Help with computing density of one variable change.

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SUMMARY

The discussion focuses on computing the density function fy(y) for the variable Y, defined as Y = X^(1/3), where X has the density function fx(x) = 1/x^2 for x > 1. The correct density function is established as fy(y) = 3y^4 for y > 1 and 0 for y ≤ 1. However, a participant points out that this answer is incorrect due to the integral of fy(y) diverging, indicating that the total probability must equal 1. The discussion emphasizes the need to derive the cumulative distribution function F(y) to correctly relate Y to X.

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sneaky666
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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
 
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hi sneaky666! :smile:

(try using the X2 and X2 icons just above the Reply box :wink:)
sneaky666 said:
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. :wink:
 
If you want the distribution function begin looking at

<br /> F(y) = P(Y \le y)<br />

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

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