= Solution
Any unbiased linear estimator has the form considered in part (b), and its error is the <linear combination of independent normal random variables>
$$
\widetilde\beta-\beta=\sum_i a_i\varepsilon_i
\sim N\left(0,\sigma^2\sum_i a_i^2\right).
$$
The <moment-generating function of a normal distribution> therefore gives, for every nonzero real $\theta$,
$$
\mathbb E_{\beta,\sigma^2}
\left[e^{\theta(\widetilde\beta-\beta)}\right]
=\exp\left(\frac{\theta^2\sigma^2}{2}\sum_i a_i^2\right).
$$
Since the <exponential function> is strictly increasing and $\theta^2\sigma^2/2>0$, minimizing this <exponential moment> is exactly the same as minimizing $\sum_i a_i^2$. Part (b) shows that the unique minimizer is $a_i=x_i/S_{xx}$. Thus, independently of the sign or magnitude of $\theta$,
$$
\boxed{\widetilde\beta
=\widehat\beta
=\frac{\sum_i x_iY_i}{\sum_i x_i^2}}.
$$
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