>   restart:
read `C:\\Tanis-Hogg\\Maple Examples\\stat.m`:

Elliot Tanis
April 19, 2006
Exercise 5.2-4
This program finds the p.d.f. of the mean of a random sample of size 2 from a Cauchy distributuion.

Let X, Y be a random sample of size 2 from a Cauchy distribution. The joint p.d.f. of X and Y is:

>   g := 1/Pi/(1 + x^2)*1/Pi/(1 + y^2);

g := 1/(Pi^2*(1+x^2)*(1+y^2))

Let W = (X + Y)/2. We first define the distribution function of W.

>   F := int(Int(g, y = 0 .. 2*w - x), x = -infinity .. infinity);

F := int(Int(1/(Pi^2*(1+x^2)*(1+y^2)),y = 0 .. 2*w-x),x = -infinity .. infinity)

The p.d.f. of W is the derivative of F with respect to w.

>   f := simplify(diff(F, w));

f := -I/Pi/(-1+w*I)/(-w+I)

>   f := factor(evalc(f));

f := 1/(Pi*(1+w^2))

>   CauchyPDF(x);

1/(Pi*(1+x^2))

>   

Note that the p.d.f. of W is the same as the p.d.f. of the Cauchy distribution.

If you would like to learn more about the Cauchy distribution using simulation, click the following.

Cauchy distribution simulations

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