Theory of Copula in Hydrology and Hydroclimatology

Abstract

This chapter starts with an introduction to copulas. The copula theory is relatively new to hydrology and hydroclimatology but has already established itself to be highly potential in frequency analysis, multivariate modeling, simulation and prediction. Development of joint distribution between multiple variables is the key to analyze utilizing the potential of copulas. The chapter starts with the mathematical theory of copulas and gradually move on to the application. If the readers are already aware of the background theory and look for application of copula theory, they can directly proceed to Sect. 10.8. Basic mathematical formulations for most commonly used copulas are discussed, and illustrative examples are provided. It will enable the readers to carry out applications to other problems. All the illustrative examples are designed with very few data points. This helps to show the calculation steps explicitly. Please note that any statistical analysis should be done with sufficiently long data. Once the readers understand the steps, computer codes can be written easily for large data sets. Example of MATLAB codes is also provided at the end.

Exercise

10.1

Check whether the following functions are valid copula functions or not?

• (a) $C(u,v)=\frac{uv}{u-v+uv}$

• (b) $C(u,v)=\sqrt{\max (u^2+v^2-1,0)}$

• (c) $C(u,v)=\max (e^{\ln (u)}+e^{\ln (v)}-1,0)$

• (d) $C(u,v)=\frac{\sqrt{u^2+v^2}}{2}$

• (e) $C(u,v)=|u+v-1|$

• (f) $C(u,v)=\frac{uv}{1-0.3(1-u)(1-v)}$

(Ans. Only (d) and (e) are not copula functions.)

10.2

Using the first 6 values of total monthly precipitation depth and mean monthly specific humidity in Table A.1 (p. 429), calculate the Kendall’s tau and Spearman’s rho.

(Ans. Kendall’s $\uptau =0.20$, Spearman’s rho $\left( \rho _s\right) =0.43$)

10.3

A location is frequently hit by cyclone. For the location, in a cyclonic event, the total rainfall depth and maximum pressure difference between eye and periphery of cyclone are assumed to be associated. For last six cyclones, the maximum pressure difference (in millibar) between the eye and periphery and total rainfall received (in cm) are (30, 70), (35, 77), (27, 75), (32, 81), (37, 87), and (25, 70). Calculate the Kendall’s tau and Spearman’s rho for the data set.

(Ans. Kendall’s $\uptau =0.69$, Spearman’s rho $\left( \rho _s\right) =0.84$)

10.4

For the data used in Exercise 10.2, total monthly precipitation depth is distributed exponentially with mean 82 mm and the specific humidity is distributed normally with mean 10.7 and standard deviation of 2. Fit following copulas to the data:

• (a) Independent Copula

• (b) Gaussian Copula

• (c) Clayton Copula

• (d) Frank Copula

(Hint: Numerical solution may be needed for fitting some of the copula functions.)  

Ans. (a):

For independent copula, $C(u,v)=\left[ \begin{array}{rrrrrr} 0.001&\quad 0&\quad 0&\quad 0&\quad 0.001&\quad 0.970\end{array}\right]$

(b):

For Gaussian copula, $\rho =0.994$, $C(u,v)=\left[ \begin{array}{rrrrrr} 0.01&\quad 0&\quad 0&\quad 0&\quad 0.02&\quad 0.97\end{array}\right]$

(c):

For Clayton copula, $\theta =0.5$, $C(u,v)=\left[ \begin{array}{rrrrrr} 0.006&\quad 0&\quad 0&\quad 0&\quad 0.009&\quad 0.970\end{array}\right]$

(d):

For Frank copula, $\theta =1.86$, $C(u,v)=\left[ \begin{array}{rrrrrr} 0.001&\quad 0&\quad 0&\quad 0&\quad 0.002&\quad 0.970\end{array}\right]$

 

10.5

For the data given in Exercise 10.3, assume that the pressure difference (in millibar) between the eye and periphery of cyclone follows normal distribution with mean 30 and standard deviation 3.2. Similarly, the rainfall is gamma distributed with $\alpha =35$ and $\beta =2.5$. Fit following copula functions over the data,

• (a) Frank Copula

• (b) Gumbel–Hougaard Copula

• (c) Clayton Copula

• (d) Ali–Mikhail–Haq Copula

 

Ans. (a):

For Frank copula, $\theta =10.968$, $C(u,v)=\left[ \begin{array}{rrrrrr} 0.1112&\quad 0.2471&\quad 0.1301&\quad 0.3455&\quad 0.5089&\quad 0.0375\end{array}\right]$

(b):

For Gumbel–Hougaard copula, $\theta =3.226$, $C(u,v)=\left[ \begin{array}{rrrrrr} 0.1103&\quad 0.2471&\quad 0.1253&\quad 0.3446&\quad 0.5090&\quad 0.0422\end{array}\right]$

(c):

For Clayton copula, $\theta =4.45$, $C(u,v)=\left[ \begin{array}{rrrrrr}0.1121&\quad 0.2471&\quad 0.1588&\quad 0.3447&\quad 0.5086&\quad 0.0583\end{array}\right]$

(d):

Ali–Mikhail–Haq copula cannot be fitted over the data.

 

10.6

From historical records, the daily rainfall depths (in mm) in two nearby cities A and B are found to have Kendall’s tau as 0.45. Fit Gumbel–Hougaard, Ali–Mikhail–Haq, and Clayton copulas between the daily rainfall depth of two cities.

(Ans. For Gumbel–Hougaard $\theta =1.818$. For Clayton copula $\theta =1.636$. Ali–Mikhail–Haq copula can not be fitted on the data.)

10.7

Select the best copula function for Example 10.4 using Kolmogorov–Smirnov statistic and Cram$\acute{\text {e}}$r-von Mises statistic.(Ans. Gaussian copula)

10.8

For a location A, monthly mean potential evaporation follows an exponential distribution with mean 4 mm/day, and mean monthly air temperature follows normal distribution with mean 25 $^{\circ }$C and standard deviation 3.7 $^{\circ }$C. Clayton copula with $\theta =0.7$ is found to be the best-fit copula. Generate the mean potential evaporation and mean monthly temperature values for a year.

(Answers may vary depending on random number generated. Refer to Sect. 10.10.1.)

10.9

Using the Frank copula fitted in Exercise 10.6, predict the daily rainfall depth (in mm) for city A, if the daily rainfall depths (in mm) for city B recorded in last week are 0, 2, 5, 20, 8, 0, and 3. Also, calculate 90% confidence interval for predictions. Assume that daily rainfall in cities A and B follows exponential distribution with mean 4.5 and 3 mm/day, respectively.

(Ans. Expected rainfall (in mm/day) for city A $=\left[ \begin{array}{*{7}{r}} 1.7&\, 3.1&\, 4.3&\, 5.2&\, 4.9&\, 1.7&3.6\end{array}\right]$ and corresponding 90% confidence interval are (0.1, 9.7), (0.2, 13.0), (0.4, 15.2), (0.5, 16.6), (0.5, 16.1), (0.1, 9.7), and (0.3, 14.0).)

10.10

For the last 5 months due to some technical problem at site A (Exercise 10.8), evaporation was not recorded. However, if the observed mean monthly air temperature (in $^{\circ }$C) for last 5 months is 24, 28, 30, 32, and 27, then using the copula function given in Exercise 10.8, calculate the expected value of mean monthly evaporation for these month along with their interquartile range (25–75% range).

(Ans. Expected evaporation (in mm/day) is 2.7, 3.9, 4.2, 4.32, and 3.7. The interquartile range of evaporation is (1.3, 5.3), (2.0, 6.9), (2.2, 7.2), (2.3, 7.4), and (1.9, 6.6).)

10.11

In the Table A.1 (p. 429), model the association of the total precipitation depth and mean monthly pressure using Frank copula. Assume that pressure is distributed normally and total monthly rainfall follows exponential distribution with mean 95 mm. Furthermore, predict the total monthly rainfall depth and 95% confidence interval if the mean monthly pressure is 960 mb.

(Hint: Numerical solution may be needed for fitting some of the copula function.)

(Ans. For Frank copula $\theta =-5$. The expected total monthly rainfall depth for 960 mb pressure is 50 mm and its 95% confidence interval is 6.2 and 182.3 mm.)

10.12

In the Table A.2 (p. 431), the locations A1 and A2 are 50 km apart, and their monthly mean sea surface temperature is assumed to associated. Assuming that SST at both the places are normally distributed, fit a Clayton copula to model the relationship of SST between these two places. Using the fitted copula, generate the monthly mean sea surface temperature for one year.

(Hint: Numerical solution may be needed for fitting some of the copula function.)

(Ans. For Clayton copula $\theta =28.21$.)