Say I have the following regression model:

$\ln\left(\dfrac{y_i}{x_{2i}}\right)=\alpha_1+\alpha_2\ln(x_{2i}) + \alpha_3\ln(x_{3i}) +e_i$

where I know the values of the regression coefficients and standard errors.

Now, I can rewrite the model as:

$\ln(y_i)=\beta_1+\beta_2\ln(x_{2i}) + \beta_3\ln(x_{3i}) +e_i$

where $\beta_1=\alpha_1$, $\beta_2=\alpha_2+1$, and $\beta_3=\alpha_3$.

How do I compute the standard errors for the new model, given that I know what they are for the original model? I feel like they are just the same? But I am not sure... if someone could explain to me that would be greatly appreciated.

  • 1
    $\begingroup$ $Var(\hat{\beta}_1) = Var(\hat{\alpha}_1 + 1) = Var(\hat{\alpha}_1)$, thus standard error shouldn't change. This is assuming the OLS estimate for $\hat{\beta} = \hat{\alpha}_1 + 1$ which I think it will be since that will also be the MLE estimate. $\endgroup$ – Greenparker May 15 '16 at 16:54
  • $\begingroup$ Yes I thought that was why, thank you for your answer! $\endgroup$ – aml3 May 15 '16 at 17:03

The answer is yes, in this case, the standard errors are the same.


Your vector of estimates $\hat{\boldsymbol{\beta}} = \left[\begin{array}{c} \hat{\beta}_1\\ \hat{\beta}_2 \\ \hat{\beta}_3 \end{array} \right]$ of the true vector $\boldsymbol{\beta}$ will be a random variable. One of the key outputs of any regression will be an estimate of the covariance matrix of estimator $\hat{\boldsymbol{\beta}}$. That is:

$$ Cov\left( \hat{\boldsymbol{\beta}} \right) \approx \Sigma \quad \quad \Sigma = \left[ \begin{array}{ccc} \Sigma_{1,1}, \Sigma_{1,2}, \Sigma_{1,3} \\ \Sigma_{2,1}, \Sigma_{2,2}, \Sigma_{2,3} \\ \Sigma_{3,1}, \Sigma_{3,2}, \Sigma_{3,3} \end{array} \right] $$

Your standard errors are the square root of the elements along the diagonal of the $\Sigma$ matrix. That is:

$$ SE = \left[ \begin{array}{c} \sqrt{\Sigma_{1,1}} \\ \sqrt{\Sigma_{2,2}} \\ \sqrt{\Sigma_{3,3}}\end{array} \right] $$

Trivially we have that covariance of $\hat{\boldsymbol{\beta}} + \left[\begin{array}{c} 0 \\ 1 \\ 0\end{array} \right]$ is the same as the covariance of $\hat{\boldsymbol{\beta}}$ (i.e. both covariance matrices are $\approx \Sigma$, and hence the standard errors will be the same as well.

Further note:

A key step in the above argument is that adding a constant to a random variable does not change the variance. Quick sketch in the univariate case:

$$ \begin{align*} Var(\tilde{x} + c) &= E[(\tilde{x} + c - E[\tilde{x} + c])^2] \\ &= E[(\tilde{x} + c - \left(E[\tilde{x}] + c\right) )^2] \\ &= E[(\tilde{x} - E[\tilde{x}] )^2] \\ &= Var(\tilde{x}) \end{align*}$$

Intuition: variance measures average square of deviations from the mean. Adding a constant to a random variable doesn't change deviations from the mean and hence doesn't change the variance.

  • $\begingroup$ I think you might want to explain why "the covariance of $\hat{\beta} + (0 \, \, 1 \, \, 0)^T$ is the same as the covariance of $\hat{\beta}$". $\endgroup$ – Greenparker May 15 '16 at 16:59

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