Value-at-Risk

Chapter 2, Page 83
Exercise 14

Use ordinary least squares to fit a linear polynomial

[2.110]

to the five points indicated in Exhibit 2.13.


k

1	1	13
2	2	10
3	3	11
4	4	8
5	5	6

Exhibit 2.13 Point set for Exercise 2.14.

Solution

Let

[s1]

We have:

[s2]

and:

[s3]

Applying [2.103], we obtain:

[s4]

and our linear polynomial is:

[s5]

This and the five points are graphed below.