Show that $T(\mathbf{X})=(\sum X_i, \sum X_i^2)$ is not complete

Question

Let $X_1, \cdots X_n \stackrel{\text{iid}}{\sim} N(\alpha \sigma, \sigma^2)$, where $\alpha$ is known, and $\sigma > 0$ is unknown. Show that the family of distributions of $$T(\mathbf{X})=(\sum X_i, \sum X_i^2)$$ is not complete.

My work:

I am getting that this family is complete with the following work.

\begin{align*}E_\sigma[g(T(X))]&=\int_{-\infty}^\infty g(T(x))\frac{1}{\sqrt{2 \pi}\sigma}\exp(\frac{-1}{2\sigma^2}(x-\alpha \sigma)^2)dx\\ &=\frac{1}{\sqrt{2\pi}\sigma}\exp(\frac{-\alpha^2}{2})\int_{-\infty}^\infty g(T(x))\exp(\frac{-x^2}{2\sigma^2} + \frac{x\alpha}{\sigma})dx\end{align*}

For this to be $0$, $\int_{-\infty}^\infty g(T(x))dx=0$, since the exponential terms can never be equal to 0. Does this imply that the family of distributions of $T(X_1,\cdots,X_n)$ is complete?

Answer 1

The argument is incorrect: it is not because $$\int_{-\infty}^\infty g(T(x))\exp(\frac{-x^2}{2\sigma^2} + \frac{x\alpha}{\sigma})\text{d}x=0$$that $g\circ T$ is necessarily zero. (The argument does not even use the specific functional form of $T$.) Furthermore, as pointed out by @whuber, the integral in your approach should be on $\mathbb R^n$ rather than $\mathbb R$.

As suggested by @whuber, the standard line of attack is to find a function of $T(X)$ that is independent from $\sigma$. What could help in this regard is to rewrite the observations as $X_i\sim\sigma Y_i$, where $Y_i\sim N(\alpha,1)$, and to notice that $$T(X)\sim(\sigma\sum_i Y_i,\sigma^2\sum_i Y_i^2)$$ to guess a transform of $T(X)$ that does not depend on $\sigma$. (Hint: $\sigma^2=(\sigma)^2$.)