Showing $\sum a_iX_i$ and $\sum X_i$ are independent iff $\sum a_i=0$ where $X_i$'s are i.i.d $N(\theta,\sigma^2)$

Question

Let $X_1, X_2,\ldots, X_n$ be iid with the distribution $N(\theta, \sigma^2),$ $-\infty < \theta < \infty$.

Prove that a necessary and sufficient condition that the statistics $Z =\sum a_iX_i$ and $Y=\sum X_i $ are independent is that $\sum a_i=0$.

My try:

I have showed that $Y=\sum X_i$ is a complete sufficient statistic.

To use Basu's theorem, since $Z$ is a linear combination of independent normal distribution then $$Z\sim N\left(\theta\sum a_i, \sigma^2\sum a_i^2\right)$$

then I said $Z$ is free of $\theta$ i.e ( an ancillary) iff $\sum a_i=0$.

Then by Basu's theorem they are independent. Is this a sufficient answer? Thank you

Answer 1

Note that $(Y,Z)$ is a bivariate normal random variable (can you say why?). So, a necessary and sufficient condition that they are independent, is $\textrm{Cov}(Y,Z) = 0$. But $\textrm{Cov}(Y,Z) = \sigma^2 \sum a_i$, and you are done.