Gujinder Singh wrote: > 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) >
Your mistake is in the above. You have not noticed or accounted for the fact that the tranformation Y=X^2 is many to one (two to one). > Now letīs 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/ . =================================================================
