Consider a random sample of size n n that follows a density probability function given by: f ( x , θ ) = 1 θ x − θ + 1 θ 1 ( 1 , + ∞ ) , θ &gt; 0where θ is unknown.

Question

Consider a random sample of size n  n that follows a density probability function given by:  f  (  x  ,  θ  )  =      1    θ        x          −                        θ          +          1                θ                        1              (      1      ,      +      ∞      )        ,      θ  &amp;gt;  0where   θ is unknown.

Averi Shaffer · Accepted Answer

Indeed you can use logarithms if you want to, and you don’t need to consider   log  ⁡  f instead of   f in order to do so. Since  f  (  x  ∣  θ  )  =      ∏          i      =      1        n    f  (      x    i    ,  θ  )  =      1          θ      n            ∏          i      =      1        n        x    i          −                        θ          +          1                θ              =      1          θ      n        exp  ⁡      (    −                  θ        +        1            θ              ∑              i        =        1            n        ln    ⁡          x      i        )      ,we have  f  (  x  ∣  θ  )  =  g  (  T  (  x  )  ∣  θ  )  h  (  x  )    ,with  T  (  x  )  =      ∑          i      =      1        n    ln  ⁡  x    .Thus   T  (  x  ) is a sufficient statistic.But you don’t have to use logarithms. You could also just write  f  (  x  ∣  θ  )  =      ∏          i      =      1        n    f  (      x    i    ,  θ  )  =      1          θ      n            ∏          i      =      1        n        x    i          −                        θ          +          1                θ              =      1          θ      n                  (              ∏                  i          =          1                n                    x        i            )              −                        θ          +          1                θ                ,which factorizes with   S  (  x  )  =      ∏          i      =      1        n        x    i  . That   S  (  x  ) and   T  (  x  ) are likewise sufficient statistics is not surprising, since one is an injective function of the other  S  (  x  )  =  exp  ⁡  (  T  (  x  )  ) and   T  (  x  )  =  log  ⁡  (  S  (  x  )  ), and an injective function of a sufficient statistic is again a sufficient statistic.

Consider a random sample of size n that follows a density probability function given by: f(x,theta)=1/theta x^(−theta+1/theta 1_(1,+oo),theta>0 where theta is unknown.

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