Counterproof for sufficient statistic as well as an example for minimal sufficient statistic 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 $\theta$?
 A: 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 $T_2(x) = T_2(y)$.
A: 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 $0$.
