Given the increase in value of a company stock with time t f

Given the increase in value of a company stock with time t follows the form $ = at/b + t Find the best fit values of a and b using a \"linearized\" least squares for the following data

Solution

Answer.

This equation is a non-linear regression, so using some math artifices we can find a linear equation that best fits.

Linear equation is:

y = a + bx

--

And we have:

y = ax/(b+x)

log y = log a - b log (x)

inverting equation we have:

(1/y) = (1/a) + (b/a)(1/x)

Thus, a plot of 1/$ against 1/t will be linear with slope b/a and an intersection with the ordinate axis 1/a.

--

$

t (yr)

1/$

1/t(yr)

(1/$)*(1/t)

(1/t)²

18

2

0,05555556

0,5

0,02777778

23

4

0,04347826

0,25

0,01086957

0,0625

26

6

0,03846154

0,16666667

0,00641026

0,02777778

28

8

0,03571429

0,125

0,00446429

0,015625

=

0,17320964

1,04166667

0,04952189

0,35590278

a = [((1/t)² x 1/$) - (t x $)]/[n(1/t)² - ((1/t))²] = (0.35590278)(0.17320964)-(0.04952189)(0.17320964)/4(0.35590278)-(0.04952189)²

a = 0.4134599769

--

b = [n(1/$*1/t) - (1/t)x(1/$)]/[n(1/t)²-((1/t))²] = 4(0.04952189)-(1.04166667)(0.17320964)/4(0.35590278)-(1.04166667)²

b = 0.0521674367

$	t (yr)	1/$	1/t(yr)	(1/$)*(1/t)	(1/t)2
18	2	0,05555556	0,5	0,02777778	0,25
23	4	0,04347826	0,25	0,01086957	0,0625
26	6	0,03846154	0,16666667	0,00641026	0,02777778
28	8	0,03571429	0,125	0,00446429	0,015625
	=	0,17320964	1,04166667	0,04952189	0,35590278