"In presence of heteroscedasticity, OLS estimators are unbiased but inefficient"

Showing the unbiased part is relatively easy. Some authors have explained the inefficiency with the help of new variance of the least square estimator. However, I was asked to show the same thing using the variance of weighted least square estimator ($\hat{\beta^*}$). I proceeded in the following way:-

Consider the model, $Y_t=\alpha+\beta X_t+u_t$ where $u_t$'s are heteroscedastic.

Suppose ${\sigma _u}^2$ be the constant variance of $u_t$'s under the homoscedastic assumption.

Let $Var(u_t)={\sigma _t}^2$ be the variance of the disturbances under the heteroscedastic assumption.

In Particular, suppose ${\sigma _t}^2=k_t {\sigma _u}^2$, $k_t$'s being some non stochastic constant weights.

Now, consider the above model in deviation form

$y_t=\beta x_t +u_t$, where $y_t=Y_t-E(Y)$, $x_t=X_t-E(X)$

$\Rightarrow \frac{y_t}{k_t}=\beta \frac{x_t}{k_t}+\frac{u_t}{k_t}$

$\Rightarrow \frac{y_t}{k_t}=\beta \frac{x_t}{k_t}+v_t$
where $v_t$ has constant variance ${\sigma _u}^2$

Now, the weighted least square estimator is $\hat{\beta^*}=\frac{\sum \frac{y_t}{k_t}\frac{x_t}{k_t}}{\sum \frac{{x_t}^2}{{k_t}^2}}$

which would eventually give us $Var(\hat{\beta^*})=\frac{{\sigma _u}^2}{\sum \frac{{x_t}^2}{{k_t}^2}}$

Now, under the heteroscedastic assumption, we have $Var(\hat{\beta})=\frac{\sum {x_t}^2{\sigma_t}^2}{(\sum {x_t}^2)^2}$

Now, $$\frac{Var(\hat{\beta^*})}{Var(\hat{\beta})}=\frac{{\sigma _u}^2}{\sum \frac{{x_t}^2}{{k_t}^2}}\times \frac{(\sum {x_t}^2)^2}{\sum {x_t}^2{\sigma_t}^2}$$
$$\Rightarrow \frac{Var(\hat{\beta^*})}{Var(\hat{\beta})}=\frac{{\sigma _u}^2}{\sum \frac{{x_t}^2}{{k_t}^2}}\times \frac{(\sum {x_t}^2)^2}{\sum {x_t}^2{k_t\sigma_u}^2}$$
$$\Rightarrow \frac{Var(\hat{\beta^*})}{Var(\hat{\beta})}=\frac{(\sum {x_t}^2)^2} {\sum \frac{{x_t}^2}{{k_t}^2} \sum {x_t}^2{k_t}}$$

However, before I could proceed further, my instructor asked me to check it again. He was insisting that I should get something like $\frac{\sum (a_t b_t)^2}{\sum {a_t}^2 {b_t}^2}$, so that I can use Cauchy-Schwartz inequality resulting in $Var(\hat{\beta^*})<Var(\hat{\beta})$. This would eventually prove the *inefficiency * part.

Since I am not getting that form, I think I have made mistakes. I would be happy if somebody can point it.

  • 1
    $\begingroup$ Divide by $\sqrt{k_t}$ and not $k_t$, $\dfrac{y_t}{\sqrt{k_t}} = \beta \dfrac{y_t}{\sqrt{k_t}} + \dfrac{u_t}{\sqrt{k_t}}$ $\endgroup$
    – Amit
    May 27, 2018 at 4:08
  • $\begingroup$ If we divide by $\sqrt{k_t}$, $\frac{u_t}{\sqrt{k_t}}$ would not have constant variance. $\endgroup$
    – user18429
    May 27, 2018 at 4:14
  • $\begingroup$ Given that variance of $u_t$ is $\sigma_t^2 = k_t \sigma_u^2$, variance of $\dfrac{u_t}{\sqrt{k_t}}$ is $\sigma^2_u$ and variance of $\dfrac{u_t}{k_t}$ is $\dfrac{\sigma^2_u}{k_t}$ $\endgroup$
    – Amit
    May 27, 2018 at 4:20
  • 1
    $\begingroup$ That's incorrect. $\text{Var}(cX) = c^2\text{Var}(X)$. $\endgroup$
    – Amit
    May 27, 2018 at 4:29
  • $\begingroup$ Sorry, I missed that point. You are correct. Now it would have constant variance. $\endgroup$
    – user18429
    May 27, 2018 at 4:34

1 Answer 1


Let $\hat{\beta}$ be the OLS estimator of $\beta$ in $$y_t = \beta x_t + u_t$$ Let $\tilde{\beta}$ be the OLS estimator of $\beta$ in $$\dfrac{y_t}{\sqrt{k_t}} = \beta \dfrac{x_t}{\sqrt{k_t}} + \dfrac{u_t}{\sqrt{k_t}}$$ $\text{Var}(\hat{\beta}) = \dfrac{\sum x_t^2\sigma_t^2}{\left(\sum x_t^2\right)^{2}} $

$\text{Var}(\tilde{\beta}) = \dfrac{\sigma^2_u}{\sum \dfrac{x_t^2}{k_t}} $

So, $\dfrac{\text{Var}(\tilde{\beta})}{\text{Var}(\hat{\beta})} = \dfrac{\left(\sum x_t^2\right)^{2}}{\sum \dfrac{x_t^2}{k_t} \sum {x_t^2}{k_t}} $.

Let $a_t = \dfrac{x_t}{\sqrt{k_t}}$ and $b_t = {x_t}{\sqrt{k_t}}$, and we can rewrite the ratio of variances as

$\dfrac{\text{Var}(\tilde{\beta})}{\text{Var}(\hat{\beta})} = \dfrac{\left(\sum a_tb_t\right)^{2}}{\sum a_t^2 \sum {b_t^2}} \leq 1 $ [By Cauchy-Schwarz inequality]


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