- #1
longrob
- 52
- 0
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| ?
[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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