Given $X_1,…,X_n$ iid. random variables which follow a $\mathcal{N}(\mu, \sigma^2)$ distribution. I already proved that
$T:=\frac{1}{n}\sum_{i=1}^n (X_i-\mu)^2$ is a minimal sufficient statistic for $\sigma^2$. Since, I want to apply the Lehmann-Scheffe Lemma, I have further to show that $T$ is complete, meaning that $\mathbb{E}_{\sigma^2}[g(T)]=0\Rightarrow g(T)=0$ for any measurable function $g$.
However, I am unable to do this, as I don't know how $T$ is distributed. Also there is a theorem which states that in exponential families the statistic is complete if the parameter space contains an open neighbourhood about the origin, but this is clearly not applicable since $\sigma^2\in (0,\infty)$. I would appreciate a little hint.
Best Answer
You can prove the completeness directly as well without using the exponential family result.
A minimal sufficient statistic for $\sigma^2$ when $X_1,\ldots,X_n$ are i.i.d $N(\mu,\sigma^2)$ with $\mu$ known is $$Z=\sum_{i=1}^n (X_i-\mu)^2$$
Since $Z/\sigma^2\sim \chi^2_n$, pdf of $Z$ is $$f_Z(z)=ce^{-z/2\sigma^2}z^{n/2-1}1_{z>0}\,,$$
where $c=c(n,\sigma^2)$ is such that $\displaystyle\int f_Z(z)\,dz=1$.
Now $\mathbb E_{\sigma^2}[g(Z)]=0$ for all $\sigma^2$ implies
$$\int_0^\infty g(z)e^{-z/2\sigma^2}z^{n/2-1}\,dz=0\quad,\forall\, \sigma^2$$
This is a one-sided Laplace transform of $g(z)z^{n/2-1}$ and by property of integral transforms, you have $$g(z)z^{n/2-1}=0\,\,,\text{ a.e.}$$
Thus implying $$g(z)=0\,\,,\text{ a.e.}$$
So $\mathbb E_{\sigma^2}[g(Z)]=0$ for all $\sigma^2$ implies $g(Z)=0$ almost everywhere for any measurable function $g$.
Hence the family of distributions is complete.