How can I find the BUE of $\theta$ in the simple linear relationship $Y_i=\theta x_i^2+\epsilon_i$?

Question

Let $Y_1,...,Y_n$ be described by the relationship $Y_i=\theta x_i^2+\epsilon_i$, where $x_1,...,x_n$ are fixed constants and $\epsilon_1,...,\epsilon_n$ are iid $N(0,\sigma^2)$. How can I find the best unbiased estimator?

my work:

I am considering using the Rao-Blackwell Theorem, where $\phi(T)=E(W|T)$ is the BUE/MVUE of $\theta$. Here, $T$ is a complete, sufficient statistic for $\theta$ and $W$ is an unbiased estimator for $\theta$.

Since $f(y|\theta)=\frac{1}{\sqrt{2\pi\sigma^2}}exp[-y_i^2/(2\sigma^2)]exp[y_ix_i\theta/(\sigma^2)]exp[-\theta^2x_i^2/(2\sigma^2)]$, we have that $T=\sum^n_{i=1}y_i$ is a complete, sufficient statistic for $\theta$ since $\{\theta y_i: y_i \in R^1\}$. However, I am struggling finding a good unbiased estimator such that I can obtain $\phi(T)$.

Answer 1

As @StubbornAtom stated, the correct complete sufficient statistic for $\theta$ is $T(\mathbf{Y})=\sum_iy_ix_i^2$, since we can form a full-rank exponential family from the random sample $Y_i \sim N(\theta x_i^2,\sigma^2)$, where $i=1,...,n$.

$ET(\mathbf{Y})=\sum \theta x_i^4$. So, if we take $\phi(\mathbf{Y})=\frac{\sum y_i x_i^2}{\sum x_i^4}$, then $E\phi(\mathbf{Y})=\theta$.

So $\phi(\mathbf{Y})$ is an unbiased estimator for $\theta$ based on the complete sufficient statistic $T(\mathbf{Y})$ for $\theta$. By Rao-Blackwell Theorem, $\phi(\mathbf{Y})$ is our best unbiased estiamtor for $\theta$.