**2. A Skew Logistic Distribution**

Here, we introduce a novel *skew logistic distribution*, which extends, in straightforward manner, the standard two parameter logistic distribution [3] and [4] (Chapter 22) by adding to it a skew parameter. The rationale for introducing the distribution is that, apart from its simple formulation, we believe that the maximum likelihood solution presented below is also simpler than those derived for other skew logistic distributions, such as the ones investigated in [8,9]. This point provides further justification for our skew logistic distribution when introducing the bi-skew logistic distribution in Section 4.

Now, let *μ* be a location parameter, *s* be a scale parameter and *λ* be a skew parameter, where *s* > 0 and 0 < *λ* < 2. Then, the probability density function of the skew logistic distribution at a value *x* of the random variable *X*, denoted as *f*(*x*; *λ*, *μ*,*s*), is given by:

$$f(x; \lambda, \mu, s) = \kappa\_{\lambda} \frac{\exp\left(-\lambda \frac{x - \mu}{s}\right)}{s \left(1 + \exp\left(-\frac{x - \mu}{s}\right)\right)^2} \tag{1}$$

noting that for clarity we write *<sup>x</sup>* <sup>−</sup> *<sup>μ</sup>* above as a shorthand for *x* − *μ* , and *κλ* is a normalisation constant, which depends on *λ*.

When *λ* = 1, the skew logistic distribution reduces to the standard logistic distribution as in [3] and [4] (Chapter 22), which is symmetric. On the other hand, when 0 < *λ* < 1, the skew logistic distribution is positively skewed, and when 1 < *λ* < 2, it is negatively skewed. So, when *λ* = 1, *κλ* = 1, and, for example, when *λ* = 0.5 or 1.5, *κλ* = 2/*π*. For simplicity, from now on, unless necessary, we will omit to mention the constant *κλ* as it will not effect any of the results.

The *skewness* of a random variable *X* [4,5], is defined as:

$$\mathbb{E}\left[\left(\frac{X-\mu}{s}\right)^3\right],$$

and thus, assuming for simplicity of exposition (due the linearity of expectations [5]) that *μ* = 0 and *s* = 1, the skewness of the skew logistic distribution, denoted by *γ*(*λ*), is given by:

$$\gamma(\lambda) = \int\_{-\infty}^{\infty} x^3 \, \frac{\exp(-\lambda x)}{s \left(1 + \exp(-x)\right)^2} \, dx. \tag{2}$$

First, we will show that letting *λ*<sup>1</sup> = *λ*, with 0 < *λ*<sup>1</sup> < 1, we have *γ*(*λ*1) > 0, that is *f*(*x*; *λ*1, 0, 1) is positively skewed. We can split the integral in (2) into two integrals for the negative part from −∞ to 0 and the positive part from 0 to ∞, noting that when *x* = 0, the expression to the right of the integral is equal to 0. Then, on setting *y* = −*x* for the negative part, and *y* = *x* for the positive part, the result follows, as by algebraic manipulation it can be shown that:

$$\frac{\exp(-\lambda\_1 y)}{\left(1 + \exp(-y)\right)^2} > \frac{\exp(\lambda\_1 y)}{\left(1 + \exp(y)\right)^2} \tag{3}$$

implying that *γ*(*λ*1) > 0 as required.

Second, in a similar fashion to above, on letting *λ*<sup>2</sup> = *λ*<sup>1</sup> + 1 = *λ*, with 1 < *λ*<sup>2</sup> < 2, it follows that *γ*(*λ*2) < 0, that is *f*(*x*; *λ*2, 0, 1) is negatively skewed. In particular, by algebraic manipulation we have that:

$$\frac{\exp\left(-\lambda\_2 y\right)}{\left(1 + \exp(-y)\right)^2} < \frac{\exp\left(\lambda\_2 y\right)}{\left(1 + \exp(y)\right)^2} \tag{4}$$

implying that *γ*(*λ*2) < 0 as required.

The cumulative distribution function of the skew logistic distribution at a value *x* of the random variable *X* is obtained by integrating *f*(*x*; *λ*, *μ*,*s*), to obtain *F*(*x*; *μ*,*s*, *λ*), which is given by:

$$F(\mathbf{x}; \lambda, \mu, s) = \kappa\_{\lambda} \exp\left(-(\lambda - 2)\frac{\mathbf{x} - \mu}{s}\right) \left(\frac{1}{\left(1 + \exp\left(\frac{\mathbf{x} - \mu}{s}\right)\right)} - 1\right) \tag{5}$$

$$\frac{\lambda - 1}{\lambda - 2} \,\_2F\_1\left(1, 2 - \lambda; 3 - \lambda; -\exp\left(\frac{\mathbf{x} - \mu}{s}\right)\right) \,\_2F\_1\left(\frac{\lambda}{s}, \frac{\lambda}{s}\right) \tag{6}$$

where <sup>2</sup>*F*1(*a*, *b*; *c*; *z*) is the *Gauss hypergemoetric function* [29] (Chapter 15); we assume *a*, *b* and *c* are positive real numbers, and that *z* is a real number extended outside the unit disk by analytic continuation [30].

The hypergeometric function has the following integral representation [29] (Chapter 15),

$$\frac{\Gamma(c)}{\Gamma(b)\Gamma(c-b)} \int\_0^1 \frac{t^{b-1}(1-t)^{c-b-1}}{(1-tz)^a} \, dt,\tag{6}$$

where *c* > *b*. Now, assuming without loss of generality that *μ* = 0 and *s* = 1, we have that:

$$\,\_2F\_1\left(1,2-\lambda;3-\lambda;-\exp(x)\right) = \left(2-\lambda\right)\int\_0^1 \frac{t^{1-\lambda}}{\left(1+t\ \exp(x)\right)}\,dt,\tag{7}$$

where *x* is a real number.

Therefore, from (7) it can be verified that: (i) <sup>2</sup>*F*1(1, 2 − *λ*; 3 − *λ*; − exp(*x*)) is monotonically decreasing with *x*, (ii) as *x* tends to plus infinity, <sup>2</sup>*F*1(1, 2 − *λ*; 3 − *λ*; − exp(*x*)) tends to 0 and (iii) as *x* tends to minus infinity, <sup>2</sup>*F*1(1, 2 − *λ*; 3 − *λ*; − exp(*x*)) tends to 1, since:

$$(2 - \lambda) \int\_0^1 t^{1 - \lambda} \, dt = 1.$$

#### **3. Maximum Likelihood Estimation for the Skew Logistic Distribution**

We now formulate the maximum likelihood estimation [31] of the parameters *μ*,*s* and *λ* of the skew logistic distribution. Let {*x*1, *x*2, ... , *xn*} be a random sample of *n* values from the density function of the skew logistic distribution in (1). Then, the log likelihood function of its three parameters is given by:

$$\ln L(\lambda, \mu, s) = -n \ln(s) - \frac{\lambda}{s} \sum\_{i=1}^{n} (\mathbf{x}\_i - \mu) - 2 \sum\_{i=1}^{n} \ln \left( 1 + \exp \left( -\frac{\mathbf{x}\_i - \mu}{s} \right) \right). \tag{8}$$

In order to solve the log likelihood function, we first partially differentiate ln *L*(*λ*, *μ*,*s*) as follows:

$$\begin{split} \frac{\partial \ln L(\lambda, \mu, s)}{\partial \lambda} &= \sum\_{i=1}^{n} \frac{\mu - \mu\_{i}}{s}, \\ \frac{\partial \ln L(\lambda, \mu, s)}{\partial \mu} &= \frac{\lambda n}{s} - \frac{2}{s} \sum\_{i=1}^{n} \frac{1}{1 + \exp\left(\frac{x\_{i} - \mu}{s}\right)} \text{ and} \\ \frac{\partial \ln L(\lambda, \mu, s)}{\partial s} &= -\frac{n}{s} + \frac{1}{s^{2}} \sum\_{i=1}^{n} (x\_{i} - \mu) \left(\lambda - \frac{2}{1 + \exp\left(\frac{x\_{i} - \mu}{s}\right)}\right). \end{split} \tag{9}$$

It is therefore implied that the maximum likelihood estimators are the solutions to the following three equations:

$$\begin{aligned} \mu &= \frac{\sum\_{i=1}^{n} x\_i}{n}, \\ \lambda &= \frac{2}{n} \sum\_{i=1}^{n} \frac{1}{1 + \exp\left(\frac{x\_i - \mu}{s}\right)} \text{ and} \\ s &= \frac{1}{n} \sum\_{i=1}^{n} (x\_i - \mu) \left(\lambda - \frac{2}{1 + \exp\left(\frac{x\_i - \mu}{s}\right)}\right), \end{aligned} \tag{10}$$

which can be solved numerically.

We observe that the equation for *μ* in (10) does not contribute to solving the maximum likelihood, since the location parameter *μ* is equal to the mean only when *λ* = 1. We thus look at an alternative equation for *μ*, which involves the mode of the skew logistic distribution.

To derive the mode of the skew logistic distribution we solve the equation,

$$\frac{\partial}{\partial x} \frac{\exp\left(-\lambda \frac{x-\mu}{s}\right)}{s\left(1+\exp\left(-\frac{x-\mu}{s}\right)\right)^2} = 0,\tag{11}$$

to obtain:

$$
\mu = x - s \, \log \left( -\frac{\lambda - 2}{\lambda} \right). \tag{12}
$$

Thus, motivated by (12) we replace the equation for *μ* in (10) with:

$$
\mu = m - s \, \log \left( -\frac{\lambda - 2}{\lambda} \right),
\tag{13}
$$

where *m* is the mode of the random sample.
