Let X 1 , … , X n be a random sample with the pdf f X i ( x | θ ) = { e i θ − x , x ≥ i θ 0 , x &lt; i θ Show that T = min i X i i is a sufficient statistic for θ.

Question

Let       X    1    ,  …  ,      X    n   be a random sample with the pdf      f                  X        i              (  x      |    θ  )  =      {                                        e                          i              θ              −              x                                ,                                  x          ≥          i          θ                                      0          ,                                  x          &amp;lt;          i          θ                        Show that   T  =      min    i              X      i        i   is a sufficient statistic for   θ.

ustalovatfog · Accepted Answer

Write                    f        (                  X          1                ,        …        ,                  X          n                                   |                         θ        )                            =                  ∏                      i            =            1                    n                          e                      i            θ            −                          X              i                                                            1                                              X              i                        ≥            i            θ                                                            =        exp        ⁡                  {          θ                                    n              (              n              +              1              )                        2                    −                      ∑                          i              =              1                        n                                X            i                    }                                      1                                              min              i                                                      X                i                            i                        ≥            θ                                                            =        exp        ⁡                  (          −                      ∑                          i              =              1                        n                                X            i                    )                          [          exp          ⁡                      {                                          θ                n                (                n                +                1                )                            2                        }                    ⋅                                    1                                                      min                i                                                              X                  i                                i                            ≥              θ                                ]                .            So, the first exponent doesn&#039;t depend on   θ, while the second term depends on   θ and the statistics used in second exponent is exactly       min    i              X      i        i  . Hence,   T  (  X  )  =      min    i              X      i        i   is indeed a sufficient statistics.

Let X_1,…,X_n be a random sample with the pdf fXi(x|theta)={e^(i theta−x),x>=i theta, 0, x<i theta Show that T=min_i X_i/i is a sufficient statistic for theta.

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