# Math Help - help with a question about change of variable

1. ## help with a question about change of variable

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

2. Originally Posted by Sneaky
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
I don't understand where most of this work has come from. You have been given a detailed approach here: http://www.mathhelpforum.com/math-he...ge-159842.html

Follow it. Show all steps. Please say which step you cannot do. It sounds harsh but I do not see that you have learned anything from all my postings to your questions.

By the way, the given answer you have posted is wrong. It should be 3/y^4.

3. your equation from that other post gets y >= 0, but to get y> 1
isnt it just
1 - integral from -inf to 1 of f(x) dx
?

4. Originally Posted by Sneaky
your equation from that other post gets y >= 0, but to get y> 1
isnt it just
1 - integral from -inf to 1 of f(x) dx
?
No it is not.

The support of X is $x > 1$. Do you understand what a cdf is?

5. ok i redid it and i am getting

integral from 1 to y^3 of 1/x^2 dx
then take the derivative and i get
3/y^4

this is not the answer that was given but i don't know how to get to that answer.

6. Originally Posted by Sneaky
ok i redid it and i am getting

integral from 1 to y^3 of 1/x^2 dx
then take the derivative and i get
3/y^4

this is not the answer that was given but i don't know how to get to that answer.
I said in post #2 that the given answer was wrong and that 3/y^4 was the correct answer. Please try to read more carefully the replies you are given.