Consider the simple linear regression model Y_i = βx_i + ϵ_i, for i = 1, …, n; where E(ϵ_i) = 0, Cov(ϵ_i, ϵ_k) = 0 if i ≠ k and Var(ϵ_i) = ${\mathrm{x}}_{\mathrm{i}}^2{\sigma }^2$. The best linear unbiased estimator of β is:

Concept:

Ley Y = β₀ +β₁x + ϵ  be a linear regression model such that ϵ ∼ N(0, σ²) then estimator 

 $\widehat{{\beta }_1}=\frac{\sum (x-\overline{x})(y-\overline{y})}{\sum (x-\overline{x}{)}^2}$ and $\widehat{{\beta }_0}=\overline{y}-\widehat{{\beta }_1}\overline{x}$

Unbiassed estimator: α is said to be unbiassed estimator of β if E(α) = β

Explanation:

Given Yi = βxi + ϵi and

E(ϵi) = 0, Cov(ϵi, ϵk) = 0 if i ≠ k and Var(ϵi) = ${\mathrm{x}}_{\mathrm{i}}^2{\sigma }^2$....(i)

⇒ $\frac{Y_i}{x_i}=\beta +\frac{{ϵ}_i}{x_i}$

var($\frac{Y_i}{x_i}$) = var($\beta +\frac{{ϵ}_i}{x_i}$) = $\frac{1}{x_i^2}var({ϵ}_i)$ = σ2 (using (i))

and E($\frac{Y_i}{x_i}$) = β + $\frac{1}{x_i}$E(ϵi) = β (as by (i) E(ϵi) = 0)

So E($\frac{1}{\mathrm{n}}{\displaystyle \sum _{\mathrm{i}=1}^{\mathrm{n}}\frac{{\mathrm{Y}}_{\mathrm{i}}}{{\mathrm{x}}_{\mathrm{i}}}}$) = E($\frac{1}{\mathrm{n}}{\displaystyle \sum _{\mathrm{i}=1}^{\mathrm{n}}(\beta +\frac{{ϵ}_{\mathrm{i}}}{{\mathrm{x}}_{\mathrm{i}}})}$) = E($\frac{1}{\mathrm{n}}(\beta \mathrm{n}+{\displaystyle \sum _{\mathrm{i}=1}^{\mathrm{n}}\frac{{ϵ}_{\mathrm{i}}}{{\mathrm{x}}_{\mathrm{i}}})}$) = E(β + $\frac{1}{\mathrm{n}}{\displaystyle \sum _{\mathrm{i}=1}^{\mathrm{n}}\frac{{ϵ}_{\mathrm{i}}}{{\mathrm{x}}_{\mathrm{i}}}}$) = β + 0 = β (by (i))

Therefore the best linear unbiased estimator of β is $\frac{1}{\mathrm{n}}{\displaystyle \sum _{\mathrm{i}=1}^{\mathrm{n}}\frac{{\mathrm{Y}}_{\mathrm{i}}}{{\mathrm{x}}_{\mathrm{i}}}}$

Option (3) is correct