A new probability model with application to heavy-tailed hydrological data

Abstract

Because of the dramatic changes that are being observed in the climatic conditions of the world, such as excess of rains, drought and huge floods, we introduce a versatile hydrologic probability model with three parameters. The proposed model is a combination of the Lomax and generalized Weibull distributions based on an exponent odd function. Main properties of the distribution are obtained, such as shapes of the probability density and hazard rate functions, quantile function, asymptotic distribution, information matrix and characterization via hazard rate function. Parameters are estimated via the maximum likelihood estimation method. Four data sets are used to compare the proposed model with a number of well-known hydrologic models. The proposed model is found to be suitable and representative for heavy-tailed hydrological data sets, with least loss of information attitude and a realistic return period.

Appendices

Appendix-A

Proof

Necessity:

If $X\sim LDGW(\lambda ,\theta ,\beta )$, with a cdf defined by Equation (1), then its hrf can be expressed as

$$\begin{aligned} h(x)=h_{{LDGW}}(x|\lambda , \theta , \beta )=\frac{e^{-\theta +\theta \left( 1-\lambda x^{\theta } \right) ^{-\frac{1}{\lambda }}} x^{-1+\theta } \beta \theta ^2 \left( 1-x^{\theta } \lambda \right) ^{-1-\frac{1}{\lambda }}}{ \left( e^{\theta }-1\right) \left( 1+\frac{e^{-\theta +\theta \left( 1-\lambda x^{\theta } \right) ^{-\frac{1}{\lambda }}}-1}{e^{\theta }-1}\right) }, \end{aligned}$$

where $x>0$, $\lambda \le 0$, $\alpha >0$ and $\beta >0$.

Now, differentiating the logarithmic form of the hrf with respect to x, we get

$$\begin{aligned}&\frac{d \ln h(x)}{dx} =\frac{\theta -1}{x}+\theta ^{2}x^{\theta -1}\left( 1-\lambda x^{\theta }\right) ^{-\frac{1}{\lambda }-1} +\left( \frac{1}{\lambda }+1\right) \frac{\lambda \theta x^{\theta -1}}{1-\lambda x^{\theta }}\\&\qquad -\frac{e^{-\theta +\theta \left( 1-\lambda x^{\theta } \right) ^{-\frac{1}{\lambda }}} x^{-1+\theta } \beta \theta ^2 \left( 1- \lambda x^{\theta } \right) ^{-1-\frac{1}{\lambda }}}{ \left( e^{\theta }-1\right) \left( 1+\frac{e^{-\theta +\theta \left( 1-\lambda x^{\theta } \right) ^{-\frac{1}{\lambda }}}-1}{e^{\theta }-1}\right) }, \frac{h^{\prime }(x)}{h(x)}-\left( \frac{1}{\lambda }+1\right) \frac{\lambda \theta x^{\theta -1}}{1-\lambda x^{\theta }}\\&\quad =\frac{\theta -1}{x}+\theta ^{2}x^{\theta -1}(1-\lambda x^{\theta })^{-\frac{1}{\lambda }-1} -\frac{e^{-\theta +\theta \left( 1-\lambda x^{\theta } \right) ^{-\frac{1}{\lambda }}} x^{-1+\theta } \beta \theta ^2 \left( 1-\lambda x^{\theta } \right) ^{-1-\frac{1}{\lambda }}}{ \left( e^{\theta }-1\right) \left( 1+\frac{e^{-\theta +\theta \left( 1-\lambda x^{\theta } \right) ^{-\frac{1}{\lambda }}}-1}{e^{\theta }-1}\right) }, \end{aligned}$$

which after some algebraic manipulations we get Equation(8).

Sufficiency:

Suppose Equation(8) holds, then it may be re-written as

$$\begin{aligned}&\frac{h^{\prime }(x)}{(h(x))^{2}} =\frac{\left( 1+\dfrac{e^{-\theta +\theta \left( 1-\lambda x^{\theta }\right) ^{-\frac{1}{\lambda }}}-1}{e^{\theta }-1}\right) ^{-\beta }}{\left[ \beta \theta ^{2} \left( 1+\dfrac{e^{-\theta +\theta \left( 1-\lambda x^{\theta }\right) ^{-\frac{1}{\lambda }}}-1}{e^{\theta }-1}\right) ^{-\beta -1} \dfrac{e^{-\theta +\theta \left( 1-\lambda x^{\theta }\right) ^{-\frac{1}{\lambda }}} x^{\theta -1}\left( 1-\lambda x^{\theta }\right) ^{-\frac{1}{\lambda }-1}}{e^{\theta }-1}\right] ^{2}} \\&\qquad \times \frac{d\left\{ \beta \theta ^{2} \left( 1+\dfrac{e^{-\theta +\theta \left( 1-\lambda x^{\theta }\right) ^{-\frac{1}{\lambda }}}-1}{e^{\theta }-1}\right) ^{-\beta -1} \dfrac{e^{\theta -\theta \left( 1-\lambda x^{\theta }\right) ^{-\frac{1}{\lambda }}} x^{\theta -1}\left( 1-\lambda x^{\theta }\right) ^{-\frac{1}{\lambda }-1}}{e^{\theta }-1}\right\} }{dx}+1. \end{aligned}$$

From the above differential equation, we have

$$\begin{aligned} h(u)=\frac{e^{-\theta +\theta \left( 1-\lambda u^{\theta }\right) ^{-\frac{1}{\lambda }}} u^{-1+\theta } \beta \theta ^2 \left( 1-\lambda u^{\theta } \right) ^{-1-\frac{1}{\lambda }}}{ \left( e^{\theta }-1\right) \left( 1+\frac{e^{-\theta +\theta \left( 1-\lambda u^{\theta }\right) ^{-\frac{1}{\lambda }}}-1}{e^{\theta }-1}\right) }. \end{aligned}$$

(10)

Integrating the Equation (10) from 0 to x we get

$$\begin{aligned} -\ln (1-F_{{LDGW}}(x|\lambda , \theta , \beta ))=-\ln \left( \left( 1+\dfrac{e^{-\theta +\theta \left( 1-\lambda x^{\theta }\right) ^{-\frac{1}{\lambda }}}-1}{e^{\theta }-1} \right) ^{-\beta }\right) , \end{aligned}$$

which after simplification yields

$$\begin{aligned} F_{{LDGW}}(x|\lambda , \theta , \beta )=1-\left( 1+\dfrac{e^{-\theta +\theta \left( 1-\lambda x^{\theta }\right) ^{-\frac{1}{\lambda }}}-1}{e^{\theta }-1}\right) ^{-\beta }. \end{aligned}$$

This completes the proof. $\square$

Appendix-B

$$\begin{aligned}&\frac{\partial \ell (\Theta )}{\partial \theta } =-n-\frac{e^{\theta } n}{e^{\theta }-1}+\frac{2 n}{\theta }+\sum _{i=1}^n \ln (x_i)+\left( 1+\frac{1}{\lambda }\right) \sum _{i=1}^n \frac{\lambda \ln (x_i) x_i^{\theta }}{1-\lambda x_i^{\theta }}\\&\qquad \qquad \quad +\theta \sum _{i=1}^n \ln (x_i) x_i^{\theta } \left( 1-\lambda x_i^{\theta }\right) {}^{-1-\frac{1}{\lambda }}+\sum _{i=1}^n \left( 1-\lambda x_i^{\theta }\right) {}^{-\frac{1}{\lambda }}-(1+\beta )\\&\qquad \quad \times \sum _{i=1}^n \frac{-\frac{e^{\theta } \left( -1+e^{-\theta +\theta \left( 1-\lambda x_i^{\theta }\right) {}^{-\frac{1}{\lambda }}}\right) }{\left( -1+e^{\theta }\right) ^2}+\frac{e^{-\theta +\theta \left( 1-\lambda x_i^{\theta }\right) {}^{-\frac{1}{\lambda }}} \left( -1+\theta \ln (x_i) x_i^{\theta } \left( 1-\lambda x_i^{\theta }\right) {}^{-1-\frac{1}{\lambda }}-\left( 1-\lambda x_i^{\theta }\right) {}^{-\frac{1}{\lambda }}\right) }{-1+e^{\theta }}}{1+\frac{e^{-\theta +\theta \left( 1-\lambda x_i^{\theta }\right) {}^{-\frac{1}{\lambda }}}-1}{e^{\theta }-1}}=0,\\&\frac{\partial \ell (\Theta )}{\partial \lambda } =\frac{1}{\lambda ^2}\sum _{i=1}^n \ln (1-\lambda x_i^{\theta })+\left( 1+\frac{1}{\lambda }\right) \sum _{i=1}^n \frac{x_i^{\theta }}{1-\lambda x_i^{\theta }}\\&\qquad \qquad + \theta \sum _{i=1}^n \left( 1-\lambda x_i^{\theta }\right) {}^{-\frac{1}{\lambda }} \left( \frac{\ln (1-\lambda x_i^{\theta })}{\lambda ^2}+\frac{x_i^{\theta }}{\lambda \left( 1-\lambda x_i^{\theta }\right) }\right) \\&\qquad \qquad -(1+\beta ) \sum _{i=1}^n -\frac{e^{-\theta +\theta \left( 1-\lambda x_i^{\theta }\right) {}^{-\frac{1}{\lambda }}} \theta \left( 1-\lambda x_i^{\theta }\right) {}^{-\frac{1}{\lambda }} \left( \frac{\ln (1-\lambda x_i^{\theta })}{\lambda ^2}+\frac{x_i^{\theta }}{\lambda \left( 1-\lambda x_i^{\theta }\right) }\right) }{\left( e^{\theta }-1\right) \left( 1+\frac{e^{-\theta +\theta \left( 1-\lambda x_i^{\theta }\right) {}^{-\frac{1}{\lambda }}}-1}{e^{\theta }-1}\right) }=0 \end{aligned}$$

and

$$\begin{aligned} \frac{\partial \ell (\Theta )}{\partial \beta }=\frac{n}{\beta }-\sum _{i=1}^n \ln \left[ 1+\frac{e^{-\theta +\theta \left( 1-\lambda x_i^{\theta }\right) {}^{-\frac{1}{\lambda }}}-1}{e^{\theta }-1}\right] =0. \end{aligned}$$

Appendix-C

See Tables 18,19, 20, 21, 22, 23, 24 and 25.

Table 18 Variances and co-variances of MLEs for data set-I to IV

Table 19 Confidence interval with level of significance $\alpha$ for data set-I

Table 20 Confidence interval with level of significance $\alpha$ for data set-II

Table 21 Confidence interval with level of significance $\alpha$ for data set-III

Table 22 Confidence interval with level of significance $\alpha$ for data set-IV

Table 23 Return level estimates $\hat{x_T}$ for T of data set-I to IV

Table 24 Return periods for some largest values of data set-I and II

Table 25 Return periods for some largest values of data set-III and IV