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Probability density: change of variable

by longrob
Tags: density, probability, variable
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longrob
#1
Mar28-10, 03:43 AM
P: 50
If x is a random variable uniformly continuously distributed on [0.1], and y=x^3, then y has the density:

[tex]\frac{1}{3}y^{-2/3}[/tex]

on [0,1]

But, if x has the same distribution, but on [-0.5, 0.5], there seems to be a problem because we have [tex]y^{-2/3}[/tex] for negative values of y. This is overcome if we use the absolute value of y, so in this case we get:

[tex]\frac{1}{3}\mid y\mid{}^{-2/3}[/tex]

on [-1/8, 1/8]

Is this correct ? It seems to be, since integrating it yields 1, but how can I justify just replacing y with |y| ?
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