# Examples of Matrix Algebra

### The average

- Typical formulation of the average: \(\frac{1}{N} \sum_{i=1}^N Y_i = \bar{Y}\)
- \(Y_{i}\) as the sum of all values, divided by the number of values \(N\)

- Can be represented in matrix form as: \[\frac{1}{N}\left(\begin{array}{c}1\\1\\\vdots\\1\end{array}\right)^{T}\left(\begin{array}{c}Y_{1}\\Y_{2}\\\vdots\\Y_{N}\end{array}\right) = \frac{1}{N}A^{T}Y\]
- \(Y\) is a one column matrix of values from which the average will be calculated
- \(A\) is a one column matrix of equal length to Y, composed entirely of ones
- Multiplying the transpose of \(A\) by \(Y\) will result in a single dot product and when \(A\) is composed entirely of ones it will simply result in the sum of all values in \(Y\) (each value in \(Y\) is multiplied by a corresponding 1 and then added to the total)
- The transpose of \(A\) allows the for the proper dimensions in order to matrix multiply \(A\) and \(Y\)

- The resulting sum (dot product) is then divided by \(N\) to obtain the average

### Variance

- Variance is defined as \(\mbox{var}(Y)=\frac{1}{N} \sum_{i=1}^N (Y_i - \bar{Y})^2\)
- \((Y_i - \bar{Y})^2\) is the squared difference between each value in \(Y\) and the mean of \(Y\)
- Each squared difference is added to a sum then divided by \(N\) to obtain the variance

- In matrix form: \[\frac{1}{N}\left(\begin{array}{c}Y_{1}-\bar{Y}\\\vdots\\Y_{N}-\bar{Y}\end{array}\right)^{T}\left(\begin{array}{c}Y_{1}-\bar{Y}\\\vdots\\Y_{N}-\bar{Y}\end{array}\right) = \frac{1}{N}R^{T}R\]
- \(R\) is a single column matrix of the difference between each value in \(Y\) and the mean of \(Y\) (aka the
**residuals**) - The dot product of \(R\) and itself results in the square of each difference, which is then summed by the dot product +One instance of \(R\) is transposed to allow for the proper dimension for matrix multiplication
- The resulting sum (dot product) is then divided by \(N\) to obtain the variance
- Recall that multiplying the transpose of one matrix by another is the
**cross product**and thus**variance is the cross product of the residuals divided by \(N\)**

- \(R\) is a single column matrix of the difference between each value in \(Y\) and the mean of \(Y\) (aka the

### Representation of linear Models

- The general forumula for a linear model is \(Y_i = \beta_0 + \beta_1 x_{1i} + ... + \beta_j x_{ji}+ \varepsilon\)
- \(Y_i\) is each actual data point to be fit in the model
- \(\beta_0\) is the a constant, base value from which all data is relative to (graphically it is the intercept)
- \(x_{ji}\) is each variable to be included in the model
- Each \(i\) corresponds to one value in data \(Y\), \(1,...,N\)
- Each \(j\) corresponds to on variable in the model, the number of variables depending on the models design, \(1,..,M\)

- \(\beta_j\) is the co-variate or coefficient modifying each variable to result in the fit of the model
- \(\varepsilon\) is the residual difference between the model and each individual data point

- In matrix notation: \[\left(\begin{array}{c}Y_1\\Y_2\\\vdots\\Y_N\end{array}\right) = \left(\begin{array}{cccc}1 & x_{1,1} & x_{1,2} & x_{1,...} & x_{1,M}\\1 & x_{2,1} & x_{2,2} & x_{2,...} & x_{2,M}\\\vdots & \vdots & \vdots & \vdots & \vdots\\1 & x_{N,1} & x_{N,2} & x_{N,...} & x_{N,M}\end{array}\right)\left(\begin{array}{c}\beta_0\\\beta_1\\\beta_2\\\vdots\\\beta_M\end{array}\right) + \left(\begin{array}{c}\varepsilon_1\\\varepsilon_2\\\vdots\\\varepsilon_N\end{array}\right)\]
- \(Y_1,..,Y_N\) represents each point of data
- The first column of 1s in the \(x\) matrix is there because values in that column will always be matrix multiplied by the value of \(\beta_0\). Since \(\beta_0\) is a constant, it should not change according to any data.
- Each remaining column in the \(X\) matrix represents a variable of interest in the model from 1 to \(M\)
- Each row in the \(X\) matrix represents the variables applied to each point of data, 1 through \(N\), in \(Y\)
- The one column \(\beta\) matrix is what is to be calculated by the linear model
- When completed, each “row” of the model will completely calculate each point in \(Y\)

- Model can be more simply denoted by: \[Y = X\beta + \varepsilon\]

### Solving linear models by least squares

- The least squares regression is the model that minimizes the square of the residuals (i.e. the difference between an actual value and the expected value based on the model, at the same point)
- The sum of squares can ordinarly be writted as \(\sum_{i=1}^N R_i^2\), where \(R\) is the residual
- In matrix notation this would be represented as \((Y-X\beta)^T(Y-X\beta)\)
- \(X\beta\) represents the dot product of all variables and their co-variates for each data point, thus representing a vector of all expected values
- \(Y\) represents the vector of all actual values
- Therefore \((Y-X\beta)\) is the vector of all residuals
- Finally, recall the crossproduct of a single matrix is the way to obtain the square of that matrix

- The first derivative of an equation allows us to determine minimums (and maximums) based on where the derivative is equal to zero
- The derivative of the matrix-form residual equation equal to zero (to find the minumum) is \(2X^T(Y-X\hat{\beta}) = 0\)
- Formula is similar to the derivative of \(f(x)^2\) being \(2f(x)f\prime(x)\)
- \(\hat{\beta}\) because \(\beta\) is being calculated

- Solving for zero results in \(2X^T(Y-X\hat{\beta})\Rightarrow X^TX\hat{\beta} = X^TY \Rightarrow \hat{\beta} = (X^TX)^{-1}X^TY\)
- Remember that dividing by a matrix requires instead multiplying by that matrix’s inverse (see last step)
- Notice that the final equation contains a two cross products

- The derivative of the matrix-form residual equation equal to zero (to find the minumum) is \(2X^T(Y-X\hat{\beta}) = 0\)
- Thus, the equation to solve a linear model is: \[\hat{\beta} = (X^TX)^{-1}X^TY\]

# Linear model R example

- Finding the regression line between father vs. son height

```
library(UsingR)
x=father.son$fheight
y=father.son$sheight
X <- cbind(1,x)
beta <- solve(t(X)%*%X)%*%t(X)%*%y
###or, because of the cross products
beta <- solve(crossprod(X))%*%crossprod(X,y)
beta
```

```
## [,1]
## 33.886604
## x 0.514093
```

- Thus, the least square line is \(f(x) = 33.89 + 0.51x\), which looks like:
- In R, the
`lm`

function is used to automatically calculate such models

```
fit <- lm(y ~ x)
summary(fit)
```

```
##
## Call:
## lm(formula = y ~ x)
##
## Residuals:
## Min 1Q Median 3Q Max
## -8.8772 -1.5144 -0.0079 1.6285 8.9685
##
## Coefficients:
## Estimate Std. Error t value Pr(>|t|)
## (Intercept) 33.88660 1.83235 18.49 <2e-16 ***
## x 0.51409 0.02705 19.01 <2e-16 ***
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
## Residual standard error: 2.437 on 1076 degrees of freedom
## Multiple R-squared: 0.2513, Adjusted R-squared: 0.2506
## F-statistic: 361.2 on 1 and 1076 DF, p-value: < 2.2e-16
```

*Notice that the `Estimate`

for the `(Intercept)`

and `x`

are quivalent to the `beta`

vector calculated manually