Given the parameter space $\Theta = (0,1] \times \{1,2\}$ we have the following distributions: If $\theta = (\tau,1)$, then $X\sim P_{\tau}$ Poisson distributed and if $\theta = (\tau,2)$ then $X\sim\operatorname{Bernoulli}(\tau)$ distributed.
How can one show, that $T= \sum_i X_i$ with a given sample $X_1, \ldots, X_n$ is not sufficient for $\theta$? Furthermore how can one contruct an example of a two-dimensional minimal sufficient statistic $(T_1, T_2)$ for $\theta$?
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What is the conditional distribution of $X_1,\ldots,X_n$ given $\displaystyle \sum_{i=1}^n X_i = 2n\text{ ?}$ It is impossible that $X_i$ has a Bernoulli distribution in that case. That implies that $$ \Pr\left( X_1 = 2 \mid \sum_{i=1}^n X_i = 2n \right) $$ depends on whether the second component of the parameter is $1$ or $2$. In particular, if the second component is $1$, then this conditional probability is $0$.
Ok I will have a go at this; first of all we know that a statistic, $T(x)$, is sufficient if and only if the likelihood decomposes as $$ f_\theta(x) = h(x)g_\theta(T(x)), $$ so lets have a go at writing out the likelihood for $\theta = (\tau,c)^T$, $$ \begin{align} f_\theta(x) &= \frac 1 {\prod_i x_i!}e^{-n\tau}\tau^{\sum_i x_i} \mathbb{1}_{c=1} + \tau^n(1-\tau)^{n-\sum_i x_i}\mathbb{1}_{c=2} \\ &= \frac 1 {\prod_i x_i!} e^{-n\tau}\tau^{T_1(x)} \mathbb{1}_{c=1} + \tau^n (1-\tau)^{n-T(x)}\mathbb{1}_{c=2}. \end{align} $$ As it stands this expression certainly isn't in the factorised form we would like it so we might try doing something like this $$ \frac 1 {\prod_{i}x_i!} \left( e^{-n\tau}\tau^{T(x)}\mathbb{1}_{c=1} + \left(\prod_i x_i!\right)\tau^n(1-\tau)^{n-T(x)} \mathbb{1}_{c=2}\right), $$ but now the second term in the product depends on the data, the important observation is that the Bernoulli random variable has support $\{ 0, 1 \}$, and the Poisson random variable has support $\mathbb{N}$, so that when $X_i$ follows the Bernoulli distribution we have $\prod x_i! = 1$, so add a new statistic $T_2(x) = \max_i{x_i}$, say, and then we have $$ \frac 1 {\prod_i x_i!} \left( e^{-n\tau}\tau^{T_1(x)}\mathbb{1}_{c=1} + \tau^n(1-\tau)^{n-T_1(x)} \mathbb{1}_{c=2}\mathbb{1}_{T_2(x) \leq 1}\right). $$ which now factors in the desired way. To argue this is also minimal sufficient it should be possible to show that for different samples $x$ and $y$ that the ratio is constant as a function of $\theta$ if and only if $T_1(x)=T_1(y)$ and $T_2(x) = T_2(y)$.