# probability distributions

• Dec 6th 2009, 10:37 PM
CarmineCortez
probability distributions
Suppose that X ~ U(-2, 2). What is the DENSITY of Y = X^4? Note that if
Z ~ U(a, b) then it has density f(z) = 1/(b - a) and cdf F(z) = (z - a)/(b - a) for a < z < b.

So I used change of variables ie.
f(y^(1/4)) = 1/4*dx/dy = (1/4)*(1/4) * y^(-3/4)

and this is wrong. any ideas what is the mistake?
• Dec 7th 2009, 12:00 AM
matheagle
Note that this is a two to one mapping, Y is defined on (0,16).
I'll derive it, but it's just like the squaring of a st normal to get a chi-square.

$F_Y(y)=P(Y\le y)=P(X^4\le y)$

here's the algebra involving the 2-1 mapping

$=P(-y^{1/4}\le X\le y^{1/4} )= F_X(y^{1/4})- F_X(-y^{1/4})$

If you differentiate, which is not necessary, you get the sum of the two solutions TIMES the jacobian.

$f_Y(y)=\Biggl(f_X(y^{1/4})+f_X(-y^{1/4})\Biggr){1\over 4y^{3/4}}$

AND when 0<y<16 we get................

$f_Y(y)=\Biggl({1\over 4}+{1\over 4}\Biggr){1\over 4y^{3/4}}={1\over 8}y^{-3/4}$

If y is outside that region you get zero.