91影视

The data are provided that consists of 14 observations on temperature at a sewage treatment plan.

a.

Following steps are taken to make time-series plot by hand:

1. Draw the Cartesian plane x-y graph, and label the x-axis by 鈥淭ime鈥� and label the y-axis by 鈥淭emperature鈥�.

2. Make a dot on every value of temperatures with respect to the time-stamp.

3. Attach two near dots with the line and then join each pair of dots to make a time-series plot.

There is no sudden shift in the given time series on temperature. There are cycles of rising and falling data values that is not repeating at any interval, so the data appears to have cyclic movements.

b.

The formula of single exponential smoothing is given as,

\({\bar x_t} = \alpha {x_t} + \left( {1 - \alpha } \right){\bar x_{t - 1}}\)

Since \({\bar x_1} = {x_1}\) and \(\alpha = 0.1\), put these values in the above formula to compute\({\bar x_2}\).

\(\begin{aligned}{{\bar x}_2} &= \alpha {x_2} + \left( {1 - \alpha } \right){{\bar x}_{2 - 1}}\\ &= \alpha {x_2} + \left( {1 - \alpha } \right){{\bar x}_1}\\ &= \left( {0.1} \right)\left( {54} \right) + \left( {1 - 0.1} \right)\left( {47} \right)\\ &= 5.4 + 42.3\\ &= 47.7\end{aligned}\)

Similarly, calculate for \({\bar x_3}\), \({\bar x_4}\)up to\({\bar x_{14}}\) by constructing a table when \(\alpha = 0.1\).

Repeat the process of obtaining table of given data when \(\alpha = 0.5\) by using simple exponential smoothing method.

Since \({\bar x_1} = {x_1}\) and \(\alpha = 0.5\), put these values in the above formula to compute \({\bar x_2}\).

\(\begin{aligned}{{\bar x}_2} &= \alpha {x_2} + \left( {1 - \alpha } \right){{\bar x}_{2 - 1}}\\ &= \alpha {x_2} + \left( {1 - \alpha } \right){{\bar x}_1}\\ &= \left( {0.5} \right)\left( {54} \right) + \left( {1 - 0.5} \right)\left( {47} \right)\\ &= 27 + 23.5\\ &= 50.5\end{aligned}\)

Similarly, calculate for \({\bar x_3}\),\({\bar x_4}\)up to \({\bar x_{14}}\) by constructing a table when \(\alpha = 0.5\).

In comparison to the second time series data, the first time series data is smoother and has less variability.

There is no sudden shift in first time series but due to high switches between the data of second time series there is drastic shift intemperatures at sewage treatment plant. Both data appears to have cyclic movements.

c.

The formula of single exponential smoothing is given by,

\({\bar x_t} = \alpha {x_t} + \left( {1 - \alpha } \right){\bar x_{t - 1}}\)

Replace \({\bar x_{t - 1}}\) by \({\bar x_{t - 1}} = \alpha {x_{t - 1}} + \left( {1 - \alpha } \right){\bar x_{t - 2}}\), we get

\(\begin{aligned}{{\bar x}_t} &= \alpha {x_t} + \left( {1 - \alpha } \right)\left( {\alpha {x_{t - 1}} + \left( {1 - \alpha } \right){{\bar x}_{t - 2}}} \right)\\ &= \alpha {x_t} + \alpha \left( {1 - \alpha } \right){x_{t - 1}} + {\left( {1 - \alpha } \right)^2}{{\bar x}_{t - 2}}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, \ldots \left( 1 \right)\end{aligned}\)

Repeat this process again by replacing \({\bar x_{t - 2}}\) by \({\bar x_{t - 2}} = \alpha {x_{t - 2}} + {\left( {1 - \alpha } \right)^2}{\bar x_{t - 3}}\) in eq(1),

\(\begin{aligned}{{\bar x}_t} &= \alpha {x_t} + \alpha \left( {1 - \alpha } \right){x_{t - 1}} + {\left( {1 - \alpha } \right)^2}\left( {\alpha {x_{t - 2}} + \left( {1 - \alpha } \right){{\bar x}_{t - 3}}} \right)\\ &= \alpha {x_t} + \alpha \left( {1 - \alpha } \right){x_{t - 1}} + \alpha {\left( {1 - \alpha } \right)^2}{x_{t - 2}} + {\left( {1 - \alpha } \right)^3}{{\bar x}_{t - 3}}\\ &= {\left( {1 - \alpha } \right)^3}{{\bar x}_{t - 3}} + \alpha \sum\limits_{k = 0}^2 {{{\left( {1 - \alpha } \right)}^k}{x_{t - k}}} \end{aligned}\)

Iterating this process and recall \({\bar x_1} = {x_1}\) yields the following formula,

\(\begin{aligned}{{\bar x}_t} &= {\left( {1 - \alpha } \right)^{t - 1}}{{\bar x}_1} + \alpha \sum\limits_{k = 0}^{t - 2} {{{\left( {1 - \alpha } \right)}^k}{x_{t - k}}} \\ &= {\left( {1 - \alpha } \right)^{t - 1}}{x_1} + \alpha \sum\limits_{k = 0}^{t - 2} {{{\left( {1 - \alpha } \right)}^k}{x_{t - k}}} \end{aligned}\)

The obtained formula suggest that the smoothing statistic depend on all the previous values of previous smoothed statistic. Thus, the value of \({\bar x_t}\) depends on all the previous values of \({x_t},{x_{t - 1}},{x_{t - 2}}, \ldots {x_1}\).

It is quite clear from the formula that by increasing the value of k term in the formula, the value of coefficient i.e.,\(\left( {1 - \alpha } \right)\)becomes zero as the value of smoothing factor ranges between 0 and 1.

Thus, the coefficient of \({x_{t - k}}\) decreases with an increase in smoothing factor \(\alpha \).

d.

Since the coefficient of \({\bar x_t}\)is \({\left( {1 - \alpha } \right)^{t - 1}}\) , by making it large would result as the same discussed in part(c) of the problem. There is an inverse relationship between the value of t and \({\bar x_t}\).

Thus, if t is large, then \({\bar x_t}\) is very insensitive to the initialization \({\bar x_1} = {x_1}\).