8
$\begingroup$

Wikipedia gives the following expression in "degrees of freedom" section, calling it the "Satterthwaite" approximation:

\begin{equation*} \text{df}\approx \frac{\operatorname{Tr}(H'H)^2}{\operatorname{Tr}(H'HH'H)} \tag{1}\label{1} \end{equation*}

where $H=X(X'X)^{-1}X'$ and $X$ is the data matrix.

How/where is this formula obtained? It's not in the references they cite.


A slightly more general definition of $H$ which may have originated this formula, is the non-parametric extension -- matrix which maps labels to predictions of the fitted model. AKA "the smoother matrix"

$$\hat{y}=Hy$$

Edit, Sep 27

Possibly related: let $x$ be a multivariate Gaussian random variable and we model $\|x\|^2$ as Chi-Squared with $k$ degrees of freedom. What's $k$? Satterthwaite approximation says to use $k=\text{df}$ from Eq \eqref{1} where $H$ consists of many samples of $x$ stacked as rows.

Excerpt from Encyclopedia of Statistical Sciences, Vol 3:

$\endgroup$
11
  • $\begingroup$ Is it $\operatorname{tr}\left(H^\mathsf T H\right) ^2$ or $\{\operatorname{tr}\left(H^\mathsf T H\right)\} ^2 ?$ $\endgroup$ Commented Sep 27, 2022 at 22:23
  • $\begingroup$ the formula is taken verbatim from wikipedia, it's safe to assume that it's equivalent to the latter, ie $(\text{trace(...)})^2$ $\endgroup$ Commented Sep 27, 2022 at 22:34
  • 3
    $\begingroup$ With the definition of $H$ given in the question $H=H^2=H'$ so the ratio in the question equals $(tr(H))^2/tr(H)=tr(H)=rank(H)=rank(X)$ $\endgroup$ Commented Sep 27, 2022 at 22:53
  • 1
    $\begingroup$ Interesting....I suspect the formula is meant for a general smoother matrix $S$, of which hat matrix $H$ is a special case $\endgroup$ Commented Sep 27, 2022 at 23:11
  • 1
    $\begingroup$ @User1865345 I've updated post with explanation of Satterthwaite approximation I found, which seems to lead to the formula in question for a different problem (where $H$ is the data matrix). Still unclear why this would come up when $H$ is the smoother matrix. $\endgroup$ Commented Sep 28, 2022 at 5:16

1 Answer 1

4
$\begingroup$

There is a Wikipedia article that is more directly about the Welch-Satterthwaite approximation.

The Wikipedia article makes the following citations to original sources

Satterthwaite, F. E. (1946), "An Approximate Distribution of Estimates of Variance Components.", I Biometrics Bulletin, 2 (6): 110–114

Welch, B. L. (1947), "The generalization of "student's" problem when several different population variances are involved.", Biometrika, 34 (1/2): 28–35


The principle behind it is to approximate a sum of squares (as used in several hypothesis tests or estimates of variance), when it is distributed as a linear sum of chi-squared distributions, by a single chi squared distribution.

The approximation applies the methods of moments. For a chi-squared distribution we have that the degrees of freedom is expressed by $k = \frac{\text{Mean}(\chi)^2}{\text{Var}(\chi)}$, and the method of moments uses sample estimates to for the mean and variance.

Let $y_i=\bar{y_i}+\epsilon_i$ be observations of true labels labels $\bar{y_i}$ corrupted with IID zero-centered noise $\epsilon_i$. For a given linear estimator

$\hat{y} = H y$

The sum of squares of the model and the residuals are

$$\begin{array}{rcl} SS_{model} &=& \Vert H (y-\bar{y}) \Vert^2 \\ SS_{residuals} &=& \Vert y- H y \Vert^2 = \Vert (I-H) (y-\bar{y}) \Vert^2 \\ \end{array}$$

If distribution of $y-\bar{y}$ has identity covariance, we have the following expressions for covariance of $\hat{y}$ and $y-\hat{y}$

$$\Sigma_{\hat{y}} = HH^T = H^2$$

$$\Sigma_{y-\hat{y}} = (I-H)(I-H)^T = I - 2H + H^2$$

the distribution of the sum of squares of a multivariate normal distribution can be seen as a sum of the independent principle components with variance equal to the eigenvalues. The sum of these eigenvalues is alse the trace of the covariance matrix.

The mean and variance will be

$$\begin{array}{rcl}\text{mean}(SS_{model}) &= &tr(HH^T) \\ \text{var}(SS_{model})& = &tr((HH^T)^T(HH^T)) = tr((HH^T)(HH^T)) \end{array}$$

and for $SS_{residuals}$ you get something similar but I won't wrote it out as it becomes a bit more complex, but that is where the different expressions come from.

When $H$ is a projection matrix (as in OLS) then $H^tH = H$. That is another source for getting different types of expressions.

$\endgroup$
4
  • 1
    $\begingroup$ So connection to "degrees of freedom of chi-squared" distribution makes sense. I'm a bit confused why this formula appears alongside $tr(H'H)$, $tr(2H – H H')$ quantities which are formulas for effective degrees of freedom in non-parametric regression where $H$ is the smoother matrix. Hence wondering if that formula is also valid in this context $\endgroup$ Commented Sep 28, 2022 at 6:01
  • $\begingroup$ Thanks, now I can see where the serendipitous connection to iterative method comes from. From statistical point of view, this gives degrees of freedom of the estimator but from minimization point of view, this gives the drop in variance when applying y=Hy transformation. $\endgroup$ Commented Sep 28, 2022 at 14:38
  • $\begingroup$ Final thoughts, tr(H)^2/tr(H^2) is an approximation to tr(2H-HH'), so your expression for $\Sigma_{y-\hat{y}}$ makes it clear why it's equivalent to "degrees of freedom". It's the drop in mean residual after applying $H$ transformation which is $d$ when $H$ is projection, at most $d$ when $H$ is a contraction $\endgroup$ Commented Sep 28, 2022 at 20:34
  • $\begingroup$ BTW, "Plane Answers to Complex Questions" gives a justification for when we can use a much simpler Tr(H) as the degrees of freedom parameter i.sstatic.net/pvXkl.png $\endgroup$ Commented Oct 12, 2022 at 0:31

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.