Difference between revisions of "Aufgaben:Exercise 3.8Z: Circle (Ring) Area"
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| − | {{quiz-Header|Buchseite= | + | {{quiz-Header|Buchseite=Theory_of_Stochastic_Signals/Exponentially_Distributed_Random_Variables |
}} | }} | ||
| − | [[File:P_ID133__Sto_Z_3_8.png|right|frame| | + | [[File:P_ID133__Sto_Z_3_8.png|right|frame|To the circular ring areas]] |
| − | + | We consider circles of different sizes: | |
| − | * | + | *The radius $r$ and the area $A$ can be thought of as interdependent random variables. |
| − | * | + | *It is assumed that the radius is restricted to the area $6 \le r \le 8$ . |
| − | In | + | In the sketch above, the area in which such circles (all with center at coordinate origin) can lie is marked in yellow. Furthermore, it can be assumed that the radius in this interval is equally distributed: |
| − | :$$f_r(r)=\left\{ \begin{array}{*{4}{c}} 0.5 & \rm f\ddot{u}r\hspace{0.2cm}{\rm 6\le \it r \le \rm 8}, | + | :$$f_r(r)=\left\{ \begin{array}{*{4}{c}} 0.5 & \rm f\ddot{u}r\hspace{0.2cm}{\rm 6\le \it r \le \rm 8}, \rm 0 & \rm else. \end{array} \right.$$ |
| − | + | From subtask '''(5)''' narrow circular rings with center radius $r$ and width $b$ are considered (lower sketch): | |
| − | * | + | *The area of such a circular ring is denoted by $R$ . |
| − | * | + | *The possible center radii $r$ are again equally distributed between $6$ and $8$. |
| − | * | + | *The circular ring width is $b = 0.1$. |
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| − | + | Hints: | |
| − | + | *The task belongs to the chapter [[Theory_of_Stochastic_Signals/Theory_of_Stochastic_Signals/Exponentially_Distributed_Random_Variables]]. | |
| − | * | + | *In particular, reference is made to the page [[Theory_of_Stochastic_Signals/Exponentially_Distributed_Random_Variables#Transformation_of_random_variables|Transformation of random variables]]. |
| − | * | ||
| − | === | + | ===Questions=== |
<quiz display=simple> | <quiz display=simple> | ||
| − | { | + | {Give the transformation characteristic curve $A = g(r)$ analytically. What is the minimum value of the random variable $A$? |
|type="{}"} | |type="{}"} | ||
$A_\text{min} \ = \ $ { 113.09 3% } | $A_\text{min} \ = \ $ { 113.09 3% } | ||
| − | { | + | {what is the maximum value of the random variable of $A$? |
|type="{}"} | |type="{}"} | ||
| − | $A_\text{max} \ = | + | $A_\text{max} \ = \ $ { 201.06 3% } |
| − | { | + | {What value $m_{ A} = {\rm E}\big[A\big]$ results for the "mean" circular area? |
|type="{}"} | |type="{}"} | ||
| − | $m_{ A} \ = | + | $m_{ A} \ = \ $ { 154.98 3% } |
| − | { | + | {Calculate the probability density function of the random variable $A$. What is the probability that the area $A> 150$ ? |
|type="{}"} | |type="{}"} | ||
| − | ${\rm Pr}(A > 150) \ = | + | ${\rm Pr}(A > 150) \ = \ $ { 54.5 3% } $\ \%$ |
| − | { | + | {What is the PDF of the random variable $R$ (area of the circular rings according to the sketch below)? What is its minimum value? Let $b = 0.1$. |
|type="{}"} | |type="{}"} | ||
| − | $R_\text{min} \ = | + | $R_\text{min} \ = \ $ { 3.77 3% } |
| − | { | + | {It continues to apply $b = 0.1$. What is the maximum value of the random variable $R$? |
|type="{}"} | |type="{}"} | ||
| − | $R_\text{max} \ = | + | $R_\text{max} \ = \ $ { 5.03 3% } |
| − | { | + | {What is the expected value of the random variable $R$ for $b = 0.1$? |
|type="{}"} | |type="{}"} | ||
| − | ${\rm E}\big[R\big] \ = | + | ${\rm E}\big[R\big] \ = \ $ { 4.4 3% } |
| − | |||
</quiz> | </quiz> | ||
| − | === | + | ===Solution=== |
{{ML-Kopf}} | {{ML-Kopf}} | ||
| − | '''(1)''' | + | '''(1)''' The equation of the circular surface is at the same time the transformation characteristic: $A = \pi \cdot r^2$. |
| − | * | + | *From this, with $r = 6$ for the minimum value: |
:$$A_\text{min} \hspace{0.15cm}\underline {= 113.09}.$$ | :$$A_\text{min} \hspace{0.15cm}\underline {= 113.09}.$$ | ||
| − | '''(2)''' | + | '''(2)''' Correspondingly, with $r = 8$ for the maximum value: |
:$$A_\text{max} \hspace{0.15cm}\underline {= 201.06}.$$ | :$$A_\text{max} \hspace{0.15cm}\underline {= 201.06}.$$ | ||
| − | '''(3)''' | + | '''(3)''' The simplest way to solve this problem is as follows: |
:$$m_{\rm A}={\rm E}\big[A\big]={\rm E}\big[g(r)\big]=\int_{ -\infty}^{+\infty}g(r)\cdot f_r(r) {\rm d}r.$$ | :$$m_{\rm A}={\rm E}\big[A\big]={\rm E}\big[g(r)\big]=\int_{ -\infty}^{+\infty}g(r)\cdot f_r(r) {\rm d}r.$$ | ||
| − | * | + | *With $g(r) = \pi \cdot r^2$ and $f_r(r) = 1/2$ in the range of $6$ ... $8$ obtains: |
:$$m_{\rm A}=\int_{\rm 6}^{\rm 8}1/2 \cdot\pi\cdot r^{\rm 2}\, {\rm d} \it r=\frac{\pi}{\rm 6}\cdot \rm ( 8^3-6^3) | :$$m_{\rm A}=\int_{\rm 6}^{\rm 8}1/2 \cdot\pi\cdot r^{\rm 2}\, {\rm d} \it r=\frac{\pi}{\rm 6}\cdot \rm ( 8^3-6^3) | ||
\hspace{0.15cm}\underline{=\rm 154.98}.$$ | \hspace{0.15cm}\underline{=\rm 154.98}.$$ | ||
| − | '''(4)''' | + | '''(4)''' The PDF of the transformed randomgrö&transform;e $A$ is: |
:$$f_A(A)=\frac{f_r(r)}{|g\hspace{0.05cm}'(r)|}\Bigg|_{r=h(y) = \sqrt{A/ \pi }}.$$ | :$$f_A(A)=\frac{f_r(r)}{|g\hspace{0.05cm}'(r)|}\Bigg|_{r=h(y) = \sqrt{A/ \pi }}.$$ | ||
| − | * | + | *In the range between $A_\text{min} {= 113.09}$ and $A_\text{max} {= 201.06}$ then holds: |
:$$f_A(A)=\frac{\rm 1/2}{\rm 2\cdot \pi\cdot\it r}\Bigg|_{\it r=\sqrt{\it A/\rm \pi}}=\frac{\rm 1}{\rm 4\cdot\sqrt{\it A\cdot\rm \pi}}.$$ | :$$f_A(A)=\frac{\rm 1/2}{\rm 2\cdot \pi\cdot\it r}\Bigg|_{\it r=\sqrt{\it A/\rm \pi}}=\frac{\rm 1}{\rm 4\cdot\sqrt{\it A\cdot\rm \pi}}.$$ | ||
| − | * | + | *The probability we are looking for is obtained by integration: |
:$${\rm Pr}(A> 150)=\int_{\rm 150}^{\it A_{\rm max}}\frac{\rm 1}{\rm 4\cdot\sqrt{\it A\cdot\rm \pi}} \; \rm d \it A= \frac{\rm 2\cdot\sqrt{\it A}}{\rm 4\cdot\sqrt{\pi}}\Big|_{\rm 150}^{\it A_{\rm max}}.$$ | :$${\rm Pr}(A> 150)=\int_{\rm 150}^{\it A_{\rm max}}\frac{\rm 1}{\rm 4\cdot\sqrt{\it A\cdot\rm \pi}} \; \rm d \it A= \frac{\rm 2\cdot\sqrt{\it A}}{\rm 4\cdot\sqrt{\pi}}\Big|_{\rm 150}^{\it A_{\rm max}}.$$ | ||
| − | * | + | *The upper limit of integration yields the value $4$ and the lower limit $3.455$. This yields the probability we are looking for: |
:$${\rm Pr}(A> 150) \hspace{0.15cm}\underline {=54.5\%}.$$ | :$${\rm Pr}(A> 150) \hspace{0.15cm}\underline {=54.5\%}.$$ | ||
| − | '''(5)''' | + | '''(5)''' For the circular ring surface $R$ holds for a given radius $r$: |
:$$R=\left (r+{b}/{\rm 2} \right)^{\rm 2}\cdot \rm\pi-\left ({\it r}-{\it b}/{\rm 2} \right)^{\rm 2}\cdot \rm\pi= \rm2\cdot\pi\cdot\it r \cdot b.$$ | :$$R=\left (r+{b}/{\rm 2} \right)^{\rm 2}\cdot \rm\pi-\left ({\it r}-{\it b}/{\rm 2} \right)^{\rm 2}\cdot \rm\pi= \rm2\cdot\pi\cdot\it r \cdot b.$$ | ||
| − | * | + | *There is thus a linear relationship between $R$ and $r$ . |
| − | * | + | *That is: $R$ is also uniformly distributed independently of the width $b$ as long as $b \ll r$ . |
| − | * | + | *For the minimum value holds: |
:$$R_{\rm min}=\rm 2\pi\cdot 6\cdot 0.1\hspace{0.15cm}\underline{\approx3.77}. $$ | :$$R_{\rm min}=\rm 2\pi\cdot 6\cdot 0.1\hspace{0.15cm}\underline{\approx3.77}. $$ | ||
| − | '''(6)''' | + | '''(6)''' Accordingly, the maximum value is: |
:$$R_{\rm max}=\rm 2\pi\cdot 8\cdot 0.1\hspace{0.15cm}\underline{\approx 5.03}.$$ | :$$R_{\rm max}=\rm 2\pi\cdot 8\cdot 0.1\hspace{0.15cm}\underline{\approx 5.03}.$$ | ||
| − | '''(7)''' | + | '''(7)''' Due to the linear relationship between $R$ and $r$ the mean radius $r = 7$ also leads to the mean circular ring surface: |
:$${\rm E}\big[R\big]=\rm 2\pi\cdot 7\cdot 0.1\hspace{0.15cm}\underline{\approx 4.4}.$$ | :$${\rm E}\big[R\big]=\rm 2\pi\cdot 7\cdot 0.1\hspace{0.15cm}\underline{\approx 4.4}.$$ | ||
| − | |||
{{ML-Fuß}} | {{ML-Fuß}} | ||
Revision as of 21:00, 11 January 2022
To the circular ring areas
We consider circles of different sizes:
- The radius $r$ and the area $A$ can be thought of as interdependent random variables.
- It is assumed that the radius is restricted to the area $6 \le r \le 8$ .
In the sketch above, the area in which such circles (all with center at coordinate origin) can lie is marked in yellow. Furthermore, it can be assumed that the radius in this interval is equally distributed:
- $$f_r(r)=\left\{ \begin{array}{*{4}{c}} 0.5 & \rm f\ddot{u}r\hspace{0.2cm}{\rm 6\le \it r \le \rm 8}, \rm 0 & \rm else. \end{array} \right.$$
From subtask (5) narrow circular rings with center radius $r$ and width $b$ are considered (lower sketch):
- The area of such a circular ring is denoted by $R$ .
- The possible center radii $r$ are again equally distributed between $6$ and $8$.
- The circular ring width is $b = 0.1$.
Hints:
- The task belongs to the chapter Theory of Stochastic Signals/Theory of Stochastic Signals/Exponentially Distributed Random Variables.
- In particular, reference is made to the page Transformation of random variables.
Questions
Solution
(1) The equation of the circular surface is at the same time the transformation characteristic: $A = \pi \cdot r^2$.
- From this, with $r = 6$ for the minimum value:
- $$A_\text{min} \hspace{0.15cm}\underline {= 113.09}.$$
(2) Correspondingly, with $r = 8$ for the maximum value:
- $$A_\text{max} \hspace{0.15cm}\underline {= 201.06}.$$
(3) The simplest way to solve this problem is as follows:
- $$m_{\rm A}={\rm E}\big[A\big]={\rm E}\big[g(r)\big]=\int_{ -\infty}^{+\infty}g(r)\cdot f_r(r) {\rm d}r.$$
- With $g(r) = \pi \cdot r^2$ and $f_r(r) = 1/2$ in the range of $6$ ... $8$ obtains:
- $$m_{\rm A}=\int_{\rm 6}^{\rm 8}1/2 \cdot\pi\cdot r^{\rm 2}\, {\rm d} \it r=\frac{\pi}{\rm 6}\cdot \rm ( 8^3-6^3) \hspace{0.15cm}\underline{=\rm 154.98}.$$
(4) The PDF of the transformed randomgrö&transform;e $A$ is:
- $$f_A(A)=\frac{f_r(r)}{|g\hspace{0.05cm}'(r)|}\Bigg|_{r=h(y) = \sqrt{A/ \pi }}.$$
- In the range between $A_\text{min} {= 113.09}$ and $A_\text{max} {= 201.06}$ then holds:
- $$f_A(A)=\frac{\rm 1/2}{\rm 2\cdot \pi\cdot\it r}\Bigg|_{\it r=\sqrt{\it A/\rm \pi}}=\frac{\rm 1}{\rm 4\cdot\sqrt{\it A\cdot\rm \pi}}.$$
- The probability we are looking for is obtained by integration:
- $${\rm Pr}(A> 150)=\int_{\rm 150}^{\it A_{\rm max}}\frac{\rm 1}{\rm 4\cdot\sqrt{\it A\cdot\rm \pi}} \; \rm d \it A= \frac{\rm 2\cdot\sqrt{\it A}}{\rm 4\cdot\sqrt{\pi}}\Big|_{\rm 150}^{\it A_{\rm max}}.$$
- The upper limit of integration yields the value $4$ and the lower limit $3.455$. This yields the probability we are looking for:
- $${\rm Pr}(A> 150) \hspace{0.15cm}\underline {=54.5\%}.$$
(5) For the circular ring surface $R$ holds for a given radius $r$:
- $$R=\left (r+{b}/{\rm 2} \right)^{\rm 2}\cdot \rm\pi-\left ({\it r}-{\it b}/{\rm 2} \right)^{\rm 2}\cdot \rm\pi= \rm2\cdot\pi\cdot\it r \cdot b.$$
- There is thus a linear relationship between $R$ and $r$ .
- That is: $R$ is also uniformly distributed independently of the width $b$ as long as $b \ll r$ .
- For the minimum value holds:
- $$R_{\rm min}=\rm 2\pi\cdot 6\cdot 0.1\hspace{0.15cm}\underline{\approx3.77}. $$
(6) Accordingly, the maximum value is:
- $$R_{\rm max}=\rm 2\pi\cdot 8\cdot 0.1\hspace{0.15cm}\underline{\approx 5.03}.$$
(7) Due to the linear relationship between $R$ and $r$ the mean radius $r = 7$ also leads to the mean circular ring surface:
- $${\rm E}\big[R\big]=\rm 2\pi\cdot 7\cdot 0.1\hspace{0.15cm}\underline{\approx 4.4}.$$
