I have this problem:
Consider the regression model $y=ax^3+\epsilon$, therefore $y_k=ax_k^3+\epsilon_k$, where $e_k \approx N(0,\sigma^2)$ for all $k=1,2,...,n$.
a) Find the estimation of $\hat{a}$.
b) Find $E(y_k)$ and $var(y_k)$.
c) Prove that the estimation $\hat{a}$ is unbiased and $var(\hat{a})=\sigma^2 \frac{\sum x_k^3}{(\sum x_k^6)^2}$.
I think I managed to derive that $\hat{a}=\frac{\sum y_k}{\sum x_k^3}$...
Then I think that $E(y_k)=E(ax_k^3+\epsilon_k)=a.E(x_k^3)+E(\epsilon_k)=a.\frac{\sum x_k^3}{n}=\overline{y}$ and
$E(\hat{a})=\frac{E(\sum y_k)}{\sum x_k^3} = \frac{\sum y_k}{\sum x_k^3}=\hat{a}$.
But I'm not able to calculate the variances. I don't really think I know what I do. How do I do it please?
a.$$ S(\alpha) = \sum (y_k - \alpha x_k^3)^2$$ $$s'(\alpha ) = -2 \sum y_kx_k^3 + 2\alpha \sum x_k^6 = 0 $$ $$\hat{\alpha} = \frac{\sum y_k x_k^3}{\sum x_k^6} $$ $$ s''(\hat{\alpha}) = 2 \sum x_k^6 >0$$ b. $$\mathbb{E} \hat{\alpha} = \frac{ \sum x_k^3 \mathbb{E} y_k}{\sum x_k^6} = \frac{ \sum x_k^3 \alpha x_k^3}{\sum x_k^6} = \alpha. $$ c.
$$ \operatorname{Var}( \hat{\alpha} ) = \frac{ \sum x_k^6 \operatorname{Var} ( y_k) }{(\sum x_k^6 )^2} = \frac{ \sum x_k^6 \operatorname{Var} (\alpha x_k + e_k) }{(\sum x_k^6 )^2}= \frac{ \sum x_k^6 \operatorname{Var} (e_k) }{(\sum x_k^6 )^2} = \frac{ \sigma^2\sum x_k^6 }{(\sum x_k^6 )^2} = \frac{ \sigma^2 }{\sum x_k^6 } $$