Problem 1

Consider the the following artificial data set and its summary of the lm command:

set.seed(111)
x1=round(sort(runif(100)), 2)
x2=round(sort(runif(100)), 2)
x3=round(sort(runif(100)), 2)
y=round(1+2*x1+3*x2+4*x3+rnorm(100),1)
summary(lm(y~x1+x2+x3))
## 
## Call:
## lm(formula = y ~ x1 + x2 + x3)
## 
## Residuals:
##      Min       1Q   Median       3Q      Max 
## -2.35695 -0.51916 -0.05192  0.70102  2.33240 
## 
## Coefficients:
##             Estimate Std. Error t value Pr(>|t|)
## (Intercept)   0.6556     0.1810   3.622  0.00047
## x1            1.8731     2.1465   0.873  0.38505
## x2           -4.8388     4.0227  -1.203  0.23199
## x3           12.4945     4.3210   2.892  0.00474
## 
## Residual standard error: 0.8695 on 96 degrees of freedom
## Multiple R-squared:  0.9148, Adjusted R-squared:  0.9121 
## F-statistic: 343.6 on 3 and 96 DF,  p-value: < 2.2e-16

Use R but not the lm command to find the following numbers from the printout:

##  [1] -4.8388 -2.3570 -0.5192 -0.0519  0.1810  0.6556  0.7010  0.8695  1.8731
## [10]  2.1464  2.3324  4.0226  4.3208 12.4945

Problem 2

Say we just found \(\pmb{\hat{\beta}}\) with k predictor variables. Now we would like to use another predictor, together with those already used. If k is large this means we have to invert a \(k+1\times k+1\) matrix. But we just inverted a very similar \(k\times k\) matrix. Show how we can make use of that.

Problem 3

Consider the following data set:

set.seed(1111)
x1=1:10
x2=sort(round(runif(10,0,10), 1))
y=round(2+3*x1+4*x2+rnorm(10),1)
cbind(y,x1,x2)
##          y x1  x2
##  [1,]  6.9  1 1.2
##  [2,] 14.3  2 1.4
##  [3,] 17.7  3 1.4
##  [4,] 31.8  4 4.1
##  [5,] 37.1  5 4.7
##  [6,] 41.0  6 5.5
##  [7,] 53.6  7 7.4
##  [8,] 59.7  8 8.8
##  [9,] 63.6  9 9.1
## [10,] 71.6 10 9.8

Use the Gram-Schmitt method to find a transformation \(\pmb{U=AX}\) such that \(\pmb{U}\) is orthogonal. Use this to verify theorem (6.3.4).