Suppose we have a NO INTERCEPT model, $$y_i=\beta x_i+e_i$$

where $e_{i}$ follows a N(0,$\sigma^2 x_i^h$), so $e_i$ is equal in distribution to $e_{0i} x_i^{\frac{h}{2}}$, where $e_{0i}$ follows a N(0,$\sigma^2$) .

I know that in order to obtain the BLUE (Best Linear Unbiased Estimator) of the $\beta$ coefficient, set $w_i = x_i^{-\frac{h}{2}}$.

$h$ can be estimated by calculating the marginal variances at each level of x. Then, using the model $\log(\hat{e}_i^2)=\log(\sigma^2)+h\log(X)$ which yields $\hat{h}$, so $w_i = x_i^{-\frac{\hat{h}}{2}}$ .

This then gives the model $$y_i w_i = \beta x_i w_i + e_{0i}$$ which gives $\hat{\beta }=\frac{\sum_{i=1}^{n}w_i^2 x_i y_i}{\sum_{i=1}^{n}w_i^2 x_i^2}$.

I am looking to find $Var(\hat{\beta })=\frac{\sum_{i=1}^{n}w_i^2 x_i^2 }{(\sum_{i=1}^{n}w_i^2 x_i^2)^2}Var(w_i y_i)=\sigma^2 \frac{\sum_{i=1}^{n}w_i^2 x_i^2 }{(\sum_{i=1}^{n}w_i^2 x_i^2)^2}$, but I'm not completely sure how to correctly estimate $\sigma^2$.

I am estimating $\sigma^2$ using the usual $s^2$ formula with the transformed data, $\frac{1}{n-1}\sum_{i=1}^{n}(w_i y_i -\hat{\beta}w_i x_i)^2$. Is this correct? Should I be dividing by n-2 because I've estimated TWO parameters ($\beta$ and $h$)?

  • $\begingroup$ Rather than trying to fix up this ad hoc two-step approach (which will be complicated and perhaps not be worth the effort, since its statistical properties are unknown), why not select a suitable estimation procedure for $(\beta,h)$--Maximum Likelihood might work well--and apply it directly? $\endgroup$ – whuber Mar 17 '15 at 23:25
  • $\begingroup$ I suppose I'm using the two step approach because I would like unbiased estimates for the variance. Not sure how to adjust the MLE estimator for the variance to correct it. $\endgroup$ – Lewkrr Apr 14 '15 at 2:11
  • $\begingroup$ My sample size is not very large, so bias is not trivial. Am I being too perfectionist? Should I just an ML solution, as Zachary suggests below? $\endgroup$ – Lewkrr Apr 14 '15 at 2:18

First off I assume you mean $e_i \sim N(0,\sigma^2|x_i|^h)$ or that $x_i$ is always positive. Otherwise you will get negative and or complex valued variances.

In practice, when we know heteroskedasticity exists but are unaware of its form we may use the two-step method of Feasible generalized least squares (FGLS). However FGLS may be inconsistent. So to @whuber 's point, given you know something about the heteroskedasticity structure, you can construct a consistent maximum likelihood estimator (MLE) for your parameters. The likelihood function would have the form $$ L=\prod_{i=1}^{n} \frac{1}{\sigma |x_i|^{h/2}}\phi \bigg(\frac{y_i-x_i'\beta}{\sigma |x_i|^{h/2}}\bigg) $$ Where $\phi$ is the standard normal pdf.
Your optimal parameter estimates may then be found by maximizing the log likelihood i.e $$ (\hat \beta,\hat \sigma, \hat h)=\arg \max_{\beta,\sigma,h}\bigg\{\sum_{i=1}^{n}-\ln(\sigma |x_i|^{h/2})+\ln\bigg(\phi \bigg(\frac{y_i-x_i'\beta}{\sigma |x_i|^{h/2}}\bigg)\bigg) \bigg\} $$ Which simplifies too $$ (\hat \beta,\hat \sigma, \hat h)=\arg \max_{\beta,\sigma,h}\bigg\{-\frac{n}{2}\ln(2\pi)-\sum_{i=1}^{n}\ln\bigg(\sigma |x_i|^{h/2}\bigg)+\frac{(y_i-x_i'\beta)^2}{\sigma^2|x_i|^h}\bigg\} $$

This is extremely useful as you can construct a hypothesis test for homoskedasticity. $$ H_0: h=0 $$ $$ H_a: h\neq 0 $$ Where the null is homoskedasticity.

$Var(\hat \beta)$ may be estimated using the fisher information equality, the negative inverse of the hessian of the above log likelihood evaluated at $(\hat \beta, \hat \sigma, \hat h)$. These estimates may be obtained using built in optimization and maximum likelihood routines for R, Matlab, Python, etc. Below is an R example.


# estimates 
    #standard errors
# maximum log likelihood

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.