Difference between revisions of "Aufgaben:Exercise 2.3: Algebraic Sum of Binary Numbers"
From LNTwww
| (4 intermediate revisions by the same user not shown) | |||
| Line 5: | Line 5: | ||
[[File:EN_Sto_A_2_3_neu.png|right|frame|Considered random generator]] | [[File:EN_Sto_A_2_3_neu.png|right|frame|Considered random generator]] | ||
A random number generator outputs a binary random number $x_\nu$ at each clock time $(\nu)$ , which can be $0$ or $1$ . | A random number generator outputs a binary random number $x_\nu$ at each clock time $(\nu)$ , which can be $0$ or $1$ . | ||
| − | *The value "1" occurs with probability $p = 0.25$ . | + | *The value "1" occurs with probability $p = 0.25$ . |
*The individual values $x_\nu$ are statistically independent of each other. | *The individual values $x_\nu$ are statistically independent of each other. | ||
| − | The binary numbers are stored in a shift register | + | The binary numbers are stored in a shift register with $I = 6$ memory cells. |
| − | At each clock instant, the contents of this shift register are shifted one place to the right and the algebraic sum $y_\nu$ of the shift register contents is formed in each case: | + | At each clock instant, the contents of this shift register are shifted one place to the right and the algebraic sum $y_\nu$ of the shift register contents is formed in each case: |
:$$y_{\nu}=\sum\limits_{i=0}^{5}x_{\nu-i}=x_{\nu}+x_{\nu-1}+\ \text{...} \ +x_{\nu-5}.$$ | :$$y_{\nu}=\sum\limits_{i=0}^{5}x_{\nu-i}=x_{\nu}+x_{\nu-1}+\ \text{...} \ +x_{\nu-5}.$$ | ||
| − | |||
| − | |||
| − | |||
| Line 21: | Line 18: | ||
Hints: | Hints: | ||
| − | *The exercise belongs to the chapter [[Theory_of_Stochastic_Signals/Binomial_Distribution| | + | *The exercise belongs to the chapter [[Theory_of_Stochastic_Signals/Binomial_Distribution|Binomial Distribution]]. |
| − | *To check your results you can use the interactive applet [[Applets: | + | *To check your results you can use the interactive HTML5/JavaScript applet [[Applets:Binomial_and_Poisson_Distribution_(Applet)|Binomial and Poisson distribution]]. |
| Line 37: | Line 34: | ||
| − | {Calculate the probability that $y$ is greater than $2$ | + | {Calculate the probability that $y$ is greater than $2$. |
|type="{}"} | |type="{}"} | ||
${\rm Pr}(y > 2) \ = \ $ { 0.169 3% } | ${\rm Pr}(y > 2) \ = \ $ { 0.169 3% } | ||
| − | {What is the mean value of the random variable $y$ | + | {What is the mean value of the random variable $y$? |
|type="{}"} | |type="{}"} | ||
$m_y \ =$ { 1.5 3% } | $m_y \ =$ { 1.5 3% } | ||
| − | {Find the | + | {Find the standard deviation of the random variable $y$. |
|type="{}"} | |type="{}"} | ||
$\sigma_y \ = \ $ { 1.061 3% } | $\sigma_y \ = \ $ { 1.061 3% } | ||
| Line 53: | Line 50: | ||
{Are the random numbers $y_\nu$ statistically independent? Justify your result. | {Are the random numbers $y_\nu$ statistically independent? Justify your result. | ||
| − | |type=" | + | |type="()"} |
- The random numbers are statistically independent. | - The random numbers are statistically independent. | ||
+ The random numbers are statistically dependent. | + The random numbers are statistically dependent. | ||
| − | {What is the conditional probability that $y_\nu | + | {What is the conditional probability that $y_\nu = \mu$ if $y_{\nu-1} = \mu$ occured previously? $(\mu = 0, \ 1, \ \text{...} \ , \ I)$. |
|type="{}"} | |type="{}"} | ||
${\rm Pr}(y_\nu = \mu \hspace{0.05cm} | \hspace{0.05cm} y_{\nu-1} = \mu ) \ = \ $ { 0.625 3% } | ${\rm Pr}(y_\nu = \mu \hspace{0.05cm} | \hspace{0.05cm} y_{\nu-1} = \mu ) \ = \ $ { 0.625 3% } | ||
| Line 67: | Line 64: | ||
===Solution=== | ===Solution=== | ||
{{ML-Kopf}} | {{ML-Kopf}} | ||
| − | '''(1)''' Each cell can contain a $0$ or a $1$ . Therefore, the sum can take all integer values between $0$ ánd $6$ : | + | '''(1)''' Each cell can contain a $0$ or a $1$ . Therefore, the sum can take all integer values between $0$ ánd $6$ : |
:$$y_{\nu}\in\{0,1,\ \text{...} \ ,6\}\hspace{0.3cm}\Rightarrow\hspace{0.3cm} | :$$y_{\nu}\in\{0,1,\ \text{...} \ ,6\}\hspace{0.3cm}\Rightarrow\hspace{0.3cm} | ||
y_{\rm max} \hspace{0.15cm} \underline{= 6}.$$ | y_{\rm max} \hspace{0.15cm} \underline{= 6}.$$ | ||
| Line 73: | Line 70: | ||
| − | '''(2)''' There is a binomial distribution. Therefore, with $p = 0.25$: | + | '''(2)''' There is a binomial distribution. Therefore, with $p = 0.25$: |
:$${\rm Pr}(y =0)=(1-p)^{\it I}=0.75^6=0.178,$$ | :$${\rm Pr}(y =0)=(1-p)^{\it I}=0.75^6=0.178,$$ | ||
:$${\rm Pr}(y=1)=\left({ I \atop {1}}\right)\cdot (1-p)^{I-1}\cdot p= \rm 6\cdot 0.75^5\cdot 0.25=0.356,$$ | :$${\rm Pr}(y=1)=\left({ I \atop {1}}\right)\cdot (1-p)^{I-1}\cdot p= \rm 6\cdot 0.75^5\cdot 0.25=0.356,$$ | ||
| Line 81: | Line 78: | ||
| − | '''(3)''' According to the general equation, the mean of the binomial distribution is: | + | '''(3)''' According to the general equation, the mean of the binomial distribution is: |
:$$m_y= I\cdot p\hspace{0.15cm} \underline{=\rm 1.5}.$$ | :$$m_y= I\cdot p\hspace{0.15cm} \underline{=\rm 1.5}.$$ | ||
| − | '''(4)''' Accordingly, for the | + | '''(4)''' Accordingly, for the standard deviation of the binomial distribution: |
:$$\sigma_y=\sqrt{ I \cdot p \cdot( 1- p)} \hspace{0.15cm} \underline{= \rm 1.061}.$$ | :$$\sigma_y=\sqrt{ I \cdot p \cdot( 1- p)} \hspace{0.15cm} \underline{= \rm 1.061}.$$ | ||
| − | '''(5)''' Correct is the <u>proposed solution 2</u>: | + | '''(5)''' Correct is the <u>proposed solution 2</u>: |
| − | *If $y_\nu = 0$, then only the values $0$ and $1$ can follow at the next time point, but not $2$, ... , $6$. | + | *If $y_\nu = 0$, then only the values $0$ and $1$ can follow at the next time point, but not $2$, ... , $6$. |
*That is: The sequence $ \langle y_\nu \rangle$ has (strong) statistical bindings. | *That is: The sequence $ \langle y_\nu \rangle$ has (strong) statistical bindings. | ||
| − | '''(6)''' The probability we are looking for is identical to the probability that the new binary symbol is equal to the symbol dropped out of the shift register. It follows that: | + | '''(6)''' The probability we are looking for is identical to the probability that the new binary symbol is equal to the symbol dropped out of the shift register. It follows that: |
:$${\rm Pr} (y_{\nu} = \mu\hspace{0.05cm}| \hspace{0.05cm} y_{\nu-{1}} = \mu) = {\rm Pr}(x_{\nu}= x_{\nu-6}). $$ | :$${\rm Pr} (y_{\nu} = \mu\hspace{0.05cm}| \hspace{0.05cm} y_{\nu-{1}} = \mu) = {\rm Pr}(x_{\nu}= x_{\nu-6}). $$ | ||
| − | *Since the symbols $x_\nu$ are statistically independent of each other, we can also write for this: | + | *Since the symbols $x_\nu$ are statistically independent of each other, we can also write for this: |
:$${\rm Pr}(x_{\nu} = x_{\nu-6}) = {\rm Pr}\big[(x_{\nu}= 1)\hspace{0.05cm}\cap\hspace{0.05cm}(x_{\nu-6}= 1)\hspace{0.05cm}\cup \hspace{0.05cm}(x_\nu=0)\hspace{0.05cm}\cap\hspace{0.05cm}(x_{\nu-6} =0)\big]= p^{2}+(1- p)^{2}=\rm 0.25^2 + 0.75^2\hspace{0.15cm} \underline{ = 0.625}. $$ | :$${\rm Pr}(x_{\nu} = x_{\nu-6}) = {\rm Pr}\big[(x_{\nu}= 1)\hspace{0.05cm}\cap\hspace{0.05cm}(x_{\nu-6}= 1)\hspace{0.05cm}\cup \hspace{0.05cm}(x_\nu=0)\hspace{0.05cm}\cap\hspace{0.05cm}(x_{\nu-6} =0)\big]= p^{2}+(1- p)^{2}=\rm 0.25^2 + 0.75^2\hspace{0.15cm} \underline{ = 0.625}. $$ | ||
Latest revision as of 14:58, 16 February 2022
Considered random generator
A random number generator outputs a binary random number $x_\nu$ at each clock time $(\nu)$ , which can be $0$ or $1$ .
- The value "1" occurs with probability $p = 0.25$ .
- The individual values $x_\nu$ are statistically independent of each other.
The binary numbers are stored in a shift register with $I = 6$ memory cells.
At each clock instant, the contents of this shift register are shifted one place to the right and the algebraic sum $y_\nu$ of the shift register contents is formed in each case:
- $$y_{\nu}=\sum\limits_{i=0}^{5}x_{\nu-i}=x_{\nu}+x_{\nu-1}+\ \text{...} \ +x_{\nu-5}.$$
Hints:
- The exercise belongs to the chapter Binomial Distribution.
- To check your results you can use the interactive HTML5/JavaScript applet Binomial and Poisson distribution.
Questions
Solution
(1) Each cell can contain a $0$ or a $1$ . Therefore, the sum can take all integer values between $0$ ánd $6$ :
- $$y_{\nu}\in\{0,1,\ \text{...} \ ,6\}\hspace{0.3cm}\Rightarrow\hspace{0.3cm} y_{\rm max} \hspace{0.15cm} \underline{= 6}.$$
(2) There is a binomial distribution. Therefore, with $p = 0.25$:
- $${\rm Pr}(y =0)=(1-p)^{\it I}=0.75^6=0.178,$$
- $${\rm Pr}(y=1)=\left({ I \atop {1}}\right)\cdot (1-p)^{I-1}\cdot p= \rm 6\cdot 0.75^5\cdot 0.25=0.356,$$
- $${\rm Pr}(y=2)=\left({ I \atop { 2}}\right)\cdot (1-p)^{I-2}\cdot p^{\rm 2}= \rm 15\cdot 0.75^4\cdot 0.25^2=0.297,$$
- $$\Rightarrow \hspace{0.3cm}{\rm Pr}(y>2)=1-{\rm Pr}(y=0)-{\rm Pr}( y=1)-{\rm Pr}( y=2)\hspace{0.15cm} \underline{=\rm 0.169}.$$
(3) According to the general equation, the mean of the binomial distribution is:
- $$m_y= I\cdot p\hspace{0.15cm} \underline{=\rm 1.5}.$$
(4) Accordingly, for the standard deviation of the binomial distribution:
- $$\sigma_y=\sqrt{ I \cdot p \cdot( 1- p)} \hspace{0.15cm} \underline{= \rm 1.061}.$$
(5) Correct is the proposed solution 2:
- If $y_\nu = 0$, then only the values $0$ and $1$ can follow at the next time point, but not $2$, ... , $6$.
- That is: The sequence $ \langle y_\nu \rangle$ has (strong) statistical bindings.
(6) The probability we are looking for is identical to the probability that the new binary symbol is equal to the symbol dropped out of the shift register. It follows that:
- $${\rm Pr} (y_{\nu} = \mu\hspace{0.05cm}| \hspace{0.05cm} y_{\nu-{1}} = \mu) = {\rm Pr}(x_{\nu}= x_{\nu-6}). $$
- Since the symbols $x_\nu$ are statistically independent of each other, we can also write for this:
- $${\rm Pr}(x_{\nu} = x_{\nu-6}) = {\rm Pr}\big[(x_{\nu}= 1)\hspace{0.05cm}\cap\hspace{0.05cm}(x_{\nu-6}= 1)\hspace{0.05cm}\cup \hspace{0.05cm}(x_\nu=0)\hspace{0.05cm}\cap\hspace{0.05cm}(x_{\nu-6} =0)\big]= p^{2}+(1- p)^{2}=\rm 0.25^2 + 0.75^2\hspace{0.15cm} \underline{ = 0.625}. $$
