Minimal sufficient statistic for normal distribution with known variance Let $X_1, ..., X_n$ be a random sample from the $N(\theta,1)$ distribution. Find a minimal sufficient statistic for $\theta$.
Now, I can find a sufficient statistic using the factorisation theorem ($\sum X_i$), but I don't think that this statistic is in fact minimal sufficient.
The question seems to imply that there exists a minimal sufficient statistic, but I'm not even sure that there is one.
MY QUESTION: How would I go about proving that there is no minimal sufficient statistic, or if there is one, what is it!?
Any hints greatly appreciated!
 A: By the factorization criterion
$$
\mathcal{L}(\theta)=\frac{1}{(2\pi)^{n/2}}\exp\{-\sum_{i=1}^nX_i^2/2 +\bar{X}_n \theta -n\theta^2/2\} 
$$
$$
  \qquad = \exp\{\bar{X}_n\theta-n\theta^2/2\}\times(2\pi)^{-n/2}\exp\{-\sum X_i^2/2\}.
$$
So $\bar{X}_n$ is sufficient statistic.
A: A sufficient statistic $Y$ is a function of $X_1,\ldots,X_n$ for which the conditional distribution of $X_1,\ldots,X_n$ given $Y$ does not depend on $\theta.$ Implicit in this is that one need not know the value of $\theta$ in order to know which function of $X_1,\ldots,X_n$ is $Y.$
A sufficient statistic $Y$ is minimal if for every function $g$ that is not one-to-one, and whose domain is the support of the random variable $Y,$ the random variable $g(Y)$ is not sufficient. I.e. no coarser statistic is sufficient.
A coarser statistic than $X_1+\cdots+X_n$ must have equal values for two different values of $X_1+\cdots+X_n.$ Suppose that for $X_1+\cdots+X_n=a$ and $X_1+\cdots+X_n=b,$ we have the same value of $Y,$ say $Y=c,$ although $a\ne b.$
Consider the conditional distribution of $X_1,\ldots,X_n$ given $Y=c.$ What is that distribution when $\theta=a/n,$ and what is it when $\theta=b/n$? You get two different distributions; therefore $Y$ is not sufficient.
