Re: Distribution of x^3?
- From: "Henri Courant Junior" <ProfoundlyGifted@xxxxxxxxx>
- Date: 14 Aug 2006 23:24:16 -0700
Ray Koopman wrote:
Because the PDF is defined as the derivative of the CDF,
and the CDF of a uniform variable is easy to work with.
X ~ Uniform[0,1].
CDF: F_X[x] = P[X <= x]
= x, 0 <= x <= 1.
Y = X^3.
CDF: F_Y[y] = P[Y <= y]
= P[X^3 <= y]
= P[X <= y^(1/3)]
= y^(1/3), 0 <= y <= 1.
PDF: f_Y[y] = dF_Y[y]/dy
= (1/3) y^(-2/3).
Wow, it was really simple. I had already read the articles you
suggested, but I was doing all sorts of crazy calculations :-(. Many
thanks, Mr. Koopman!!
.
- References:
- Distribution of x^3?
- From: ProfoundlyGifted
- Re: Distribution of x^3?
- From: Lou Thraki
- Re: Distribution of x^3?
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- Re: Distribution of x^3?
- From: Ray Koopman
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