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Math Help - exponential family help

  1. #1
    Senior Member Danneedshelp's Avatar
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    exponential family help

    Q: Let Y_{1},...,Y_{n} denote a random sample from the probability density function
    f(y|\theta)=\theta\\y^{\theta\\-1} for 0<y<1, \theta>0 and 0 otherwise.
    a) Show that this density function is in the (one-parameter) exponential family and that \sum_{i=1}^{n}-ln(Y_{i}) is sufficient for \theta (see previous exercise).
    b) If W_{i}=-ln(Y_{i}), show that W_{i} has an exponential distribution with mean \frac{1}{\theta}.

    Previews exercise mentioned in (a):

    1) Suppose that Y_{1},...,Y_{n} is a random sample from a probability density in the (one-parameter) exponential family so that

    f(y|\theta)=a(\theta)b(y)e^{-[c(\theta)d(y)]}I_{[a,b]}(y), where I_{[a,b]}(y) is the indicator function and a and b do not depend on \theta.

    I think I am supposed to define each function in the density directly above for the orginal question; however, I am having a hard time doing so. Some direction would be great. Thanks.
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  2. #2
    MHF Contributor harish21's Avatar
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    Quote Originally Posted by Danneedshelp View Post
    Q: Let Y_{1},...,Y_{n} denote a random sample from the probability density function
    f(y|\theta)=\theta\\y^{\theta\\-1} for 0<y<1, \theta>0 and 0 otherwise.
    a) Show that this density function is in the (one-parameter) exponential family and that \sum_{i=1}^{n}-ln(Y_{i}) is sufficient for \theta (see previous exercise).
    b) If W_{i}=-ln(Y_{i}), show that W_{i} has an exponential distribution with mean \frac{1}{\theta}.

    Previews exercise mentioned in (a):

    1) Suppose that Y_{1},...,Y_{n} is a random sample from a probability density in the (one-parameter) exponential family so that

    f(y|\theta)=a(\theta)b(y)e^{-[c(\theta)d(y)]}I_{[a,b]}(y), where I_{[a,b]}(y) is the indicator function and a and b do not depend on \theta.

    I think I am supposed to define each function in the density directly above for the orginal question; however, I am having a hard time doing so. Some direction would be great. Thanks.
    You can say that the distribution f(y,\theta) belongs to a one parameter exponential family if you can express its pdf in this form:

    f(y,\theta) = a(\theta) \times g(y) \times exp(b(\theta) \times R(y))

    so your pdf is :
     \theta\\y^{\theta\\-1}

    = \theta \frac{y^{\theta}}{y}

    = \theta \times \frac{1}{y} \times {y^{\theta}}

    = \theta \times \frac{1}{y} \times exp(\theta \times log(y))

    compare the above system with f(y,\theta) = a(\theta) \times g(y) \times exp(b(\theta) \times R(y)), you can say that:

     a(\theta) = \theta

     g(y) = \frac{1}{y}

     b(\theta) = \theta

    and  R(y) = log(y)



    and

    \sum_{i=1}^{n} R(y) is a complete sufficient statistic for  \theta
    Last edited by harish21; March 24th 2010 at 12:49 PM.
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