Let X 1 , X 2 Poisson random variables with mean μ. Is T = X 1 − X 2 a sufficient statistic?

Question

Let       X    1    ,      X    2   Poisson random variables with mean   μ. Is   T  =      X    1    −      X    2   a sufficient statistic?

mentest91k99 · Accepted Answer

There&#039;s no need to keep separate values       x    1   and       x    2  : If they differ, the probability is zero, which is obviously independent of   μ; so only the case       X    1    =      X    2    =  x is of interest. In that case the numerator is                    μ                  2          x                                      e                          −          2          μ                            x              !        2            and the denominator is      ∑          k      =      0        ∞                      μ                  2          k                                      e                          −          2          μ                            k              !        2              =            e              −      2      μ            I    0    (  2  μ  )    ,where       I    0   is a modified Bessel function of the first kind. Thus  P  (      X    1    =      X    2    =  x  ∣  T  =  0  )  =            μ              2        x                    x              !        2                    I        0            (      2      μ      )          ,which depends on   μ.

Let X_1,X_2 Poisson random variables with mean mu. Is T=X_1−X_2 a sufficient statistic?

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