Difference between revisions of "Aufgaben:Exercise 3.12: Cauchy Distribution"
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− | [[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 (PDF) of the Cauchy distribution is given as follows: |
:fx(x)=12π⋅11+(x/2)2. | :fx(x)=12π⋅11+(x/2)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. | ||
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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"]]. |
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<quiz display=simple> | <quiz display=simple> | ||
− | {What is the distribution function Fx(r)? What is the probability that $x | + | {What is the cumulative distribution function (CDF) Fx(r)? What is the probability that $|x|<2$? |
|type="{}"} | |type="{}"} | ||
Pr(|x|<2) = { 50 3% } % | Pr(|x|<2) = { 50 3% } % | ||
− | {What is the probability that $x | + | {What is the probability that $|x|>4$? |
|type="{}"} | |type="{}"} | ||
Pr(|x|>4) = { 29.6 3% } % | Pr(|x|>4) = { 29.6 3% } % | ||
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===Solution=== | ===Solution=== | ||
{{ML-Kopf}} | {{ML-Kopf}} | ||
− | '''(1)''' Comparing the given PDF with the general equation in the theory part, we see that the parameter λ=2 | + | '''(1)''' Comparing the given PDF with the general equation in the theory part, we see that the parameter is λ=2. |
*From this follows (after integration over the PDF): | *From this follows (after integration over the PDF): | ||
:Fx(r)=12+1π⋅arctan(r/2). | :Fx(r)=12+1π⋅arctan(r/2). | ||
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:Fx(r=−2)=12+1π⋅arctan(−1)=12−1π⋅π4=0.25. | :Fx(r=−2)=12+1π⋅arctan(−1)=12−1π⋅π4=0.25. | ||
− | *The probability we are looking for is given by the difference | + | *The probability we are looking for is given by the difference: |
:Pr(|x|<2)=0.75−0.25=50%_. | :Pr(|x|<2)=0.75−0.25=50%_. | ||
− | '''(2)''' According to the result of the subtask '''(1)''' | + | '''(2)''' According to the result of the subtask '''(1)''' ⇒ Fx(r=4)=0.5+1/π=0.852. |
− | *Thus, for the "complementary" probability Pr(x>4)=0.148. | + | *Thus, for the "complementary" probability: 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: |
:Pr(|x|>4)=29.6%_. | :Pr(|x|>4)=29.6%_. | ||
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\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 σx→∞. |
− | *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 14:24, 3 February 2022
The probability density function (PDF) of the Cauchy distribution is given as follows:
- fx(x)=12π⋅11+(x/2)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 λ=2.
- From this follows (after integration over the PDF):
- Fx(r)=12+1π⋅arctan(r/2).
- In particular.
- Fx(r=+2)=12+1π⋅arctan(1)=12+1π⋅π4=0.75,
- Fx(r=−2)=12+1π⋅arctan(−1)=12−1π⋅π4=0.25.
- The probability we are looking for is given by the difference:
- Pr(|x|<2)=0.75−0.25=50%_.
(2) According to the result of the subtask (1) ⇒ Fx(r=4)=0.5+1/π=0.852.
- Thus, for the "complementary" probability: Pr(x>4)=0.148.
- For symmetry reasons, the probability we are looking for is twice as large:
- Pr(|x|>4)=29.6%_.
(3) All proposed solutions are true:
- For the variance of the Cauchy distribution holds namely:
- σ2x=12π∫+∞−∞x21+(x/2)2dx.
- 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 σx→∞.
- "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.