Consider the following model with the usual OLS assumptions: $\epsilon_i$ are uncorrelated random variables with mean zero and constant variance $\sigma^2$.
$$y_i=\beta+ 2 \beta x_i+\epsilon_i$$
$(a)$Derive the least squares estimator for $\beta$
$(b)$Compute $\mathsf{E}(\hat{\beta})$
$(c)$ Compute $\mathsf{Var}(\hat{\beta})$
Attempted Solutions:
$(a)$ Let $S(\beta)=\sum(y_i-(\beta+2\beta x_i))^2$
Then set
$$\frac{\partial S}{\partial \beta}=2\sum(y_i-\beta-2\beta x_i)(-1-2x_i)=0$$
From here, I get that
$$\begin{align*}
\frac{\partial S}{\partial \beta}
&=2\sum(y_i-\beta-2\beta x_i)(-1-2x_i)\\\\
&=2\sum(-y_i+\beta+4\beta x_i-2y_i x_i+4\beta x_i^2)\\\\
&=2\left(\sum-y_i-2y_ix_i+\beta\sum(1+4x_i+4x_i^2)\right)
\end{align*}$$
Thus,
$$\hat{\beta}=\frac{\sum y_i+2y_ix_i}{\sum1+4x_i+4x_i^2}$$
Checking that this is, in fact, a minimum,
$$\frac{\partial}{\partial\beta}\left(2\sum-y_i-2y_ix_i+2\beta\sum1+4x_i+4x_i^2\right)=2\gt0 \text{ }\checkmark$$
Is that valid?
For $(b)$ and $(c)$ I am stumped. I think I need to figure out what can be treated as constants within my estimate and what needs to be treated as random variables.
My first thought for $(b)$ is to note that
$$\mathsf E\left(\frac{\sum y_i+2y_ix_i}{\sum1+4x_i+4x_i^2}\right)=\mathsf E\left(\frac{\sum (\beta+2\beta x_i+\epsilon_i)+2x_i(\beta+2\beta x_i+\epsilon_i)}{\sum1+4x_i+4x_i^2}\right)$$
but I don't see how that would help me.
Edit:
I have
$$\begin{align*}
\mathsf{Var}\left(\frac{\sum y_i+2y_ix_i}{\sum1+4x_i+4x_i^2}\right)
&=\mathsf{Var}\left(\frac{\sum (\beta+2\beta x_i+\epsilon_i)+2x_i(\beta+2\beta x_i+\epsilon_i)}{\sum1+4x_i+4x_i^2}\right)\\\\
&=\mathsf{Var}\left(\frac{\sum \epsilon_i+2x_i\epsilon_i+\sum\beta+4\beta x_i +2\beta x_i^2}{\sum1+4x_i+4x_i^2}\right)\\\\
&=\mathsf{Var}\left(\frac{\sum \epsilon_i+2x_i\epsilon_i}{\sum1+4x_i+4x_i^2}+\frac{\sum\beta+4\beta x_i +2\beta x_i^2}{\sum1+4x_i+4x_i^2}\right)\\\\
&=\mathsf{Var}\left(\frac{\sum \epsilon_i+2x_i\epsilon_i}{\sum1+4x_i+4x_i^2}\right)\\\\
&=\mathsf{Var}\left(\frac{\sum\epsilon_i(1+2x_i)}{\sum(1+2x_i)^2}\right)
\end{align*}$$
but I cannot isolate the $\epsilon_i$, the only random variable remaining.
Best Answer
To finish off, write $$\operatorname{Var}\left(\frac{\sum\epsilon_it_i}{\sum t_i^2}\right)\stackrel{(1)}=\frac{\operatorname{Var}(\sum\epsilon_it_i)}{(\sum t_i^2)^2}\stackrel{(2)}=\frac{\sum\sigma^2t_i^2}{(\sum t_i^2)^2}\stackrel{(3)}=\frac{\sigma^2\sum t_i^2}{(\sum t_i^2)^2}\stackrel{(4)}=\frac{\sigma^2}{\sum t_i^2}.$$ I've written $t_i:=1+2x_i$ for brevity. Step (1) is the rule $\operatorname{Var}(cX)=c^2\operatorname{Var}(X)$. Step (2) is using the uncorrelatedness of the $\epsilon$'s. Step (3) is where you can pull out the constant $\sigma^2$ from the summation. Step (4) is cancelling a common factor from top and bottom.