Joint distribution simplification in minimal sufficient statistics proof

Question

My notes introduce the concept of minimal sufficient statistics as follows:

Definition

A sufficient statistic $T(\mathbf{Y})$ is called a minimal sufficient statistic if it is a function of any other sufficient statistic.

Remark

Except for several very special examples, a minimal sufficient statistic always exists.

Assume the existence of a minimal sufficient statistic and consider partitioning the sample space $\Omega$, where $\mathbf{y}_1, \mathbf{y}_2 \in \Omega$ are assigned to the same equivalence class iff the likelihood ratio $L(\theta; \mathbf{y})/L(\theta,\mathbf{y})$ does not depend on $\theta$.

Define a statistic $T(\mathbf{Y})$ in such a way that $T(\mathbf{y}_1) = T(\mathbf{y_2})$ if $\mathbf{y}_1$ and $\mathbf{y}_2$ belong to the same equivalence class and $T(\mathbf{y}_1) \not= T(\mathbf{y_2})$ otherwise.

Therorem 2

The statistic $T(\mathbf{Y})$ defined above is the minimal sufficient statistic for $\theta$.

Proof of theorem 2 for the discrete case

First we show that $T(\mathbf{Y})$ is sufficient.

$$\begin{align} P_\theta (\mathbf{Y} = \mathbf{y} \vert T(\mathbf{Y}) = t) &= \dfrac{P(\mathbf{Y} = \mathbf{y}, T(\mathbf{Y}) = t)}{P(T(\mathbf{Y}) = t)} \\ &= \dfrac{P(\mathbf{Y} = \mathbf{y})}{\sum_{\mathbf{y}_i : T(\mathbf{y}_i) = t} P(\mathbf{Y} = \mathbf{y}_i)} \\ &= \dfrac{L(\theta; \mathbf{y})}{\sum_{\mathbf{y}_i : T(\mathbf{y}_i) = t} L(\theta; \mathbf{y})} \\ &= \dfrac{1}{\sum_{\mathbf{y}_i : T(\mathbf{y}_i) = t} \dfrac{L(\theta; \mathbf{y}_i)}{L(\theta; \mathbf{y})}} \end{align}$$

Since $T(\mathbf{y}) = T(\mathbf{y}_i) = t$, all $\mathbf{y}_i$ and $\mathbf{y}$ belong to the same equivalence class induced by $T(\mathbf{y})$ and, therefore, the likelihood ratios $L(\theta; \mathbf{y}_i)/L(\theta; \mathbf{y})$ do not depend on $\theta$.

What is the reasoning that leads to the changes in the numerator and denominator of $\dfrac{P(\mathbf{Y} = \mathbf{y}, T(\mathbf{Y}) = t)}{P(T(\mathbf{Y}) = t)} = \dfrac{P(\mathbf{Y} = \mathbf{y})}{\sum_{\mathbf{y}_i : T(\mathbf{y}_i) = t} P(\mathbf{Y} = \mathbf{y}_i)}$? In the numerator, the joint PMF becomes a univariate PMF, and in the denominator, the PMF is changed from $P(T(\mathbf{Y}) = t)$ to $P(\mathbf{Y} = \mathbf{y}_i)$ and a sum is inserted.

Answer 1

There is an assumption that $t=T(y)$, otherwise the equation is not true.

To see that the assumption is needed, suppose $t \ne T(y)$, then $P_\theta(Y=y,T(Y)=t)=0$ but $P(Y=y)$ can be positive.

$$P(Y=y, T(Y)=t)=P(Y=y)$$ since we already know that $T(y)=t$.

Also $$P(T(Y)=t)=\sum_{y_i:T(y_i)=t}P(Y=y_i)$$

since \begin{align}\{\omega: T(Y(\omega))=t\}&=\bigsqcup_{y_i:T(y_i)=t}\{\omega: Y(\omega)=y_i , T(Y(\omega))=t\} \\ &=\bigsqcup_{y_i:T(y_i)=t}\{\omega: Y(\omega)=y_i \}\end{align} the right hand side forms a partition for the set on the left and we used the law of total probability.