# Probability density: change of variable

1. Mar 28, 2010

### longrob

If x is a random variable uniformly continuously distributed on [0.1], and y=x^3, then y has the density:

$$\frac{1}{3}y^{-2/3}$$

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 $$y^{-2/3}$$ for negative values of y. This is overcome if we use the absolute value of y, so in this case we get:

$$\frac{1}{3}\mid y\mid{}^{-2/3}$$

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| ?

Last edited: Mar 28, 2010