Show that Statistic is complete

Solution 1:

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.