Aufgaben:Exercise 3.12: Cauchy Distribution: Difference between revisions
From LNTwww
No edit summary |
No edit summary |
||
| 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|PDF of Cauchy distribution]] | ||
The probability density function 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. | |||
| Line 15: | Line 15: | ||
Hints: | |||
* | *The exercise belongs to the chapter [[Theory_of_Stochastic_Signals/Further_Distributions|Further Distributions]]. | ||
* | *In particular, reference is made to the page [[Theory_of_Stochastic_Signals/Further_Distributions#Cauchy_PDF|Cauchy PDF]] . | ||
=== | ===Questions=== | ||
<quiz display=simple> | <quiz display=simple> | ||
{ | {What is the distribution function $F_x(r)$? What is the probability that $x$ is smaller than $2$ by a small amount? | ||
|type="{}"} | |type="{}"} | ||
${\rm Pr} (|x| < 2) \ = \ $ | ${\rm Pr} (|x| < 2) \ = \ $ { 50 3% } $ \ \%$ | ||
{ | {What is the probability that $x$ is greater in amount than $4$? | ||
|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 $\lambda= 2$ is. | ||
* | *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 of. | ||
:$${\rm Pr} (|x| < 2) = 0.75 - 0.25 \hspace{0.15cm}\underline{=50\%}.$$ | :$${\rm Pr} (|x| < 2) = 0.75 - 0.25 \hspace{0.15cm}\underline{=50\%}.$$ | ||
'''(2)''' | '''(2)''' According to the result of the subtask '''(1)''' is $F_x ( r = 4 ) = 0.5 + 1/\pi = 0.852$. | ||
* | *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 große $x$ the integrand yields the constant value $4$. Therefore the integral diverges. | ||
* | *With $\sigma_x \to \infty$ however, even Chebyshev's inequality does not provide an evaluable bound. | ||
* | *Natural" random variables (physically interpretable) can never be cauchy distributed, otherwise they would have to have infinite power. | ||
* | *On the other hand, an "artificial" (or mathematical) random variable (example: the quotient of two zero mean quantities) is not subject to this restriction. | ||
{{ML-Fuß}} | {{ML-Fuß}} | ||
Revision as of 23:57, 12 January 2022

The probability density function 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 page Cauchy PDF .
Questions
Solution
(1) Comparing the given PDF with the general equation in the theory part, we see that the parameter $\lambda= 2$ is.
- 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 of.
- $${\rm Pr} (|x| < 2) = 0.75 - 0.25 \hspace{0.15cm}\underline{=50\%}.$$
(2) According to the result of the subtask (1) is $F_x ( r = 4 ) = 0.5 + 1/\pi = 0.852$.
- 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 große $x$ the integrand yields the constant value $4$. Therefore the integral diverges.
- With $\sigma_x \to \infty$ however, even Chebyshev's inequality does not provide an evaluable bound.
- Natural" random variables (physically interpretable) can never be cauchy distributed, otherwise they would have to have infinite power.
- On the other hand, an "artificial" (or mathematical) random variable (example: the quotient of two zero mean quantities) is not subject to this restriction.