I need to define a design matrix $X$ so that it fits this scenario:
For $1\leq i< j \leq 5$, where $i$ are competitors and $j$ are different competitors. Their score is $y_{ij} =$ score of player $i$ minus score of player $j$.
$$Y_{ij} =\beta_i – \beta_j+\varepsilon_{ij}.$$
$\beta_i,\ldots,\beta_5$ are unknown parameters. $\varepsilon \sim N(0,\sigma^2I)$.
The $X$ matrix has to represent only 4 matches: $y_{12}$, $y_{34}$, $y_{25}$, $y_{15}$. and then be usable in general linear form $y=X\beta+\varepsilon$.
A couple challenges:
(a) I've never done this before. I am thinking I should use zeros, ones and negative ones. Am I right in putting ones and negative ones in the two parameter columns for a match, and zeros in the parameter columns that aren't playing in the match?
(b) How do I decide if a couple comparisons can be made, $\beta_1 – \beta_2$ and $\beta_1 – \beta_3$?
$X =$ in the row order $y_{12}$, $y_{34}$, $y_{25}$, $y_{15}$
$$
\begin{bmatrix}
1 & -1 & 0 & 0 & 0 \\
0 & 0 & 1 & -1 & 0 \\
0 & 1 & 0 & 0 & -1 \\
1 & 0 & 0 & 0 & -1 \\
\end{bmatrix}
$$
I further have $E(y)$ =
$$
\begin{bmatrix}
\mu_1 \\
\mu_2 \\
\mu_3 \\
\mu_4 \\
\end{bmatrix}
$$
I know of 2 ways to see if it's estimable. The first is by seeing if a matrix $A$ exists such that $Ae(Y) = AX\beta.$ The other way is the way I need to practice, so where I look to you: By seeing if there is a given linear combination of the elements of beta that can be written as a linear combination of the elements of E(y).
Best Answer
The design matrix you have set up is not quite right. Since this is a homework problem, I'll give you hints and help you through the problem. The key to setting up your design matrix involves, understanding what Y represents. In this case, is represents, the difference between player $i$'s score and player $j$'s score. So, to get started set up the matrices (in doing so, I'm ignoring the error terms $\epsilon_{ij}$, but remember those are important!) accordingly:
$\begin{bmatrix} y_{12}\\ y_{34}\\ y_{25}\\ y_{15}\\ \end{bmatrix}=X\begin{bmatrix} \beta_1\\ \beta_2\\ \beta_3\\ \beta_4\\ \end{bmatrix}=\begin{bmatrix} \beta_1-\beta_2\\ \beta_3-\beta_4\\ \beta_2-\beta_5\\ \beta_1-\beta_5\\ \end{bmatrix}$
The $X$ matrix is a 4 $\times$5 matrix, where each column is represented by a $\beta$. So you need to ask yourself, what should the elements of $X$ be so that when you multiply $X\beta$ you get the differences of $\beta$s shown on the right hand sight of the matrix equation above. The elements should consists of only 1, -1, and 0. Once you do this correction, it should be fairly straight forward to answer your remaining questions.