# Is a $t$-statistic still $t$-distributed under linearly transforming the original OLS model?

For an OLS model $$y=X\beta +\epsilon$$ it is well known that any of the estimated coefficients, say $$\widehat\beta_i$$, divided by its estimated standard error $$s_{\widehat\beta_i}$$, must follow a $$t$$-distribution whose degree of freedom equal is $$n-p$$ where $$p$$ is the number of explanatory variables (including intercept).

Now, suppose we linearly transform the model by a non-singular matrix $$C$$: \begin{align} y &= X\beta + \epsilon\\ &=XC C^{-1}\beta + \epsilon\\ &= XC\color{red}{\gamma} + \epsilon \end{align} Then in this new model where the coeffcients to find are $$\gamma$$, we can find an OLS estimator $$\widehat\gamma$$ which is again a BLUE. But the question is, is there any way we can tell about the distribution of $$\widehat\gamma$$'s $$t$$-statistic $$\widehat\gamma_i/s_{\widehat\gamma_i}$$? For example, I believe it is still $$t$$-distributed, but how can we prove it? And what will be its dof?

Thanks!

• Are you cared about the asymptotic distribution? – RScrlli Nov 1 '18 at 17:14

The (exact) $$t$$ distribution arises not from the OLS itself but from the combination of OLS, multivariate normal vector $$\epsilon$$ with covariance matrix of $$\sigma ^ 2I$$, and by assuming that the null hypothesis of $$\gamma_i = 0$$ is true.
In such a case, the OLS vector of $$\hat{\gamma}$$ or $$\hat{\beta}$$ is a linear transformation of $$y$$, i.e., $$\hat{\beta} = (X'X)^{-1}X'y$$, hence it is multivariate by itself and the marginal distribution of each estimator $$\hat{\beta}_i$$ is $$N(0, \sigma_{i}^2)$$. The residuals vector $$e$$ is also a linear transformation of $$y$$, i.e., $$(I - H)y$$, hence it is also follows a multivariate distribution and then you can show that $$\hat{\sigma}_i^2 \sim \sigma_i ^ 2\chi^2(n-p)$$, hence the ratio $$\hat{\beta}_i/\hat{\sigma}_i$$ by definition follows $$t$$ distribution with $$n-p$$ dof. The explicit derivations can be found on wikipedia or any standard book on linear regression.
Having said that, regressing on $$XC$$ and $$\gamma = X^{-1}\beta$$ induces no change to this properties. Namely, you transformed the data $$X$$ and accordingly the coefficients vector $$\beta$$, you induced no change to the assumptions on $$\epsilon$$ or the estimation method (OLS), hence same properties will follow from the same reasoning. The only difference is that the BLUE property is now w.r.t. $$\gamma$$ given $$XC$$ and not w.r.t. $$\beta$$ given $$X$$.
If no normality assumption on $$\epsilon$$ is made then the normal distribution of $$\hat{\beta}_i$$ is merely an approximation for any finite sample size $$n$$, hence I would not bother at all with student's $$t$$ and will stick to the normal approximation.
• Thanks! But why does $\hat{\beta}_i/\hat{\sigma}_i$ by definition follow a $t$-distribution with dof=1? Actually I think we need to know the dependencies between $\hat{\beta}_i$ and $\hat{\sigma}_i$, and the dof should be $n-p$ where $p$ is the dof of the model, right? – Vim Nov 2 '18 at 0:44
• 1) Sure, $n-p$ dof. Corrected. 2) $t(k) = Z/\sqrt{\chi^2(k)/k}$. Where $\hat{\beta}_i$ is normal and $\hat{\sigma}_i = \hat{\sigma}_{\epsilon}g(X)$, then the ratio is standard normal divided by square root of chi square etc. The independence stems from the fact that the residuals that construct $\hat{\sigma}^2_{\epsilon}$ are orthogonal to the columns space of $X$. – V. Vancak Nov 2 '18 at 9:27