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In light of this question : Proof that the coefficients in an OLS model follow a t-distribution with (n-k) degrees of freedom

I would love to understand why

$$ F = \frac{(\text{TSS}-\text{RSS})/(p-1)}{\text{RSS}/(n-p)},$$

where $p$ is the number of model parameters and $n$ the number of observations and $TSS$ the total variance, $RSS$ the residual variance, follows an $F_{p-1,n-p}$ distribution.

I must admit I have not even attempted to prove it as I wouldn't know where to start.

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  • $\begingroup$ Christoph Hanck and Francis has given a very good answer already. If you still have difficulties in understanding the proof of f test for linear regression, try to checkout teamdable.github.io/techblog/… . I wrote the blog post about the proof of the ftest for linear regression. It is written in Korean but it may not be a problem because almost all of it is math formula. I hope it would help if you still have difficulties in understanding the proof of f test for linear regression. $\endgroup$
    – Taeho Oh
    Aug 5, 2019 at 1:34
  • $\begingroup$ While this link may answer the question, it is better to include the essential parts of the answer here and provide the link for reference. Link-only answers can become invalid if the linked page changes. - From Review $\endgroup$
    – mkt
    Aug 5, 2019 at 5:36
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    $\begingroup$ Ultimately, there are only three fundamental things to know here. The first is that sums of squares of zero-mean Normal variables are multiples of chi-squared distributions. This is often taken as the definition of a chi-square distribution. The second is that TSS-RSS and RSS are independent; this is a matter of linear algebra, which expresses them as functions of uncorrelated Normal variables. The third is that a ratio of independent chi-squared variables has (up to a constant multiple) an $F$ distribution: indeed, you can take this as a definition of $F.$ $\endgroup$
    – whuber
    Feb 4, 2022 at 16:11

2 Answers 2

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Let us show the result for the general case of which your formula for the test statistic is a special case. In general, we need to verify that the statistic can be, according to the characterization of the $F$ distribution, be written as the ratio of independent $\chi^2$ r.v.s divided by their degrees of freedom.

Let $H_{0}:R^\prime\beta=r$ with $R$ and $r$ known, nonrandom and $R:k\times q$ has full column rank $q$. This represents $q$ linear restrictions for (unlike in OPs notation) $k$ regressors including the constant term. So, in @user1627466's example, $p-1$ corresponds to the $q=k-1$ restrictions of setting all slope coefficients to zero.

In view of $Var\bigl(\hat{\beta}_{\text{ols}}\bigr)=\sigma^2(X'X)^{-1}$, we have \begin{eqnarray*} R^\prime(\hat{\beta}_{\text{ols}}-\beta)\sim N\left(0,\sigma^{2}R^\prime(X^\prime X)^{-1} R\right), \end{eqnarray*} so that (with $B^{-1/2}=\{R^\prime(X^\prime X)^{-1} R\}^{-1/2}$ being a "matrix square root" of $B^{-1}=\{R^\prime(X^\prime X)^{-1} R\}^{-1}$, via, e.g., a Cholesky decomposition) \begin{eqnarray*} n:=\frac{B^{-1/2}}{\sigma}R^\prime(\hat{\beta}_{\text{ols}}-\beta)\sim N(0,I_{q}), \end{eqnarray*} as \begin{eqnarray*} Var(n)&=&\frac{B^{-1/2}}{\sigma}R^\prime Var\bigl(\hat{\beta}_{\text{ols}}\bigr)R\frac{B^{-1/2}}{\sigma}\\ &=&\frac{B^{-1/2}}{\sigma}\sigma^2B\frac{B^{-1/2}}{\sigma}=I \end{eqnarray*} where the second line uses the variance of the OLSE.

This, as shown in the answer that you link to (see also here), is independent of $$d:=(n-k)\frac{\hat{\sigma}^{2}}{\sigma^{2}}\sim\chi^{2}_{n-k},$$ where $\hat{\sigma}^{2}=y'M_Xy/(n-k)$ is the usual unbiased error variance estimate, with $M_{X}=I-X(X'X)^{-1}X'$ is the "residual maker matrix" from regressing on $X$.

So, as $n'n$ is a quadratic form in normals, \begin{eqnarray*} \frac{\overbrace{n^\prime n}^{\sim\chi^{2}_{q}}/q}{d/(n-k)}=\frac{(\hat{\beta}_{\text{ols}}-\beta)^\prime R\left\{R^\prime(X^\prime X)^{-1}R\right\}^{-1}R^\prime(\hat{\beta}_{\text{ols}}-\beta)/q}{\hat{\sigma}^{2}}\sim F_{q,n-k}. \end{eqnarray*} In particular, under $H_{0}:R^\prime\beta=r$, this reduces to the statistic \begin{eqnarray} F=\frac{(R^\prime\hat{\beta}_{\text{ols}}-r)^\prime\left\{R^\prime(X^\prime X)^{-1}R\right\}^{-1}(R^\prime\hat{\beta}_{\text{ols}}-r)/q}{\hat{\sigma}^{2}}\sim F_{q,n-k}. \end{eqnarray}

For illustration, consider the special case $R^\prime=I$, $r=0$, $q=2$, $\hat{\sigma}^{2}=1$ and $X^\prime X=I$. Then, \begin{eqnarray} F=\hat{\beta}_{\text{ols}}^\prime\hat{\beta}_{\text{ols}}/2=\frac{\hat{\beta}_{\text{ols},1}^2+\hat{\beta}_{\text{ols},2}^2}{2}, \end{eqnarray} the squared Euclidean distance of the OLS estimate from the origin standardized by the number of elements - highlighting that, since $\hat{\beta}_{\text{ols},2}^2$ are squared standard normals and hence $\chi^2_1$, the $F$ distribution may be seen as an "average $\chi^2$ distribution.

In case you prefer a little simulation (which is of course not a proof!), in which the null is tested that none of the $k$ regressors matter - which they indeed do not, so that we simulate the null distribution.

enter image description here

We see very good agreement between the theoretical density and the histogram of the Monte Carlo test statistics.

library(lmtest)
n <- 100
reps <- 20000
sloperegs <- 5 # number of slope regressors, q or k-1 (minus the constant) in the above notation
critical.value <- qf(p = .95, df1 = sloperegs, df2 = n-sloperegs-1) 
# for the null that none of the slope regrssors matter

Fstat <- rep(NA,reps)
for (i in 1:reps){
  y <- rnorm(n)
  X <- matrix(rnorm(n*sloperegs), ncol=sloperegs)
  reg <- lm(y~X)
  Fstat[i] <- waldtest(reg, test="F")$F[2] 
}

mean(Fstat>critical.value) # very close to 0.05

hist(Fstat, breaks = 60, col="lightblue", freq = F, xlim=c(0,4))
x <- seq(0,6,by=.1)
lines(x, df(x, df1 = sloperegs, df2 = n-sloperegs-1), lwd=2, col="purple")

To see that the versions of the test statistics in the question and the answer are indeed equivalent, note that the null corresponds to the restrictions $R'=[0\;\;I]$ and $r=0$.

Let $X=[X_1\;\;X_2]$ be partitioned according to which coefficients are restricted to be zero under the null (in your case, all but the constant, but the derivation to follow is general). Also, let $\hat{\beta}_{\text{ols}}=(\hat{\beta}_{\text{ols},1}^\prime,\hat{\beta}_{\text{ols},2}')'$ be the suitably partitioned OLS estimate.

Then, $$ R'\hat{\beta}_{\text{ols}}=\hat{\beta}_{\text{ols},2} $$ and $$ R^\prime(X^\prime X)^{-1}R\equiv\tilde D, $$ the lower right block of \begin{align*} (X^TX)^{-1}&=\left( \begin{array} {c,c} X_1'X_1&X_1'X_2 \\ X_2'X_1&X_2'X_2\end{array} \right)^{-1}\\&\equiv\left( \begin{array} {c,c} \tilde A&\tilde B \\ \tilde C&\tilde D\end{array} \right) \end{align*} Now, use results for partitioned inverses to obtain $$ \tilde D=(X_2'X_2-X_2'X_1(X_1'X_1)^{-1}X_1'X_2)^{-1}=(X_2'M_{X_1}X_2)^{-1} $$ where $M_{X_1}=I-X_1(X_1'X_1)^{-1}X_1'$.

Thus, the numerator of the $F$ statistic becomes (without the division by $q$) $$ F_{num}=\hat{\beta}_{\text{ols},2}'(X_2'M_{X_1}X_2)\hat{\beta}_{\text{ols},2} $$ Next, recall that by the Frisch-Waugh-Lovell theorem we may write $$ \hat{\beta}_{\text{ols},2}=(X_2'M_{X_1}X_2)^{-1}X_2'M_{X_1}y $$ so that \begin{align*} F_{num}&=y'M_{X_1}X_2(X_2'M_{X_1}X_2)^{-1}(X_2'M_{X_1}X_2)(X_2'M_{X_1}X_2)^{-1}X_2'M_{X_1}y\\ &=y'M_{X_1}X_2(X_2'M_{X_1}X_2)^{-1}X_2'M_{X_1}y \end{align*}

It remains to show that this numerator is identical to $\text{RSSR}-\text{USSR}$, the difference in restricted and unrestricted sum of squared residuals.

Here, $$\text{RSSR}=y'M_{X_1}y$$ is the residual sum of squares from regressing $y$ on $X_1$, i.e., with $H_0$ imposed. In your special case, this is just $TSS=\sum_i(y_i-\bar y)^2$, the residuals of a regression on a constant.

Again using FWL (which also shows that the residuals of the two approaches are identical), we can write $\text{USSR}$ (SSR in your notation) as the SSR of the regression $$ M_{X_1}y\quad\text{on}\quad M_{X_1}X_2 $$

That is, \begin{eqnarray*} \text{USSR}&=&y'M_{X_1}'M_{M_{X_1}X_2}M_{X_1}y\\ &=&y'M_{X_1}'(I-P_{M_{X_1}X_2})M_{X_1}y\\ &=&y'M_{X_1}y-y'M_{X_1}M_{X_1}X_2((M_{X_1}X_2)'M_{X_1}X_2)^{-1}(M_{X_1}X_2)'M_{X_1}y\\ &=&y'M_{X_1}y-y'M_{X_1}X_2(X_2'M_{X_1}X_2)^{-1}X_2'M_{X_1}y \end{eqnarray*}

Thus,

\begin{eqnarray*} \text{RSSR}-\text{USSR}&=&y'M_{X_1}y-(y'M_{X_1}y-y'M_{X_1}X_2(X_2'M_{X_1}X_2)^{-1}X_2'M_{X_1}y)\\ &=&y'M_{X_1}X_2(X_2'M_{X_1}X_2)^{-1}X_2'M_{X_1}y \end{eqnarray*}

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    $\begingroup$ Thanks. I don't know if it's considered hand holding at this point but how do you go from your sum of squared betas to an expression that contains sum of squares? $\endgroup$ Jan 27, 2017 at 17:17
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    $\begingroup$ @user1627466, I added a derivation of the equivalence of the two formulae. $\endgroup$ Jan 28, 2017 at 10:51
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@ChristophHanck has provided a very comprehensive answer, here I will add a sketch of proof on the special case OP mentioned. Hopefully it's also easier to follow for beginners.

A random variable $Y\sim F_{d_1,d_2}$ if $$Y=\frac{X_1/d_1}{X_2/d_2},$$ where $X_1\sim\chi^2_{d_1}$ and $X_2\sim\chi^2_{d_2}$ are independent. Thus, to show that the $F$-statistic has $F$-distribution, we may as well show that $c\text{ESS}\sim\chi^2_{p-1}$ and $c\text{RSS}\sim\chi^2_{n-p}$ for some constant $c$, and that they are independent.

In OLS model we write $$y=X\beta+\varepsilon,$$ where $X$ is a $n\times p$ matrix, and ideally $\varepsilon\sim N_n(\mathbf{0}, \sigma^2I)$. For convenience we introduce the hat matrix $H=X(X^TX)^{-1}X^{T}$ (note $\hat{y}=Hy$), and the residual maker $M=I-H$. Important properties of $H$ and $M$ are that they are both symmetric and idempotent. In addition, we have $\operatorname{tr}(H)=p$ and $HX=X$, these will come in handy later.

Let us denote the matrix of all ones as $J$, the sum of squares can then be expressed with quadratic forms: $$\text{TSS}=y^T\left(I-\frac{1}{n}J\right)y,\quad\text{RSS}=y^TMy,\quad\text{ESS}=y^T\left(H-\frac{1}{n}J\right)y.$$ Note that $M+(H-J/n)+J/n=I$. One can verify that $J/n$ is idempotent and $\operatorname{rank}(M)+\operatorname{rank}(H-J/n)+\operatorname{rank}(J/n)=n$. It follows from this then that $H-J/n$ is also idempotent and $M(H-J/n)=0$.

We can now set out to show that $F$-statistic has $F$-distribution (search Cochran's theorem for more). Here we need two facts:

  1. Let $x\sim N_n(\mu,\Sigma)$. Suppose $A$ is symmetric with rank $r$ and $A\Sigma$ is idempotent, then $x^TAx\sim\chi^2_r(\mu^TA\mu/2)$, i.e. non-central $\chi^2$ with d.f. $r$ and non-centrality $\mu^TA\mu/2$. This is a special case of Baldessari's result, a proof can also be found here.
  2. Let $x\sim N_n(\mu,\Sigma)$. If $A\Sigma B=0$, then $x^TAx$ and $x^TBx$ are independent. This is known as Craig's theorem.

Since $y\sim N_n(X\beta,\sigma^2I)$, we have $$\frac{\text{ESS}}{\sigma^2}=\left(\frac{y}{\sigma}\right)^T\left(H-\frac{1}{n}J\right)\frac{y}{\sigma}\sim\chi^2_{p-1}\left((X\beta)^T\left(H-\frac{J}{n}\right)X\beta\right).$$ However, under null hypothesis $\beta=\mathbf{0}$, so really $\text{ESS}/\sigma^2\sim\chi^2_{p-1}$. On the other hand, note that $y^TMy=\varepsilon^TM\varepsilon$ since $HX=X$. Therefore $\text{RSS}/\sigma^2\sim\chi^2_{n-p}$. Since $M(H-J/n)=0$, $\text{ESS}/\sigma^2$ and $\text{RSS}/\sigma^2$ are also independent. It immediately follows then $$F = \frac{(\text{TSS}-\text{RSS})/(p-1)}{\text{RSS}/(n-p)}=\frac{\dfrac{\text{ESS}}{\sigma^2}/(p-1)}{\dfrac{\text{RSS}}{\sigma^2}/(n-p)}\sim F_{p-1,n-p}.$$

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  • $\begingroup$ shouldn't the null hypothesis be $\beta = [1 \space\space \mathbf{0} ]^T$ ? $\endgroup$
    – abhishek
    Sep 12, 2023 at 14:10

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