6
$\begingroup$

In grad school, I was always taught the general linear model $$\mathbf{y} = \mathbf{X}\boldsymbol\beta + \boldsymbol\epsilon\tag{1}$$ where $\mathbf{y}$ is a vector, $\mathbf{X}$ is some matrix, $\boldsymbol\beta$ is a parameter vector, and $\boldsymbol\epsilon$ is a vector of error terms satisfying $\mathbb{E}[\boldsymbol\epsilon] = \mathbf0$.

However, in Izenman's Modern Multivariate Statistical Techniques, the following model is used instead to explain multivariate linear regression, as well as the reduced-rank regression model: $$\mathbf{y} = \boldsymbol\mu + \mathbf{C}\mathbf{x}+\boldsymbol\epsilon\tag{2}$$ where $\mathbf{y}, \boldsymbol\mu, \boldsymbol\epsilon \in \mathbb{R}^s$, $\mathbf{C}$ is a $s \times r$ matrix, and $\mathbf{x} \in \mathbb{R}^r$.

At first, this seems confusing - but I think I understand now why Izenman does this: it's so that the covariance matrix of $\begin{bmatrix}\mathbf{x} \\ \mathbf{y}\end{bmatrix}$ is defined. Note that in this case, $\mathbf{C}$ is taken to be a matrix of parameters and $\mathbf{x}$ is the design matrix (vector?).

Is there a way to rewrite $(2)$ in the form of $(1)$? I imagine $(2)$ could be reformulated to look like $(1)$ somehow, but I can't get it to work.

$\endgroup$
3
  • 3
    $\begingroup$ It's more likely because Izenman's definition of $x$ does not include a constant term, so he needs to subtract the mean from $y$ in order to have the mean of $\epsilon = 0$. The way to change it into the first form is to add a column of $1$s to $x$ and expand the coefficient matrix to include one more term.. $\endgroup$
    – jbowman
    Oct 18, 2017 at 13:44
  • $\begingroup$ @jbowman Yes, but it's not only this: see my answer. $\endgroup$
    – amoeba
    Dec 12, 2017 at 10:10
  • $\begingroup$ In Inzemann (1975) the matrices $X$ and $Y$ are turned sideways. GLM would have $N \times r$, one row per case, but the original RRR paper models Y as $r \times N$. Same with X. So rather than $X\beta$ in RRR $CX$. For me, that's confusing. It all works out, but notation is difficult comparing documents. $\endgroup$
    – pauljohn32
    Feb 13, 2019 at 10:56

1 Answer 1

4
+50
$\begingroup$

In usual multiple regression the response variable $y$ is 1-dimensional so for each sample we can write an equation $$y = \boldsymbol \beta^\top \mathbf x + \epsilon,\tag{1a}$$ where $\mathbf x$ is an $r$-dimensional vector of predictors and $\boldsymbol\beta$ is the vector of regression coefficients. If the sample size is $n$, we can combine $n$ such equations into one, stacking all $y_i$ into one vector $\mathbf y$ and all $\mathbf x_i$, as rows, into one data matrix $\mathbf X$. This yields the form that you gave as Equation 1: $$\mathbf y = \mathbf X\boldsymbol\beta + \boldsymbol\epsilon.\tag{1b}$$

Your Equation 2 describes a single sample of multivariate regression where response variable $\mathbf y$ is a vector. Using the similar notation as above, we could write for each sample $$\mathbf y = \mathbf B^\top\mathbf x + \boldsymbol\epsilon,\tag{2a}$$ with the only difference to your Equation 2 being that here the intercept is included in $\mathbf x$, i.e. the first element of $\mathbf x$ is always equal to $1$, and so $\mathbf B = [\boldsymbol \mu, \mathbf C^\top]$ stacked together. Here we can also combine $n$ such equations together by using data matrices: $$\mathbf Y = \mathbf {XB} + \mathbf E.\tag{2b}$$

The key point and probably the source of confusion for you was that $\mathbf y$ in Equations (1b) and (2a) above are very different things! In (1b) it denotes an $n$-dimensional data vector comprising $n$ one-dimensional sample points while in (2a) it denotes an $s$-dimensional response vector which is one single $s$-dimensional sample point.

$\endgroup$
1
  • 1
    $\begingroup$ I should've figured this out from what I wrote in the question! Multivariate as opposed to multiple. $\endgroup$ Dec 12, 2017 at 14:21

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.