Showing the sample mean is a sufficient statistics from an exponential distribution





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Suppose that the lifelengths ( in thousands of hours) of light bulbs are distributed Exponential($theta$), where $theta>0$ is unknown. If we observe $overline x = 5.2$ for a sample of $20$ light bulbs, record a representative likelihood function. Why is it that we only need to observe the sample average to obtain a representative likelihood?




The likelihood is pretty straightforward to find:
$$L(theta mid x_1,ldots,x_{20})=prod_{i=1}^{20}theta e^{-theta overline x} = theta^{20}e^{-20 theta overline x}$$



I am having an issue showing this is a sufficient statistic however, as I cannot seem to factor it in the form of $h(overline x)g_theta (T(overline x))$






likelihood sufficient-statistics factorisation-theorem







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    $begingroup$



    Suppose that the lifelengths ( in thousands of hours) of light bulbs are distributed Exponential($theta$), where $theta>0$ is unknown. If we observe $overline x = 5.2$ for a sample of $20$ light bulbs, record a representative likelihood function. Why is it that we only need to observe the sample average to obtain a representative likelihood?




    The likelihood is pretty straightforward to find:
    $$L(theta mid x_1,ldots,x_{20})=prod_{i=1}^{20}theta e^{-theta overline x} = theta^{20}e^{-20 theta overline x}$$



    I am having an issue showing this is a sufficient statistic however, as I cannot seem to factor it in the form of $h(overline x)g_theta (T(overline x))$






    likelihood sufficient-statistics factorisation-theorem







    share|cite|improve this question









    New contributor




    hkj447 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
    Check out our Code of Conduct.







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      1












      1








      1





      $begingroup$



      Suppose that the lifelengths ( in thousands of hours) of light bulbs are distributed Exponential($theta$), where $theta>0$ is unknown. If we observe $overline x = 5.2$ for a sample of $20$ light bulbs, record a representative likelihood function. Why is it that we only need to observe the sample average to obtain a representative likelihood?




      The likelihood is pretty straightforward to find:
      $$L(theta mid x_1,ldots,x_{20})=prod_{i=1}^{20}theta e^{-theta overline x} = theta^{20}e^{-20 theta overline x}$$



      I am having an issue showing this is a sufficient statistic however, as I cannot seem to factor it in the form of $h(overline x)g_theta (T(overline x))$






      likelihood sufficient-statistics factorisation-theorem







      share|cite|improve this question









      New contributor




      hkj447 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
      Check out our Code of Conduct.







      $endgroup$





      Suppose that the lifelengths ( in thousands of hours) of light bulbs are distributed Exponential($theta$), where $theta>0$ is unknown. If we observe $overline x = 5.2$ for a sample of $20$ light bulbs, record a representative likelihood function. Why is it that we only need to observe the sample average to obtain a representative likelihood?




      The likelihood is pretty straightforward to find:
      $$L(theta mid x_1,ldots,x_{20})=prod_{i=1}^{20}theta e^{-theta overline x} = theta^{20}e^{-20 theta overline x}$$



      I am having an issue showing this is a sufficient statistic however, as I cannot seem to factor it in the form of $h(overline x)g_theta (T(overline x))$






      likelihood sufficient-statistics factorisation-theorem




      likelihood sufficient-statistics factorisation-theorem






      share|cite|improve this question









      New contributor




      hkj447 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
      Check out our Code of Conduct.











      share|cite|improve this question









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      edited 4 hours ago









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          $begingroup$

          You've already done what you think you haven't done.



          You should write it as $h(x_1,ldots,x_{20}) g_theta( (x_1 + cdots + x_{20})/20),$ or in other words as $h(x_1,ldots,x_{20}) g_theta(T(x_1,ldots, x_{20})).$



          What's going on is camouflaged by the fact that $h(x_1,ldots,x_{20}) = 1$ regardless of the value of $(x_1,ldots,x_{20}).$



          That often happens. (One situation where it does not happen is with an i.i.d. sample from the family of Poisson distributions.)






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            $begingroup$

            You've already done what you think you haven't done.



            You should write it as $h(x_1,ldots,x_{20}) g_theta( (x_1 + cdots + x_{20})/20),$ or in other words as $h(x_1,ldots,x_{20}) g_theta(T(x_1,ldots, x_{20})).$



            What's going on is camouflaged by the fact that $h(x_1,ldots,x_{20}) = 1$ regardless of the value of $(x_1,ldots,x_{20}).$



            That often happens. (One situation where it does not happen is with an i.i.d. sample from the family of Poisson distributions.)






            share|cite|improve this answer









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              2












              $begingroup$

              You've already done what you think you haven't done.



              You should write it as $h(x_1,ldots,x_{20}) g_theta( (x_1 + cdots + x_{20})/20),$ or in other words as $h(x_1,ldots,x_{20}) g_theta(T(x_1,ldots, x_{20})).$



              What's going on is camouflaged by the fact that $h(x_1,ldots,x_{20}) = 1$ regardless of the value of $(x_1,ldots,x_{20}).$



              That often happens. (One situation where it does not happen is with an i.i.d. sample from the family of Poisson distributions.)






              share|cite|improve this answer









              $endgroup$
















                2












                2








                2





                $begingroup$

                You've already done what you think you haven't done.



                You should write it as $h(x_1,ldots,x_{20}) g_theta( (x_1 + cdots + x_{20})/20),$ or in other words as $h(x_1,ldots,x_{20}) g_theta(T(x_1,ldots, x_{20})).$



                What's going on is camouflaged by the fact that $h(x_1,ldots,x_{20}) = 1$ regardless of the value of $(x_1,ldots,x_{20}).$



                That often happens. (One situation where it does not happen is with an i.i.d. sample from the family of Poisson distributions.)






                share|cite|improve this answer









                $endgroup$



                You've already done what you think you haven't done.



                You should write it as $h(x_1,ldots,x_{20}) g_theta( (x_1 + cdots + x_{20})/20),$ or in other words as $h(x_1,ldots,x_{20}) g_theta(T(x_1,ldots, x_{20})).$



                What's going on is camouflaged by the fact that $h(x_1,ldots,x_{20}) = 1$ regardless of the value of $(x_1,ldots,x_{20}).$



                That often happens. (One situation where it does not happen is with an i.i.d. sample from the family of Poisson distributions.)







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered 4 hours ago









                Michael HardyMichael Hardy

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