# Factorisation Theorem for Sufficiency

1. Feb 21, 2014

### Coolster7

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.

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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)$\sum$(xi) - n

you set h(x) =1 here and g(x) = (t,θ) = θn(1-θ)t - n where t = $\sum$(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.

2. Feb 21, 2014

### h6ss

The idea is to have h(x) not depending on the parameter. Whether it's a constant or not.

3. Feb 21, 2014

### Coolster7

Ah I see, thanks for this I understand now.. it's simple really.

4. Feb 21, 2014

### Coolster7

Actually just one more question. What about n? Would n need to be sufficient for the parameter or is it treated as constant/number?

5. Feb 24, 2014

Anyone?