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eban science forum beginner
Joined: 09 Apr 2006
Posts: 3
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Posted: Sun Apr 09, 2006 8:49 pm Post subject:
Expectation/Integral problem
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Hi,
Almost to embarrassed to ask, but I cannot find the solution in my calculus (binmore) or statistics (Gourieroux) books.
I want to express the expectation as integrals with cumulative distribution functions in the integrands.:
/ inf
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EX = | x dF(x)
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/ -inf
This should be equal to:
/ 0 / inf
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EX = - | F(x) dx + | (1-F(x)) dx
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/ -inf / 0
This should be easily proved with integration by parts, is what my book says.
I can think of:
absolute integrability:
/ 0 / inf
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EX = - | xf(x)dx + | xf(x)dx
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/ -inf / 0
integration by parts:
/ 0
| 0 |
= x.F(x) | - | F(x)dx
| -Inf |
/ -inf
/ Inf
| Inf |
+ x.F(x) | - | F(x)dx
| 0 |
/ 0
As lim(inf) F(x)=1, the second line becomes
/ inf
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| (1-F(x)) dx
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/ 0
which is the second part of what i am supposed to find.
In the first part I am stuck with -lim(-inf)x.F(x) which is not necessarily zero, however.
What have i done wrong ???????
Thanks for tips
Etty |
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eban science forum beginner
Joined: 09 Apr 2006
Posts: 3
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Posted: Sun Apr 09, 2006 8:54 pm Post subject:
Re: Expectation/Integral problem
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In article <4439733c$0$11064$e4fe514c@news.xs4all.nl>,
etx@pubara.net (eban) writes:
Sorry, F is the cdf, f is the pdf, inf is infinite |
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A.G.McDowell science forum beginner
Joined: 17 Mar 2005
Posts: 4
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Posted: Sat Apr 15, 2006 6:04 am Post subject:
Re: Expectation/Integral problem
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In article <4439733c$0$11064$e4fe514c@news.xs4all.nl>, eban
<etx@pubara.net> writes
| Quote: | Hi,
Almost to embarrassed to ask, but I cannot find the solution in my calculus
(binmore) or statistics (Gourieroux) books.
I want to express the expectation as integrals with cumulative distribution
functions in the integrands.:
/ inf
|
EX = | x dF(x)
|
/ -inf
This should be equal to:
/ 0 / inf
| |
EX = - | F(x) dx + | (1-F(x)) dx
| |
/ -inf / 0
This should be easily proved with integration by parts, is what my book says.
Here goes: |
Integral(-inf, inf) xF' dx = Integral(-inf, 0) xF' dx +
Integral(0, inf) xF' dx
The first part is the easy one, by integration by parts
Integral(-inf, 0) xF' dx = (0)F(0) -(-inf)F(-inf) -
Integral(-inf, 0) F dx
and the first two terms drop off because 0F(0) = 0 and F(-inf) = 0
The second part is tricky because xF(x) goes to infinity as x goes to
infinity, so we rely on the fact that (1-F(x))' = -F' and therefore
Integral(0, inf) xF' dx = -Integral(0, inf) x(1-F(x)) dx which
integrates by parts to
-(-inf)(1-1) + 0(1-F(0)) + Integral(0, inf)(1-F(x)) dx
and the first two terms again drop off
(and I have blithely taken 0 * inf = 0, which I get away with if I know
the expectation actually exists, and don't in cases such as F(x) =
1/sqrt(-x) for all x < -1000)
--
A.G.McDowell |
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eban science forum beginner
Joined: 09 Apr 2006
Posts: 3
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Posted: Sat Apr 15, 2006 6:51 pm Post subject:
Re: Expectation/Integral problem
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Thanks,
You solve the 'tricky' part in a neater way than I had done.
My biggest problem was the inf*0 limit. I now see it is coming from the convergence of the integral.
It feels like the case of the limit of the terms of a converging series going to zero. I shall look for a comparable theorem for integrals.
Etty
In article <ZVeBhKAB0IQEFwor@mcdowella.demon.co.uk>,
"A.G.McDowell" <mcdowella@nospam.co.uk> writes:
| Quote: | In article <4439733c$0$11064$e4fe514c@news.xs4all.nl>, eban
etx@pubara.net> writes
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EX = - | F(x) dx + | (1-F(x)) dx
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/ -inf / 0
(and I have blithely taken 0 * inf = 0, which I get away with if I know
the expectation actually exists, and don't in cases such as F(x) =
1/sqrt(-x) for all x < -1000) |
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