Is sample variance a complete statistic for variance of a normal distribution if the mean is known?

Question

Suppose $X \sim N(\mu,\sigma^2)$. I know that $T(X)=(\bar X, S^2)$ is a complete sufficient statistic for $\mu, \sigma^2$ if $\mu, \sigma^2$ are unknown. But if $\mu$ is known, is $S^2$ still a complete statistic of $\sigma^2$?

Answer 1

$$S^2=\frac{1}{n-1}\Sigma_i (X_i-\overline{X}_n)^2$$

being

$$\Sigma_i (X_i-\overline{X}_n)^2=\Sigma_i (X_i-\mu_0)^2-n(\overline{X}_n-\mu_0)^2$$

Let's define the following funcion

$$g(S^2)=\Sigma_i (X_i-\mu_0)^2-n^2(\overline{X}_n-\mu_0)^2$$

where, $\forall \sigma^2$

$$\mathbb{E}[g(S^2)]=n\sigma^2-n^2\frac{\sigma^2}{n}=0$$

but evidently

$$P[g(S^2)=0] \neq 1$$

this shows that $S^2$ is not a complete statistic for $\sigma^2$ when $\mu=\mu_0$ is a known value