I need to show that if the stochastic variable X is Cauchy(0,1) then X^2 is Fisher(1,1). Now Let Y=X^2
By the transformation theorem we have the density function f_Y(y) = f_X(sqrt[y]) * 1/(2*sqrt[y]) = (1/Pi) * 1/(2*sqrt[y](1+y)) = (a) Now let�s see what the density function for Y is if we let it be Fisher(1,1) f_Y(y)=1/(Gamma[1/2])^2 * 1/(sqrt[y](1+y)) = (1/Pi)* 1/(sqrt[y](1+y)) = (b) Clearly (a) != (b). Did i make a mistake somewhere or could it be that X^2 is not Fisher(1,1) distributed ? . . ================================================================= Instructions for joining and leaving this list, remarks about the problem of INAPPROPRIATE MESSAGES, and archives are available at: . http://jse.stat.ncsu.edu/ . =================================================================
