= Solution
Write the linear mean model as
$$
Y_i=\theta x_i+\varepsilon_i,
\qquad
\mathbb E\varepsilon_i=0,
\qquad
\operatorname{var}(\varepsilon_i)=\theta x_i.
$$
The errors are independent but <heteroscedastic>. Minimizing the unweighted residual sum of squares
$$
Q(\theta)=\sum_{i=1}^n(Y_i-\theta x_i)^2
$$
gives the <normal equation>
$$
0=Q'(\theta)=-2\sum_i x_i(Y_i-\theta x_i).
$$
Thus, provided $\sum_i x_i^2>0$,
$$
\boxed{\widehat\theta_{LS}
=\frac{\sum_i x_iY_i}{\sum_i x_i^2}}.
$$
Since $\mathbb EY_i=\theta x_i$,
$$
\mathbb E\widehat\theta_{LS}
=\frac{\theta\sum_i x_i^2}{\sum_i x_i^2}=\theta.
$$
\b[Hence it is an <unbiased estimator>. This is the least-squares part of the <estimators for a Poisson exposure model>.]
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