Difference between revisions of "Aufgaben:Exercise 1.1: Entropy of the Weather"

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{{quiz-Header|Buchseite=Informationstheorie/Gedächtnislose Nachrichtenquellen
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{{quiz-Header|Buchseite=Information_Theory/Discrete_Memoryless_Sources
 
}}
 
}}
  
[[File:Inf_A_1_1_vers2.png|right|frame|Five different binary sources]]
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[[File:EN_Inf_A_1_1_v2.png|right|frame|Five different binary sources]]
 
A weather station queries different regions every day and receives a message   $x$  back as a response in each case, namely
 
A weather station queries different regions every day and receives a message   $x$  back as a response in each case, namely
  
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:$$H =  p_{\rm B} \cdot {\rm log}_2\hspace{0.1cm}\frac{1}{p_{\rm B}} + p_{\rm G} \cdot {\rm log}_2\hspace{0.1cm}\frac{1}{p_{\rm G}}$$
 
:$$H =  p_{\rm B} \cdot {\rm log}_2\hspace{0.1cm}\frac{1}{p_{\rm B}} + p_{\rm G} \cdot {\rm log}_2\hspace{0.1cm}\frac{1}{p_{\rm G}}$$
  
with the  "Logarithm dualis"
+
with the  base-2 logarithm
:$${\rm log}_2\hspace{0.1cm}p=\frac{{\rm lg}\hspace{0.1cm}p}{{\rm lg}\hspace{0.1cm}2}\hspace{0.3cm} \left ( =  {\rm ld}\hspace{0.1cm}p \right ) \hspace{0.05cm}.$$
+
:$${\rm log}_2\hspace{0.1cm}p=\frac{{\rm lg}\hspace{0.1cm}p}{{\rm lg}\hspace{0.1cm}2}.$$
 
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.
 
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.
  
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*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.
 
*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  [https://en.wikipedia.org/wiki/Greta_Thunberg Greta]  was just starting school.  We leave it to you to rename   "Paradise"  tonbsp;  "Hell".  
+
*The task was designed at a time when  [https://en.wikipedia.org/wiki/Greta_Thunberg Greta]  was just starting school.  We leave it to you to rename   "Paradise"  to   "Hell".  
  
  
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{What is the entropy  (that is:  the average information content)  $H_{\rm P}$  of the file  „paradise”?  Interpret the result.
+
{What is the entropy  (that is:  the average information content)  $H_{\rm P}$  of the file  "paradise"?  Interpret the result.
 
|type="{}"}
 
|type="{}"}
 
$H_{\rm P}\ =  \ $ { 0.211 3% } $\ \rm bit/enquiry$
 
$H_{\rm P}\ =  \ $ { 0.211 3% } $\ \rm bit/enquiry$
  
  
{Which statements could be true for the file  "Unknown&"?
+
{Which statements could be true for the file  "Unknown"?
 
|type="[]"}
 
|type="[]"}
 
+ Events  $\rm B$  and  $\rm G$  are approximately equally probable.
 
+ Events  $\rm B$  and  $\rm G$  are approximately equally probable.
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  \hspace{0.15cm}  
 
  \hspace{0.15cm}  
 
\underline {= 0.211\,{\rm bit/enquiry}}\hspace{0.05cm}.$$
 
\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 greater.
+
*Although  (more precisely:  because)  event  $\rm B$  occurs less frequently than  $\rm G$, its contribution to entropy is much greater.
  
  
 
'''(6)'''&nbsp; Statements <u>1 and 3</u> are correct:
 
'''(6)'''&nbsp; Statements <u>1 and 3</u> are correct:
*$\rm B$&nbsp; and&nbsp; $\rm G$&nbsp; are indeed equally probable in the&nbsp; "unknown"&nbsp; file: &nbsp; the&nbsp; $60$&nbsp; symbols shown divide into&nbsp;  $30$&nbsp;times&nbsp; $\rm B$&nbsp; and&nbsp; $30$&nbsp;times&nbsp; $\rm G$.  
+
*$\rm B$&nbsp; and&nbsp; $\rm G$&nbsp; are indeed equally probable in the&nbsp; "unknown"&nbsp; file: &nbsp; The&nbsp; $60$&nbsp; symbols shown divide into&nbsp;  $30$&nbsp;times&nbsp; $\rm B$&nbsp; and&nbsp; $30$&nbsp;times&nbsp; $\rm G$.  
 
*However, there are now strong statistical ties within the temporal sequence.&nbsp; Long periods of good weather are usually followed by many bad days in a row.
 
*However, there are now strong statistical ties within the temporal sequence.&nbsp; Long periods of good weather are usually followed by many bad days in a row.
 
*Because of this statistical dependence within the&nbsp; $\rm B/G$&nbsp; sequence&nbsp; $H_\text{U} = 0.722 \; \rm bit/enquiry$&nbsp; is smaller than&nbsp; $H_\text{M} = 1 \; \rm bit/enquiry$.  
 
*Because of this statistical dependence within the&nbsp; $\rm B/G$&nbsp; sequence&nbsp; $H_\text{U} = 0.722 \; \rm bit/enquiry$&nbsp; is smaller than&nbsp; $H_\text{M} = 1 \; \rm bit/enquiry$.  

Latest revision as of 12:59, 10 August 2021

Five different binary sources

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

(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.