Exercise 3.12: Cauchy Distribution
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The probability density function 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 page Cauchy PDF .
Questions
Solution
(1) Comparing the given PDF with the general equation in the theory part, we see that the parameter λ=2 is.
- 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 of.
- Pr(|x|<2)=0.75−0.25=50%_.
(2) According to the result of the subtask (1) is 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 große x the integrand yields the constant value 4. Therefore the integral diverges.
- With σx→∞ 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.