Let X &#x223C; B e r ( n 1 ; p ) , Y &#x223C; B e r ( n 2 ; p 2 ) , where X and Y are independent. Find a minimal sufficient statistic T and, using a nontrivial function, show that it is not complete.

Question

Let     X    &amp;#x223C;    B    e    r    (          n      1        ;    p    )    ,    Y    &amp;#x223C;    B    e    r    (          n      2        ;          p      2        )  , where     X   and     Y   are independent. Find a minimal sufficient statistic     T   and, using a nontrivial function, show that it is not complete.

Ostrakodec3 · Accepted Answer

First write the likelihood function:  f  (  x  ,  y      |    p  )  =  [      p    x    (  1  −  p      )          1      −      x        ]  [      p          2      y        (  1  −      p    2        )          1      −      y        ]From this use the Lehmann-Scheffe Theorem. Need to find T such that  T  (      x    1    ,      y    1    )  =  T  (      x    2    ,      y    2    )  ⇔            f      (              x        1            ,              y        1                    |            p      )              f      (              x        2            ,              y        2                    |            p      )        =  c  (      x    1    ,      y    1    ,      x    2    ,      y    2    )To this end, let&#039;s compute the likelihood ratio:                                          f            (                          x              1                        ,                          y              1                                      |                        p            )                                f            (                          x              2                        ,                          y              2                                      |                        p            )                                              =                                            p                                                x                  1                                                      (            1            −            p                          )                              1                −                                  x                  1                                                                    p                              2                                  y                  1                                                      (            1            −                          p              2                                      )                              1                −                                  y                  1                                                                                        p                                                x                  2                                                      (            1            −            p                          )                              1                −                                  x                  2                                                                    p                              2                                  y                  2                                                      (            1            −                          p              2                                      )                              1                −                                  y                  2                                                                                                      =                  p                      (                          x              1                        −                          x              2                        )                          (        1        −        p                  )                                    x              2                        −                          x              1                                                p                      2            (                          y              1                        −                          y              2                        )                          (        1        −                  p          2                          )                                    y              2                        −                          y              1                                          It can be seen then that the ratio does not depend on   p if and only if       x    1    =      x    2    ,      y    1    =      y    2  ,      x    1    =      x    2    ,      y    1    =      y    2  . By the Lehmann-Scheffe, this implies that   T  (  X  ,  Y  )  =  (  X  ,  Y  ) is the minimal sufficient statistic.

Let X ~ Ber(n1; p), Y ~ Ber(n2; p^2), where X and Y are independent. Find a minimal sufficient statistic T and, using a nontrivial function, show that it is not complete.

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