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

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 $$\phi(T)$$.

• Complete sufficient for $\theta$ is $\sum x_i^2y_i$. If you take the expectation you will get the answer as expected in a regression without intercept. Must have been answered here before. May 4, 2020 at 5:34
• May 4, 2020 at 5:52
• @StubbornAtom Thank you for linking that related post. I have posted my work as an answer below. I appreciate it! May 4, 2020 at 23:54

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 $$\theta$$.