Chapter 1, Problem 82SE

a.

To determine

Construct a time series plot for temperature using the given data.

Comment on the pattern of obtained time series plot.

Answer to Problem 82SE

Output obtained from MINITAB is given below:

The obtained time series plot represents a cyclic pattern of temperature.

Explanation of Solution

Given info:

The data represents the observed value of response variable x at time t. The observed values are $x_t$. Exponential smoothing is done to produce a smoothed version of the series. The value of smoothing constant is $\alpha$ and the smoothing equation is ${\overline{x}}_t=\alpha x_t+\left(1-\alpha \right){\overline{x}}_{t-1}$ for $t=2,3,\mathrm{...},n$. Additionally it is given that ${\overline{x}}_1=x_1$.

Calculation:

Software Procedure:

Step-by-step procedure to draw the time series plot for temperature using the MINITAB software:

• Choose Graph > Time Series Plot.
•  Choose Simple, and then click OK.
• In Series, enter the column of Temperature.
• Click OK.

Observation:

Time series plot shows that the highest temperature is at time 3 and the lowest temperature is at time 6 and 7. From the graph it can be concluded that temperature oscillates with the change in time. Overall the plot represents a cyclic pattern of temperature.

b.

To determine

Find the smoothed value ${\overline{x}}_t$ at time t for $\alpha =0.1$ and $\alpha =0.5$.

Answer to Problem 82SE

Smoothed values ${\overline{x}}_t$ at time t for $\alpha =0.1$are as follows:

Time t	$x_t$	${\overline{x}}_t$
2	47	47
3	54	47.7
4	53	48.2
5	50	48.4
6	46	48.2
7	46	48
8	47	47.9
9	50	48.1
10	51	48.4
11	50	48.5
12	46	48.3
13	52	48.6
14	50	48.8
15	50	48.9

Smoothed values ${\overline{x}}_t$ at time t for $\alpha =0.5$are as follows:

Time t	$x_t$	${\overline{x}}_t$
2	47	47
3	54	50.5
4	53	51.8
5	50	50.9
6	46	48.4
7	46	47.2
8	47	47.1
9	50	48.6
10	51	49.8
11	50	49.9
12	46	47.9
13	52	50
14	50	50
15	50	50

Explanation of Solution

Calculation:

The exponential smoothing equation is ${\overline{x}}_t=\alpha x_t+\left(1-\alpha \right){\overline{x}}_{t-1}$ for $t=2,3,\mathrm{...},n$.

Smoothed values ${\overline{x}}_t$ at time t for $\alpha =0.1$are as follows:

Here, to calculate the smoothed value ${\overline{x}}_t$ at time 2 the smoothed value at time 1 is necessary.

That is, ${\overline{x}}_{t-1}$ is necessary. The value of ${\overline{x}}_1$ is unknown.

Moreover, it is given that ${\overline{x}}_1=x_1$.

Hence, the smoothed value ${\overline{x}}_t$ at time 2 for $\alpha =0.1$ remains 47.

Thus, the smoothed value at time 2 is ${\overline{x}}_2=47$.

The smoothed value ${\overline{x}}_t$ at time 3 is calculated as follows:

$\begin{array}{c}{\overline{x}}_3\text{ =}\alpha x_3+\left(1-\alpha \right){\overline{x}}_2\\ =0.1\times 54+\left(1-0.1\right)\times 47\\ =5.4+42.3\\ =47.7\end{array}$

Thus, the smoothed value at time 3 is ${\overline{x}}_3=47.7$.

Similarly, smoothed values for the remaining times are given below:

Time t	$x_t$	${\overline{x}}_t$
2	47	47
3	54	47.7
4	53	48.2
5	50	48.4
6	46	48.2
7	46	48
8	47	47.9
9	50	48.1
10	51	48.4
11	50	48.5
12	46	48.3
13	52	48.6
14	50	48.8
15	50	48.9

Smoothed values ${\overline{x}}_t$ at time t for $\alpha =0.5$are as follows:

Here, to calculate the smoothed value ${\overline{x}}_t$ at time 2 the smoothed value at time 1 is necessary.

That is, ${\overline{x}}_{t-1}$ is necessary. The value of ${\overline{x}}_1$ is unknown.

Moreover, it is given that ${\overline{x}}_1=x_1$.

Hence, the smoothed value ${\overline{x}}_t$ at time 2 for $\alpha =0.5$ remains 47.

Thus, the smoothed value at time 2 is ${\overline{x}}_2=47$.

The smoothed value ${\overline{x}}_t$ at time 3 is calculated as follows:

$\begin{array}{c}{\overline{x}}_3\text{ =}\alpha x_3+\left(1-\alpha \right){\overline{x}}_2\\ =0.5\times 54+\left(1-0.5\right)\times 47\\ =27+23.5\\ =50.5\end{array}$

Thus, the smoothed value at time 3 is ${\overline{x}}_3=50.5$.

Similarly, smoothed values for the remaining times are given below:

Time t	$x_t$	${\overline{x}}_t$
2	47	47
3	54	50.5
4	53	51.8
5	50	50.9
6	46	48.4
7	46	47.2
8	47	47.1
9	50	48.6
10	51	49.8
11	50	49.9
12	46	47.9
13	52	50
14	50	50
15	50	50

c.

To determine

Find the number of values of $x_t,x_{t-1},\mathrm{...},x_1$ that depend on ${\overline{x}}_t$ when ${\overline{x}}_{t-1}=\alpha x_{t-1}+\left(1-\alpha \right){\overline{x}}_{t-2}$ is substituted in place of ${\overline{x}}_{t-1}$ in the smoothing equation ${\overline{x}}_t$ and ${\overline{x}}_{t-2}$ in terms of $x_{t-2}$ and so on

Find the change in the coefficient of $x_{t-k}$ with the increase in k.

Answer to Problem 82SE

The value of ${\overline{x}}_t$ depends on all the values of $x_t,x_{t-1},\mathrm{...},x_1$.

The coefficient on $x_{t-k}$ decreases as k increases.

Explanation of Solution

Calculation:

The exponential smoothing equation is ${\overline{x}}_t=\alpha x_t+\left(1-\alpha \right){\overline{x}}_{t-1}$ for $t=2,3,\mathrm{...},n$.

Substituting the value, ${\overline{x}}_{t-1}=\alpha x_{t-1}+\left(1-\alpha \right){\overline{x}}_{t-2}$ in the exponential smoothing equation is ${\overline{x}}_t$,

$\begin{array}{c}{\overline{x}}_t=\alpha x_t+\left(1-\alpha \right)\left(\alpha x_{t-1}+\left(1-\alpha \right){\overline{x}}_{t-2}\right)\\ =\alpha x_t+\alpha \left(1-\alpha \right)x_{t-1}+{\left(1-\alpha \right)}^2{\overline{x}}_{t-2}\end{array}$

Substituting the value, ${\overline{x}}_{t-2}=\alpha x_{t-2}+\left(1-\alpha \right){\overline{x}}_{t-2}$ in the above exponential smoothing equation is ${\overline{x}}_t$,

$\begin{array}{c}{\overline{x}}_t=\alpha x_t+\alpha \left(1-\alpha \right)x_{t-1}+{\left(1-\alpha \right)}^2{\overline{x}}_{t-2}\\ =\alpha x_t+\alpha \left(1-\alpha \right)x_{t-1}+{\left(1-\alpha \right)}^2\left(\alpha x_{t-2}+\left(1-\alpha \right){\overline{x}}_{t-2}\right)\\ =\alpha x_t+\alpha \left(1-\alpha \right)x_{t-1}+\alpha {\left(1-\alpha \right)}^2{x}_{t-2}+{\left(1-\alpha \right)}^3{\overline{x}}_{t-2}\end{array}$

Continuing the same computational procedure till the value of ${\overline{x}}_3$.

${\overline{x}}_3=\alpha {\left(1-\alpha \right)}^{t-3}{x}_3+{\left(1-\alpha \right)}^{t-2}{\overline{x}}_2$

Now, the equation reduces as follows:

$\begin{array}{c}{\overline{x}}_3=\alpha {\left(1-\alpha \right)}^{t-3}{x}_3+{\left(1-\alpha \right)}^{t-2}{\overline{x}}_2\\ =\alpha {\left(1-\alpha \right)}^{t-3}{x}_3+{\left(1-\alpha \right)}^{t-2}\left(\alpha x_2+\left(1-\alpha \right){\overline{x}}_1\right)\\ =\alpha {\left(1-\alpha \right)}^{t-3}{x}_3+\alpha {\left(1-\alpha \right)}^{t-2}{x}_2+{\left(1-\alpha \right)}^{t-1}{\overline{x}}_1\end{array}$’

The value of ${\overline{x}}_1$ is set as $x_1$. That is, ${\overline{x}}_1=x_1$.

Now, the exponential smoothing equation ${\overline{x}}_t$ is as follows:

${\overline{x}}_t=\alpha x_t+\alpha \left(1-\alpha \right)x_{t-1}+\alpha {\left(1-\alpha \right)}^2{x}_{t-2}+\mathrm{...}+\alpha {\left(1-\alpha \right)}^{t-3}{x}_3+\alpha {\left(1-\alpha \right)}^{t-2}{x}_2+{\left(1-\alpha \right)}^{t-1}{x}_1$

Here, from the above obtained equation it is seen that the value of ${\overline{x}}_t$ depends on all the values of $x_t,x_{t-1},\mathrm{...},x_1$.

Here, form the equation it can be said that the coefficient of $x_{t-k}$ is ${\left(1-\alpha \right)}^{t-k}$.

Moreover, the smoothing constant $\alpha$ lies between 0 and 1.

That is, $0<\alpha <1$.

Hence the value of ${\left(1-\alpha \right)}^{t-k}$ decreases as k increases.

Therefore, the coefficient on $x_{t-k}$ decreases as k increases.

c.

To determine

Explain the sensitivity of the initialization of ${\overline{x}}_1=x_1$ for large t in the equation ${\overline{x}}_t$ obtained in part (c).

Answer to Problem 82SE

The smoothed series ${\overline{x}}_t$ is not very sensitive for the initialization of ${\overline{x}}_1=x_1$ for large t.

Explanation of Solution

Calculation:

From part (c), the exponential smoothing equation is,

${\overline{x}}_t=\alpha x_t+\alpha \left(1-\alpha \right)x_{t-1}+\alpha {\left(1-\alpha \right)}^2{x}_{t-2}+\mathrm{...}+\alpha {\left(1-\alpha \right)}^{t-3}{x}_3+\alpha {\left(1-\alpha \right)}^{t-2}{x}_2+{\left(1-\alpha \right)}^{t-1}{x}_1$

The substitution of of ${\overline{x}}_1$ is as follows:

${\overline{x}}_1={\left(1-\alpha \right)}^{t-1}{x}_1$

Here, form the equation it can be said that the coefficient of $x_1$ is ${\left(1-\alpha \right)}^{t-1}$.

Moreover, the smoothing constant $\alpha$ lies between 0 and 1.

That is, $0<\alpha <1$.

Hence the value of ${\left(1-\alpha \right)}^{t-1}$ is very small for large t.

Therefore, the smoothed series ${\overline{x}}_t$ is not very sensitive for the initialization of ${\overline{x}}_1=x_1$ for large t.