My question is essentially identical to this In linear regression, how to prove the equivalence of F-test and t-test? - it was (I believe humbly) erroneously marked as a duplicate of this Prove F test is equal to T test squared. These questions are not duplicates, as linear regression is a generalisation of ANOVA and the calculations are quite different, involving more linear algebra.

My question is also identical to ESL Exercise 3.1. In a multivariate linear regression, the square of the t-statistic for $H_0: \beta_j=0$ is $$\frac{\hat{\beta_j}^2}{\hat{\sigma}^2 (X^T X)^{-1}_{jj}}$$

This should coincide with the F-statistic comparing the full model with the model that leaves out the variable $X_j$. This quickly reduces to proving

$$RSS_0 - RSS = \frac{\hat{\beta_j}^2}{(X^T X)^{-1}_{jj}}$$

I have seen this problem solved with Lagrange multipliers (eg here https://yuhangzhou88.github.io/ESL_Solution/ESL-Solution/3-Linear-Methods-for-Regression/ex3-01/), but it should be able to be solved purely with linear algebra.

For example, $RSS= y^T(I-H)y$ where $H=X(X^TX)^{-1}X^T$ is the hat matrix and the projection onto the column space of $X$. Similarly, $RSS_0 = y^T(I-H_0)y$. On the other hand, $\hat{\beta}=(X^TX)^{-1}X^Ty$ so $\hat{\beta}_j = Row_j((X^TX)^{-1}) X^Ty$. So

$$ \hat{\beta}_j^2 = y^T X Row_j((X^TX)^{-1})^T Row_j((X^TX)^{-1}) X^T y $$

Thus we can remove $y$ entirely and just need to prove the following

$$X Row_j((X^TX)^{-1})^T Row_j((X^TX)^{-1}) X^T = (X^T X)^{-1}_{jj} (H - H_0). $$

That is, the desired equality is now something entirely expressible in terms of $X$, but I find it quite intractable. It reminds me of Schur complement/inversion, which I have never really understood, so if you use that, an explanation of that would be appreciated too.

Edit: I have found quite a nice proof of this ultimately by careful use of orthogonal decompositions, but I'm still wondering if this has something to do with Schur inversion.

  • $\begingroup$ I have removed the additional query on Cook's D, which would warrant ab additional question, if OP wants. $\endgroup$ Feb 10 at 4:35

1 Answer 1


More appealing (than Lagrangian multipliers), intuitive from a geometrical point of view is to resort to perpendicular projection operators.

It is apt to work in a more general setting: let us consider for the linear model $$\mathbf Y = \mathbf X\boldsymbol\beta +\boldsymbol{\varepsilon},$$ the hypothesis $$\mathrm H_0: \mathbf \Lambda^\top\boldsymbol{\beta}= \mathbf 0.\tag 1\label 1$$

By $\eqref 1,~ \boldsymbol\beta \in \mathcal C(\mathbf \Lambda)^{\perp}=:\mathcal C(\mathbf U)\implies \boldsymbol\beta = \mathbf U\boldsymbol\gamma.$ Using this, we can get the reduced model as $\mathbf Y = \mathbf X_0\boldsymbol\gamma +\boldsymbol{\varepsilon}, ~~\mathbf X_0:= \mathbf {XU}.$

We know the numerator of the F-statistic is expressed as $$\mathbf Y^\top(\mathbf M-\mathbf M_0)\mathbf Y/\operatorname{rank}(\mathbf M-\mathbf M_0),\tag 2\label 2$$ where $\mathbf M,~ \mathbf M_0$ are respectively the perpendicular projection operators onto $\mathcal C(\mathbf X) $ and $\mathcal C(\mathbf X_0) .$ It is a routine calculation to check $\mathcal C(\mathbf M-\mathbf M_0)= \mathcal C(\mathbf X_0)^{\perp}_{\mathcal C(\mathbf X)}.$

Assume the linear constraint in $\eqref 1$ is estimable (If not, we would consider the estimable part of $\eqref 1: \mathbf \Lambda_0^\top\boldsymbol{\beta}= \mathbf 0,$ where $\mathcal C(\boldsymbol\Lambda_0):= \mathcal C(\boldsymbol\Lambda)\cap \mathcal C\left(\mathbf X^\top\right);$ the rationale behind it being that $\mathcal C(\boldsymbol\Lambda_0)\subset \mathcal C(\boldsymbol\Lambda)\implies \mathbf \Lambda_0^\top\boldsymbol{\beta}= \mathbf 0$ and $\mathcal C(\boldsymbol\Lambda_0)\subset\mathcal C\left(\mathbf X^\top\right)$ means $\mathbf \Lambda_0^\top\boldsymbol{\beta}$ is estimable; both constraints induce the same reduced model). Then $\mathbf \Lambda = \mathbf X^\top\mathbf P,$ for some $\mathbf P.$

The goal is to re-write $\eqref 2$ in terms of $\mathbf \Lambda$ and $\hat{\boldsymbol\beta}.$ For that, the approach would be to express $\mathbf M-\mathbf M_0$ in terms of $\mathbf \Lambda.$

Result $1.$ $$\mathcal C(\mathbf M-\mathbf M_0) =\mathcal C(\mathbf{MP} ).$$

From above $\mathcal C(\mathbf M-\mathbf M_0)= \mathcal C(\mathbf {XU})^{\perp}_{\mathcal C(\mathbf X)}.$ If $\mathbf x\in \mathcal C(\mathbf {XU})^{\perp}_{\mathcal C(\mathbf X)},$ then $\mathbf X^\top\mathbf x\perp \mathcal C(\mathbf U)\implies \mathbf X^\top\mathbf x\in \mathcal C(\mathbf \Lambda).$ Now $$\mathbf x = \mathbf{ Mx} = \mathbf X\left(\mathbf X^\top \mathbf X\right)^{-}\mathbf X^\top\mathbf x\in \mathcal C(\mathbf{MP} ).$$ Conversely, $\mathbf x\in \mathcal C(\mathbf{MP})\implies \mathbf x= \mathbf{MPb},$ and $$\mathbf x^\top\mathbf {XU} = \mathbf b^\top\underbrace{\mathbf P^\top\underbrace{\mathbf M\mathbf X}_{\mathbf{MX}= \mathbf X}}_{\mathbf P^\top\mathbf X = \mathbf \Lambda^\top }\mathbf U = 0, $$ as $\mathcal C(\mathbf U)\equiv \mathcal C(\mathbf \Lambda)^{\perp}.$


Result $2.$ $$\operatorname{rank}(\mathbf \Lambda) =\operatorname{rank}\left(\mathbf M_{\mathbf{ MP}}\right). $$

It suffices to check $\mathcal C(\mathbf X^\top\mathbf P) = \mathcal C(\mathbf P^\top\mathbf M)$ as this would imply $\operatorname{rank}(\mathbf X^\top\mathbf P) = \operatorname{rank}\left(\left(\mathbf P^\top\mathbf M\right)^\top\right)= \operatorname{rank}\left(\mathbf{ MP}\right)=\operatorname{rank}\left(\mathbf M_{\mathbf{ MP}}\right) . $ Now, $\mathbf X^\top\mathbf P\mathbf z = 0\iff \mathbf P\mathbf z\perp \mathcal C(\mathbf X) = \mathcal C(\mathbf M).$ This means $\mathcal C(\mathbf X^\top\mathbf P)^\perp = \mathcal C(\mathbf P^\top\mathbf M)^\perp.$


Using these two results, we can re-write $\eqref 2$ as

$$\begin{align}\frac{\mathbf Y^\top(\mathbf M-\mathbf M_0)\mathbf Y}{\operatorname{rank}(\mathbf M-\mathbf M_0)} &= \frac{\mathbf Y^\top\mathbf M_\mathbf{MP}\mathbf Y}{\operatorname{rank}(\mathbf \Lambda)}\\&=\frac{\mathbf Y^\top\mathbf{MP}\left(\mathbf P^\top\mathbf{MP}\right)^{-}\mathbf P^\top\mathbf{M Y}}{\operatorname{rank}(\mathbf \Lambda)} \\&=\frac{\hat{\boldsymbol \beta}^\top\mathbf{X}^\top\mathbf {P}\left(\mathbf P^\top\mathbf X\left(\mathbf X^\top\mathbf X\right)^{-}\mathbf X^\top\mathbf{P}\right)^{-}\mathbf P^\top\mathbf{X}\hat{\boldsymbol\beta}}{\operatorname{rank}(\mathbf \Lambda)}\\&=\frac{\hat{\boldsymbol \beta}^\top\mathbf{\Lambda}\left(\mathbf \Lambda^\top\left(\mathbf X^\top\mathbf X\right)^{-}\mathbf \Lambda\right)^{-}\mathbf{ \Lambda}^\top\hat{\boldsymbol\beta}}{\operatorname{rank}(\mathbf \Lambda)}\tag 3,\label 3\end{align}$$

which is the desired form.

If $\eqref 1$ is of single degree of freedom hypothesis, say, $\mathrm H_0: \boldsymbol \lambda^\top\boldsymbol{\beta}= 0,$ then $\eqref 3$ becomes $\frac{\left(\boldsymbol\lambda^\top\hat{\boldsymbol\beta}\right)^2}{\left(\boldsymbol\lambda^\top\left(\mathbf X^\top\mathbf X\right)^{-}\boldsymbol\lambda\right)}.$

$\rm[II]$ also has somewhat similar treatment using perpendicular projection operators specifically for multiple regression problems.


$\rm[I]$ Plane Answers to Complex Questions: Theory of Linear Models, Ronald Christensen, Springer Science$+$Business, $2011,$ sec. $3.2-3.3.$

$\rm[II]$ Linear Regression Analysis, George A. F. Seber, Alan J. Lee, John Wiley & Sons, $2003,$ sec. $3.8.2, ~4.3.2.$

  • 1
    $\begingroup$ I really appreciate the explicitness of the mathematics in this answer (+1). $\endgroup$
    – Galen
    Feb 10 at 5:01
  • $\begingroup$ Thanks, this looks similar in scope to the orthogonal projections proof that I found, but certainly in a bit of a different direction. I'm still curious if there is a relation with Schur's inversion. $\endgroup$ Feb 18 at 5:40
  • $\begingroup$ By Schur's inversion, are you talking about the Schur formulas for calculating inverse of partitioned matrices? AFAIR, it is not directly needed in the derivation of the result. $\endgroup$ Feb 19 at 2:25

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