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Thread: Distribution function of an exponential random variable

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    Distribution function of an exponential random variable




    Let Y ⇠ Exp(2) be the size of an insurance loss. If the insurance company pays min(Y, 4), what is the variance of that payment?

    I know that lambda =2

    If I put that in the pdf I get: 2*e^(-2x). I don't understand how to use the information given by min(Y, 4)
    Last edited by endlessend25; 11-14-2016 at 01:51 PM.

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    Re: Distribution function of an exponential random variable

    Instead of integrating your pdf from 0 to infinity, split into two integrals (0 to 4, and 4 to infinity). You figure out what to put for your integrand.
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    Re: Distribution function of an exponential random variable

    If i get you:
    \int_{0}^{4}2e^{-2x}dx + \int_{4}^{\propto } 4 dx
    Last edited by endlessend25; 11-14-2016 at 02:01 PM.

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    Re: Distribution function of an exponential random variable


    OK clearly your 2nd integral is divergent, because you forget the cdf for x>4.

    I would use Var(X) = E[X^2] - E^2[X], but that's because I forgot my cool equations for insurance payments.
    All things are known because we want to believe in them.

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