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Chapter 5: Q.5.16 (page 218)

A standard Cauchy random variable has density function

f(x)=1π1+x2−∞<x<∞

Show that if X is a standard Cauchy random variable, then 1/X is also a standard Cauchy random variable.

Short Answer

Expert verified

−∫−∞∞ 1π1+y2dy is also the probability density function of a standard Cauchy's distribution. So, 1xis also standard Cauchy's distribution.

Step by step solution

01

Given information

A standard Cauchy random variable has density function

f(x)=1π1+x2−∞<x<∞

02

Solution

The probability density function of Cauchy's distribution will be,

f(x)=1π1+x2,−∞<x<∞

Now let's calculate,

y=1x⇒x=1y

We need to Differentiate on both sides

dx=−1y2dy

The probability density function of x is,

f(x)=∫−∞∞ 1π1+x2dx

f(x)=∫−∞∞ 1π1+x2dx

=∫−∞∞ 1π1+1y2−dyy2

=∫−∞∞ 1π1+1y2−dyy2

=∫−∞∞ y2π1+y2−dyy2

=−∫−∞∞ 1π1+y2dy

03

Final answer

−∫−∞∞ 1π1+y2dyis also the probability density function of a standard Cauchy's distribution. So,1xis also standard Cauchy's distribution.

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