
#1
Mar2810, 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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