Difference between revisions of "Aufgaben:Exercise 3.12: Cauchy Distribution"
From LNTwww
m (Text replacement - "„" to """) |
|||
| (7 intermediate revisions by 2 users not shown) | |||
| Line 1: | Line 1: | ||
| − | {{quiz-Header|Buchseite= | + | {{quiz-Header|Buchseite=Theory_of_Stochastic_Signals/Further_Distributions |
}} | }} | ||
| − | [[File:P_ID207__Sto_A_3_12.png|right|frame| | + | [[File:P_ID207__Sto_A_3_12.png|right|frame|Cauchy PDF]] |
| − | + | The probability density function $\rm (PDF)$ of the Cauchy distribution is given as follows: | |
:$$f_x(x)=\frac{\rm 1}{\rm 2 \pi}\cdot \frac{\rm 1}{\rm 1+ (\it x/\rm 2)^{\rm 2}}.$$ | :$$f_x(x)=\frac{\rm 1}{\rm 2 \pi}\cdot \frac{\rm 1}{\rm 1+ (\it x/\rm 2)^{\rm 2}}.$$ | ||
| − | + | From the graph you can already see the extremely slow decay of the PDF course. | |
| − | + | Hints: | |
| − | + | *The exercise belongs to the chapter [[Theory_of_Stochastic_Signals/Further_Distributions|"Further Distributions"]]. | |
| − | + | *In particular, reference is made to the section [[Theory_of_Stochastic_Signals/Further_Distributions#Cauchy_PDF|"Cauchy PDF"]]. | |
| − | |||
| − | |||
| − | * | ||
| − | * | ||
| − | === | + | ===Questions=== |
<quiz display=simple> | <quiz display=simple> | ||
| − | { | + | {What is the cumulative distribution function $\rm (CDF)$ $F_x(r)$? What is the probability that $|x|<2$? |
|type="{}"} | |type="{}"} | ||
| − | ${\rm Pr} (|x| < 2) \ = \ $ | + | ${\rm Pr} (|x| < 2) \ = \ $ { 50 3% } $ \ \%$ |
| − | { | + | {What is the probability that $|x|>4$? |
|type="{}"} | |type="{}"} | ||
${\rm Pr} (|x| > 4) \ = \ $ { 29.6 3% } $ \ \%$ | ${\rm Pr} (|x| > 4) \ = \ $ { 29.6 3% } $ \ \%$ | ||
| − | { | + | {Which of the following statements are true for the Cauchy distribution? |
|type="[]"} | |type="[]"} | ||
| − | + | + | + The Cauchy distribution has an infinitely large variance. |
| − | + | + | + The Chebyshev inequality makes no sense here. |
| − | + | + | + A random variable that can be measured in nature is never Cauchy distributed. |
</quiz> | </quiz> | ||
| − | === | + | ===Solution=== |
{{ML-Kopf}} | {{ML-Kopf}} | ||
| − | '''(1)''' | + | '''(1)''' Comparing the given PDF with the general equation in the theory part, we see that the parameter is $\lambda= 2$. |
| − | * | + | *From this follows (after integration over the PDF): |
:$$F_x ( r ) =\frac{1}{2} + \frac{\rm 1}{\rm \pi}\cdot \rm arctan(\it r/\rm 2).$$ | :$$F_x ( r ) =\frac{1}{2} + \frac{\rm 1}{\rm \pi}\cdot \rm arctan(\it r/\rm 2).$$ | ||
| − | * | + | *In particular. |
:$$F_x ( r = +2 ) =\frac{1}{2} + \frac{\rm 1}{\rm \pi}\cdot \rm arctan(1)=\frac{1}{2} + \frac{\rm 1}{\rm \pi} \cdot \frac{\rm \pi}{4 }=0.75,$$ | :$$F_x ( r = +2 ) =\frac{1}{2} + \frac{\rm 1}{\rm \pi}\cdot \rm arctan(1)=\frac{1}{2} + \frac{\rm 1}{\rm \pi} \cdot \frac{\rm \pi}{4 }=0.75,$$ | ||
:$$F_x ( r = -2 ) =\frac{1}{2} + \frac{\rm 1}{\rm \pi}\cdot \rm arctan(-1)=\frac{1}{2} - \frac{\rm 1}{\rm \pi} \cdot \frac{\rm \pi}{4 }=0.25.$$ | :$$F_x ( r = -2 ) =\frac{1}{2} + \frac{\rm 1}{\rm \pi}\cdot \rm arctan(-1)=\frac{1}{2} - \frac{\rm 1}{\rm \pi} \cdot \frac{\rm \pi}{4 }=0.25.$$ | ||
| − | * | + | *The probability we are looking for is given by the difference: |
:$${\rm Pr} (|x| < 2) = 0.75 - 0.25 \hspace{0.15cm}\underline{=50\%}.$$ | :$${\rm Pr} (|x| < 2) = 0.75 - 0.25 \hspace{0.15cm}\underline{=50\%}.$$ | ||
| − | '''(2)''' | + | '''(2)''' According to the result of the subtask '''(1)''' ⇒ $F_x ( r = 4 ) = 0.5 + 1/\pi = 0.852$. |
| − | * | + | *Thus, for the "complementary" probability: ${\rm Pr} (x > 4)= 0.148$. |
| − | * | + | *For symmetry reasons, the probability we are looking for is twice as large: |
:$${\rm Pr} (|x| >4) \hspace{0.15cm}\underline{ = 29.6\%}.$$ | :$${\rm Pr} (|x| >4) \hspace{0.15cm}\underline{ = 29.6\%}.$$ | ||
| − | '''(3)''' <u> | + | '''(3)''' <u>All proposed solutions</u> are true: |
| − | * | + | *For the variance of the Cauchy distribution holds namely: |
:$$\sigma_x^{\rm 2}=\frac{1}{2\pi}\int_{-\infty}^{+\infty} | :$$\sigma_x^{\rm 2}=\frac{1}{2\pi}\int_{-\infty}^{+\infty} | ||
\hspace{-0.15cm} | \hspace{-0.15cm} | ||
\frac{\it x^{\rm 2}}{\rm 1+(\it x/\rm 2)^{\rm 2}} \,\,{\rm d}x.$$ | \frac{\it x^{\rm 2}}{\rm 1+(\it x/\rm 2)^{\rm 2}} \,\,{\rm d}x.$$ | ||
| − | * | + | *For large $x$ the integrand yields the constant value $4$. Therefore the integral diverges. |
| − | * | + | *Chebyshev's inequality does not provide an evaluable bound, even with $\sigma_x \to \infty$. |
| − | * | + | *"Natural" random variables (physically interpretable) can never be cauchy distributed, otherwise they would have an infinite power. |
| − | * | + | *On the other hand, an "artificial" (or mathematical) random variable is not subject to this restriction. Example: '''The quotient of two zero mean quantities'''. |
{{ML-Fuß}} | {{ML-Fuß}} | ||
| − | [[Category:Theory of Stochastic Signals: Exercises|^3.7 | + | [[Category:Theory of Stochastic Signals: Exercises|^3.7 Further Distributions^]] |
Latest revision as of 14:24, 3 February 2022
Cauchy PDF
The probability density function $\rm (PDF)$ of the Cauchy distribution is given as follows:
- $$f_x(x)=\frac{\rm 1}{\rm 2 \pi}\cdot \frac{\rm 1}{\rm 1+ (\it x/\rm 2)^{\rm 2}}.$$
From the graph you can already see the extremely slow decay of the PDF course.
Hints:
- The exercise belongs to the chapter "Further Distributions".
- In particular, reference is made to the section "Cauchy PDF".
Questions
Solution
(1) Comparing the given PDF with the general equation in the theory part, we see that the parameter is $\lambda= 2$.
- From this follows (after integration over the PDF):
- $$F_x ( r ) =\frac{1}{2} + \frac{\rm 1}{\rm \pi}\cdot \rm arctan(\it r/\rm 2).$$
- In particular.
- $$F_x ( r = +2 ) =\frac{1}{2} + \frac{\rm 1}{\rm \pi}\cdot \rm arctan(1)=\frac{1}{2} + \frac{\rm 1}{\rm \pi} \cdot \frac{\rm \pi}{4 }=0.75,$$
- $$F_x ( r = -2 ) =\frac{1}{2} + \frac{\rm 1}{\rm \pi}\cdot \rm arctan(-1)=\frac{1}{2} - \frac{\rm 1}{\rm \pi} \cdot \frac{\rm \pi}{4 }=0.25.$$
- The probability we are looking for is given by the difference:
- $${\rm Pr} (|x| < 2) = 0.75 - 0.25 \hspace{0.15cm}\underline{=50\%}.$$
(2) According to the result of the subtask (1) ⇒ $F_x ( r = 4 ) = 0.5 + 1/\pi = 0.852$.
- Thus, for the "complementary" probability: ${\rm Pr} (x > 4)= 0.148$.
- For symmetry reasons, the probability we are looking for is twice as large:
- $${\rm Pr} (|x| >4) \hspace{0.15cm}\underline{ = 29.6\%}.$$
(3) All proposed solutions are true:
- For the variance of the Cauchy distribution holds namely:
- $$\sigma_x^{\rm 2}=\frac{1}{2\pi}\int_{-\infty}^{+\infty} \hspace{-0.15cm} \frac{\it x^{\rm 2}}{\rm 1+(\it x/\rm 2)^{\rm 2}} \,\,{\rm d}x.$$
- For large $x$ the integrand yields the constant value $4$. Therefore the integral diverges.
- Chebyshev's inequality does not provide an evaluable bound, even with $\sigma_x \to \infty$.
- "Natural" random variables (physically interpretable) can never be cauchy distributed, otherwise they would have an infinite power.
- On the other hand, an "artificial" (or mathematical) random variable is not subject to this restriction. Example: The quotient of two zero mean quantities.
