The Method of Least Squares Lecture 05

 The Method of Least Squares 

Lecture 05

Introduction

The method of least squares is applicable when there is a strong association between time series observations. demonstrating a trend towards increment or decrement as time increases. This method is suitable for a reliable forecast value.

The principle of least squares is used when the trend is linear or non-linear.

Linear Trend

If time is treated as an independent variable, a mathematical equation in the shape of a straight line, a parabola, or an exponential can be used to calculate the secular trend. Let's assume that the secular trend is linear. Then the equation of the least squares linear trend would be

Y^t = a + b t

The following two normal equations are used to estimate the parameters of the above-stated model.

∑Yt = na + b ∑t

∑t Yt = a ∑t + b ∑t²

If the time variable (t) is coded by taking the time deviation from the middle point of the time period.

then ∑t = 0

a = ∑Yt / n

b = ∑t Yt / ∑t²

When the secular trend of a time series graph appears to be a curve, then a parabola of second or third degree or a curve of some other type, such as the exponential, modified exponential, etc., is to be used.

The second degree of parabola

Y^t = a + b t + c t² 

The following three normal equations are used to estimate the parameters of the above-stated model.

 na + b ∑t + c ∑t² = ∑Yt

 a ∑t + b ∑t² + c ∑t³ = ∑t Yt

a ∑ t² + b ∑t³ +  c ∑t⁴ = ∑t² Yt

The simple criterion is that the curve will be best fit, for which ∑(Yt - Y^t)² is minimum.

Practice Question

Determine the trend line by straight-line equation from the following data.

Year	1991	1992	1993	1994	1995	1996	1997
Sale (00,000,000)	21	24	28	18	20	16	25

 Solution: Assign 0 to 1994

a = ∑Yt / n

a = 152 / 7

a = 21.71

b = ∑t Yt / ∑t²

b = -12 / 28

b = -0.4285

The estimated model

Y^t = a + b t

Y^t = 21.71- 0.4285 t

Practice Question

Fit a second-degree equation to the following data and find the trend line.

Year	1985	1986	1987	1988	1989
Values	125	160	150	135	170

Solution:

Year	Yt	t	t Yt	t²	t² Yt	t³	t⁴
1985	125	-2	-250	4	500	-8	16
1986	160	-1	-160	1	160	-1	1
1987	150	0	0	0	0	0	0
1988	135	+1	135	1	135	+1	1
1989	170	+2	340	4	680	+8	16
	740	0	65	10	1475	0	34

 na + b ∑t + c ∑t² = ∑Yt

 a ∑t + b ∑t² + c ∑t³ = ∑t Yt

a ∑ t² + b ∑t³ +  c ∑t⁴ = ∑t² Yt