Back to the coin flipping example. For the three assumptions, we denote respectively $$\begin{array}{rl}& X_{++}\sim \mathrm{Binom}(n,\theta ),\\ & X_{i+}\sim \mathrm{Binom}(n_i,{\theta }_i),1\le i\le 48,\\ & X_{i,j}\sim \mathrm{Bernoulli}({\theta }_{i,t}),1\le i\le 48,1\le j\le n_i.\end{array}$$
Check for ${\displaystyle T(X)=X_{++}}$, we have $$p_{\theta }(x)=\prod _{i=1}^{48}\prod _{j=1}^{n_i}{\theta }^{x_i}(1-\theta {)}^{n_i-x_i}={\theta }^{X_{++}}(1-\theta {)}^{n-X_{++}},$$ so $$\begin{array}{rl}& {\mathbb{P}}_{\theta }(X=x|X_{++}=t)=\frac{{\mathbb{P}}_{\theta }(X=x,X_{++}=t)}{{\mathbb{P}}_{\theta }(X_{++}=t)}\\ =& \frac{\mathbf{1}\{X_{++}=t\}{\theta }^t(1-\theta {)}^{n-t}}{(\genfrac{}{}{0ex}{}{n}{t}){\theta }^t(1-\theta {)}^{n-t}}={(\genfrac{}{}{0ex}{}{n}{t})}^{-1}\mathbf{1}\{X_{++}=t\}.\end{array}$$
This expression is irrelevant to ${\displaystyle \theta }$, so ${\displaystyle T(X)=X_{++}}$ is sufficient for model 2.

The proof can also be seen in here. For this note, we anly check the discrete case. WLOG assume ${\displaystyle \mu }$ is counting measure.

• Assume factorization ${\displaystyle p_{\theta }(x)=g_{\theta }(T(x))h(x)}$ exists. Then we have $$\begin{array}{rl}{\mathbb{P}}_{\theta }(X=x|T(X)=t)& =\frac{{\mathbb{P}}_{\theta }(X=x,T(X)=t)}{{\mathbb{P}}_{\theta }(T(X)=t)}\\ & =\frac{g_{\theta }(t)h(x)\mathbf{1}\{T(x)=t\}}{g_{\theta }(t)\sum _{z:T(z)=t}h(z)}\\ & =\frac{h(x)\mathbf{1}\{T(x)=t\}}{\sum _{z:T(z)=t}h(z)}\end{array}$$
  does not depend on ${\displaystyle \theta }$.
• If ${\displaystyle T(X)}$ is sufficient, construct $$g_{\theta }(t)={\mathbb{P}}_{\theta }(T(X)=t),h(x)=\mathbb{P}(X=x|T(X)=T(x)),$$ here ${\displaystyle h(x)}$ does not depend on ${\displaystyle \theta }$. Then $${\mathbb{P}}_{\theta }(X=x)=g_{\theta }(T(X))h(x).$$

• Exponential family exactly fits the expression. So it is sufficient.
• Normal distribution family: ${\displaystyle X_1,\cdots ,X_n\stackrel{\mathrm{i}.\mathrm{i}.\mathrm{d}}{\sim }\mathcal{N}(\theta ,1)=\frac{1}{\sqrt{2\pi }}{\mathrm{e}}^{-\frac{(x-\theta {)}^2}{2}}}$. Then $$p_{\theta }(x)=\prod _{i=1}^n{\mathrm{e}}^{-\frac{(x_i-\theta {)}^2}{2}}=\underset{g_{\theta }\left(\sum _{i=1}^n{x}_i\right)}{\underbrace{\left({\mathrm{e}}^{\theta \sum _{i=1}^n{x}_i-\frac{n{\theta }^2}{2}}\right)}}\cdot \underset{h(x)}{\underbrace{2{\pi }^{-\frac{n}{2}}\left({\mathrm{e}}^{-\frac{1}{2}\sum _{i=1}^n{x}_i^2}\right)}}.$$
  So ${\displaystyle T(X)=\sum _{i=1}^n{X}_i}$.
• Poisson family: ${\displaystyle X_1,\cdots ,X_n\stackrel{\mathrm{i}.\mathrm{i}.\mathrm{d}}{\sim }\mathrm{Poisson}(\theta )}$. So $$p_{\theta }(x)=\prod _{i=1}^n\frac{{\theta }^{x_i}{\mathrm{e}}^{-\theta }}{x_i!}=\underset{g_{\theta }\left(\sum _{i=1}^n{x}_i\right)}{\underbrace{{\theta }^{\sum _{i=1}^n{x}_i}{\mathrm{e}}^{-n\theta }}}\cdot \underset{h(x)}{\underbrace{{(\prod _{i=1}^n{x}_i!)}^{-1}}}.$$ So ${\displaystyle T(X)=\sum _{i=1}^n{X}_i}$.
• Uniform location family: ${\displaystyle X_1,\cdots ,X_n\stackrel{\mathrm{i}.\mathrm{i}.\mathrm{d}}{\sim }\mathrm{Uniform}(\theta ,\theta +1)}$. So $$\begin{array}{rl}{p}_{\theta }(x)& =\prod _{i=1}^n\mathbf{1}\{\theta \le x_i\le \theta +1\}\\ & =\mathbf{1}\{\theta \le \underset{i}{min}{X}_i\}\cdot \mathbf{1}\{\underset{i}{max}{X}_i\le \theta +1\}.\end{array}$$ So ${\displaystyle T(X)=(\underset{i}{min}{X}_i,\underset{i}{max}{X}_i)}$.

By factorization theorem we have $$p_{\theta }(x)=g_{\theta }(T(x))h(x)=(g_{\theta }\circ f)(S(x))h(x),$$ showing that ${\displaystyle S(X)}$ is sufficient.

• First, show ${\displaystyle T(X)}$ is sufficient. ${\displaystyle \mathrm{\forall }x:T(x)=t}$, we have $${\mathbb{P}}_{\theta }(X=x|T(x)=t)=\frac{p_{\theta }(x)}{\sum _{z:T(z)=t}{p}_{\theta }(z)}={\left(\sum _{z:T(z)=t}\frac{p_{\theta }(z)}{p_{\theta }(x)}\right)}^{-1}$$ does not depend on ${\displaystyle \theta }$.
• Then show ${\displaystyle T(X)}$ is minimal. Assume another sufficient statistic ${\displaystyle S(X)}$. Suppose ${\displaystyle S(x)=S(y)=s}$, then ${\displaystyle x\equiv py}$, so by assumption of the theorem ${\displaystyle T(x)=T(y)}$. Assume ${\displaystyle f(s)=T(x)}$.
  For any other ${\displaystyle z}$ with ${\displaystyle S(z)=s}$, we must also have ${\displaystyle T(z)=T(x)=f(s)}$. So ${\displaystyle T(x)=f(S(x))}$.

${\displaystyle X_1,\cdots ,X_n\stackrel{\mathrm{i}.\mathrm{i}.\mathrm{d}}{\sim }p-\theta (x)=\frac{1}{2}{\mathrm{e}}^{-|x-\theta |}}$. Then ${\displaystyle l(\theta ;x)=-\sum _{i=1}^n|x-\theta |-n\log 2}$. The image is a combination of linear segments. So ${\displaystyle x\equiv py⟺(x_{(1)},\cdots ,x_{(n)})=(y_{(1)},\cdots ,y_{(n)})}$. So ${\displaystyle S(X)=(X_{(1)},\cdots ,X_{(n)})}$ is minimal sufficient.

By theorem, we only need to show ${\displaystyle x\equiv py\Rightarrow T(x)=T(y)}$. By this argument, we need to show ${\displaystyle l(\cdot ;x)=l(\cdot ;y)+c_{xy}\Rightarrow T(x)=T(y)}$.
Now, $$l(\eta ;x)-l(\eta ;y)={\eta }^{\mathrm{T}}(T(x)-T(y)).$$ If ${\displaystyle T(x)-T(y)\ne 0}$, we can always find ${\displaystyle \eta ,\xi \in \mathrm{\Xi }}$ s.t. ${\displaystyle {\eta }^{\mathrm{T}}(T(x)-T(y))\ne {\xi }^{\mathrm{T}}(T(x)-T(y))}$ (this is undesirable, because ${\displaystyle c_{xy}}$ should be irrelevant of ${\displaystyle \eta }$), so ${\displaystyle T(x)=T(y)}$. Q.E.D.