The skew-symmetric-Laplace-uniform distribution (2024)

\nameRaju.K. Lohot^a and V. U. Dixit^b CONTACT Raju.K. Lohot. Email: rajulohot.92@gmail.com^aDepartment of Statistics, SVKM’s Mithibai College of Arts, Chauhan Institute of Science & Amrutben Jivanlal College of Commerce and Economics, Vile Parle (W), Mumbai, Maharashtra, India; ^bDepartment of Statistics, University of Mumbai, Vidyanagari, Santacruz (E), Mumbai, Maharashtra, India

Abstract

Laplace distribution is popular in the field of economics and finance. Still, data sets often show a lack of symmetry and a tendency of being bounded from either side of their support. In view of this, we introduce a new family of skew distribution using the skewing mechanism of Azzalini, (1985), namely, skew-symmetric-Laplace-uniform distribution (SSLUD). Here uniform distribution is used not only to introduce skewness in Laplace distribution but also to restrict distribution support on one side of the real line. This paper provides a comprehensive description of the essential distributional properties of SSLUD. Estimators of the parameter are obtained using the method of moments and the method of maximum likelihood. The finite sample and asymptotic properties of these estimators are studied using simulation. It is observed that the maximum likelihood estimator is better than the moment estimator through a simulation study. Finally, an application of SSLUD to real-life data on the daily percentage change in the price of NIFTY 50, an Indian stock market index, is presented.

keywords:

Estimation; Indian stock market index; one side bounded support distribution; simulation; skew-symmetric-Laplace-uniform distribution

{jelcode}

C10, C13

{amscode}

62E10, 62F10

1 Introduction

Symmetry is something which we try to seek naturally in everything, but not everything in the world is symmetric. So expecting symmetry in everything is unrealistic. In statistics, most classical procedures assume some kind of symmetry. However, the absence of symmetry is much more common in many data sets. In particular, much interest has been shown recently in a family of distributions called “Skew-symmetric distributions”. Let $f$ be a density function symmetric about zero, and $K$ an absolutely continuous distribution function such that the corresponding density function $K\,^{\prime}$ is symmetric about zero. Then, Azzalini’s form of skew-symmetric density function for any real $\lambda$ , as mentioned in Azzalini, (1985), is given as

2\,f(x)\,K(\lambda x).

(1)

Arnold and Lin, (2004) studied a special case using $K$ as the cumulative distribution function (cdf) of $f$ in (1). Nadarajah and Kotz, (2003) introduced the skew-symmetric-normal distribution family by replacing $f$ with $\phi$ , the probability density function (pdf) of the standard normal distribution in (1). Further, they studied various skew-symmetric distributions by choosing $K$ as the cdf of normal, Student’s t, Laplace, logistic, and uniform distributions. Nadarajah, (2009) introduced and studied the skew logistic distribution considering $f$ and $K$ as pdf and cdf of logistic distribution, respectively in (1).

When $f$ and $K$ are the density and distribution functions of the Laplace distribution in (1), respectively, it is called a skew-Laplace distribution. Aryal and Rao, (2005) studied some properties of truncated skew-Laplace distribution, and Kozubowski and Nolan, (2008) showed that a skew-Laplace distribution is infinitely divisible. Further, Nekoukhou and Alamatsaz, (2012) introduced a more general family of skew-Laplace distributions by considering $f$ as a standard Laplace pdf, $K$ as an arbitrary symmetric cdf, and $w$ as any odd continuous function in place of $\lambda x$ in (1). That is,

e^{-\lvert x\rvert}\,F(w(x)).

(2)

Recently much interest has been shown in the construction of flexible parametric classes of distributions that exhibit skewness and kurtosis, which is different from the normal distribution. While much of classical statistical analysis is based on Gaussian distributional assumptions, statistical modeling with the Laplace distribution has gained importance in many applied fields. The motivation originates from data sets, including environmental, financial, and biomedical ones, which often do not follow the normal law. Models based on the Laplace distributions are popular in economics and finance; see Zeckhauser and Thompson, (1970); Rachev and SenGupta, (1993); Rydén etal., (1998); Theodossiou, (1998); Kotz etal., (2001); Kozubowski and Podgórski, (2001). They are rapidly becoming distributions of the first choice whenever “something” with heavier than normal tail is observed in the data. The interesting characteristic has often bound on its support from either end along with skew nature. i.e., data is positively skewed but bounded below or negatively skewed but bounded above. For example, consider the scenario of family income, which is typically positively skewed and bounded below by a certain amount. In this paper, by considering interesting applications of Laplace distribution, the need for skewness and restriction on the support of variable of interest, skew-symmetric-Laplace-uniform distribution (SSLUD) is introduced. Here, we consider $f$ as the standard Laplace density function and $K$ as a distribution function of Uniform $(-\theta,\theta)$ in (1). It provides a more flexible model representing the data as adequately as possible. Thus, we can expect this to be useful in more practical situations. The standard Laplace pdf is

f(x)=\frac{1}{2}\,e^{-\lvert x\rvert},\quad x\in\mathbb{R}.

(3)

The distribution function of Uniform $(-\theta,\theta)$ where $\theta>0$ is

K(x)=\begin{cases}\;0&\text{if}\ x<-\theta,\\\displaystyle\;\frac{x+\theta}{2\theta}&\text{if}\ -\theta\leqslant x<\theta,%\\\;1&\text{if}\ x\geqslant\theta.\end{cases}

(4)

Thus, the density function of SSLUD is

2 Moment generating function, characteristic function, and moments

Here, we derive the moment generating function and the characteristic function of r. v. $X$ having pdf given in (6). The moment generating function (MGF) is $M_{X}(t)=E(e^{tX})$ . By using (6), one obtains

M_{X}(t)=\frac{1}{2\mu}\left\{\frac{-1+e^{-\mu(1+t)}}{(1+t)^{2}}+\frac{1-e^{-%\mu(1-t)}}{(1-t)^{2}}\right\}+\frac{1}{(1-t^{2})}\ ,\quad\text{for}\ t<1.

(8)

The corresponding characteristic function defined by $\phi_{X}(t)=E(e^{itX})$ is given as

\phi_{X}(t)=\frac{1}{2\mu}\left\{\frac{-1+e^{-\mu(1+it)}}{(1+it)^{2}}+\frac{1-%e^{-\mu(1-it)}}{(1-it)^{2}}\right\}+\frac{1}{(1+t^{2})}\ ,\quad\text{for}\ it<1,

(9)

where $i=\sqrt{-1}$ is the complex imaginary unit.

The moments of a probability distribution are a collection of descriptive constants used for measuring its properties. Here, we derive the expression of the first four raw moments of $X$ . They are as follows.

\begin{split}\mu_{1}^{{}^{\prime}}&=\frac{2}{\mu}-\left(1+\frac{2}{\mu}\right)%e^{-\mu}\,,\\\mu_{2}^{{}^{\prime}}&=2\,,\\\mu_{3}^{{}^{\prime}}&=\frac{24}{\mu}-e^{-\mu}\left[\mu^{2}+6\mu+18+\frac{24}{%\mu}\right]\,,\\\mu_{4}^{{}^{\prime}}&=24\,.\end{split}

(10)

We see that $\mu_{2r}^{{}^{\prime}}=(2r)!$ for $r=1,2,\ldots$ and corresponding central moments can be obtained using these raw moments but can not be simplified further. Note that, expressions given in (10) are valid only for $\mu>0$ . If $\mu<0$ , one must replace $\mu$ by $-\mu$ in each of these expressions; in addition, the expressions for the odd order moments must be multiplied by -1.

The skew-symmetric-Laplace-uniform distribution (4)

Figure 2 illustrates the behavior of the four measures E( $X$ ), Var( $X$ ), Skewness( $X$ ) and Kurtosis( $X$ ) for $\mu=-10,\ldots,10$ . Mean and skewness are decreasing functions of $\mu$ over the range $(-\infty,0)$ and $(0,\infty)$ , while variance and kurtosis are even functions of $\mu$ . The variance strictly decreases as $\mu$ moves from $-\infty$ to 0 and increases as $\mu$ moves from 0 to $\infty$ .

3 Mode and median

Mode is the value of the r. v. $X$ at which pdf $g(x)$ is maximum. When $\mu>0$ , $g^{\prime}(x)$ is,

g^{\prime}(x)=\begin{cases}\displaystyle\;e^{x}\left(\frac{x+1}{2\mu}+\frac{1}%{2}\right)&\text{if}\ -\mu\leqslant x<0,\\\displaystyle\;e^{-x}\ \left(\frac{1-x}{2\mu}-\frac{1}{2}\right)&\text{if}\ 0%\leqslant x<\mu,\\\displaystyle\;-\,e^{-x}\ &\text{if}\ x\geqslant\mu.\end{cases}

(11)

It is clear from (11) that the function $g(x)$ is increasing in $[-\mu,0)$ and decreasing in $[\mu,\infty)$ . Hence, mode $M_{0}$ of (6) lies in the interval $[0,\mu]$ . Accordingly, we equate $g^{\prime}(x)$ to zero and solve for x. Thus, the value of $M_{0}$ is $M_{0}=1-\mu$ for $\displaystyle\frac{1}{2}\leqslant\mu<1$ . But when $\mu<\displaystyle\frac{1}{2}$ , the function $g(x)$ increases in $[-\mu,\mu)$ and decreases in $[\mu,\infty)$ . Hence, $M_{0}=\mu$ . Similarly, when $\mu>1$ , the function $g(x)$ increases in $[-\mu,0)$ and decreases in $[0,\infty)$ . Hence, $M_{0}=0$ . On similar lines, one can derive the expression of $M_{0}$ for $\mu<0$ . Thus, combining these two expressions of $M_{0}$ , we get

M_{0}=\begin{cases}\;\mu&\text{if}\ \displaystyle 0<|\mu|<\frac{1}{2},\\\;\text{sign}(\mu)-\mu&\text{if}\ \displaystyle\frac{1}{2}\leqslant|\mu|<1,\\\;0&\text{if}\ |\mu|\geqslant 1.\end{cases}

(12)

The median $M$ of (6) is the value of r. v. $X$ such that $G(M)=\frac{1}{2}$ . Thus, for $\mu>0$ using (7b),

M=\begin{cases}\;\text{Solution of the equation,}&\text{if}\ \displaystyle G(0%)>\frac{1}{2},\\\;e^{M}(M-1+\mu)+e^{-\mu}-\mu=0\vspace{0.35 cm}\\\;\text{Solution of the equation,}&\text{if}\ \displaystyle G(0)\leqslant\frac%{1}{2}<G(\mu),\\\;e^{-M}(-M-1-\mu)+e^{-\mu}+\mu=0\vspace{0.35 cm}\\\;\ln 2&\text{if}\ \displaystyle G(\mu)\leqslant\frac{1}{2},\par\end{cases}

(13)

where $\displaystyle G(0)=\frac{1}{2}+\frac{e^{-\mu}-1}{2\mu}$ and $\displaystyle G(\mu)=1-e^{-\mu}$ . Table 1 represents values of the median of (6) for different positive values of $\mu$ using the Newton-Raphson iterative procedure in R-software. If $\mu<0$ , one can obtain the median $M$ on similar lines using (7a).

$\mu$	0.25	0.5	0.75	1	1.25	1.5
$M$	0.6931472	0.6931472	0.6920484	0.6681079	0.6273646	0.5811654

4 Hazard rate function

The reliability function $R(x)=1-G(x)$ for $\mu>0$ is obtained using (7b) as,

R(x)=\begin{cases}\;1&\text{if}\ x<-\mu,\vspace{0.25 cm}\\\displaystyle\;1-\frac{e^{x}}{2\mu}(x+\mu-1)-\frac{e^{-\mu}}{2\mu}&\text{if}\ %-\mu\leqslant x<0,\vspace{0.25 cm}\\\displaystyle\;-\frac{e^{-\mu}}{2\mu}+\frac{e^{-x}}{2\mu}(x+\mu+1)&\text{if}\ %0\leqslant x<\mu,\vspace{0.25 cm}\\\displaystyle\;e^{-x}&\text{if}\ x\geqslant\mu.\end{cases}

(14)

The hazard rate function is an important quantity, characterizing life phenomena. After some simple steps, one can get the hazard function $\displaystyle h(x)=\frac{g(x)}{R(x)}$ for $\mu>0$ as follows.

h(x)=\begin{cases}\;0&\text{if}\ x<-\mu,\vspace{0.25 cm}\\\displaystyle\;\left[-1+\frac{1+(2\mu-e^{-\mu})e^{-x}}{x+\mu}\right]^{-1}&%\text{if}\ -\mu\leqslant x<0,\vspace{0.25 cm}\\\displaystyle\;\left[1+\frac{1-e^{(x-\mu)}}{x+\mu}\right]^{-1}&\text{if}\ 0%\leqslant x<\mu,\vspace{0.25 cm}\\\;1&\text{if}\ x\geqslant\mu.\end{cases}

(15)

One can easily check that $h(x)$ is increasing function of $x$ for $\mu<0$ as well as for $\mu>0$ . Hence, $SSLUD(\mu)$ is increasing failure rate (IFR) distribution.

5 Mean deviation

The amount of scatter in a population is evidently measured to some extent by the totality of deviations from the mean and median. These are known as the mean deviation about the mean and the mean deviation about the median, respectively. Mean deviation about an arbitrary real number ‘ $a$ ’ is defined by $\eta_{a}=E\lvert X-a\rvert$ .

It leads to expression as

\eta_{a}=\begin{cases}\displaystyle\;-\left(1+\frac{2}{\mu}\right)e^{-\mu}+%\left(\frac{2}{\mu}-a\right)&\text{if}\ a<-\mu,\vspace{0.35cm}\\\displaystyle\;\left(\frac{a}{\mu}\right)e^{-\mu}+\left(\frac{a}{\mu}-\frac{2}%{\mu}+1\right)e^{a}+\left(\frac{2}{\mu}-a\right)&\text{if}\ -\mu\leqslant a<0,%\vspace{0.35 cm}\\\displaystyle\;\left(\frac{a}{\mu}\right)e^{-\mu}+\left(\frac{a}{\mu}+\frac{2}%{\mu}+1\right)e^{-a}+\left(a-\frac{2}{\mu}\right)&\text{if}\ 0\leqslant a<\mu,%\vspace{0.35cm}\\\displaystyle\;\left(1+\frac{2}{\mu}\right)e^{-\mu}+2e^{-a}+\left(a-\frac{2}{%\mu}\right)&\text{if}\ a\geqslant\mu.\end{cases}

(16)

To obtain mean deviation about mean and mean deviation about median, ‘ $a$ ’ in the above expression can be replaced by mean and median, respectively.

6 Entropy

The entropy of a random variable $X$ measures the variation of uncertainty. The Rènyi entropy of order $\alpha$ is

H_{\alpha}=\frac{1}{1-\alpha}\,\log_{2}\left\{\int g^{\alpha}(x)dx\right\},\ %\ \alpha>0\ ,\ \alpha\neq 1,

(17)

where $g(x)$ is pdf of random variable $X$ . By using (6), one can write

\int g^{\alpha}(x)dx=I_{1}+I_{2}+\frac{e^{-\alpha x}}{\alpha},

where $I_{1}=\int\displaylimits_{-\mu}^{0}e^{\alpha x}\left(\frac{x}{2\mu}+\frac{1}{2%}\right)^{\alpha}dx=\left(\frac{1}{\alpha}\right)\left(\frac{1}{2}\right)^{%\alpha}\left[\sum_{j=0}^{\alpha}\left(\frac{-1}{\mu\alpha}\right)^{j}\frac{%\alpha!}{(\alpha-j)!}-\left(\frac{-1}{\mu\alpha}\right)^{\alpha}e^{-\mu\alpha}\right]$ and $I_{2}=\int\displaylimits_{0}^{\mu}e^{-\alpha x}\left(\frac{x}{2\mu}+\frac{1}{2%}\right)^{\alpha}dx=\sum_{j=0}^{\alpha}\frac{\alpha!}{(\alpha-j)!(2\mu\alpha)^%{j}}\left[\frac{2^{-(\alpha-j)}-e^{-\mu\alpha}}{\alpha}\right].$

Therefore,

\begin{split}\int g^{\alpha}(x)dx=&\frac{1}{\alpha}\sum_{j=0}^{\alpha}\frac{%\alpha!}{(\alpha-j)!}\left[\frac{2^{-(\alpha-j)}(1+(-1)^{j})-e^{-\mu\alpha}}{(%2\mu\alpha)^{j}}\right]\\&+\frac{e^{-\mu\alpha}}{\alpha}\left[1-\left(\frac{-1}{2\mu\alpha}\right)^{%\alpha}\right].\end{split}

(18)

One can obtain the Rènyi entropy of order $\alpha$ by substituting (18) in (17).

The Shannon entropy function is the particular case of (17) for $\alpha\uparrow 1$ , and it is $H=E[-\log_{2}g(X)]$ , where $g(x)$ is pdf of random variable $X$ . Using this definition, after some simplification we get,

\begin{split}H=&\frac{1}{\ln 2}-\int\displaylimits_{0}^{\mu}\frac{xe^{-x}}{2%\mu}\ \log_{2}\left[\frac{\left(\displaystyle\frac{x}{2\mu}+\frac{1}{2}\right)%}{\left(\displaystyle\frac{-x}{2\mu}+\frac{1}{2}\right)}\right]dx\\&-\int\displaylimits_{0}^{\mu}\frac{e^{-x}}{2}\ \log_{2}\left[\left(\frac{-x}{%2\mu}+\frac{1}{2}\right)\left(\frac{x}{2\mu}+\frac{1}{2}\right)\right]dx.\end{split}

(19)

Since the above integration is cumbersome, we numerically evaluate $H$ for different values of $\mu$ using R-software. Figure 3 represents a graph of $\mu$ $(\mu>0)$ versus $H$ .

The skew-symmetric-Laplace-uniform distribution (5)

7 Estimation

Here, we first consider simulating values of a random variable $X$ with the pdf (6) using the inverse transformation technique. Let $r$ be a random number between zero and one.The generator to generate a random sample is

X=\begin{cases}\;\text{Solution of the equation,}&\text{if}\ 0\leqslant r<G(0)%,\\\;e^{x}(x-1+\mu)+e^{-\mu}-2r\mu=0\vspace{0.35 cm}\\\;\text{Solution of the equation,}&\text{if}\ G(0)\leqslant r<G(\mu),\\\;e^{-x}(-x-1-\mu)+e^{-\mu}+2(1-r)\mu=0\vspace{0.35 cm}\\\;-\ln(1-r)&\text{if}\ G(\mu)\leqslant r\leqslant 1.\par\end{cases}

(20)

One can use the Newton-Raphson method to solve the equation in (20) and generate a random sample from $SSLUD(\mu)$ given in (6).

Now, we consider the estimation of $\mu$ by the method of moments and the method of maximum likelihood. To estimate unknown parameter $\mu$ , we have to consider both the cases $\mu<0$ and $\mu>0$ together. Suppose $x_{1},...,x_{n}$ is an observed random sample of size ‘ $n$ ’ from (6). For the method of moments estimation, after equating sample mean $\overline{x}$ to the first population raw moment of (6), one obtains the equation

\overline{x}=\begin{cases}\displaystyle\;\frac{2}{\mu}+\left(1-\frac{2}{\mu}%\right)e^{\mu}&\quad\text{if}\ \mu<0,\\\displaystyle\;\frac{2}{\mu}-\left(1+\frac{2}{\mu}\right)e^{-\mu}&\quad\text{%if}\ \mu>0.\end{cases}

(21)

From Figure 2, we see that $\mu_{1}^{{}^{\prime}}$ decreases from 0 to -1 when $-\infty<\mu<0$ and it decreases from 1 to 0 when $0<\mu<\infty$ , i.e., always $-1<\mu_{1}^{{}^{\prime}}<1$ . Therefore, if $\overline{x}<-1$ or $\overline{x}>1$ for a particular sample, then (21) will not have an exact solution. As per Figure 2, $\mu$ corresponds to the closest value of $\mu_{1}^{{}^{\prime}}$ to $\overline{x}$ if $\overline{x}<-1$ or $\overline{x}>1$ is a value close to zero. But as per parameter space, $\mu$ can not take the value zero. Hence, we define the moment estimator $\tilde{\mu}$ of $\mu$ as $-10^{-5}$ if $\overline{x}<-1$ and $10^{-5}$ if $\overline{x}>1$ . Thus, the moment estimator $\tilde{\mu}$ of $\mu$ is obtained as follows.

\tilde{\mu}=\begin{cases}\;-10^{-5}&\quad\text{if}\ \ \overline{x}\leqslant-1,%\vspace{0.35 cm}\\\;\text{Solution of the equation,}&\quad\text{if}\ \ -1<\overline{x}<0,\\\;\displaystyle\frac{2}{\mu}+\left(1-\frac{2}{\mu}\right)e^{\mu}-\overline{x}=%0\vspace{0.35 cm}\\\;\text{Solution of the equation,}&\quad\text{if}\ \ 0\leqslant\overline{x}<1,%\\\;\displaystyle\frac{2}{\mu}-\left(1+\frac{2}{\mu}\right)e^{-\mu}-\overline{x}%=0\vspace{0.35 cm}\\\;10^{-5}&\quad\text{if}\ \ \overline{x}\geqslant 1.\end{cases}

(22)

We consider the estimation of $\mu$ by the method of maximum likelihood in the following. Let $x_{(1)},x_{(2)},\ldots,x_{(n)}$ be the order statistics of given sample. Suppose $\mu<0$ and ‘ $r_{1}$ ’ denotes the number of observations less than $\mu$ such that $-\infty<x_{(1)}<x_{(2)}<\ldots<x_{(r_{1})}\leqslant\mu\leqslant x_{(r_{1}+1)}<%\ldots<x_{(n)}<-\mu<\infty$ , i.e. $-\infty<\mu<\min(0,\,-x_{(n)})$ where $r_{1}=0,1,2,\ldots,n$ . Similarly, suppose $\mu>0$ and ‘ $r_{2}$ ’ denotes the number of observations lying in the interval $[-\mu,\mu]$ such that $-\infty<-\mu<x_{(1)}<x_{(2)}<\ldots<x_{(r_{2})}\leqslant\mu\leqslant x_{(r_{2}%+1)}<\ldots<x_{(n)}<\infty$ , i.e. $\max(0,\,-x_{(1)})<\mu<\infty$ where $r_{2}=0,1,2,\ldots,n$ . Hence, the log-likelihood function of $\mu$ is written as

l=\begin{cases}\;l_{1}=\displaystyle-\sum\limits_{i=1}^{n}\lvert x_{(i)}\rvert%+\sum\limits_{i=r_{1}+1}^{n}ln\left(\frac{x_{(i)}+\mu}{2\mu}\right)&\text{if}%\ -\infty<\mu<\min\{0,\,-x_{(n)}\},\\\;l_{2}=\displaystyle-\sum\limits_{i=1}^{n}\lvert x_{(i)}\rvert+\sum\limits_{i%=1}^{r_{2}}ln\left(\frac{x_{(i)}+\mu}{2\mu}\right)&\text{if}\ \max\{-x_{(1)},%\,0\}<\mu<\infty.\end{cases}

(23)

In the following, we give a step-wise procedure for computation of the MLE $\hat{\mu}$ of $\mu$ .

Step 1:
Numerically maximize $l_{1}$ over the range $(-a,\min\{0,-x_{(n)}\})$ . Suppose the maximum value of $l_{1}$ is $\hat{l}_{1}$ which is attained at $\hat{\mu}_{1}$ , say, where ‘ $a$ ’ is a sufficiently large positive number chosen for computation purposes.
Step 2:
Numerically maximize $l_{2}$ over the range $(\max\{-x_{(1)},0\},a)$ . Suppose the maximum value of $l_{2}$ is $\hat{l}_{2}$ which is attained at $\hat{\mu}_{2}$ , say.

Step 3:

MLE $\hat{\mu}$ of $\mu$ is

\displaystyle\hat{\mu}=

\displaystyle\begin{cases}\;\hat{\mu}_{1}\quad\text{if}\ \hat{l}_{1}>\hat{l}_{%2},\\\;\hat{\mu}_{2}\quad\text{otherwise}.\end{cases}

(24)

Finite sample properties of $\tilde{\mu}$ and $\hat{\mu}$ are studied using simulation, and computations are done using R- software. Table 2 and Table 3 presents bias and MSE of $\tilde{\mu}$ and $\hat{\mu}$ for $n=100(100)1000$ and for $\mu=-1.5,-0.75,-0.25,0.25,0.75,1.5$ . We see that bias and MSE decrease as sample size $n$ increases for both MLE $\hat{\mu}$ and moment estimator $\tilde{\mu}$ , with few exceptions only for bias. Further, the MSE of $\hat{\mu}$ is always less than the corresponding MSE of $\tilde{\mu}$ . Also, one can observe that sign of bias of MLE $\hat{\mu}$ is opposite to the sign of parameter $\mu$ . As parameter $\mu$ approaches zero from any side, MSE and magnitude of bias of $\hat{\mu}$ decrease. But, no such observation in the case of the moment estimator $\tilde{\mu}$ . To check the asymptotic nature of the distribution of $\hat{\mu}$ and $\tilde{\mu}$ using simulation, we plotted observed densities for various values of the sample size $n$ . We observe that as $n$ increases, the distribution of both $\hat{\mu}$ and $\tilde{\mu}$ converges to the normal distribution, but the rate of convergence to normal distribution seems to be much higher for $\hat{\mu}$ than $\tilde{\mu}$ . Thus, based on all the above results, we conclude that MLE is better than the moment estimator of $\mu$ for SSLUD( $\mu$ ).

$\mu$	$n$	MLE		Moment estimator
$\mu$	$n$	Bias	MSE	Bias	MSE
-1.5	100	0.06024381	0.045176654	0.0003797204	0.50541564
	200	0.03906522	0.021357856	0.0157055747	0.26786436
	300	0.02933963	0.012989993	-0.0008644177	0.16465504
	400	0.02375948	0.009209691	-0.0085926100	0.12896987
	500	0.01766114	0.007451091	-0.0065899772	0.10033671
	600	0.01808905	0.006370478	0.0075580998	0.08311324
	700	0.01456689	0.005214147	-0.0003676365	0.06853064
	800	0.01492506	0.004456640	-0.0164346584	0.05661172
	900	0.01381575	0.003861085	-0.0079539753	0.05188121
	1000	0.01378036	0.003294051	0.0052817371	0.04687188
-0.75	100	0.032511901	0.0134889603	-0.02488011	0.37927153
	200	0.023375017	0.0062341973	0.02678563	0.24807994
	300	0.022614234	0.0040753483	0.05442608	0.18159122
	400	0.013782428	0.0026481775	0.02407104	0.14540491
	500	0.014188578	0.0019942288	0.04011415	0.12661616
	600	0.010531113	0.0016580024	0.02741234	0.10965835
	700	0.009567891	0.0014142729	0.02247448	0.09215384
	800	0.008969822	0.0012107034	0.03299546	0.08416938
	900	0.008833928	0.0010258905	0.02604549	0.07738691
	1000	0.007307065	0.0009473139	0.01597401	0.06532169
-0.25	100	0.026910685	0.0043941552	-0.20148295	0.28701317
	200	0.014633591	0.0016442189	-0.12561277	0.18340413
	300	0.011298482	0.0010134272	-0.11008763	0.14648063
	400	0.008403715	0.0007178242	-0.07115405	0.11693260
	500	0.006866262	0.0005193217	-0.05475621	0.10104767
	600	0.006577965	0.0004359336	-0.04101810	0.09227325
	700	0.005574953	0.0003294627	-0.03931742	0.08363312
	800	0.005196938	0.0002914057	-0.02245561	0.07867760
	900	0.004681950	0.0002552058	-0.03305608	0.07551096
	1000	0.003797666	0.0002160883	-0.01393033	0.06746986

$\mu$	$n$	MLE		Moment estimator
$\mu$	$n$	Bias	MSE	Bias	MSE
0.25	100	-0.026141601	0.0040926127	0.19908173	0.30299408
	200	-0.015706993	0.0017378035	0.11979148	0.18277572
	300	-0.010537695	0.0009629724	0.10281278	0.15122684
	400	-0.009388154	0.0006823647	0.07138595	0.11273918
	500	-0.007654403	0.0005241705	0.06670007	0.10613451
	600	-0.006541976	0.0004273222	0.04710248	0.09317236
	700	-0.005775066	0.0003498924	0.03681952	0.08264223
	800	-0.005424833	0.0003076071	0.02613340	0.07893736
	900	-0.004687041	0.0002641379	0.02308814	0.07288723
	1000	-0.004587965	0.0002188929	0.02168387	0.07098162
0.75	100	-0.039403641	0.0141163154	0.02381151	0.36120747
	200	-0.023215667	0.0057148979	-0.01903135	0.24011612
	300	-0.018010206	0.0037173032	-0.04910507	0.18917471
	400	-0.013004298	0.0026953723	-0.02435556	0.13988301
	500	-0.011937422	0.0020292707	-0.04849263	0.13064004
	600	-0.010423756	0.0016294899	-0.02277927	0.10673985
	700	-0.009229675	0.0013339162	-0.03201923	0.09343322
	800	-0.009286969	0.0011661790	-0.02381294	0.08053673
	900	-0.008850638	0.0010360152	-0.02366172	0.07657251
	1000	-0.008957869	0.0009469547	-0.01562601	0.06498910
1.5	100	-0.05884554	0.047383765	-0.013127745	0.54926083
	200	-0.03000957	0.020339855	-0.006090958	0.25400260
	300	-0.03186303	0.013144354	-0.010153344	0.16976755
	400	-0.02284458	0.009625618	-0.002482506	0.13057787
	500	-0.02293975	0.007759937	-0.008421723	0.09775602
	600	-0.01841574	0.005969112	-0.009321518	0.08234397
	700	-0.01724714	0.005152509	-0.004056375	0.06780765
	800	-0.01197158	0.004282729	-0.008411236	0.06338525
	900	-0.01484989	0.004060138	-0.009773787	0.05273530
	1000	-0.01228560	0.003408502	0.003389947	0.04925195

8 Application

In this section, we present the application of skew-symmetric-Laplace-uniform distribution for modeling daily percentage change in the price of NIFTY 50, an Indian stock market index. Further, we have fitted and compared the proposed distribution $SSLUD(\mu)$ with normal distribution $N(\theta,\sigma^{2})$ , Laplace distribution $L(\theta,\beta)$ , and skew-Laplace distribution $SL(\lambda)$ for percentage change data. Here, $SL(\lambda)$ refers to a special case of skew-Laplace distribution using $f$ and $K$ as pdf and cdf of standard Laplace distribution in (1). The NIFTY 50 is a benchmark Indian stock market index representing the weighted average of 50 of the largest Indian companies listed on the National Stock Exchange (NSE). It is one of the two leading stock indices used in India. The daily price of NIFTY 50 quoted in the National Stock Exchange of India Ltd. is available at https://in.investing.com/indices/s-p-cnx-nifty-historical-data and is selected for the current study. We consider the daily percentage change $Y_{t}$ on day t given by $\displaystyle Y_{t}=\frac{X_{t}\,-\,X_{t-1}}{X_{t-1}}\times 100$ , where $X_{t}$ denotes the price of NIFTY 50 on day t. This transformed data covering the period $16^{th}$ December 2021 to $13^{th}$ April 2022 (82 working days) is as follows :
0.16, - 1.53, - 2.18, 0.94, 1.10, 0.69, - 0.40, 0.49, 0.86, - 0.11, - 0.06, 0.87, 1.57, 1.02, 0.67, - 1.00, 0.38, 1.07, 0.29, 0.87, 0.25, - 0.01, 0.29, - 1.07, - 0.96, - 1.01, - 0.79, - 2.66, 0.75, - 0.97, - 0.05, 1.39, 1.37, 1.16, - 1.24, - 0.25, - 1.73, 0.31, 1.14, 0.81, - 1.31, - 3.06, 3.03, - 0.17, - 0.10, - 0.16, - 0.40, - 0.67, - 0.17, - 4.78, 2.53, 0.81, - 1.12, - 0.65, - 1.53, - 2.35, 0.95, 2.07, 1.53, 0.21, 1.45, - 1.23, 1.87, 1.84, - 0.98, 1.16, - 0.40, - 0.13, - 0.40, 0.40, 0.60, 1.00, - 0.19, 1.18, 2.17, - 0.53, - 0.83, - 0.94, 0.82, - 0.62, - 0.82, - 0.31.

Mean, variance, and skewness for the above data is 0.027, 1.671, and - 0.639 respectively.The Wald-Wolfowitz runs test for randomness of $Y_{t}$ yields a p-value of 0.076, justifying the assumption of independence of the $Y_{t}$ values. We consider fitting the proposed skew-symmetric-Laplace-uniform distribution $SSLUD(\mu)$ along with normal distribution $N(\theta,\sigma^{2})$ , Laplace distribution $L(\theta,\beta)$ , and skew-Laplace distribution $SL(\lambda)$ to the data on percentage change. Using R-software, the MLE of the parameters and hence, the estimated value of log-likelihood are obtained. Akaike’s Information Criteria (AIC) and Bayesian Information Criteria (BIC) are used for model comparison. Table 4 shows that the proposed $SSLUD(\mu)$ provides the best fit for the data set which is very close to $SL(\lambda)$ in terms of BIC. But in terms of AIC, N( $\theta,\sigma^{2}$ ) seems to be better than $SSLUD(\mu)$ and the best among the four distributions.

Distribution	MLEs	ln $L$	AIC	BIC
$SSLUD(\mu)$	$\hat{\mu}$ = 62.38674	- 138.7604	279.5207	281.9274
$SL(\lambda)$	$\hat{\lambda}$ = -6.247468e-05	- 138.7782	279.5564	281.9631
$L(\theta,\beta)$	$\hat{\theta}$ = - 0.03, $\hat{\beta}$ = 0.9990244	- 138.7580	281.5161	286.3295
$N(\theta,\sigma^{2})$	$\hat{\theta}$ = 0.02682927, $\hat{\sigma}^{2}$ = 1.650275	- 136.9081	277.8162	282.6296

The skew-symmetric-Laplace-uniform distribution (6)

For $SSLUD(\mu)$ , MLE of $\mu$ is $\hat{\mu}$ =62.38674 which is relatively high, and by definition of $g(x)$ in (6) for a large value of $\mu$ , SSLUD approaches to Laplace distribution. But, from the histogram in Figure 4 and the value of skewness for $Y_{t}$ , one can observe that data is negatively skewed. It might be due to a single parameter in the proposed distribution unable to give the best fit to the data. By changing the data location, significant change observed in SSLUD’s curve in terms of location, scale, and shape. So, by observation, one can choose an appropriate change in location such that the value of $\hat{\mu}$ is significantly small for possible better fitting of data using the proposed distribution. Through the trial and error method, here we consider a change as - 0.8 and define transformed daily percentage change in Nifty 50 index price, $Z_{t}=Y_{t}-0.8$ which gives $\hat{\mu}=-2.589259$ , a significantly small value. A possible generalization of proposed distribution with additional location parameter to avoid hindrance to employ it is under consideration, in order to make it more flexible and apt to catch the features present in real data.

Table 5 shows the MLEs, estimated log-likelihood, AIC, and BIC by fitting the distributions mentioned above to $Z_{t}$ . The graphical representation of the results is given in Figure 5. It is clear from Table 5 that the proposed $SSLUD(\mu)$ provides the best fit for the data set in terms of both AIC and BIC, but close to $SL(\lambda)$ . The plot of observed and expected densities presented in Figure 5 also confirms our findings. Thus, $SSLUD(\mu)$ is better for modeling daily percentage change in the price of NIFTY 50 in comparison to normal distribution $N(\theta,\sigma^{2})$ and Laplace distribution $L(\theta,\beta)$ , and one good alternative to skew-Laplace distribution $SL(\lambda)$ .

Distribution	MLEs	ln $L$	AIC	BIC
$SSLUD(\mu)$	$\hat{\mu}$ = - 2.589259	- 136.8343	275.6685	278.0752
$SL(\lambda)$	$\hat{\lambda}$ = - 0.6988722	- 137.0020	276.0040	278.4107
$L(\theta,\beta)$	$\hat{\theta}$ = - 0.83, $\hat{\beta}$ = 0.9990244	- 138.7580	281.5161	286.3295
$N(\theta,\sigma^{2})$	$\hat{\theta}$ = - 0.7731707, $\hat{\sigma}^{2}$ = 1.650275	- 136.9081	277.8162	282.6296

The skew-symmetric-Laplace-uniform distribution (7)

Declarations

Conflict of interest

The authors declare that they have no conflict of interest to disclose.

Ethics approval and consent to participate

Not applicable.

Consent for publication

Both authors have agreed to submit and publish this paper.

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