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Factorisation Theorem for Sufficiency

  1. Feb 21, 2014 #1
    I'm having trouble with applying this theorem to likelihood functions in order to obtain a sufficiency statistic for the relevant variables.


    The factorisation theorem being:

    Under certain regularity conditions;

    T(X) is sufficient for θ ⇔ f(x|θ) = h(x)g(t(x),θ)

    for some functions h and g.


    The main problem I'm having is when to allow h(x) = 1.

    For example in the exponential distribution you get a likelihood function: f(x|θ) = θn(1-θ)[itex]\sum[/itex](xi) - n

    you set h(x) =1 here and g(x) = (t,θ) = θn(1-θ)t - n where t = [itex]\sum[/itex](xi)


    However, in the Poisson distribution you get a likelihood function: 1/(x1!x2!) x e-2θθx1+x2.

    here you set h(x) = 1/(x1!x2!) and not h(x) = 1.

    Is this because h(x) has to be a constant or involving just x?


    So say for example you had a likelihood function:

    f(x|θ) = xnσ-2ne-0.5nx2σ-2

    using the Factorisation method would you let h(x) = 1 with g(x, σ) = f(x|θ)

    and say x is sufficient for σ

    OR would you let h(x) = xn and g(x, σ) = σ-2ne-0.5nx2σ-2

    and say x is sufficient for σ

    Note: Obviously there is the same outcome for the sufficiency statistic, but in a different problem this may not be the case.

    Can anyone help me please?
  2. jcsd
  3. Feb 21, 2014 #2
    The idea is to have h(x) not depending on the parameter. Whether it's a constant or not.
  4. Feb 21, 2014 #3
    Ah I see, thanks for this I understand now.. it's simple really.
  5. Feb 21, 2014 #4
    Actually just one more question. What about n? Would n need to be sufficient for the parameter or is it treated as constant/number?
  6. Feb 24, 2014 #5

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