- #1
Coolster7
- 14
- 0
I'm having trouble with applying this theorem to likelihood functions in order to obtain a sufficiency statistic for the relevant variables.
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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)
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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?
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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?
_________________________________________________________________________________________
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?