# Why does t-statistic for arbitrary parameter follow t distribution?

I understand that given a standard normal variable $$Z$$ and a $$\chi^2$$ random variable $$V$$ with $$\upsilon$$ degrees of freedom that \begin{align*} T := \frac{Z}{\sqrt{V/\upsilon}} \end{align*} follows a $$t$$-distribution with $$\upsilon$$ degrees of freedom (assuming $$Z$$ and $$V$$ are independent). From this one may derive (source) the ubiquitous $$t$$-statistic \begin{align}\tag{1} \frac{\bar{X} - \mu_0}{S/\sqrt{n}} \sim t_{n-1} \end{align} from \begin{align*} Z = \frac{\bar{X} - \mu_0}{\sigma/\sqrt{n}} \sim \mathcal{N}(0,1) \quad \text{and} \quad V = \frac{(n-1)S^2}{\sigma^2}\sim \chi^2_{n-1}. \end{align*}

However, when learning about hypothesis testing for regression population coefficients, I learned that given $$\hat{\beta}$$, an estimator of an arbitrary parameter $$\beta$$, one also uses the $$t$$-statistic \begin{align}\tag{2} t_{\hat{\beta}} = \frac{\hat{\beta} - \beta_0}{s.e.(\hat{\beta})}, \end{align} where $$s.e.(\hat{\beta})$$ is the standard error, i.e. the standard deviation of the sampling distribution of $$\hat{\beta}$$.

Whilst this looks similar to equation (1), I don't know why it should necessarily follow a $$t$$-distribution. I don't believe one would derive this fact in a similar manner to above: if our estimator $$\hat{\beta}$$ is not the sample mean then the random variable \begin{align*} \tilde{Z} = \frac{\hat{\beta} - \beta_0}{s.e.(\hat{\beta})} \end{align*} will not necessarily be standard normal (the central limit theorem can only be invoked for the sample mean).

Therefore my question is, does someone have a source or a proof to show that equation (2) follows a $$t$$-distribution given that $$\hat{\beta}$$ is an arbitrary parameter and not necessarily the sample mean? Everything online seems to focus on the specific case of the sample mean.

• We have posts here that prove this for ordinary least squares regression. The proofs are all simple, deriving from an assumed Normal distribution of errors. The sample mean is a special case (it's OLS with only an intercept) but serves as a model for the general case. In other kinds of regression these statistics tend not to have Student t distributions but they often are asymptotically Normal.
– whuber
Jun 28 at 12:47
• @hamster230: My guess is that the point you might be missing is that the t-statistic of a coefficient in an OLS has a t-distribution under the assumption that the null hypothesis is TRUE. So this means that the true coefficient is $\beta$. In that case, one can show ( as whuber pointed out it's probably in many threads ) that the numerator is normal and the standard deviation is chi-squared and they are independent etc. If the null hypothesis is not true, then the distribution would be non-central t. Jun 28 at 13:32

$$\hat{\beta}$$ is an arbitrary parameter and not necessarily the sample mean

The behaviour is the same as a sample mean.

• The sample mean is a linear sum of the observations $$y_i$$

$$\bar{y} = \sum_{i=1}^n a_{i} y_i \qquad \text{with} \, a_i = \frac{1}{n}$$

• Ordinary least squares regression estimates the parameters also as a linear sum of the observations $$y_i$$

$$\beta_j = \sum_{i=1}^n a_{ji}y_i \qquad \text{with} \, a_{ji} = \left[(X^TX)^{-1}X^T\right]_{ji}$$

So when these observations $$y_i$$ are normal distributed then $$\beta_j$$ will also be normal distributed just like $$\bar{y}$$ is considered to be normal distributed. (or if the $$y_i$$ are not approximately normal distributions then we might often still consider the linear sum as approximately normal distributed)

A geometrical viewpoint of OLS is that you are splitting the n-dimensional space of observations into two orthogonal sub-spaces as illustrated in the image below (from the question Why are the residuals in $\mathbb{R}^{n-p}$? ).

The error distribution of the observations can then also be considered to $$y_i$$ split into two parts. One part for the estimate(s) and one part for the residuals.

In the case of the estimate of the mean this sub-space will be a single diagonal line.