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- SigmundFraud
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Expectation
Saw this, thought it's half fun:Suppose there is a PDF [$]f(X)[$] and a CDF [$]F(x) = Pr[\text{X}<x][$] then what is [$]\mathbb{E}[F(\text{X})][$]?
Expectation
EBal got the answer, here's the reason: X has the same distribution as F^{-1}(U), where U is uniformly distributed over [0,1], so E[F(X)] = E[F(F^{-1}(U))] = E = 1/2, since F is continuous by assumption.
Expectation
Alternatively, you can write down the expectation as the integral of the product of F(x) and f(x) and apply integration by parts.
Last edited by bearish on November 15th, 2013, 11:00 pm, edited 1 time in total.
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- Ultraviolet
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Expectation
Alternatively 
[$]dF(x)/dx = \rho(x)[$][$]E[F(X)] = \int_{-\infty}^{\infty} F(x)\rho(x) dx = \int_0^1 F(x) dF(x) = \int_0^1 t dt = 1/2[$]
[$]dF(x)/dx = \rho(x)[$][$]E[F(X)] = \int_{-\infty}^{\infty} F(x)\rho(x) dx = \int_0^1 F(x) dF(x) = \int_0^1 t dt = 1/2[$]
Last edited by Ultraviolet on November 15th, 2013, 11:00 pm, edited 1 time in total.
Expectation
Clearly, this cat can be skinned in all sorts of ways. kinnally's answer actually contains the intuition, but it is hard to improve on EBal's brevity.
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- Cuchulainn
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Expectation
QuoteOriginally posted by: bearish[...] it is hard to improve on EBal's brevity.You could try removing the question mark. 0.5
Last edited by Cuchulainn on November 16th, 2013, 11:00 pm, edited 1 time in total.
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- Traden4Alpha
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Expectation
QuoteOriginally posted by: CuchulainnQuoteOriginally posted by: bearish[...] it is hard to improve on EBal's brevity.You could try removing the question mark. 0.5And the zero..5
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- Cuchulainn
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Expectation
QuoteOriginally posted by: Traden4AlphaQuoteOriginally posted by: CuchulainnQuoteOriginally posted by: bearish[...] it is hard to improve on EBal's brevity.You could try removing the question mark. 0.5And the zero..5Yes, how sloppy.
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- FritzJacob
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Expectation
QuoteOriginally posted by: CuchulainnQuoteOriginally posted by: Traden4AlphaQuoteOriginally posted by: CuchulainnQuoteOriginally posted by: bearish[...] it is hard to improve on EBal's brevity.You could try removing the question mark. 0.5And the zero..5Yes, how sloppy.Yeah....he could have answered under 1 minute....took 32.
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- Cuchulainn
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Expectation
QuoteOriginally posted by: FritzJacobQuoteOriginally posted by: CuchulainnQuoteOriginally posted by: Traden4AlphaQuoteOriginally posted by: CuchulainnQuoteOriginally posted by: bearish[...] it is hard to improve on EBal's brevity.You could try removing the question mark. 0.5And the zero..5Yes, how sloppy.Yeah....he could have answered under 1 minute....took 32. Maybe he wanted to check the round-off errors.
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- Traden4Alpha
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Expectation
QuoteOriginally posted by: CuchulainnQuoteOriginally posted by: FritzJacobQuoteOriginally posted by: CuchulainnQuoteOriginally posted by: Traden4AlphaQuoteOriginally posted by: CuchulainnQuoteOriginally posted by: bearish[...] it is hard to improve on EBal's brevity.You could try removing the question mark. 0.5And the zero..5Yes, how sloppy.Yeah....he could have answered under 1 minute....took 32. Maybe he wanted to check the round-off errors.0.25^(1.0/2.0)
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- Cuchulainn
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Expectation
QuoteOriginally posted by: UltravioletAlternatively [$]dF(x)/dx = \rho(x)[$][$]E[F(X)] = \int_{-\infty}^{\infty} F(x)\rho(x) dx = \int_0^1 F(x) dF(x) = \int_0^1 t dt = 1/2[$]Is it possible to get the answer using geometry?
[$]dF(x)/dx = \rho(x)[$][$]E[F(X)] = \int_{-\infty}^{\infty} F(x)\rho(x) dx = \int_0^1 F(x) dF(x) = \int_0^1 t dt = 1/2[$]Is it possible to get the answer using geometry? -
- FritzJacob
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Expectation
QuoteOriginally posted by: CuchulainnQuoteOriginally posted by: UltravioletAlternatively [$]dF(x)/dx = \rho(x)[$][$]E[F(X)] = \int_{-\infty}^{\infty} F(x)\rho(x) dx = \int_0^1 F(x) dF(x) = \int_0^1 t dt = 1/2[$]Is it possible to get the answer using geometry?When F(x) is plotted against its own "probability density" pdf_F(x), we get a rectangle.E[F(x)] = half the area of this rectangle = 0.5*1*1= 0.5
[$]dF(x)/dx = \rho(x)[$][$]E[F(X)] = \int_{-\infty}^{\infty} F(x)\rho(x) dx = \int_0^1 F(x) dF(x) = \int_0^1 t dt = 1/2[$]Is it possible to get the answer using geometry?When F(x) is plotted against its own "probability density" pdf_F(x), we get a rectangle.E[F(x)] = half the area of this rectangle = 0.5*1*1= 0.5
Last edited by FritzJacob on November 17th, 2013, 11:00 pm, edited 1 time in total.
Expectation
here is a symmetry argument: simply consider two i.i.d. random variables labeled as X and x, what is the probability of X < x?note this indeed highlights what kinnally said above that the CDF needs to be continuous for the answer to be 1/2, i.e., the PDF has no point mass (Dirac delta) thus Pr(X=x) would be zero