Suppose we have an Ordinary Least Squares model where we have $k$ coefficients in our regression model, $$\mathbf{y}=\mathbf{X}\mathbf{\beta} + \mathbf{\epsilon}$$

where $\mathbf{\beta}$ is an $(k\times1)$ vector of coefficients, $\mathbf{X}$ is the design matrix defined by

$$\mathbf{X} = \begin{pmatrix} 1 & x_{11} & x_{12} & \dots & x_{1\;(k-1)} \\ 1 & x_{21} & \dots & & \vdots \\ \vdots & & \ddots & & \vdots \\ 1 & x_{n1} & \dots & \dots & x_{n\;(k-1)} \end{pmatrix}$$ and the errors are IID normal, $$\mathbf{\epsilon} \sim \mathcal{N}\left(\mathbf{0},\sigma^2 \mathbf{I}\right) \;.$$

We minimize the sum-of-squared-errors by setting our estimates for $\mathbf{\beta}$ to be $$\mathbf{\hat{\beta}}= (\mathbf{X^T X})^{-1}\mathbf{X}^T \mathbf{y}\;. $$

An unbiased estimator of $\sigma^2$ is $$s^2 = \frac{\left\Vert \mathbf{y}-\mathbf{\hat{y}}\right\Vert ^2}{n-p}$$ where $\mathbf{\hat{y}} \equiv \mathbf{X} \mathbf{\hat{\beta}}$ (ref).

The covariance of $\mathbf{\hat{\beta}}$ is given by $$\operatorname{Cov}\left(\mathbf{\hat{\beta}}\right) = \sigma^2 \mathbf{C}$$ where $\mathbf{C}\equiv(\mathbf{X}^T\mathbf{X})^{-1}$ (ref) .


How can I prove that for $\hat\beta_i$, $$\frac{\hat{\beta}_i - \beta_i} {s_{\hat{\beta}_i}} \sim t_{n-k}$$ where $t_{n-k}$ is a t-distribution with $(n-k)$ degrees of freedom, and the standard error of $\hat{\beta}_i$ is estimated by $s_{\hat{\beta}_i} = s\sqrt{c_{ii}}$.

My attempts

I know that for $n$ random variables sampled from $x\sim\mathcal{N}\left(\mu, \sigma^2\right)$, you can show that $$\frac{\bar{x}-\mu}{s/\sqrt{n}} \sim t_{n-1} $$ by rewriting the LHS as $$\frac{ \left(\frac{\bar x - \mu}{\sigma/\sqrt{n}}\right) } {\sqrt{s^2/\sigma^2}}$$ and realizing that the numertor is a standard normal distribution, and the denominator is square root of a Chi-square distribution with df=(n-1) and divided by (n-1) (ref). And therefore it follows a t-distribution with df=(n-1) (ref).

I was unable to extend this proof to my question...

Any ideas? I'm aware of this question, but they don't explicitly prove it, they just give a rule of thumb, saying "each predictor costs you a degree of freedom".

  • $\begingroup$ Because $\hat\beta_i$ is a linear combination of jointly Normal variables, it has a Normal distribution. Therefore all you need do are (1) establish that $\mathbb{E}(\hat\beta_i)=\beta_i$; (2) show that $s_{\hat\beta_i}^2$ is an unbiased estimator of $\text{Var}(\hat\beta_i)$; and (3) demonstrate the degrees of freedom in $s_{\hat\beta_i}$ is $n-k$. The latter has been proven on this site in several places, such as stats.stackexchange.com/a/16931. I suspect you already know how to do (1) and (2). $\endgroup$ – whuber Oct 1 '14 at 15:27

Since $$\begin{align*} \hat\beta &= (X^TX)^{-1}X^TY \\ &= (X^TX)^{-1}X^T(X\beta + \varepsilon) \\ &= \beta + (X^TX)^{-1}X^T\varepsilon \end{align*}$$ we know that $$\hat\beta-\beta \sim \mathcal{N}(0,\sigma^2 (X^TX)^{-1})$$ and thus we know that for each component $k$ of $\hat\beta$, $$\hat\beta_k -\beta_k \sim \mathcal{N}(0, \sigma^2 S_{kk})$$ where $S_{kk}$ is the $k^\text{th}$ diagonal element of $(X^TX)^{-1}$. Thus, we know that $$z_k = \frac{\hat\beta_k -\beta_k}{\sqrt{\sigma^2 S_{kk}}} \sim \mathcal{N}(0,1).$$

Take note of the statement of the Theorem for the Distribution of an Idempotent Quadratic Form in a Standard Normal Vector (Theorem B.8 in Greene):

If $x\sim\mathcal{N}(0,I)$ and $A$ is symmetric and idempotent, then $x^TAx$ is distributed $\chi^2_{\nu}$ where $\nu$ is the rank of $A$.

Let $\hat\varepsilon$ denote the regression residual vector and let $$M=I_n - X(X^TX)^{-1}X^T \text{,}$$ which is the residual maker matrix (i.e. $My=\hat\varepsilon$). It's easy to verify that $M$ is symmetric and idempotent.

Let $$s^2 = \frac{\hat\varepsilon^T \hat\varepsilon}{n-p}$$ be an estimator for $\sigma^2$.

We then need to do some linear algebra. Note these three linear algebra properties:

  • The rank of an idempotent matrix is its trace.
  • $\operatorname{Tr}(A_1+A_2) = \operatorname{Tr}(A_1) + \operatorname{Tr}(A_2)$
  • $\operatorname{Tr}(A_1A_2) = \operatorname{Tr}(A_2A_1)$ if $A_1$ is $n_1 \times n_2$ and $A_2$ is $n_2 \times n_1$ (this property is critical for the below to work)

So $$\begin{align*} \operatorname{rank}(M) = \operatorname{Tr}(M) &= \operatorname{Tr}(I_n - X(X^TX)^{-1}X^T) \\ &= \operatorname{Tr}(I_n) - \operatorname{Tr}\left( X(X^TX)^{-1}X^T) \right) \\ &= \operatorname{Tr}(I_n) - \operatorname{Tr}\left( (X^TX)^{-1}X^TX) \right) \\ &= \operatorname{Tr}(I_n) - \operatorname{Tr}(I_p) \\ &=n-p \end{align*}$$

Then $$\begin{align*} V = \frac{(n-p)s^2}{\sigma^2} = \frac{\hat\varepsilon^T\hat\varepsilon}{\sigma^2} = \left(\frac{\varepsilon}{\sigma}\right)^T M \left(\frac{\varepsilon}{\sigma}\right). \end{align*}$$

Applying the Theorem for the Distribution of an Idempotent Quadratic Form in a Standard Normal Vector (stated above), we know that $V \sim \chi^2_{n-p}$.

Since you assumed that $\varepsilon$ is normally distributed, then $\hat\beta$ is independent of $\hat\varepsilon$, and since $s^2$ is a function of $\hat\varepsilon$, then $s^2$ is also independent of $\hat\beta$. Thus, $z_k$ and $V$ are independent of each other.

Then, $$\begin{align*} t_k = \frac{z_k}{\sqrt{V/(n-p)}} \end{align*}$$ is the ratio of a standard Normal distribution with the square root of a Chi-squared distribution with the same degrees of freedom (i.e. $n-p$), which is a characterization of the $t$ distribution. Therefore, the statistic $t_k$ has a $t$ distribution with $n-p$ degrees of freedom.

It can then be algebraically manipulated into a more familiar form.

$$\begin{align*} t_k &= \frac{\frac{\hat\beta_k -\beta_k}{\sqrt{\sigma^2 S_{kk}}}}{\sqrt{\frac{(n-p)s^2}{\sigma^2}/(n-p)}} \\ &= \frac{\frac{\hat\beta_k -\beta_k}{\sqrt{S_{kk}}}}{\sqrt{s^2}} = \frac{\hat\beta_k -\beta_k}{\sqrt{s^2 S_{kk}}} \\ &= \frac{\hat\beta_k -\beta_k}{\operatorname{se}\left(\hat\beta_k \right)} \end{align*}$$

  • $\begingroup$ Also a side question: for the Theorem for the Distribution of an Idempotent Quadratic Form in a Standard Normal Vector, don't we also need $A$ to be symmetric? Unfortunately, I don't have Greene, so I can't see the proof although I saw that Wikipedia had the same form as you. However, a counter example seems to be the idempotent matrix $A=\begin{pmatrix}1&1\\0&0\end{pmatrix}$ which leads to $x_1^2+x_1 x_2$ which is not Chi-Squared since it could take on negative values... $\endgroup$ – Garrett Oct 1 '14 at 9:17
  • 1
    $\begingroup$ @Garrett My apologies, $A$ should be both symmetric and idempotent. A proof is provided as Theorem 3 in this document: www2.econ.iastate.edu/classes/econ671/hallam/documents/… Luckily, $M$ is symmetric as well as idempotent. $\endgroup$ – Blue Marker Oct 1 '14 at 13:29
  • 1
    $\begingroup$ $A$ is merely a matrix representation of a quadratic form. Every quadratic form has a symmetric representation, so the requirement of symmetry of $A$ is implicit in the statement of the theorem. (People do not use asymmetric matrices to represent quadratic forms.) Thus the quadratic form $(x_1,x_2)\to x_1^2+x_1x_2$ is uniquely represented by the matrix $A=\begin{pmatrix}1&1/2\\1/2&0\end{pmatrix}$ which is not idempotent. $\endgroup$ – whuber Oct 1 '14 at 15:33
  • $\begingroup$ this is a great explanation for nonrandom x. Can you explain the deriviation if x is assumed to be random? $\endgroup$ – denizen of the north Feb 13 '18 at 14:07
  • $\begingroup$ Why does $\epsilon\sim N(0,\sigma^2)$ imply $\hat{\beta}$ is independent of $\hat{\epsilon}$? Not quite following there. $\endgroup$ – Glassjawed Oct 25 '18 at 15:59

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.