.164 + 0.047 \hspace{0.15cm} \underline {= 0.211\,{\rm bit/enquiry}}\hspace{0.05cm}.$$
• Although  (more precisely:  because)  event  $\rm B$  occurs less frequently than  $\rm G$, its contribution to entropy is much greater.

(6)  Statements 1 and 3 are correct:

• $\rm B$  and  $\rm G$  are indeed equally probable in the  "unknown"  file:   The  $60$  symbols shown divide into  $30$ times  $\rm B$  and  $30$ times  $\rm G$.
• However, there are now strong statistical ties within the temporal sequence.  Long periods of good weather are usually followed by many bad days in a row.
• Because of this statistical dependence within the  $\rm B/G$  sequence  $H_\text{U} = 0.722 \; \rm bit/enquiry$  is smaller than  $H_\text{M} = 1 \; \rm bit/enquiry$.
• $H_\text{M}$  is at the same time the maximum for  $M = 2$   ⇒   the last statement is certainly wrong.