MVUE for a function of variance of Normal Distribution Let $X_1, X_2, ..., X_n$ be a random sample from a $N(\theta_1,\theta_2)$ distribution. Find the uniformly minimum variance unbiased estimator of $3{\theta_2}^2$.
Using factorization theorem, I found that $T(X) = (\sum X_i^2, \sum X_i)$ is a sufficient statistic. Now, how do I check if the statistic is complete?
 A: lets $X_i \sim N(\mu,\sigma^2)$.
We want to show 
$(\sum X_i, \sum X^2_i)$ is complete for $(\mu,\sigma^2)$.
It is enough to show $(\bar{X} , S=\sum (X_i-\bar{X})^2)$ is complete. We know $\bar{X}$ and $S$ are independent and  $\bar{X}\sim N(\mu,\frac{\sigma^2}{n})$, $S\sim Gamma(\frac{n-1}{2},2\sigma^2)$.
We should show if $\forall (\mu,\sigma^2)$
$$E(g(\bar{X} , S))=0 \Rightarrow P(g(\bar{X} , S)=0)=1$$.
$$0=E(g(\bar{X} , S))=\int_0^\infty \int_{-\infty}^{+\infty} g(\bar{x} , s) f(\bar{x})f(s) d\bar{x} \, ds$$
$$=\frac{1}{\Gamma(\frac{n-1}{2})(\sigma^2)^{\frac{n-1}{2}}}\int_0^\infty \left(\int_{-\infty}^{+\infty} g(\bar{x} , s) f(\bar{x})s^{\frac{n-1}{2}-1} e^{-\frac{s}{\sigma^2}} d\bar{x} \,\right) ds$$
$$=\frac{1}{\Gamma(\frac{n-1}{2})(\sigma^2)^{\frac{n-1}{2}}}\int_0^\infty \left(\int_{-\infty}^{+\infty} g(\bar{x} , s) f(\bar{x})s^{\frac{n-1}{2}-1}  d\bar{x} \,\right)e^{-\frac{s}{\sigma^2}} ds$$
$$=\frac{1}{\Gamma(\frac{n-1}{2})(\sigma^2)^{\frac{n-1}{2}}}\int_0^\infty \left(h(s)\right)e^{-\frac{s}{\sigma^2}} ds$$
The above is a  Laplace transform of $h(s)$, which implies
$h(s)=0$, a.e.
So
$$0=\int_{-\infty}^{+\infty} g(\bar{x} , s) f(\bar{x}) d\bar{x}$$
$$=\int_{-\infty}^{+\infty} g(\bar{x} , s) \frac{1}{\sqrt{2\pi \frac{\sigma^2}{n}}} e^{-\frac{n}{2\sigma^2}(\bar{x}-\mu)^2} d\bar{x}$$
$$=\int_{-\infty}^{+\infty} \left(g(\bar{x} , s) \frac{1}{\sqrt{2\pi \frac{\sigma^2}{n}}} e^{-\frac{n}{2\sigma^2}\bar{x}^2} e^{-\frac{n}{2\sigma^2}\mu^2} \right) e^{\frac{n}{2\sigma^2}2\bar{x}\mu} d\bar{x}$$
The above is a Two-sided_Laplace_transform.
So $g(\bar{x} , s)=0$ a.e.
