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).

  • $\begingroup$ Where you say "score of player $i-j$" (which seems to idendity one player with the number that is the difference between the numbers $i$ and $j$), might you have meant "score of player $i$ minus score of player $j$"? $\qquad$ $\endgroup$ – Michael Hardy Jan 24 '16 at 19:59
  • $\begingroup$ Should this be understood as meaning that if two players compete then the data available to you is only the difference between the numbers of points they scored? E.g. if player $1$ gets $13$ points and player $2$ gets $17$, then we learn only that player $2$ got $4$ points more than player $1$? $\qquad$ $\endgroup$ – Michael Hardy Jan 24 '16 at 20:09
  • $\begingroup$ I've seen this question before as a homework problem. Please add the self-study tag. $\endgroup$ – StatsStudent Jan 24 '16 at 20:33
  • $\begingroup$ Note that $y_{12}$ is properly coded as y_{12}, and $\varepsilon\sim N(0,\sigma^2 I)$ as \varepsilon\sim N(0,\sigma^2 I). I edited accordingly. $\qquad$ $\endgroup$ – Michael Hardy Jan 24 '16 at 20:41

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.

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  • $\begingroup$ I saw that mistake myself and just updated the matrix, thanks! I'm still overlooking how to know if $\beta_1 - \beta_3$ for example, as a question, is estimable. I'm practicing beyond my homework here, trying to memorize ways to figure out estimability. I read that if a given linear combination of the elements of beta can be written as a linear combination of the elements of E(y), then it's estimable. $\endgroup$ – Jennifer Cooke Jan 24 '16 at 21:10
  • $\begingroup$ OK. Great work! What is the definition of "estimable?" You should start with the definition and see how you might be able to determine the estimability of $\beta_1-\beta_3$ from the definition. Try this for a bit and post back if you have trouble and I'll assist. $\endgroup$ – StatsStudent Jan 24 '16 at 21:26
  • $\begingroup$ If I try writing $\beta_1 - \beta_3$ I get stuck because my brain is switching over to the $\mu$ vector, and there's 4 $\mu$ and 5 $\beta$ and I can't figure out a way to write an expected value statement that makes sense, like $E(y) = $? what linear combination of elements can I write of the $\mu$'s would work, since the beta and the mu don't match up? I'd start by writing $E(y) = \mu_1 - \mu_3$ but then I don't understand how to write a linear combination so I stop here. $\endgroup$ – Jennifer Cooke Jan 24 '16 at 21:53
  • $\begingroup$ I think you are making this too hard by writing things in terms of $\mu$. Don't do that. Instead, you simply need to answer the question, is there a matrix $C$ such that $CXB$=$\beta_1-\beta_3$? If you can provide the values for C to get $\beta_1-\beta_3$, you have shown that $\beta_1-\beta_3$ is estimable. $\endgroup$ – StatsStudent Jan 24 '16 at 23:09
  • $\begingroup$ But I'm not understanding $\beta_1 - \beta_3.$ What is this, do I take the values of these columns of the X matrix and subtract? I don't think this is what it means, and I'm trying to both ask and answer my own questions. Sorry if I sound a little frustrated. I'm at the end of my own understanding. $\endgroup$ – Jennifer Cooke Jan 25 '16 at 0:47

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