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[Statistics] Factorisation theorem proof

  • Thread starter Heidrun
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  • #1
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Hello. I have a question about a step in the factorization theorem demonstration.

1. Homework Statement

Here is the theorem (begins end of page 1), it is not my course but I have almost the same demonstration : http://math.arizona.edu/~jwatkins/sufficiency.pdf
Screenshot of it:
591799factorization.png


Homework Equations


Could someone please explain me how to justify the first equality of that step?
891254factorization2.png


The Attempt at a Solution


I think a possible justification is because the sample is a sufficient statistic but it feels like it's not enough/not the right justification
 

Answers and Replies

  • #2
Stephen Tashi
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Could someone please explain me how to justify the first equality of that step?
891254factorization2.png
The first equality is [itex] f_{T(X)}(t|\theta) = \sum_{\tilde{x}:T(\tilde{x}) = t} f_X(\tilde{x}|\theta) [/itex]. [itex]\ [/itex] Is that what you're asking about ?
 
  • #3
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The first equality is [itex] f_{T(X)}(t|\theta) = \sum_{\tilde{x}:T(\tilde{x}) = t} f_X(\tilde{x}|\theta) [/itex]. [itex]\ [/itex] Is that what you're asking about ?
Yes exactly. I don't see why we can write it.
 
  • #4
Stephen Tashi
Science Advisor
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That equality is due to the definitions associated with the notation being used plus the fact that the probability of an event can be expressed as the sum of the probabilities of events that partition it into mutually exclusive sets.

[itex] f_{T(X)}(t|\theta) [/itex] denotes the probability density of the discrete random variable [itex] T(X) [/itex] on the probability space "[itex] X [/itex] given [itex] \theta [/itex]".

The event [itex] T(X) = t [/itex] in that probability space is exactly the event that [itex] X [/itex] takes on some value that makes [itex] T(X) = t. [/itex] The notation "[itex]\tilde{x}: T(\tilde{x}) = t [/itex] denotes that event expressed in terms of a variable [itex] \tilde{x} [/itex]. The notation [itex] f_X(\tilde{X}|\theta) [/itex] says we are assigning probability to that event using the probability density function defined on the probability space of "[itex] X [/itex] given [itex] \theta [/itex]".

For example, if event [itex] A [/itex] can be partititoned into mutually exclusive events [itex] A1, A2[/itex] then [itex] p(A) = p(A1) + p(A2) [/itex]. If [itex] A [/itex] is the event [itex] T(X) = 4 [/itex] and this can be partitioned into the mutually exclusive events [itex] X = 2 [/itex] and [itex] X = -2 [/itex] then [itex] f_{T(X)} (4) = f_X(2) + f_X(-2) [/itex].
 
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  • #5
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Allright thanks a lot!
 

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