2
$\begingroup$

I was reading Functional Data Analysis with R and MATLAB (Ramsay et. al., 2009) and they presented the following (functional) ANOVA model (but I'm ignoring the functional aspect; it doesn't appear necessary in general):

$$X_{nj} = \mu + \sum_{j^{*} = 1}^J \mu_{j^{*}} \mathbb{I}_{\{j^{*} = j\}}(j) + \epsilon_{nj}$$

(where $\mathbb{I}$ is the indicator function and there are $J$ populations). We add the constraint $\sum_{j = 1}^J \mu_j = 0$. According to Ramsay et. al., this model can be estimated if we add a fictitious 0 observation to our sample. Then they create a design matrix with the following properties:

  1. One column consists of 1's except for the final entry corresponding to the zero entry, which is zero (I interpret this as the "intercept" column).
  2. For the remaining $J$ columns of the design matrix, the $j^{\text{th}}$ column's row is 1 if the observation is a member of the $j^{\text{th}}$ population; otherwise, it's zero. However, for the zero observation, each of these columns' entries are 1. (I interpret these as population membership columns, and apparently the zero observation is a member of all populations.)

Why does this work? I tried examining the design matrix but the linear algebra is rough and trying to just do the linear algebra (finding inverses, matrix multiplication, etc.) does not feel enlightening.

Specifically, I would like to extend this application to, say, a linear regression model with a trend. I would like to estimate the model:

$$X_{tj} = \beta_{0 \bullet} + \beta_{1 \bullet} (t - 1) + \sum_{j^{*}=1}^J \left(\beta_{0 j^{*}} + \beta_{1 j^{*}}(t-1)\right)\mathbb{I}_{\{j^{*} = j\}}(j) + \epsilon_{tj}.$$

($t$ should be interpreted to be time since start.) Add the constraints $\sum_{j =1}^J \beta_{0j} = 0$ and $\sum_{j=1}^J \beta_{1j} = 0$. How would this trick of adding a zero observation extend to this case?

I would absolutely love any references on this topic you can recommend. Thanks.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.