Help with computing density of one variable change.

[sneaky666]

Let X have density function fx(x)=1/x^2 for x>1, otherwise fx(x)=0.
Let Y=X^(1/3). Compute the density function fy(y) for Y.

my attempt
------------
P(X<=1)=0
P(Y^3<=1)=P(Y<=1)=0

P(1<x<inf)=P(1<Y^3<inf)
=1-P(Y<1)

i am stuck here.


The answer to this is
fy(y) = 3y^4 for y>1 and 0 for y<= 1

 

[tiny-tim]

hi sneaky666!

(try using the X² and X₂ icons just above the Reply box )

Let X have density function fx(x)=1/x^2 for x>1, otherwise fx(x)=0.
Let Y=X^(1/3). Compute the density function fy(y) for Y.
…
The answer to this is
fy(y) = 3y^4 for y>1 and 0 for y<= 1

No, that answer is wrong: ∫₁^∞ 3y⁴ dy = ∞, and it should be 1.

 

[statdad]

If you want the distribution function begin looking at

$$F(y) = P(Y \le y)$$

and use the definition of $$Y$$ to get this into a form that involves the distribution of $$X$$.