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- ralfbuesser
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Weibull distribution conditional expectation
I look for the expected value of a Weibull distribution conditional on the duration t < x, where x is some censoring bound. Formally,[$]E(t|t<x) = \int_0^x t f(t) dt = \int_0^x \gamma \alpha t^\alpha exp(-\gamma t^\alpha) dt [$].I have tried with integration by parts but this does not seem to lead anywhere.Many thanks for your suggestions.Ralf
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- ralfbuesser
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Weibull distribution conditional expectation
Yep, was thinking of that, because the unconditional expectation can be expressed by a gamma-fct. Still, I can't see the way there and hoped to get some hint (I think normally you go via mgf but dont think that makes sense in the conditional case).ThanksRalf
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- Cuchulainn
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Weibull distribution conditional expectation
You can solve it numerically as an ODE, but you probably want a nice 'closed' solution?
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- ralfbuesser
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Weibull distribution conditional expectation
yes, if there is one ...
Weibull distribution conditional expectation
QuoteOriginally posted by: ralfbuesserYep, was thinking of that, because the unconditional expectation can be expressed by a gamma-fct. Still, I can't see the way there and hoped to get some hint (I think normally you go via mgf but dont think that makes sense in the conditional case).ThanksRalfHint: [$]t=\left(\frac{u}{\gamma}\right)^{1/\alpha}[$]
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- ralfbuesser
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Weibull distribution conditional expectation
[$](1/\gamma)^{1/\alpha}\delta(1+1/\alpha,\gamma x^\alpha)[$] it must be then, with [$]\delta[$] being the lower incomplete gamma fct. Thanks much for help!
Last edited by ralfbuesser on July 1st, 2014, 10:00 pm, edited 1 time in total.