[Statistics] Factorisation theorem proof

[Heidrun]

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:

Homework Equations

Could someone please explain me how to justify the first equality of that step?

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

 

[Stephen Tashi]

Could someone please explain me how to justify the first equality of that step?

The first equality is f_{T(X)}(t|\theta) = \sum_{\tilde{x}:T(\tilde{x}) = t} f_X(\tilde{x}|\theta). \ Is that what you're asking about ?

 

[Heidrun]

The first equality is f_{T(X)}(t|\theta) = \sum_{\tilde{x}:T(\tilde{x}) = t} f_X(\tilde{x}|\theta). \ Is that what you're asking about ?

Yes exactly. I don't see why we can write it.

 

[Stephen Tashi]

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.

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

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

For example, if event A can be partititoned into mutually exclusive events A1, A2 then p(A) = p(A1) + p(A2). If A is the event T(X) = 4 and this can be partitioned into the mutually exclusive events X = 2 and X = -2 then f_{T(X)} (4) = f_X(2) + f_X(-2).

 

[Heidrun]

Allright thanks a lot!