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CTK Exchange
august
guest
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Sep-10-08, 07:23 AM (EST) |
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"function centroid as mean"
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Prove that, assuming that p(x) is continuous and p(x)>=0 for all x,
if
* Integrate<p(x), {x,-Infinity,Infinity}> = 1
and
* mu = Integrate<x*p(x), {x,-Infinity,Infinity}>
then
* Integrate<p(x), {x,-Infinity,mu}> = 0.5 |
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august
guest
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Sep-10-08, 09:32 AM (EST) |
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4. "RE: function centroid as mean"
In response to message #1
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You're right: if p(x) = (e^(-x^2)+(1/4)e^(-2(x-0.5)^2)-(1/4)e^(-2(x+0.5)^2))/1.772453851, then p(x) is skewed to the right, and mu = 0.1767766953 Integrate<p(x) dx, {-Infinity,mu}> = 0.4832595709 != 0.5
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august
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Sep-10-08, 09:12 AM (EST) |
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2. "case: p(x) symmetric"
In response to message #0
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Here is a proof for the special case that p(x) is symmetric. If
p(x) is symmetric about the line x=x_0 then
p(x_0 - x) = p(x_0 + x).Integrate<x p(x), {x,-Infinity,Infinity}>
=Integrate<x p(x), {-Infinity,x_0}> + Integrate<x p(x), {x_0,Infinity}>
=Integrate<(x_0 + y) p(x_0 + y), {y,-Infinity,0}> + Integrate<(x_0 + y) p(x_0 + y), {y,0,Infinity}>
= -Integrate<(x_0 - y) p(x_0 + y), {y,Infinity,0}> + Integrate<(x_0 + y) p(x_0 + y), {y,0,Infinity}>
=Integrate<(x_0 - y) p(x_0 + y), {y,0,Infinity}> + Integrate<(x_0 + y) p(x_0 + y), {y,0,Infinity}>
=Integrate<2 x_0 p(x_0 + y), {y,0,Infinity}>
=2 x_0 Integrate<p(x_0 + y), {y,0,Infinity}>
= x_0 {Integrate<p(x_0 + y), {y,0,Infinity}> + Integrate<p(x_0 + y), {y,0,Infinity}>}
= x_0 {Integrate<p(x_0 - y), {y,-Infinity,0}> + Integrate<p(x_0 + y), {y,0,Infinity}>}
= x_0 {Integrate<p(x_0 + y), {y,-Infinity,0}> + Integrate<p(x_0 + y), {y,0,Infinity}>}
= x_0 Integrate<p(x_0 + y), {y,-Infinity,Infinity}>
= x_0
thus
* mu = x_0
and
2 Integrate<p(x_0 + y), {y,0,Infinity}> = Integrate<p(x_0 + y),{y,-Infinity,Infinity}> = 1
therefore
Integrate<p(mu+y), {y,-Infinity,0}> = 0.5, i.e. letting mu+y=x, then
* Integrate<p(x), {x,-Infinity,mu}> = 0.5 |
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