I have the following problem on my Statistics Problem Set:
Consider the model \begin{equation*} y_{i}=x_{i}^{\prime }\beta +u_{i}, \end{equation*} where $\left( x_{i}^{\prime },u_{i}\right) $ are iid with $u_{i}|x_{i}$ have the density $f\left( u\right) \in C^{2}$ (with support $-\infty <u<\infty $ ). Assume that \begin{eqnarray*} E\left( U\right) &=&\int_{-\infty }^{\infty }uf\left( u\right) =0\text{ and} \\ V\left( U\right) &=&E\left( U^{2}\right) =\int_{-\infty }^{\infty }u^{2}f\left( u\right) =\sigma ^{2}. \end{eqnarray*}
a)Use transformation of variables to show that the (conditional) pdf of $ y_{i}|x_{i}$ is given by $g\left( y_{i}|x_{i}\right) =f\left( y_{i}-x_{i}^{\prime }\beta \right) $.
b) Find the likelihood of $y=\left( y_{1},...,y_{n}\right) $ conditional on $X=\left( x_{1},...,x_{n}\right) ^{\prime }$.
c) State the Gauss-Markov theorem.
d) We will show in item (f) that the asymptotic variance of$\sqrt{n}(% \widetilde{\beta }-\beta ^{\ast })$ can be smaller than the asymptotic variance of $\sqrt{n}(\widehat{\beta }-\beta ^{\ast })$, where $\widetilde{% \beta }$ is the MLE and $\widehat{\beta }$ is the OLS estimator. Explain why this result does not contradict the Gauss-Markov theorem.
e) Find the asymptotic variance of$\sqrt{n}(\widehat{\beta }-\beta ^{\ast })$.
f) Show algebraically that (i) the asymptotic variance of$\sqrt{n}(% \widetilde{\beta }-\beta ^{\ast })$ is no larger than the asymptotic variance of $\sqrt{n}(\widehat{\beta }-\beta ^{\ast })$; and (ii) give a necessary and sufficient condition on the density $f\left( u\right) $ for the asymptotic variance of $\widehat{\beta }$ and $\widetilde{\beta }$ to be the same.
I'm fine with this problem until item f).
I was able to do it all, including computing the asymptotic distributions:
$\sqrt{n}(\hat{\beta} - \beta ^{\ast }) \xrightarrow{d} N(0, \sigma^2(x^T x)^{-1})$
$\sqrt{n}(\tilde{\beta} - \beta ^{\ast }) \xrightarrow{d}N\Bigg(0, \Bigg[\mathbb{E}\Big(x_i \bigg(\frac{d \log(f(u))}{du}\bigg)^2 x_i'\Big)\Bigg]^{-1}\Bigg)$
Indeed, if $f(u)$ is Gaussian, then
$\mathbb{E}\Bigg(\Big(\frac{d\log(f(u))}{du}\Big)^2 \Bigg) = \mathbb{E}\Big(\frac{u^2}{\sigma^4} \Big) = 1/\sigma^2$,
which would take us to the previous case of OLS. I would feel happy answering at least the first part of item f), which asks us to compare these asymptotic variances. However, I don't see how to do it because we don't have so much structure on $f$. Any ideas on how to it? Thanks in advance!