Difference between revisions of "Aufgaben:Exercise 3.12: Cauchy Distribution"
From LNTwww
| (3 intermediate revisions by 2 users not shown) | |||
| Line 3: | Line 3: | ||
}} | }} | ||
| − | [[File:P_ID207__Sto_A_3_12.png|right|frame|PDF | + | [[File:P_ID207__Sto_A_3_12.png|right|frame|Cauchy PDF]] |
| − | The probability density function of the Cauchy distribution is given as follows: | + | 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. | From the graph you can already see the extremely slow decay of the PDF course. | ||
| − | |||
| − | |||
| − | |||
| − | |||
Hints: | Hints: | ||
| − | *The exercise belongs to the chapter [[Theory_of_Stochastic_Signals/Further_Distributions|Further Distributions]]. | + | *The exercise belongs to the chapter [[Theory_of_Stochastic_Signals/Further_Distributions|"Further Distributions"]]. |
| − | *In particular, reference is made to the | + | *In particular, reference is made to the section [[Theory_of_Stochastic_Signals/Further_Distributions#Cauchy_PDF|"Cauchy PDF"]]. |
| Line 25: | Line 21: | ||
<quiz display=simple> | <quiz display=simple> | ||
| − | {What is the distribution function $F_x(r)$? What is the probability that $x | + | {What is the cumulative distribution function $\rm (CDF)$ $F_x(r)$? What is the probability that $|x|<2$? |
|type="{}"} | |type="{}"} | ||
${\rm Pr} (|x| < 2) \ = \ $ { 50 3% } $ \ \%$ | ${\rm Pr} (|x| < 2) \ = \ $ { 50 3% } $ \ \%$ | ||
| − | {What is the probability that $x | + | {What is the probability that $|x|>4$? |
|type="{}"} | |type="{}"} | ||
${\rm Pr} (|x| > 4) \ = \ $ { 29.6 3% } $ \ \%$ | ${\rm Pr} (|x| > 4) \ = \ $ { 29.6 3% } $ \ \%$ | ||
| Line 46: | Line 42: | ||
===Solution=== | ===Solution=== | ||
{{ML-Kopf}} | {{ML-Kopf}} | ||
| − | '''(1)''' Comparing the given PDF with the general equation in the theory part, we see that the parameter $\lambda= 2$ | + | '''(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): | *From this follows (after integration over the PDF): | ||
:$$F_x ( r ) =\frac{1}{2} + \frac{\rm 1}{\rm \pi}\cdot \rm arctan(\it r/\rm 2).$$ | :$$F_x ( r ) =\frac{1}{2} + \frac{\rm 1}{\rm \pi}\cdot \rm arctan(\it r/\rm 2).$$ | ||
| Line 54: | Line 50: | ||
:$$F_x ( r = -2 ) =\frac{1}{2} + \frac{\rm 1}{\rm \pi}\cdot \rm arctan(-1)=\frac{1}{2} - \frac{\rm 1}{\rm \pi} \cdot \frac{\rm \pi}{4 }=0.25.$$ | :$$F_x ( r = -2 ) =\frac{1}{2} + \frac{\rm 1}{\rm \pi}\cdot \rm arctan(-1)=\frac{1}{2} - \frac{\rm 1}{\rm \pi} \cdot \frac{\rm \pi}{4 }=0.25.$$ | ||
| − | *The probability we are looking for is given by the difference | + | *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)''' According to the result of the subtask '''(1)''' | + | '''(2)''' According to the result of the subtask '''(1)''' ⇒ $F_x ( r = 4 ) = 0.5 + 1/\pi = 0.852$. |
| − | *Thus, for the "complementary" probability ${\rm Pr} (x > 4)= 0.148$. | + | *Thus, for the "complementary" probability: ${\rm Pr} (x > 4)= 0.148$. |
| − | *For symmetry reasons, the probability we are looking for is twice as large: | + | *For symmetry reasons, the probability we are looking for is twice as large: |
:$${\rm Pr} (|x| >4) \hspace{0.15cm}\underline{ = 29.6\%}.$$ | :$${\rm Pr} (|x| >4) \hspace{0.15cm}\underline{ = 29.6\%}.$$ | ||
| Line 69: | Line 65: | ||
\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 | + | *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 | + | *"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 | + | *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}} | ||
Latest revision as of 13:24, 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.
