The Method of Least Squares: NonLinear Trend

Secular Trend — Nonlinear

When a straight line does not describe accurately the long-term movement of a time series, then one might detect some curvature and decide to fit a curve instead of a straight line. The most commonly used curve, to describe the nonlinear secular trend in a time series, are:

1. Exponential curve, and
2. Second-degree parabola

1) Exponential (Nonlinear) Curve:

The exponential curve describes the trend (nonlinear) in a time series that changes by a constant percentage rate. The equation of the curve is $\hat{y} = ab^x$

Taking logarithm, we get the linear form $log\, \hat{y}=log\, a + (log\,b)x$

The method of least squares give the normal equations as:

\begin{align*}
\sum log\, y & = n\, log\, a + log\, b \sum x\\
\sum log\, y & = n\, log\, a \sum x + log\, b \sum x^2
\end{align*}

However, if $\sum x=0$ the normal equations becomes

\begin{align*}
\sum log\,y & = n\, log a\\
\sum x log\, y &= log\, b \sum x^2
\end{align*}

The values of $log\, a$ and $log\, b$ are

\begin{align*}
log\, a &=\frac{\sum log\, y}{n}\\
log\, b&= \frac{\sum x log\, y}{\sum x^2}
\end{align*}

Taking $antilog$ of of $log\, a$ and $log\, b$, we get the values of $a$ and $b$.

Question: The population of a country for the years 1911 to 1971 in ten yearly intervals in millions is 5.38, 7.22, 9.64, 12.70, 17.80, 24.02, and 31.34. (i) Fit a curve of the type $\hat{y}=ab^x$ to this time series and find the trend values, (ii) Forecast the population for the year 1991.

solution

(i) We have $\overline{t}=\frac{(1991+1971)}{2}=1941$. Let $x=\frac{t-\overline{t}}{10}=\frac{5-1941}{10}$ so that coded year number $x$ is measured in a unit of 10 years.

Year $t$	Population $y$	Coded year $x=\frac{x-1941}{10}$	$log y$	$x log\, y$	$x^2$	$\hat{y}=13.029(1.345)^x$
1911	5.38	-3	0.73078	-2.19234	9	5.355
1921	7.22	-2	0.85854	-1.71708	4	7.202
1931	9.64	-1	0.98408	-0.98498	1	9.687
1941	12.70	0	1.10380	0	0	13.029
1951	17.808	1	1.25042	1.25042	1	17.524
1961	24.02	2	1.38057	2.76114	4	23.570
1971	31.34	3	1.49610	4.48830	9	31.701

The least squares exponential curve is $\hat{y} = ab^x$

Taking logarithm, $log\, \hat{y} = log a + (log\, b)x$

since $\sum x=0$, therefore

\begin{align*}
log\, a &= \frac{\sum log\, y}{n} = \frac{7.80429}{7}=1.1149\\
log\, b &= \frac{\sum x log\, y}{\sum x^2} = \frac{3.60636}{28}=0.12880\\
a &= antilog(1.1149)=13.029\\
b &= antilog(0.1288)=1.345\\
\hat{y} &=13.029 (1.345)^x,\quad \text{with origin at 1941 and unit of $x$ as 10 years}
\end{align*}

(ii) For $t=1941$ we have $x=\frac{t-1941}{10}= \frac{1991-1994}{10}=5$. Putting $x=5$, in the least squares exponentail curve, we have
$\hat{y} = 13.029 (1.345)^5 = 57.348$ millions

2) Second Degree Parabola (Nonlinear)

It describes the trend (nonlinear) in a time series where a change in the amount of change is constant per unit time. The quadratic (parabolic) trend can be described by equation

\begin{align*}
\hat{y} = a + bx + cx^2
\end{align*}

The method of least squares gives the normal equations as

\begin{align*}
\sum y &= na + b\sum x + c \sum x^2\\
\sum xy &= a\sum x + b\sum x^2 + c \sum x^3\\
\sum x^2y &= a \sum x^2 + b\sum x^3 + c\sum x^4
\end{align*}

However if $\sum x = 0 \sum x^3$ then the normal equation reduces to

\begin{align*}
\sum y &= na + c\sum x^2\\
\sum xy &= b\sum x^2\\
\sum x^2 y &= a \sum x^2 + c \sum x^4\\
& \text{the values of $a$, $b$, and $c$ can be found as}\\
c &= \frac{n \sum x^2 y – (\sum x^2)(\sum y)}{n \sum x^2 -(\sum x^2)^2}\\
a&=\frac{\sum y – c\sum x^2}{n}\\
b&= \frac{\sum xy}{\sum x^2}
\end{align*}

Question: Given the following time series

Year	1931	1933	1935	1937	1939	1941	1943	1945
Price Index	96	87	91	102	108	139	307	289

1. Fit a second degree parabola taking the origin at 1938.
2. Find the trend values
3. What would have been the equation of parabola if origin were at 1933

Solution

(i)

Year $t$	Price index $y$	Coded Year $x=t-1938$	$x^2$	$x^4$	$xy$	$x^2y$	Trend values $y=110.2+15.48x+2.01 x^2$
1931	96	-7	49	2401	-672	4704	100.33
1933	87	-5	25	625	-435	2175	83.05
1935	91	-3	9	81	-273	819	81.85
1937	102	-1	1	1	-102	102	96.73
1939	108	1	1	1	108	108	127.69
1941	139	3	9	81	417	1251	174.73
1943	307	5	25	625	1535	7675	237.85
1945	289	7	49	2401	2023	14161	317.05
Total	1219	0	168	6216	2601	30995

(ii) Different trend values are already computed in the above table.

\begin{align*}
\hat{y} &= a + b x + c x^2\\
c &= \frac{n\sum x^2 y-(\sum x^2)(\sum y)}{n \sum x^4 -(\sum x^2)^2} =\frac{8(30995)-(168)(1219)}{8(6126)-(168)^2}=2.01\\
a &= \frac{\sum y – a \sum x^2}{n}=\frac{1219-(2.01)(168)}{8}=119.2\\
b &= \frac{\sum xy}{\sum x^2}=\frac{2601}{168} = 15.48\\
\hat{y} &= 110.2 + 15.48x + 2.01^2,\quad \text{with origin at the year 1938}
\end{align*}

For different values of $x$, the trend values are obtained in table.

For shifting the origin at 1933, replace $x$ by $(x-5)$

\begin{align*}
\hat{y} &= 110.2 + 15.48(x-5)+2.01(x-5)^2\\
&= 110.2 + 15.48(x-5)+2.01(x^2 -10x + 25)\\
&= 110.2 + 15.48x -77.4 + 2.01x^2 – 20.1x + 50.25\\
&= 83.05 -4.62x + 2.01x^2, \quad \text{with origin at the year 1933}
\end{align*}

Merits of Least Squares:

• The method of least squares gives the most satisfactory measurement of the secular trend in a time series when the distribution of the deviations is approximately normal.
• The least-squares estimates are unbiased estimates of the parameters.
• The method can be used when the trend is linear, exponential, or quadratic.

Demerits of Least Squares:

• The least-squares method gives too much weight to extremely large deviations from the trend
• The least-squares line is the best only for the period to which it has reference.
• The elimination or addition for a few or more time periods may change its position.

Learn about Method of Least Squares (Linear Trend)