Aufgaben:Exercise 4.3: Algebraic and Modulo Sum: Difference between revisions
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{{quiz-Header|Buchseite= | {{quiz-Header|Buchseite=Theory_of_Stochastic_Signals/Two-Dimensional_Random_Variables | ||
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[[File:EN_Sto_A_4_3.png|right|frame| | [[File:EN_Sto_A_4_3.png|right|frame|Algebraic sum and modulo 2 sum]] | ||
A "clocked" random number generator returns a sequence $\langle x_\nu \rangle$ of binary random numbers. | |||
* | *It is assumed that the binary numbers $0$ and $1$ occur with equal probabilities and that the individual random numbers do not depend on each other. | ||
* | *The random numbers $ x_\nu \in \{0, 1\}$ are entered into the first memory location of a shift register and shifted down one digit with each clock pulse. | ||
Two new random sequences $\langle a_\nu \rangle$ and $\langle m_\nu \rangle$ are formed from the contents of the three-digit shift register. Here denotes: | |||
* $a_\nu$ | * $a_\nu$ the <i>algebraic sum</i>: | ||
:$$a_\nu=x_\nu+x_{\nu-1}+x_{\nu-2},$$ | :$$a_\nu=x_\nu+x_{\nu-1}+x_{\nu-2},$$ | ||
*$m_\nu$ | *$m_\nu$ the <i>modulo 2 sum</i>: | ||
:$$m_\nu=x_\nu\oplus x_{\nu-1}\oplus x_{\nu-2}.$$ | :$$m_\nu=x_\nu\oplus x_{\nu-1}\oplus x_{\nu-2}.$$ | ||
This fact is shown again in the following table: | |||
[[File:P_ID254__Sto_A_4_3Tab.png|left|frame| | [[File:P_ID254__Sto_A_4_3Tab.png|left|frame|Table for moment calculation]] | ||
<br><br><br><br><br><br> | <br><br><br><br><br><br> | ||
Hints: This exercise belongs to the chapter [[Theory_of_Stochastic_Signals/Two-Dimensional_Random_Variables|Two-Dimensional Random Variables]]. | |||
<br clear=all> | <br clear=all> | ||
=== | ===Questions=== | ||
<quiz display=simple> | <quiz display=simple> | ||
{ | {Calculate the probabilities of the random variable $m_\nu$. What is the probability that the modulo-2 sum is equal to $0$ ? | ||
|type="{}"} | |type="{}"} | ||
${\rm Pr}(m_\nu = 0) \ = \ $ | ${\rm Pr}(m_\nu = 0) \ = \ $ { 0.5 3% } | ||
{ | {Are there statistical dependencies within the sequence $\langle m_\nu \rangle$? | ||
|type="()"} | |type="()"} | ||
+ | + The sequence elements $m_\nu$ are statistically independent. | ||
- | - There are statistical bindings within the sequence $\langle m_\nu \rangle$. | ||
{ | {Determine the joint PDF $f_{xm}(x_\nu, m_\nu)$. Based on the result, evaluate the following statements (true or not). | ||
|type="[]"} | |type="[]"} | ||
- | - The random variables $x_\nu$ and $m_\nu$ are statistically dependent. | ||
+ | + The random variables $x_\nu$ and $m_\nu$ are statistically independent. | ||
- | - The random variables $x_\nu$ and $m_\nu$ are correlated. | ||
+ | + The random variables $x_\nu$ and $m_\nu$ are uncorrelated. | ||
{ | {Do statistical dependencies exist within the sequence $\langle a_\nu \rangle$ ? | ||
|type="()"} | |type="()"} | ||
- | - The sequence elements $a_\nu$ are statistically independent. | ||
+ | + There are statistical bindings within the sequence $\langle a_\nu \rangle$. | ||
{ | {Determine the 2D PDF $f_{am}(a_\nu, m_\nu)$ and the correlation coefficient $\rho_{am}$. Which of the following statements are true? | ||
|type="[]"} | |type="[]"} | ||
+ | + The random variables $a_\nu$ and $m_\nu$ are statistically dependent. | ||
- | - The random variables $a_\nu$ and $m_\nu$ are statistically independent. | ||
- | - The random variables $a_\nu$ and $m_\nu$ are correlated. | ||
+ | + The random variables $a_\nu$ and $m_\nu$ are uncorrelated. | ||
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</quiz> | </quiz> | ||
=== | ===Solution=== | ||
{{ML-Kopf}} | {{ML-Kopf}} | ||
'''(1)''' | '''(1)''' It can be seen from the table on the information page that for the modulo 2 sum, the two values $0$ and $1$ have equal probability: | ||
:$${\rm Pr}(m_\nu = 0) = {\rm Pr}(m_\nu = 1)\hspace{0.15cm}\underline{=0.5}.$$ | :$${\rm Pr}(m_\nu = 0) = {\rm Pr}(m_\nu = 1)\hspace{0.15cm}\underline{=0.5}.$$ | ||
'''(2)''' | '''(2)''' The table shows that for each preassignment ⇒ $( x_{\nu-1}, x_{\nu-2}) = (0,0), (0,1), (1,0), (1,1)$ the values $m_\nu = 0$ respectively $m_\nu = 1$ are equally likely. | ||
* | *Expressed differently: ${\rm Pr}(m_{\nu}\hspace{0.05cm}|\hspace{0.05cm}m_{\nu-1}) = {\rm Pr}( m_{\nu}).$ | ||
* | *This exactly matches the definition of "statistical independence" ⇒ <u>Answer 1</u>. | ||
[[File:P_ID224__Sto_A_4_3_c.png|right|frame|2D | [[File:P_ID224__Sto_A_4_3_c.png|right|frame|2D PDF of $x$ and $m$]] | ||
'''(3)''' | '''(3)''' Correct are <u>the second and the last suggested solutions</u>. | ||
* | *The 2D PDF consists of four Dirac functions, each with weight $1/4$. | ||
* | *One obtains this result, for example, by evaluating the table on the data page. | ||
* | *Since $f_{xm}(x_\nu, m_\nu)$ is equal to the product $f_{x}(x_\nu) \cdot f_{m}(m_\nu)$ the quantities $x_\nu$ and $m_\nu$ are statistically independent. | ||
* | *Statistically independent random variables, however, are also linearly statistically independent, so they are certainly uncorrelated. | ||
'''(4)''' | '''(4)''' Within the sequence $\langle a_\nu \rangle$ of algebraic sum there are statistical bindings ⇒ <u>Answer 2</u>. | ||
* | *You can see this because the unconditional probability $ {\rm Pr}( a_{\nu} = 0) =1/8$ is, | ||
* | *while, for example ${\rm Pr}(a_{\nu} = 0\hspace{0.05cm}|\hspace{0.05cm}a_{\nu-1} = 3) =0$ holds. | ||
[[File:P_ID225__Sto_A_4_3_e.png|right|frame|2D | [[File:P_ID225__Sto_A_4_3_e.png|right|frame|2D PDF of $a$ and $m$]] | ||
'''(5)''' | '''(5)''' Correct are <u>the first and the last suggested solutions</u>: | ||
* | *As in the subtask '''(3)''' there are again four Dirac functions, but this time not with equal momentum weights $1/4$. | ||
* | *The two-dimensional PDF thus cannot be written as a product of the two marginal probability densities. | ||
* | *But this means that statistical bindings must exist between $a_\nu$ and $m_\nu$ . | ||
* | *For the joint expected value, one obtains: | ||
:$${\rm E}\big[a\cdot m \big] = \rm \frac{1}{8}\cdot 0 \cdot 0 +\frac{3}{8}\cdot 2 \cdot 0 +\frac{3}{8}\cdot 1 \cdot 1 + \frac{1}{8}\cdot 3 \cdot 1 = \frac{3}{4}.$$ | :$${\rm E}\big[a\cdot m \big] = \rm \frac{1}{8}\cdot 0 \cdot 0 +\frac{3}{8}\cdot 2 \cdot 0 +\frac{3}{8}\cdot 1 \cdot 1 + \frac{1}{8}\cdot 3 \cdot 1 = \frac{3}{4}.$$ | ||
* | *With the linear means ${\rm E}\big[a \big] = 1.5$ and ${\rm E}[m] = 0.5$ it thus follows for the covariance: | ||
:$$\mu_{am}= {\rm E}\big[ a\cdot m \big] - {\rm E}\big[ a \big]\cdot {\rm E} \big[ m \big] = \rm 0.75-1.5\cdot 0.5 = \rm 0.$$ | :$$\mu_{am}= {\rm E}\big[ a\cdot m \big] - {\rm E}\big[ a \big]\cdot {\rm E} \big[ m \big] = \rm 0.75-1.5\cdot 0.5 = \rm 0.$$ | ||
* | *Thus, the correlation coefficient $\rho_{am}= 0$. That is, The dependencies present are nonlinear. | ||
* | *The quantities $a_\nu$ and $m_\nu$ are statistically dependent, but still uncorrelated. | ||
{{ML-Fuß}} | {{ML-Fuß}} | ||
Revision as of 16:14, 19 January 2022

A "clocked" random number generator returns a sequence $\langle x_\nu \rangle$ of binary random numbers.
- It is assumed that the binary numbers $0$ and $1$ occur with equal probabilities and that the individual random numbers do not depend on each other.
- The random numbers $ x_\nu \in \{0, 1\}$ are entered into the first memory location of a shift register and shifted down one digit with each clock pulse.
Two new random sequences $\langle a_\nu \rangle$ and $\langle m_\nu \rangle$ are formed from the contents of the three-digit shift register. Here denotes:
- $a_\nu$ the algebraic sum:
- $$a_\nu=x_\nu+x_{\nu-1}+x_{\nu-2},$$
- $m_\nu$ the modulo 2 sum:
- $$m_\nu=x_\nu\oplus x_{\nu-1}\oplus x_{\nu-2}.$$
This fact is shown again in the following table:

Hints: This exercise belongs to the chapter Two-Dimensional Random Variables.
Questions
Solution
- $${\rm Pr}(m_\nu = 0) = {\rm Pr}(m_\nu = 1)\hspace{0.15cm}\underline{=0.5}.$$
(2) The table shows that for each preassignment ⇒ $( x_{\nu-1}, x_{\nu-2}) = (0,0), (0,1), (1,0), (1,1)$ the values $m_\nu = 0$ respectively $m_\nu = 1$ are equally likely.
- Expressed differently: ${\rm Pr}(m_{\nu}\hspace{0.05cm}|\hspace{0.05cm}m_{\nu-1}) = {\rm Pr}( m_{\nu}).$
- This exactly matches the definition of "statistical independence" ⇒ Answer 1.

(3) Correct are the second and the last suggested solutions.
- The 2D PDF consists of four Dirac functions, each with weight $1/4$.
- One obtains this result, for example, by evaluating the table on the data page.
- Since $f_{xm}(x_\nu, m_\nu)$ is equal to the product $f_{x}(x_\nu) \cdot f_{m}(m_\nu)$ the quantities $x_\nu$ and $m_\nu$ are statistically independent.
- Statistically independent random variables, however, are also linearly statistically independent, so they are certainly uncorrelated.
(4) Within the sequence $\langle a_\nu \rangle$ of algebraic sum there are statistical bindings ⇒ Answer 2.
- You can see this because the unconditional probability $ {\rm Pr}( a_{\nu} = 0) =1/8$ is,
- while, for example ${\rm Pr}(a_{\nu} = 0\hspace{0.05cm}|\hspace{0.05cm}a_{\nu-1} = 3) =0$ holds.

(5) Correct are the first and the last suggested solutions:
- As in the subtask (3) there are again four Dirac functions, but this time not with equal momentum weights $1/4$.
- The two-dimensional PDF thus cannot be written as a product of the two marginal probability densities.
- But this means that statistical bindings must exist between $a_\nu$ and $m_\nu$ .
- For the joint expected value, one obtains:
- $${\rm E}\big[a\cdot m \big] = \rm \frac{1}{8}\cdot 0 \cdot 0 +\frac{3}{8}\cdot 2 \cdot 0 +\frac{3}{8}\cdot 1 \cdot 1 + \frac{1}{8}\cdot 3 \cdot 1 = \frac{3}{4}.$$
- With the linear means ${\rm E}\big[a \big] = 1.5$ and ${\rm E}[m] = 0.5$ it thus follows for the covariance:
- $$\mu_{am}= {\rm E}\big[ a\cdot m \big] - {\rm E}\big[ a \big]\cdot {\rm E} \big[ m \big] = \rm 0.75-1.5\cdot 0.5 = \rm 0.$$
- Thus, the correlation coefficient $\rho_{am}= 0$. That is, The dependencies present are nonlinear.
- The quantities $a_\nu$ and $m_\nu$ are statistically dependent, but still uncorrelated.