Sufficient statistics when mean is known

Question

I want to find sufficient statistics for $N(0,\sigma^2)$. I know that $\sum_{i=1}^{n}x_i^2$ is a sufficient statistics by factorization Theorem. I wonder if $\frac{1}{n-1}\sum_{i=1}^n(x_i-\bar{X})^2$ which is sample variance is sufficient or not. It seems that it is not sufficient because I couldn't factorize the density but I couldn't prove it.

Answer 1

Define $T=\sum X_i^2$ and $U=\frac{1}{n-1}\sum (X_i-\bar{X})^2=\frac{\sum X_i^2 -n \bar{X}^2}{n-1}$.

If you want to show a statistic is not a sufficient statistic , you can compare it with minimal sufficient statistic. Use the fact that a minimal sufficient statistic is a function of any sufficient statistic.

It is obvious that $T=\sum X_i^2$ is a minimal sufficient statistic for $\sigma^2$. Since $T$ is minimal sufficient statistic, so it is a function of any sufficient statistic. It is enough to show that $T$ is not a function of $U$.

$T$ is a function of $U$ if $U(a_1)=U(a_2 )$ $\Rightarrow T(a_1)=T(a_2)$. So it is enough to find two points that $U(a_1)= U(a_2)$ but $T(a_1)\neq T(a_2)$ , and hence $T$ is not a function of $U$ and hence $U$ is not a sufficient statistic.

$a_1=(x_1=1,x_2=1, \cdots ,x_n=1)$

$a_2=(x_1=0,x_2=0, \cdots ,x_n=0)$

So $0=U(a_1)=U(a_2 )$ but $1=T(a_1)\neq 0=T(a_2 )$