Time Series Analysis

Abstract

Hydroclimatic variables such as rainfall intensity, streamflow, air temperature vary with space and time, due to different hydrological/climatic phenomena/processes. As these processes are continuously evolving over time, studying the interdependence in hydroclimatic data with proper consideration of temporal information may lead to better insight into the governing processes. Observations of any variable, recorded in chronological order, represent a time series. A time series is generally assumed to consist of deterministic components (results can be predicted with certainty) and stochastic components (results cannot be predicted with certainty as the outcome depends on chance). Analysis of time series helps to get an insight of the time series that in turn may enhance the prediction of the hydroclimatic processes/variables. The objective of this chapter is to introduce different types of time series analysis techniques. This requires an understanding of time series analysis techniques and time series properties like stationarity, homogeneity, periodicity, which is the subject matter of this chapter.

Exercise

9.1

The annual evapotranspiration (in cm/year) for a basin in last 20 years are 61.04, 58.71, 60.02, 60.36, 62.65, 64.17, 62.82, 64.41, 64.6, 63.45, 65.35, 64.65, 67.37, 66.27, 68.39, 66.77, 68.24, 68.04, 66.53, and 68.02.

Check the evapotranspiration data for any trend using (a) Mann–Kendall test and (b) Kendall tau test. Use 5% level of significance.

(Ans. At 5% significance level null hypothesis of no trend is rejected for Mann–Kendall. However, in the Kendall tau test null hypothesis of no trend cannot be rejected at 5% significance level.)

9.2

The monthly average atmospheric pressure (in mb) measured at surface level for 24 consecutive months are

963.65, 965.03, 961.18, 959.43, 957.68, 953.42, 950.11, 952.44, 952.25, 956.88, 963.66, 963.36, 965.56, 964.5, 963.66, 960.91, 956.9, 952.18, 950.71, 952.54, 951.43, 955.06, 959.01, and 962.60. Find the autocorrelation and partial autocorrelation functions at lags 0, 1, 2, and 3.

• Ans. Autocorrelation function at lag 0, 1, 2, and 3 are 1, 0.782, 0.414, and 0.008 respectively.

  Partial autocorrelation function at lag 0, 1, 2, and 3 are 1, 0.807, $-0.617$, and $-0.481$ respectively.

9.3

For the data, provided in Exercise 9.1, find the autocorrelation and partial autocorrelation coefficient at lags 0, 1 and 2. Find the 95% confidence limit for the ACF and PACF at lag 2.

• Ans. Autocorrelation function at lag 0, 1, and 2 are 1, 0.784, and 0.678 respectively.

  Partial autocorrelation function at lag 0, 1, and 2 are 1, 0.852 and 0.489 respectively.

  95% confidence interval of ACF and PACF is $[-0.462, 0.462]$.

9.4

Streamflow at a section for 30 consecutive days is shown in the following table

Day	Flow ($\times 1000$ m$^3$/s)	Day	Flow ($\times 1000$ m$^3$/s)	Day	Flow ($\times 1000$ m$^3$/s)
1	14.12	11	14.45	21	12.50
2	22.05	12	12.72	22	16.10
3	22.34	13	13.67	23	17.40
4	20.07	14	12.58	24	9.48
5	21.15	15	9.33	25	8.41
6	19.82	16	9.67	26	9.33
7	20.65	17	10.65	27	10.40
8	23.57	18	14.47	28	12.62
9	22.19	19	11.02	29	15.30
10	18.32	20	9.82	30	13.84

From historical records, streamflow is found to follow gamma distribution. The rating curve for the section is given by:

$$\begin{aligned} Q=59.5(G-5)^2 \end{aligned}$$

where Q is streamflow in m$^3$/s and G is river stage at section in meters. Calculate the river stage at the section and check whether river stage follows normal distribution at 5% level of significance or not.

(Ans. The river stage for the section (in m) are 20.4, 24.3, 24.4, 23.4, 23.9, 23.3, 23.6, 24.9, 24.3, 22.5, 20.6, 19.6, 20.2, 19.5, 17.5, 17.7, 18.4, 20.6, 18.6, 17.8, 19.5, 21.4, 22.1, 17.6, 16.9, 17.5, 18.2, 19.6, 21.0, and 20.3. At 5% significance level, the null hypothesis of river stage follows normal distribution cannot be rejected.)

9.5

Fit AR(1) and AR(2) model on the river stage data given in Exercise 9.4. What percentage of variance in the river stage time series is explained by these two models? Calculate Akaike Information Criteria for the models and suggest the best model.

• Ans. For AR(1) model $\Phi _1=0.79$, $R^2=0.626$, $\text {AIC}=187.77$

  For AR(2) model $\Phi _1=0.98$ and $\Phi _2=-0.24$, $R^2=0.647$, $\text {AIC}=189.21$

  Hence, out of AR(1) and AR(2), AR(1) is better model.

9.6

Soil moisture is usually found to have high memory component. Using a sensor the surface soil moisture was recorded daily at a location for 60 days. For this time series, PACF at successive lags from 0 to 4 are 1, 0.56, 0.41, 0.15, and 0.11 and corresponding ACF are 1, 0.85, 0.62, 0.25, and 0.12. Suggest the appropriate order of AR model and find the parameters of selected AR model. Check the AR parameters for model stationarity.

(Ans. On the basis of significance of PACF function highest order of AR model is 2. The AR(2) parameters are $\Phi _1=1.164$ and $\Phi _2=-0.370$. The AR(2) model is stationary.)

9.7

For a location, monthly average zonal wind is found to follow a moving average model. From a monthly average zonal wind time series record of length 35, the ACF function at lags 0 to 5 are found as 1, 0.45, 0.35, 0.25, 0.15, and 0.08. Suggest an appropriate order for MA model and find corresponding parameters. Check the invertibility of the selected model.

(Ans. On the basis of significance of ACF, MA(1) is an appropriate model. The parameter for MA(1) model is $-0.627$. The MA(1) is invertible.)

9.8

The parameters of AR(2) model are $\Phi _1=0.77$ and $\Phi _2=-0.25$. Calculate the ACF till lag 2 for the corresponding time series.

(Ans. $\rho _1=0.616$ and $\rho _2=0.224$)

9.9

For a MA(2) model fitted on time series X(t), if parameters are $\theta _1=0.57$ and $\theta _2=0.36$, calculate the PACF and ACF up to lag 2 for the time series X(t).

(Ans. $\rho _1=-0.251$, $\rho _2=-0.247$, $\varphi _1=-0.251$, and $\varphi _2=-0.331$)

9.10

Considering the following ARMA model,

$$\begin{aligned} X(t)=0.63X(t-1)-0.45X(t-2)+\varepsilon (t)-0.58\varepsilon (t-1)+0.21\varepsilon (t-2) \end{aligned}$$

Check the invertibility and stationarity of the model.

(Ans. The model is stationary but not invertible.)

9.11

At a location, the daily air temperature follows the ARMA(2,1) model given below,

$$\begin{aligned} X(t)=0.7X(t-1)+0.2X(t-2)+\varepsilon (t)+0.7\varepsilon (t-1) \end{aligned}$$

If the air temperature recorded in the last week (in ${}^\circ$C) was 16.5, 15.2, 18.2, 16.3, 19.4, 17.8, and 15.7, then forecast air temperature and their 95% confidence limit for next three days. Assume that the variance of residual is unity. Further, update the forecast for remaining two days, if the temperature on eighth day is recorded as 14.5${}^\circ$C.

(Ans. Forecasted temperatures (in ${}^\circ$C) for next three days are 15.7, 14.1, and 13.0 respectively. Their confidence intervals are (13.7, 17.7), (10.8, 17.5), and (9.0, 17.1) respectively. The update forecasts for next two days (in ${}^\circ$C) are 12.4 and 11.6, respectively.)

9.12

For the monthly average atmospheric pressure at surface data provided in Exercise 9.2, check the data for any seasonality (periodicity of 12 months) at 5% level of significance.

(Ans. Data is seasonal at 5% level of significance.)

9.13

For AR(2) model developed in Exercise 9.5, check that the residual series is white noise at 5% level of significance. A series is called white noise when it is independent and normally distributed with zero mean.

(Ans. The residual is white noise at 5% level of significance.)

9.14

Decompose the annual evapotranspiration time series provided in Exercise 9.1 into its Haar MRSWT components up to level 2. [Hint: Code written in Box 1.9 may be used]

Ans. The decomposed series are

$a_2$ :

[120.1, 120.9, 123.6, 125.0, 127.0, 128.0, 127.6, 128.9, 129.0, 130.4, 131.8, 133.3, 134.4, 134.8, 135.7, 134.8, 135.4, 131.8, 127.2, 123.9]

$d_2$ :

[−0.3, −2.1, −3.2, −2.0, −0.2, −1.0, −0.4, 0.1, −1.0, −1.6, −1.8, −1.3, −0.8, −0.2, −0.6, 0.2, 0.9, 2.8, 7.4, 5.2]

$d_1$ :

[1.6, −0.9, −0.2, −1.6, −1.1, 1.0, −1.1, −0.1, 0.8, −1.3, 0.5, −1.9, 0.8, −1.5, 1.1, −1.0, 0.1, 1.1, −1.1, 4.9]