# [Statistics] Factorisation theorem proof

1. Jul 29, 2016

### Heidrun

Hello. I have a question about a step in the factorization theorem demonstration.

1. The problem statement, all variables and given/known data

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:

2. Relevant equations
Could someone please explain me how to justify the first equality of that step?

3. 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

2. Jul 29, 2016

### Stephen Tashi

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 ?

3. Jul 30, 2016

### Heidrun

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

4. Jul 30, 2016

### 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 $f_{T(X)} (4) = f_X(2) + f_X(-2)$.

5. Jul 30, 2016

### Heidrun

Allright thanks a lot!