Full article: Further characterization based on conditional expectations: new and extended findings

Formulae display: $MathJax Logo$ ?Mathematical formulae have been encoded as MathML and are displayed in this HTML version using MathJax in order to improve their display. Uncheck the box to turn MathJax off. This feature requires Javascript. Click on a formula to zoom.

ABSTRACT

On the basis of conditional expectation of a random variable function, we present few characterization findings in this study. For a given function g, $g (x) = E (D (X) | X \geq x)$ , we present necessary and sufficient conditions for characterization results in terms of a single function $D (x)$ . Some of these findings are completely novel, while others are expansions of previously published characterizations.

KEYWORDS:

1. Introduction

Many scholars have explored probability distribution characterizations. According to [Citation1], a characterization theorem occurs in probability and statistics when a certain distribution is the only one that fits a specified property. Furthermore, a characterization, according to [Citation2], is a specific statistical or distributional attribute of a statistic or statistics that uniquely defines the associated stochastic model. Some academics suggested that before applying a probability distribution to real-world data, it should first be characterized under certain conditions. Several kinds of characterizations have been investigated for instance characterizations based on hazard functions, conditional expectations, truncated moments, order statistics and record values. Several authors, including [Citation3–8], have explored characterizations based on conditional expectations. In general, see, for example, for a survey or further information on these characterization subjects [Citation1, Citation2, Citation9–19] and references therein, to name a few, however there are numerous others.

We present some new and extended characterizations based on conditional expectations, motivated by the importance of probability distribution characterizations. The results of characterization are expressed in terms of a single function expression $H$ of X, say $H (x)$ .

2. Main results

In this paper, we present characterizations based on the following conditional expectations in the form $E (D (X) | X \geq x)$ where $D$ is a mathematical expression of the function $H$ : (1) $\begin{aligned} E ({(H (X))}^{a} | X \geq x) \\ = b^{a}_{2} F_{1} (a, a - r; a - r + 1; 1 - \frac{b}{H (x)}), \end{aligned}$ (1) (2) $\begin{aligned} E ({(H (X))}^{a} | X \geq x) = r + s {(H (x))}^{a}, \end{aligned}$ (2) (3) $\begin{aligned} E ({(H (X))}^{a} | X \geq x) = r {(H (x))}^{a} + s {(H (x))}^{a - 1}, \end{aligned}$ (3) (4) $\begin{aligned} E (H (X) | X \geq x) = r {(H (x))}^{a} + H (x), \end{aligned}$ (4) (5) $\begin{aligned} E ({(H (X))}^{a} e^{[c {(H (X))}^{b}]} | X \geq x) \\ = r + s {(H (x))}^{a} e^{[c {(H (x))}^{b}]}, \end{aligned}$ (5) (6) $\begin{aligned} E ({[{(H (X))}^{b} \ln {(H (X))}^{c}]}^{a} | X \geq x) \\ = r + s {[{(H (x))}^{b} \ln {(H (x))}^{c}]}^{a}, \end{aligned}$ (6) such that $H$ is a continuous and function that is differentiable on the interval $(L, U)$ .

We are expanding certain works founded by [Citation20–22] as presented in Propositions 2.1–2.4, respectively, by using Equations (Equation1(1) $\begin{aligned} E ({(H (X))}^{a} | X \geq x) \\ = b^{a}_{2} F_{1} (a, a - r; a - r + 1; 1 - \frac{b}{H (x)}), \end{aligned}$ (1) )–(Equation4(4) $\begin{aligned} E (H (X) | X \geq x) = r {(H (x))}^{a} + H (x), \end{aligned}$ (4) ). On the other side, as far as we know, the Equations (Equation5(5) $\begin{aligned} E ({(H (X))}^{a} e^{[c {(H (X))}^{b}]} | X \geq x) \\ = r + s {(H (x))}^{a} e^{[c {(H (x))}^{b}]}, \end{aligned}$ (5) ) and (Equation6(6) $\begin{aligned} E ({[{(H (X))}^{b} \ln {(H (X))}^{c}]}^{a} | X \geq x) \\ = r + s {[{(H (x))}^{b} \ln {(H (x))}^{c}]}^{a}, \end{aligned}$ (6) ) introduce new characterizations based on conditional expectations, as shown in the corresponding Propositions 2.5 and 2.6, respectively.

Proposition 2.1

Let $X : Ω \to (L, U)$ be a random variable that is continuous with cdf F. Let $H (x)$ be function that is differentiable on $(L, U)$ with $lim_{x \to L^{+}} H (x) = H (L)$ and $lim_{x \to U^{-}} H (x) = b$ if $a > r$ and $H (x) > \frac{b}{2}$ . Then, $\begin{aligned} E ({(H (X))}^{a} | X \geq x) \\ = b^{a}_{2} F_{1} (a, a - r; a - r + 1; 1 - \frac{b}{H (x)}) \end{aligned}$ if and only if (7) $H (x) = \frac{b}{1 + (\frac{b - H (L)}{H (L)}) {[1 - F_{X} (x)]}^{(\frac{1}{a - r})}}$ (7) where $x \in (L, U)$ and $_{2} F_{1}$ is the Gaussian hypergeometric function defined as $_{2} F_{1} (a, b; c; z) = \sum_{i = 0}^{\infty} \frac{{(a)}_{i} {(b)}_{i}}{{(c)}_{i}} \frac{z^{i}}{i!}$ , see Equation (15.1.1) at page 556 in [Citation23].

Proof.

From Equation (Equation1(1) $\begin{aligned} E ({(H (X))}^{a} | X \geq x) \\ = b^{a}_{2} F_{1} (a, a - r; a - r + 1; 1 - \frac{b}{H (x)}), \end{aligned}$ (1) ), we have $\begin{aligned} \int_{x}^{U} {(H (t))}^{a} f_{X} (t) d t \\ = b^{a} (1 - F_{X} (x))_{2} F_{1} \\ \times (a, a - r; a - r + 1; 1 - \frac{b}{H (x)}) \end{aligned}$ Using derivatives from both sides in relation to x, we obtain $\begin{aligned} [b^{a}_{2} F_{1} (a, a - r; a - r + 1; 1 - \frac{b}{H (x)}) - {(H (x))}^{a}] \\ \times f_{X} (x) = (\frac{a b^{(a + 1)} (a - r)}{(a - r + 1)}) \\ \times (\frac{H^{'} (x)}{{(H (x))}^{2}}) (1 - F_{X} (x))_{2} F_{1} \\ \times (a + 1, a - r + 1; a - r + 2; 1 - \frac{b}{H (x)}), \end{aligned}$ where $\frac{d}{d x}_{2} F_{1} (a, b; c; q (x)) = (\frac{ab}{c}) q^{'} (x)_{2} F_{1} (a + 1, b + 1; c + 1; q (x))$ , see Equation (15.2.1) at page 557 in [Citation23].

From which we have $\begin{aligned} \frac{f_{X} (x)}{1 - F_{X} (x)} = (\frac{a b^{(a + 1)} (a - r)}{(a - r + 1)}) \\ \times \frac{\begin{matrix} H^{'} (x)_{2} F_{1} (a + 1, a - r + 1; a - r + 2; 1 - \frac{b}{H (x)}) \end{matrix}}{\begin{matrix} {(H (x))}^{2} [b^{a}_{2} F_{1} (a, a - r; a - r + 1; 1 - \frac{b}{H (x)}) \\ - {(H (x))}^{a}] \end{matrix}} \end{aligned}$ Integrating both sides from L to x, $\begin{aligned} \int_{L}^{x} \frac{f_{X} (t)}{1 - F_{X} (t)} d t = (\frac{a b^{(a + 1)} (a - r)}{(a - r + 1)}) \\ \times \int_{L}^{x} {\frac{\begin{matrix} H^{'} (t)_{2} F_{1} \\ (a + 1, a - r + 1; a - r + 2; 1 - \frac{b}{H (t)}) \end{matrix}}{\begin{matrix} {(H (t))}^{2} \\ [b^{a}_{2} F_{1} (a, a - r; a - r + 1; 1 - \frac{b}{H (t)}) \\ - {(H (t))}^{a}] \end{matrix}}} d t . \end{aligned}$ Let $u = 1 - F_{X} (t), v = (\frac{1}{H (t)})$ , then

(8) $\begin{aligned} \int_{1 - F_{X} (x)}^{1} \frac{d u}{u} \\ = (\frac{a b^{(a + 1)} (a - r)}{(a - r + 1)}) \times \int_{(\frac{1}{H (x)})}^{(\frac{1}{H (L)})} \\ \times {\frac{_{2} F_{1} (a + 1, a - r + 1; a - r + 2; 1 - bv)}{[b^{a}_{2} F_{1} (a, a - r; a - r + 1; 1 - bv) - {(\frac{1}{v})}^{a}]}} d v \\ \ln (1) - \ln (1 - F_{X} (x)) = b (a - r) \\ \times \int_{(\frac{1}{H (x)})}^{(\frac{1}{H (L)})} {\frac{\begin{matrix} [bv_{2} F_{1} (1, 1 - r; a - r + 1; 1 - bv) - 1] \end{matrix}}{\begin{matrix} (bv - 1) [{(bv)}^{a}_{2} F_{1} \\ (a, a - r; a - r + 1; 1 - bv) - 1] \end{matrix}}} d v, \end{aligned}$ (8) where $_{2} F_{1} (a + 1, a - r + 1; a - r + 2; 1 - bv) = \frac{(a - r + 1) [bv_{2} F_{1} (1, 1 - r; a - r + 1; 1 - bv) - 1]}{a (bv - 1) {(bv)}^{a}}$ by using Equations (1.110), (9.131.1) and (9.137.11) at pages 25, 1008 and 1010, respectively in [Citation24]. Therefore, $\begin{aligned} - \ln (1 - F_{X} (x)) \\ = b (a - r) \int_{(\frac{1}{H (x)})}^{(\frac{1}{H (L)})} \\ \times {\frac{[bv_{2} F_{1} (1, 1 - r; a - r + 1; 1 - bv) - 1]}{\begin{matrix} (bv - 1) \\ [bv_{2} F_{1} (1, 1 - r; a - r + 1; 1 - bv) - 1] \end{matrix}}} d v, \end{aligned}$ where $[{(bv)}^{a}_{2} F_{1} (a, a - r; a - r + 1; 1 - bv) - 1] =$ $[bv_{2} F_{1} (1, 1 - r; 1 + a - r; 1 - bv) - 1],$ see Equation (9.131.1) at page 1008 in [Citation24]. From which we have $- \ln (1 - F_{X} (x)) = b (a - r) \int_{(\frac{1}{H (x)})}^{(\frac{1}{H (L)})} \frac{1}{(bv - 1)} d v .$ Let w = bv−1 then after simplification, we have (9) $\begin{aligned} F_{X} (x) = 1 - \exp {(a - r) \\ [\ln [(\frac{b - H (x)}{H (x)}) (\frac{H (L)}{b - H (L)})]]} \Rightarrow \\ F_{X} (x) = 1 - {[(\frac{b - H (x)}{H (x)}) (\frac{H (L)}{b - H (L)})]}^{(a - r)} \Rightarrow \\ (\frac{b - H (L)}{H (L)}) {[1 - F_{X} (x)]}^{(\frac{1}{(a - r)})} = (\frac{b}{H (x)} - 1) \\ ∴ H (x) = \frac{b}{1 + (\frac{b - H (L)}{H (L)}) {[1 - F_{X} (x)]}^{(\frac{1}{(a - r)})}} \end{aligned}$ (9) Conversely, if Equation (Equation7(7) $H (x) = \frac{b}{1 + (\frac{b - H (L)}{H (L)}) {[1 - F_{X} (x)]}^{(\frac{1}{a - r})}}$ (7) ) holds, then $\begin{aligned} E ({(H (X))}^{a} | X \geq x) \\ = \int_{x}^{U} {\frac{1 + (\frac{b - H (L)}{H (L)}) {[1 - F_{X} (t)]}^{(\frac{1}{a - r})}}{b}}^{(- a)} \\ \times \frac{f_{X} (t)}{1 - F_{X} (x)} d t \\ = \frac{b^{a}}{1 - F_{X} (x)} \sum_{i = 0}^{\infty} (\binom{(- a)}{i}) {(\frac{b - H (L)}{H (L)})}^{i} \\ \times \int_{x}^{U} {[1 - F_{X} (t)]}^{(\frac{i}{a - r})} f_{X} (t) d t . \end{aligned}$ Let $u = 1 - F_{X} (t)$ , we have $\begin{aligned} E ({(H (X))}^{a} | X \geq x) \\ = \frac{b^{a}}{1 - F_{X} (x)} \sum_{i = 0}^{\infty} (\frac{a - r}{a + i - r}) (\binom{(- a)}{i}) \\ \times {(\frac{b - H (L)}{H (L)})}^{i} {{[u]}^{(\frac{a + i - r}{a - r})} |}_{0}^{1 - F_{X} (x)} \\ E ({(H (X))}^{a} | X \geq x) \\ = b^{a} \sum_{i = 0}^{\infty} (\frac{a - r}{a + i - r}) (\binom{(- a)}{i}) {(\frac{b - H (L)}{H (L)})}^{i} \\ \times {[1 - F_{X} (x)]}^{(\frac{i}{a - r})} \end{aligned}$ Now, (10) $(\binom{m}{n}) = \frac{m (m - 1) \dots (m - n + 1)}{n!} .$ (10) We obtain $\begin{aligned} E ({(H (X))}^{a} | X \geq x) \\ = b^{a} \sum_{i = 0}^{\infty} (\frac{a - r}{a + i - r}) \\ \times (\frac{(- a) (- a - 1) \dots (- a - i + 1)}{i!}) \\ \times {(\frac{b - H (L)}{H (L)})}^{i} {[1 - F_{X} (x)]}^{(\frac{i}{a - r})} \\ = b^{a} \sum_{i = 0}^{\infty} (\frac{a - r}{a + i - r}) \\ \times (\frac{{(- 1)}^{i} (a) (a + 1) \dots (a + i - 1)}{i!}) \\ \times {(\frac{b - H (L)}{H (L)})}^{i} {[1 - F_{X} (x)]}^{(\frac{i}{a - r})} \end{aligned}$ Also, (11) ${(m)}_{n} = m (m + 1) \dots (m + n - 1) .$ (11) From which we have $\begin{aligned} E ({(H (X))}^{a} | X \geq x) \\ = b^{a} \sum_{i = 0}^{\infty} (\frac{a - r}{a + i - r}) (\frac{{(- 1)}^{i} {(a)}_{i}}{i!}) \\ \times {(\frac{b - H (L)}{H (L)})}^{i} {[1 - F_{X} (x)]}^{(\frac{i}{a - r})} \\ = b^{a} \sum_{i = 0}^{\infty} (\frac{a - r}{a + i - r}) (\frac{{(- 1)}^{i} {(a)}_{i}}{i!} \frac{{(a - r)}_{i}}{{(a - r)}_{i}}) \\ \times {(\frac{b - H (L)}{H (L)})}^{i} {[1 - F_{X} (x)]}^{(\frac{i}{a - r})} \end{aligned}$ Moreover, (12) ${(m + 1)}_{n} = \frac{(m + n) {(m)}_{n}}{m} .$ (12) Hence, $\begin{aligned} E ({(H (X))}^{a} | X \geq x) \\ = b^{a} \sum_{i = 0}^{\infty} (\frac{{(- 1)}^{i} {(a)}_{i}}{i!} \frac{{(a - r)}_{i}}{{(a - r + 1)}_{i}}) \\ \times {(\frac{b - H (L)}{H (L)})}^{i} {[1 - F_{X} (x)]}^{(\frac{i}{a - r})} \\ = b^{a}_{2} F_{1} (a, a - r; a - r + 1; - (\frac{b - H (L)}{H (L)}) \\ {[1 - F_{X} (x)]}^{(\frac{1}{a - r})}), \end{aligned}$ where the Gaussian hypergeometric function defined as $_{2} F_{1} (a, b; c; z) = \sum_{i = 0}^{\infty} \frac{{(a)}_{i} {(b)}_{i}}{{(c)}_{i}} \frac{z^{i}}{i!}$ . $\begin{aligned} E ({(H (X))}^{a} | X \geq x) \\ = b^{a}_{2} F_{1} (a, a - r; a - r + 1; - (\frac{b - H (L)}{H (L)}) \\ {[1 - F_{X} (x)]}^{(\frac{1}{a - r})}), \end{aligned}$ By using Equation (Equation9(9) $\begin{aligned} F_{X} (x) = 1 - \exp {(a - r) \\ [\ln [(\frac{b - H (x)}{H (x)}) (\frac{H (L)}{b - H (L)})]]} \Rightarrow \\ F_{X} (x) = 1 - {[(\frac{b - H (x)}{H (x)}) (\frac{H (L)}{b - H (L)})]}^{(a - r)} \Rightarrow \\ (\frac{b - H (L)}{H (L)}) {[1 - F_{X} (x)]}^{(\frac{1}{(a - r)})} = (\frac{b}{H (x)} - 1) \\ ∴ H (x) = \frac{b}{1 + (\frac{b - H (L)}{H (L)}) {[1 - F_{X} (x)]}^{(\frac{1}{(a - r)})}} \end{aligned}$ (9) ). $\begin{aligned} E ({(H (X))}^{a} | X \geq x) \\ = b^{a}_{2} F_{1} (a, a - r; a - r + 1; 1 - \frac{b}{H (x)}), \end{aligned}$

Remark 2.1

If r = 1 and using Equations (Equation10(10) $(\binom{m}{n}) = \frac{m (m - 1) \dots (m - n + 1)}{n!} .$ (10) ) and (Equation11(11) ${(m)}_{n} = m (m + 1) \dots (m + n - 1) .$ (11) ) then from Equation (Equation1(1) $\begin{aligned} E ({(H (X))}^{a} | X \geq x) \\ = b^{a}_{2} F_{1} (a, a - r; a - r + 1; 1 - \frac{b}{H (x)}), \end{aligned}$ (1) ) after simplification and using Equation (1.110) at page 25 in [Citation24], we obtain: $b^{a}_{2} F_{1} (a, a - 1; a; 1 - \frac{b}{H (x)}) = b {(H (x))}^{a - 1} .$
Taking, for example, r = 1, $a = b = δ > r$ and ${lim}_{x \to L^{+}} H (x) = H (L) = δ / 2 > 1 / 2$ then we obtain equation (44) in Proposition 10 in [Citation22].

Proposition 2.2

Let $X : Ω \to (L, U)$ be a random variable that is continuous with cdf F. Let $H (x)$ be a function that is differentiable on $(L, U)$ with $lim_{x \to L^{+}} H (x) = H (L)$ and $lim_{x \to U^{-}} H (x) = {\frac{r}{(1 - s)}}^{(\frac{1}{a})}$ if 0<s<1 and $r > (1 - s) {(H (x))}^{a}$ . Then, $E ({(H (X))}^{a} | X \geq x) = r + s {(H (x))}^{a}$ if and only if (13) $H (x) = {\frac{[\frac{r + (s - 1) {(H (L))}^{a}}{{[1 - F_{X} (x)]}^{(\frac{s - 1}{s})}}] - r}{(s - 1)}}^{(\frac{1}{a})}$ (13) where $x \in (L, U)$ .

Proof.

If Equation (Equation2(2) $\begin{aligned} E ({(H (X))}^{a} | X \geq x) = r + s {(H (x))}^{a}, \end{aligned}$ (2) ) holds, we have $\begin{aligned} \int_{x}^{U} {(H (t))}^{a} \frac{f_{X} (t)}{1 - F_{X} (x)} d t = r + s {(H (x))}^{a} \Rightarrow \\ \int_{x}^{U} {(H (t))}^{a} f_{X} (t) d t = (1 - F_{X} (x)) [r + s {(H (x))}^{a}] \end{aligned}$ Using derivatives from both sides in relation to x, we obtain $\begin{aligned} [r + (s - 1) {(H (x))}^{a}] f_{X} (x) \\ = a s (1 - F_{X} (x)) {(H (x))}^{a - 1} H^{'} (x) \end{aligned}$ from which we have $\frac{f_{X} (x)}{1 - F_{X} (x)} = \frac{a s {(H (x))}^{a - 1} H^{'} (x)}{[r + (s - 1) {(H (x))}^{a}]}$ Integrating both sides from L to x, $\int_{L}^{x} \frac{f_{X} (t)}{1 - F_{X} (t)} d t = s \int_{L}^{x} \frac{a {(H (t))}^{a - 1} H^{'} (t)}{[r + (s - 1) {(H (t))}^{a}]} d t .$ Let $u = 1 - F_{X} (t)$ and $v = [r + (s - 1) {(H (t))}^{a}]$ , then (14) $\begin{aligned} {\ln (u) |}_{1 - F_{X} (x)}^{1} & = (\frac{s}{s - 1}) {\ln (v) |}_{r + (s - 1) {(H (L))}^{a}}^{r + (s - 1) {(H (x))}^{a}} \Rightarrow \\ - \ln (1 - F_{X} (x)) = (\frac{s}{s - 1}) \\ \times \ln [\frac{r + (s - 1) {(H (x))}^{a}}{r + (s - 1) {(H (L))}^{a}}] \Rightarrow \\ F_{X} (x) & = 1 - {[\frac{r + (s - 1) {(H (x))}^{a}}{r + (s - 1) {(H (L))}^{a}}]}^{(\frac{s}{1 - s})} \\ \times \Rightarrow \\ r + (s - 1) {(H (x))}^{a} & = [r + (s - 1) {(H (L))}^{a}] \\ \times {[1 - F_{X} (x)]}^{(\frac{1 - s}{s})} \Rightarrow \\ H (x) & = {\frac{[\frac{r + (s - 1) {(H (L))}^{a}}{{[1 - F_{X} (x)]}^{(\frac{s - 1}{s})}}] - r}{(s - 1)}}^{(\frac{1}{a})} \end{aligned}$ (14) Conversely, if Equation (Equation13(13) $H (x) = {\frac{[\frac{r + (s - 1) {(H (L))}^{a}}{{[1 - F_{X} (x)]}^{(\frac{s - 1}{s})}}] - r}{(s - 1)}}^{(\frac{1}{a})}$ (13) ) holds, then $\begin{aligned} E ({(H (X))}^{a} | X \geq x) \\ = \frac{1}{(s - 1) [1 - F_{X} (x)]} \\ \times \int_{x}^{U} {[\frac{r + (s - 1) {(H (L))}^{a}}{{[1 - F_{X} (t)]}^{(\frac{s - 1}{s})}}] - r} f_{X} (t) d t . \end{aligned}$ Let $u = 1 - F_{X} (x)$ , we have

$\begin{aligned} E ({(H (X))}^{a} | X \geq x) \\ = \frac{1}{(s - 1) [1 - F_{X} (x)]} \\ \times \int_{0}^{1 - F_{X} (x)} {[\frac{r + (s - 1) {(H (L))}^{a}}{{[u]}^{(\frac{s - 1}{s})}}] - r} d u \\ = \frac{1}{(s - 1) [1 - F_{X} (x)]} \\ \times {s [r + (s - 1) {(H (L))}^{a}] {{[u]}^{(\frac{1}{s})} |}_{0}^{1 - F_{X} (x)} - r {u |}_{0}^{1 - F_{X} (x)}} \\ = \frac{1}{(s - 1)} {s [r + (s - 1) {(H (L))}^{a}] {[1 - F_{X} (x)]}^{(\frac{1 - s}{s})} - r} \end{aligned}$ Using Equation (Equation14(14) $\begin{aligned} {\ln (u) |}_{1 - F_{X} (x)}^{1} & = (\frac{s}{s - 1}) {\ln (v) |}_{r + (s - 1) {(H (L))}^{a}}^{r + (s - 1) {(H (x))}^{a}} \Rightarrow \\ - \ln (1 - F_{X} (x)) = (\frac{s}{s - 1}) \\ \times \ln [\frac{r + (s - 1) {(H (x))}^{a}}{r + (s - 1) {(H (L))}^{a}}] \Rightarrow \\ F_{X} (x) & = 1 - {[\frac{r + (s - 1) {(H (x))}^{a}}{r + (s - 1) {(H (L))}^{a}}]}^{(\frac{s}{1 - s})} \\ \times \Rightarrow \\ r + (s - 1) {(H (x))}^{a} & = [r + (s - 1) {(H (L))}^{a}] \\ \times {[1 - F_{X} (x)]}^{(\frac{1 - s}{s})} \Rightarrow \\ H (x) & = {\frac{[\frac{r + (s - 1) {(H (L))}^{a}}{{[1 - F_{X} (x)]}^{(\frac{s - 1}{s})}}] - r}{(s - 1)}}^{(\frac{1}{a})} \end{aligned}$ (14) ), we get $\begin{aligned} E ({(H (X))}^{a} | X \geq x) \\ = \frac{1}{(s - 1)} {(s - 1) r + s (s - 1) {(H (x))}^{a}} \\ = r + s {(H (x))}^{a} . \end{aligned}$

Remark 2.2

As special cases see, for example, Proposition 2.5 page 22–23 in [Citation21].

$H (L) = 1, a = 1, r = 0$ and $s = δ$
$H (L) = 1, a = δ, r = 0$ and $s = ϵ$
$H (L) = 0, a = 1, r = ϵ$ and $s = 1 - ϵ$

Remark 2.3

Taking, for example, a = 1, r = c, s = 1−c and ${lim}_{x \to L^{+}} H (x) = H (L) = 1 - c$ , we obtain Equation (1) of Proposition 2.1 in [Citation20].

Proposition 2.3

Let $X : Ω \to (L, U)$ be a random variable that is continuous with cdf F. Let $H (x)$ be a function that is differentiable on $(L, U)$ with ${lim}_{x \to L^{+}} H (x) = H (L) > 0$ . Then, for $r \geq 1, s > (1 - r) H (L)$ and $a > 1,$ $E ({(H (X))}^{a} | X \geq x) = r {(H (x))}^{a} + s {(H (x))}^{a - 1}$ implies (15) $\begin{aligned} F_{X} (x) = {\begin{cases} 1 - {(\frac{H (L)}{H (x)})}^{(a - 1)} \\ {(\frac{(r - 1) H (L) + s}{(r - 1) H (x) + s})}^{(\frac{(a + r - 1)}{(r - 1)})} & if r > 1 \\ 1 - {(\frac{H (L)}{H (x)})}^{(a - 1)} e^{(\frac{a}{s}) (H (L) - H (x))} & if r = 1 \end{cases} \end{aligned}$ (15) where $x \in (L, U)$ .

Proof.

For r>1 and from Equation (Equation3(3) $\begin{aligned} E ({(H (X))}^{a} | X \geq x) = r {(H (x))}^{a} + s {(H (x))}^{a - 1}, \end{aligned}$ (3) ), we have $\begin{aligned} \int_{x}^{U} {(H (t))}^{a} f_{X} (t) d t = [1 - F_{X} (x)] \\ \times [r {(H (x))}^{a} + s {(H (x))}^{a - 1}] \end{aligned}$ Using derivatives from both sides in relation to x, we obtain $\begin{aligned} f_{X} (x) {(H (x))}^{a - 1} [(r - 1) H (x) + s] = H^{'} (x) [1 - F_{X} (x)] \\ \times {(H (x))}^{a - 2} [a r H (x) + (a - 1) s] \end{aligned}$ from which we have $\begin{aligned} \frac{f_{X} (x)}{1 - F_{X} (x)} & = \frac{H^{'} (x) [a r H (x) + (a - 1) s]}{H (x) [(r - 1) H (x) + s]} \Rightarrow \\ \frac{f_{X} (x)}{1 - F_{X} (x)} & = H^{'} (x) {\frac{a r}{[(r - 1) H (x) + s]} \\ + \frac{(a - 1) s}{H (x) [(r - 1) H (x) + s]}} \Rightarrow \\ \frac{f_{X} (x)}{1 - F_{X} (x)} & = H^{'} (x) {\frac{a r}{[(r - 1) H (x) + s]} + \frac{(a - 1)}{H (x)} \\ - \frac{(a - 1) (r - 1)}{[(r - 1) H (x) + s]}} \Rightarrow \\ \frac{f_{X} (x)}{1 - F_{X} (x)} & = H^{'} (x) {\frac{a + r - 1}{[(r - 1) H (x) + s]} + \frac{(a - 1)}{H (x)}} \end{aligned}$ Integrating both sides from L to x, $\begin{aligned} \int_{L}^{x} \frac{f_{X} (t)}{1 - F_{X} (t)} d t \\ = \int_{L}^{x} {\frac{(a + r - 1) H^{'} (t)}{[(r - 1) H (t) + s]} + \frac{(a - 1) H^{'} (t)}{H (t)}} d t . \end{aligned}$ Let $u = 1 - F_{X} (t), v = (r - 1) H (t) + sandw = H (t)$ , then for $r \neq 1$

(16) $\begin{aligned} \int_{1 - F_{X} (x)}^{1} \frac{d u}{u} = (\frac{(a + r - 1)}{(r - 1)}) \int_{(r - 1) H (L) + s}^{(r - 1) H (x) + s} \frac{d v}{v} \\ + (a - 1) \int_{H (L)}^{H (x)} \frac{d w}{w} \Rightarrow \\ \ln (1) - \ln (1 - F_{X} (x)) = (\frac{(a + r - 1)}{(r - 1)}) \\ \times [\ln ((r - 1) H (x) + s) \\ - \ln ((r - 1) H (L) + s)] + (a - 1) \\ \times [\ln (H (x)) - \ln (H (L))] \Rightarrow \\ \ln (1 - F_{X} (x)) \\ = \ln {{(\frac{H (L)}{H (x)})}^{(a - 1)} {(\frac{(r - 1) H (L) + s}{(r - 1) H (x) + s})}^{(\frac{(a + r - 1)}{(r - 1)})}} \Rightarrow \\ F_{X} (x) = 1 - {(\frac{H (L)}{H (x)})}^{(a - 1)} {(\frac{(r - 1) H (L) + s}{(r - 1) H (x) + s})}^{(\frac{(a + r - 1)}{(r - 1)})} . \end{aligned}$ (16) For r = 1, $\begin{aligned} lim_{r \to 1} F_{X} (x) & = 1 - {(\frac{H (L)}{H (x)})}^{(a - 1)} lim_{r \to 1} \\ \times {(\frac{H (L) - H (x)}{H (x) + \frac{s}{(r - 1)}} + 1)}^{(\frac{(a + r - 1)}{(r - 1)})} . \end{aligned}$ Let $n = a / (r - 1)$ , then $\begin{aligned} lim_{r \to 1} F_{X} (x) = lim_{n \to \infty} F_{X} (x) = 1 - {(\frac{H (L)}{H (x)})}^{(a - 1)} \\ lim_{n \to \infty} {(1 + \frac{H (L) - H (x)}{H (x) + n \frac{s}{a}})}^{(1 + n)} \\ = 1 - {(\frac{H (L)}{H (x)})}^{(a - 1)} \\ \times lim_{n \to \infty} {(1 + \frac{(\frac{a}{s}) (H (L) - H (x))}{(\frac{a}{s}) H (x) + n})}^{n} . \end{aligned}$ Let $m = (\frac{a}{s}) H (x) + n$ , then $\begin{aligned} lim_{r \to 1} F_{X} (x) \\ = lim_{n \to \infty} F_{X} (x) = lim_{m \to \infty} F_{X} (x) = 1 \\ - {(\frac{H (L)}{H (x)})}^{(a - 1)} \\ \times lim_{m \to \infty} {(1 + \frac{(\frac{a}{s}) (H (L) - H (x))}{m})}^{m - (\frac{a}{s}) H (x)} \\ = 1 - {(\frac{H (L)}{H (x)})}^{(a - 1)} \\ \times lim_{m \to \infty} {(1 + \frac{(\frac{a}{s}) (H (L) - H (x))}{m})}^{m} . \end{aligned}$ Therefore, (17) $\begin{aligned} lim_{r \to 1} F_{X} (x) & = lim_{n \to \infty} F_{X} (x) = lim_{m \to \infty} F_{X} (x) \\ = 1 - {(\frac{H (L)}{H (x)})}^{(a - 1)} e^{(\frac{a}{s}) (H (L) - H (x))}, \end{aligned}$ (17) where $lim_{m \to \infty} {(1 + \frac{y}{m})}^{m} = e^{y}$ .

For r>1 and from Equation (Equation16(16) $\begin{aligned} \int_{1 - F_{X} (x)}^{1} \frac{d u}{u} = (\frac{(a + r - 1)}{(r - 1)}) \int_{(r - 1) H (L) + s}^{(r - 1) H (x) + s} \frac{d v}{v} \\ + (a - 1) \int_{H (L)}^{H (x)} \frac{d w}{w} \Rightarrow \\ \ln (1) - \ln (1 - F_{X} (x)) = (\frac{(a + r - 1)}{(r - 1)}) \\ \times [\ln ((r - 1) H (x) + s) \\ - \ln ((r - 1) H (L) + s)] + (a - 1) \\ \times [\ln (H (x)) - \ln (H (L))] \Rightarrow \\ \ln (1 - F_{X} (x)) \\ = \ln {{(\frac{H (L)}{H (x)})}^{(a - 1)} {(\frac{(r - 1) H (L) + s}{(r - 1) H (x) + s})}^{(\frac{(a + r - 1)}{(r - 1)})}} \Rightarrow \\ F_{X} (x) = 1 - {(\frac{H (L)}{H (x)})}^{(a - 1)} {(\frac{(r - 1) H (L) + s}{(r - 1) H (x) + s})}^{(\frac{(a + r - 1)}{(r - 1)})} . \end{aligned}$ (16) ), (18) $\begin{aligned} ((r - 1) H (x) + s) {(H (x))}^{(\frac{(1 - a) (1 - r)}{(a + r - 1)})} \\ = ((r - 1) H (L) + s) \\ \times {(H (L))}^{(\frac{(1 - a) (1 - r)}{(a + r - 1)})} {[1 - F_{X} (x)]}^{(\frac{(1 - r)}{(a + r - 1)})} \Rightarrow \\ \times (H (x) + \frac{s}{(r - 1)}) {(H (x))}^{(\frac{(1 - a) (1 - r)}{(a + r - 1)})} \\ = (H (L) + \frac{s}{(r - 1)}) \\ \times {(H (L))}^{(\frac{(1 - a) (1 - r)}{(a + r - 1)})} {[1 - F_{X} (x)]}^{(\frac{(1 - r)}{(a + r - 1)})} \end{aligned}$ (18) It is difficult to obtain $H (x)$ analytically in closed form from Equation (Equation18(18) $\begin{aligned} ((r - 1) H (x) + s) {(H (x))}^{(\frac{(1 - a) (1 - r)}{(a + r - 1)})} \\ = ((r - 1) H (L) + s) \\ \times {(H (L))}^{(\frac{(1 - a) (1 - r)}{(a + r - 1)})} {[1 - F_{X} (x)]}^{(\frac{(1 - r)}{(a + r - 1)})} \Rightarrow \\ \times (H (x) + \frac{s}{(r - 1)}) {(H (x))}^{(\frac{(1 - a) (1 - r)}{(a + r - 1)})} \\ = (H (L) + \frac{s}{(r - 1)}) \\ \times {(H (L))}^{(\frac{(1 - a) (1 - r)}{(a + r - 1)})} {[1 - F_{X} (x)]}^{(\frac{(1 - r)}{(a + r - 1)})} \end{aligned}$ (18) ) but we will get it numerically as follows: $\begin{aligned} Let δ & = (H (L) + \frac{s}{(r - 1)}) {(H (L))}^{(\frac{(1 - a) (1 - r)}{(a + r - 1)})} \\ \times {[1 - F_{X} (x)]}^{(\frac{(1 - r)}{(a + r - 1)})}, \\ b & = (\frac{(1 - a) (1 - r)}{(a + r - 1)}), \\ ϑ & = \frac{s}{(r - 1)} \end{aligned}$ and $y = H (x) .$ Therefore, from Equation (Equation18(18) $\begin{aligned} ((r - 1) H (x) + s) {(H (x))}^{(\frac{(1 - a) (1 - r)}{(a + r - 1)})} \\ = ((r - 1) H (L) + s) \\ \times {(H (L))}^{(\frac{(1 - a) (1 - r)}{(a + r - 1)})} {[1 - F_{X} (x)]}^{(\frac{(1 - r)}{(a + r - 1)})} \Rightarrow \\ \times (H (x) + \frac{s}{(r - 1)}) {(H (x))}^{(\frac{(1 - a) (1 - r)}{(a + r - 1)})} \\ = (H (L) + \frac{s}{(r - 1)}) \\ \times {(H (L))}^{(\frac{(1 - a) (1 - r)}{(a + r - 1)})} {[1 - F_{X} (x)]}^{(\frac{(1 - r)}{(a + r - 1)})} \end{aligned}$ (18) ), we have $(y + ϑ) y^{b} = δ$ . Now, let $\begin{aligned} φ (y) & = (y + ϑ) y^{b} - δ = 0. \\ So, φ^{'} (y) & = (y + b (ϑ + y)) y^{(b - 1)} . \end{aligned}$ By using the Newton-Raphson method as follows: $\begin{aligned} y_{n + 1} = y_{n} - \frac{φ (y_{n})}{φ^{'} (y_{n})} . \end{aligned}$ We obtain the root z such that $φ (z) = 0$ , i.e. $H (x) = z$ .

For r = 1 and from Equation (Equation17(17) $\begin{aligned} lim_{r \to 1} F_{X} (x) & = lim_{n \to \infty} F_{X} (x) = lim_{m \to \infty} F_{X} (x) \\ = 1 - {(\frac{H (L)}{H (x)})}^{(a - 1)} e^{(\frac{a}{s}) (H (L) - H (x))}, \end{aligned}$ (17) ), $\begin{aligned} {(H (x))}^{(1 - a)} e^{- (\frac{a}{s}) H (x)} \\ = {(H (L))}^{(1 - a)} e^{- (\frac{a}{s}) H (L)} [1 - F_{X} (x)] \Rightarrow \\ {{(H (x))}^{(1 - a)} e^{- (\frac{a}{s}) H (x)}}^{(\frac{1}{(1 - a)})} \\ = {{(H (L))}^{(1 - a)} e^{- (\frac{a}{s}) H (L)} [1 - F_{X} (x)]}^{(\frac{1}{(1 - a)})} \Rightarrow \\ (\frac{a}{s (a - 1)}) H (x) e^{(\frac{a}{s (a - 1)}) H (x)} = (\frac{a}{s (a - 1)}) \\ \times H (L) e^{(\frac{a}{s (a - 1)}) H (L)} {[1 - F_{X} (x)]}^{(\frac{1}{(1 - a)})} \Rightarrow \\ (\frac{a}{s (a - 1)}) H (x) = W ((\frac{a}{s (a - 1)}) \\ \times H (L) e^{(\frac{a}{s (a - 1)}) H (L)} {[1 - F_{X} (x)]}^{(\frac{1}{(1 - a)})}), \end{aligned}$ where $W (\cdot)$ is the Lambert function. Hence, $\begin{aligned} H (x) & = (\frac{s (a - 1)}{a}) W ((\frac{a}{s (a - 1)}) \\ \times H (L) e^{(\frac{a}{s (a - 1)}) H (L)} {[1 - F_{X} (x)]}^{(\frac{1}{(1 - a)})}) \end{aligned}$

Remark 2.4

Taking, for example, $(L, U) = (a, b)$ , r = 1, s = c and $H (L) = γ > 0$ , we obtain Equation (7) of Proposition 2.3 in [Citation20].

Remark 2.5

Taking, for example, $(L, U) = (a, b)$ , r = 1, s = 1−c and $H (L) = γ > 0$ , we obtain Equation (11) of Proposition 2.5 in [Citation20].

Proposition 2.4

Let $X : Ω \to (L, U)$ be a random variable that is continuous with cdf F. Let $H (x)$ be a function that is differentiable on $(L, U)$ with ${lim}_{x \to L^{+}} H (x) = H (L) > 0$ and ${lim}_{x \to U^{-}} H (x) = \infty$ . Then, for $r > 0 anda > 1,$ $E (H (X) | X \geq x) = r {(H (x))}^{a} + H (x)$ implies (19) $\begin{aligned} F_{X} (x) & = 1 - {(\frac{H (L)}{H (x)})}^{a} \\ \times \exp [\frac{{(H (L))}^{1 - a} - {(H (x))}^{1 - a}}{r (1 - a)}] \end{aligned}$ (19) where $x \in (L, U)$ .

Proof.

From Equation (Equation4(4) $\begin{aligned} E (H (X) | X \geq x) = r {(H (x))}^{a} + H (x), \end{aligned}$ (4) ), we have $\int_{x}^{U} H (t) f_{X} (t) d t = [1 - F_{X} (x)] [r {(H (x))}^{a} + H (x)]$ Using derivatives from both sides in relation to x, we obtain $\begin{aligned} r f_{X} (x) {(H (x))}^{a} & = [1 - F_{X} (x)] \\ \times [a r {(H (x))}^{a - 1} H^{'} (x) + H^{'} (x)] \end{aligned}$ from which we have $\frac{f_{X} (x)}{1 - F_{X} (x)} = \frac{[a r {(H (x))}^{a - 1} H^{'} (x) + H^{'} (x)]}{r {(H (x))}^{a}}$ Integrating both sides from L to x, $\begin{aligned} \int_{L}^{x} \frac{f_{X} (t)}{1 - F_{X} (t)} d t \\ = \int_{L}^{x} {\frac{[a r {(H (t))}^{a - 1} H^{'} (t) + H^{'} (t)]}{r {(H (t))}^{a}}} d t . \end{aligned}$ Let $u = 1 - F_{X} (t)$ and $v = H (t)$ , then $\begin{aligned} \ln (1) - \ln (1 - F_{X} (x)) = a [\ln (H (x)) - \ln (H (L))] \\ + (\frac{1}{r (1 - a)}) [{(H (x))}^{1 - a} - {(H (L))}^{1 - a}] \Rightarrow \\ \ln (1 - F_{X} (x)) = - \ln {{(\frac{H (x)}{H (L)})}^{a} \exp {(\frac{1}{r (1 - a)}) \\ \times [{(H (x))}^{1 - a} - {(H (L))}^{1 - a}]}} \Rightarrow \\ F_{X} (x) = 1 - {(\frac{H (L)}{H (x)})}^{a} \\ \times \exp [\frac{{(H (L))}^{1 - a} - {(H (x))}^{1 - a}}{r (1 - a)}] \Rightarrow \\ {(H (x))}^{- a} \exp (- \frac{{(H (x))}^{1 - a}}{r (1 - a)}) \\ = \exp (- \frac{{(H (L))}^{1 - a}}{r (1 - a)}) {(H (L))}^{- a} [1 - F_{X} (x)] \Rightarrow \\ (\frac{1}{ra}) {(H (x))}^{1 - a} \exp (\frac{{(H (x))}^{1 - a}}{ra}) = (\frac{1}{ra}) \\ \times \exp (\frac{{(H (L))}^{1 - a}}{ra}) {(H (L))}^{1 - a} {[1 - F_{X} (x)]}^{(\frac{a - 1}{a})} \Rightarrow \\ (\frac{1}{ra}) {(H (x))}^{1 - a} = W ((\frac{1}{ra}) \\ \exp (\frac{{(H (L))}^{1 - a}}{ra}) {(H (L))}^{1 - a} {[1 - F_{X} (x)]}^{(\frac{a - 1}{a})}), \end{aligned}$ where $W (\cdot)$ is the Lambert function. Therefore, $\begin{aligned} {(H (x))}^{1 - a} & = ra W ((\frac{1}{ra}) \exp (\frac{{(H (L))}^{1 - a}}{ra}) \\ \times {(H (L))}^{1 - a} {[1 - F_{X} (x)]}^{(\frac{a - 1}{a})}) . \end{aligned}$ Hence, $\begin{aligned} H (x) & = {ra W ((\frac{1}{ra}) \exp (\frac{{(H (L))}^{1 - a}}{ra}) \\ \times {(H (L))}^{1 - a} {[1 - F_{X} (x)]}^{(\frac{a - 1}{a})})}^{\frac{1}{(1 - a)}} \end{aligned}$

Remark 2.6

Taking, for example, a = 2, r = 1 and $H (L) = γ > 0$ , we obtain Equation (15) of Proposition 2.7 in [Citation20].

Proposition 2.5

Let $X : Ω \to (L, U)$ be a random variable that is continuous with cdf F. Let $H (x)$ be a function that is differentiable on $(L, U)$ with $lim_{x \to L^{+}} H (x) = H (L)$ and $lim_{x \to U^{-}} H (x) = {(\frac{a}{bc}) W ((\frac{bc}{a}) {[\frac{r}{(1 - s)}]}^{(\frac{b}{a})})}^{(\frac{1}{b})}$ . Then, for $0 < s < 1, 0 \leq r < 1 - s and a, b, c > 0,$ $\begin{aligned} E ({(H (X))}^{a} e^{[c {(H (X))}^{b}]} | X \geq x) \\ = r + s {(H (x))}^{a} e^{[c {(H (x))}^{b}]} \end{aligned}$ implies (20) $\begin{aligned} H (x) & = {(\frac{a}{bc}) W ((\frac{bc}{a}) \\ \times {[\frac{\begin{array}{l} [r + (s - 1) {(H (L))}^{a} \\ e^{[c {(H (L))}^{b}]}] \\ {[1 - F_{X} (x)]}^{(\frac{1 - s}{s})} - r \end{array}}{(s - 1)}]}^{(\frac{b}{a})})}^{(\frac{1}{b})} \end{aligned}$ (20) where $x \in (L, U)$ and $W (\cdot)$ is the Lambert function.

Proof.

From Equation (Equation5(5) $\begin{aligned} E ({(H (X))}^{a} e^{[c {(H (X))}^{b}]} | X \geq x) \\ = r + s {(H (x))}^{a} e^{[c {(H (x))}^{b}]}, \end{aligned}$ (5) ), we have $\begin{aligned} \int_{x}^{U} {(H (t))}^{a} e^{[c {(H (t))}^{b}]} f_{X} (t) d t \\ = [1 - F_{X} (x)] [r + s {(H (x))}^{a} e^{[c {(H (x))}^{b}]}] \end{aligned}$ Using derivatives from both sides in relation to x, we obtain $\begin{aligned} f_{X} (x) [r + (s - 1) {(H (x))}^{a} e^{[c {(H (x))}^{b}]}] \\ = s [1 - F_{X} (x)] H^{'} (x) {(H (x))}^{a - 1} \\ \times e^{[c {(H (x))}^{b}]} [a + b c {(H (x))}^{b}] \end{aligned}$ from which we have $\begin{aligned} \frac{f_{X} (x)}{1 - F_{X} (x)} = \frac{\begin{array}{l} s H^{'} (x) {(H (x))}^{a - 1} e^{[c {(H (x))}^{b}]} \\ [a + b c {(H (x))}^{b}] \end{array}}{[r + (s - 1) {(H (x))}^{a} e^{[c {(H (x))}^{b}]}]} \end{aligned}$ Integrating both sides from L to x, $\begin{aligned} \int_{L}^{x} \frac{f_{X} (t)}{1 - F_{X} (t)} d t \\ = s \int_{L}^{x} {\frac{\begin{array}{l} H^{'} (t) {(H (t))}^{a - 1} e^{[c {(H (t))}^{b}]} \\ [a + b c {(H (t))}^{b}] \end{array}}{[r + (s - 1) {(H (t))}^{a} e^{[c {(H (t))}^{b}]}]}} d t . \end{aligned}$ Let $u = 1 - F_{X} (t)$ and $v = [r + (s - 1) {(H (t))}^{a} e^{[c {(H (t))}^{b}]}]$ , then $\begin{aligned} \ln (1) - \ln (1 - F_{X} (x)) = (\frac{s}{s - 1}) \\ \times {\ln [r + (s - 1) {(H (x))}^{a} e^{[c {(H (x))}^{b}]}] \\ - \ln [r + (s - 1) {(H (L))}^{a} e^{[c {(H (L))}^{b}]}]} \Rightarrow \\ F_{X} (x) = 1 - {\frac{[r + (s - 1) {(H (x))}^{a} e^{[c {(H (x))}^{b}]}]}{[r + (s - 1) {(H (L))}^{a} e^{[c {(H (L))}^{b}]}]}}^{(\frac{s}{1 - s})} \\ \Rightarrow [r + (s - 1) {(H (L))}^{a} e^{[c {(H (L))}^{b}]}] {[1 - F_{X} (x)]}^{(\frac{1 - s}{s})} \\ = r + (s - 1) {(H (x))}^{a} e^{[c {(H (x))}^{b}]} \Rightarrow \\ {(H (x))}^{a} e^{[c {(H (x))}^{b}]} \\ = \frac{\begin{array}{l} [r + (s - 1) {(H (L))}^{a} e^{[c {(H (L))}^{b}]}] \\ {[1 - F_{X} (x)]}^{(\frac{1 - s}{s})} - r \end{array}}{(s - 1)} . \end{aligned}$ Let $δ = \frac{\begin{array}{l} [r + (s - 1) {(H (L))}^{a} e^{[c {(H (L))}^{b}]}] \\ {[1 - F_{X} (x)]}^{(\frac{1 - s}{s})} - r \end{array}}{(s - 1)} .$ Then, $\begin{aligned} {(H (x))}^{a} e^{[c {(H (x))}^{b}]} & = δ \Rightarrow (\frac{bc}{a}) {(H (x))}^{b} \\ = W ((\frac{bc}{a}) δ^{(\frac{b}{a})}), \end{aligned}$ where $W (\cdot)$ is the Lambert function. Hence, $\begin{aligned} H (x) = {(\frac{a}{bc}) W ((\frac{bc}{a}) \\ \times {[\frac{\begin{array}{l} [r + (s - 1) {(H (L))}^{a} e^{[c {(H (L))}^{b}]}] \\ {[1 - F_{X} (x)]}^{(\frac{1 - s}{s})} - r \end{array}}{(s - 1)}]}^{(\frac{b}{a})})}^{(\frac{1}{b})} . \end{aligned}$

Proposition 2.6

Let $X : Ω \to (L, U)$ be a random variable that is continuous with cdf F. Let $H (x)$ be a function that is differentiable on $(L, U)$ with $lim_{x \to L^{+}} H (x) = H (L) > 0$ and $lim_{x \to U^{-}} H (x) = \exp {(\frac{1}{b}) W ((\frac{b}{c}) {[\frac{r}{(1 - s)}]}^{(\frac{1}{a})})}$ . Then, for $0 < s < 1, 0 \leq r < 1 - s anda, b, c > 0,$ $\begin{aligned} E ({[{(H (X))}^{b} \ln {(H (X))}^{c}]}^{a} | X \geq x) \\ = r + s {[{(H (x))}^{b} \ln {(H (x))}^{c}]}^{a} \end{aligned}$ implies (21) $\begin{aligned} H (x) \\ = \exp {\frac{W ((\frac{b}{c}) {[\frac{\begin{array}{l} [r + (s - 1) \\ {[{(H (L))}^{b} \ln {(H (L))}^{c}]}^{a}] \\ {[1 - F_{X} (x)]}^{(\frac{1 - s}{s})} - r \end{array}}{(s - 1)}]}^{(\frac{1}{a})})}{b}} \end{aligned}$ (21) where $x \in (L, U)$ and $W (\cdot)$ is the Lambert function.

Proof.

From Equation (Equation6(6) $\begin{aligned} E ({[{(H (X))}^{b} \ln {(H (X))}^{c}]}^{a} | X \geq x) \\ = r + s {[{(H (x))}^{b} \ln {(H (x))}^{c}]}^{a}, \end{aligned}$ (6) ), we have $\begin{aligned} \int_{x}^{U} {[{(H (t))}^{b} \ln {(H (t))}^{c}]}^{a} f_{X} (t) d t \\ = [1 - F_{X} (x)] [r + s {[{(H (x))}^{b} \ln {(H (x))}^{c}]}^{a}] \end{aligned}$ Using derivatives from both sides in relation to x, we obtain $\begin{aligned} f_{X} (x) [r + (s - 1) {[{(H (x))}^{b} \ln {(H (x))}^{c}]}^{a}] \\ = a c s [1 - F_{X} (x)] H^{'} (x) {(H (x))}^{b - 1} \\ \times {[{(H (x))}^{b} \ln {(H (x))}^{c}]}^{a - 1} [1 + b \ln (H (x))] \end{aligned}$ from which we have $\begin{aligned} \frac{f_{X} (x)}{1 - F_{X} (x)} = \frac{\begin{array}{l} a c s H^{'} (x) {(H (x))}^{b - 1} \\ {[{(H (x))}^{b} \ln {(H (x))}^{c}]}^{a - 1} \\ [1 + b \ln (H (x))] \end{array}}{[r + (s - 1) {[{(H (x))}^{b} \ln {(H (x))}^{c}]}^{a}]} \end{aligned}$ Integrating both sides from L to x, $\begin{aligned} \int_{L}^{x} \frac{f_{X} (t)}{1 - F_{X} (t)} d t = a c s \\ \times \int_{L}^{x} {\frac{\begin{array}{l} H^{'} (t) {(H (t))}^{b - 1} \\ {[{(H (t))}^{b} \ln {(H (t))}^{c}]}^{a - 1} \\ [1 + b \ln (H (t))] \end{array}}{[r + (s - 1) {[{(H (t))}^{b} \ln {(H (t))}^{c}]}^{a}]}} d t . \end{aligned}$ Let $u = 1 - F_{X} (t) and v = [r + (s - 1) [{(H (t))}^{b} \ln {(H (t))}^{c}]^{a}]$ , then $\begin{aligned} \ln (1) - \ln (1 - F_{X} (x)) \\ = (\frac{s}{s - 1}) {\ln [r + (s - 1) {[{(H (x))}^{b} \ln {(H (x))}^{c}]}^{a}] \\ - \ln [r + (s - 1) {[{(H (L))}^{b} \ln {(H (L))}^{c}]}^{a}]} \Rightarrow \\ F_{X} (x) = 1 - {\frac{\begin{array}{l} [r + (s - 1) \\ {[{(H (x))}^{b} \ln {(H (x))}^{c}]}^{a}] \end{array}}{\begin{array}{l} [r + (s - 1) \\ {[{(H (L))}^{b} \ln {(H (L))}^{c}]}^{a}] \end{array}}}^{(\frac{s}{1 - s})} \Rightarrow \\ {[{(H (x))}^{b} \ln {(H (x))}^{c}]}^{a} \\ = \frac{\begin{array}{l} [r + (s - 1) {[{(H (L))}^{b} \ln {(H (L))}^{c}]}^{a}] \\ {[1 - F_{X} (x)]}^{(\frac{1 - s}{s})} - r \end{array}}{(s - 1)} . \end{aligned}$ Let $\begin{aligned} δ = \frac{\begin{array}{l} [r + (s - 1) {[{(H (L))}^{b} \ln {(H (L))}^{c}]}^{a}] \\ {[1 - F_{X} (x)]}^{(\frac{1 - s}{s})} - r \end{array}}{(s - 1)} . \end{aligned}$ Then, $\begin{aligned} {[{(H (x))}^{b} \ln {(H (x))}^{c}]}^{a} & = δ \Rightarrow b \ln (H (x)) \\ = W ((\frac{b}{c}) δ^{(\frac{1}{a})}), \end{aligned}$ where $W (\cdot)$ is the Lambert function. Hence, $\begin{aligned} H (x) \\ = \exp {\frac{W ((\frac{b}{c}) {[\frac{\begin{array}{l} [r + (s - 1) \\ {[{(H (L))}^{b} \ln {(H (L))}^{c}]}^{a}] \\ {[1 - F_{X} (x)]}^{(\frac{1 - s}{s})} - r \end{array}}{(s - 1)}]}^{(\frac{1}{a})})}{b}} . \end{aligned}$

3. Applications

As illustrations and without loss of generality, we apply three of these characterizations as follows:

Using Proposition 2.1 where $r = 1, a = b = δ > 1, {lim}_{x \to L +} H (x) = H (L) = \frac{δ}{2} > 1 / 2$ and $(L, U) = (0, \infty)$ , we have
1. If $F (x)$ is the cdf of TL-G distribution (see [Citation25]) given by $\begin{aligned} F (x) & = {(G (x) (2 - G (x)))}^{α} \\ = {(1 - {(1 - G (x))}^{2})}^{α}, \end{aligned}$ then Proposition 2.1 (using Equation (Equation7(7) $H (x) = \frac{b}{1 + (\frac{b - H (L)}{H (L)}) {[1 - F_{X} (x)]}^{(\frac{1}{a - r})}}$ (7) )) gives a characterization of TL-G distribution as follows: $\begin{aligned} H (x) & = δ [1 + (1 - (1 - \\ {{{(1 - G (x))^{2})}^{α})}^{(\frac{1}{δ - 1})}]}^{- 1} . \end{aligned}$
2. If $F (x)$ is the cdf of AGT-G distribution (see [Citation22]) given by $\begin{aligned} F (x) & = (1 + λ) [1 - {(1 - G (x))}^{α}] \\ - λ {[1 - {(1 - G (x))}^{α}]}^{2} \\ = 1 - {(1 - G (x))}^{α} (1 - λ \\ + λ {(1 - G (x))}^{α}), \end{aligned}$ then Proposition 2.1 (using Equation (Equation7(7) $H (x) = \frac{b}{1 + (\frac{b - H (L)}{H (L)}) {[1 - F_{X} (x)]}^{(\frac{1}{a - r})}}$ (7) )) gives a characterization of AGT-G distribution as follows: $\begin{aligned} H (x) & = δ [1 + ({(1 - G (x))}^{α} (1 - λ \\ {{+ λ {(1 - G (x))}^{α}))}^{(\frac{1}{δ - 1})}]}^{- 1} . \end{aligned}$
Using Proposition 2.3 where $r = 1, a = δ$ and $lim_{x \to L +} H (x) = H (L) = 1$ , we get
1. If s = 1−c and $H (x) = x$ where $(L, U) = (1, \infty)$ and $lim_{x \to U -} H (x) = \infty$ then $F (x) = 1 - x^{(1 - δ)} e^{(\frac{δ}{1 - c}) (1 - x)},$
2. If s = c and $H (x) = 1 / x$ where $(L, U) = (1, \infty)$ and $lim_{x \to U -} H (x) = 0$ then $F (x) = 1 - x^{(δ - 1)} e^{(\frac{δ}{c}) (1 - \frac{1}{x})},$
3. If s = c and $H (x) = e^{- x}$ where $(L, U) = (0, \infty)$ and $lim_{x \to U -} H (x) = 0$ then $F (x) = 1 - \exp {(δ - 1) x + (\frac{δ}{c}) (1 - e^{- x})},$
4. If s = 1−c and $H (x) = e^{x}$ where $(L, U) = (0, \infty)$ and $lim_{x \to U -} H (x) = \infty$ then $\begin{aligned} F (x) & = 1 - \exp {(1 - δ) x \\ + (\frac{δ}{1 - c}) (1 - e^{x})}, \end{aligned}$
Using Proposition 2.4 where $r = 1, a = 2$ and $lim_{x \to L +} H (x) = H (L) = 1$ , we get
1. If $H (x) = x$ where $(L, U) = (1, \infty)$ and $lim_{x \to U -} H (x) = \infty$ then $F (x) = 1 - x^{- 2} e^{(\frac{1}{x} - 1)},$
2. If $H (x) = e^{x}$ where $(L, U) = (0, \infty)$ and $lim_{x \to U -} H (x) = \infty$ then $F (x) = 1 - e^{(e^{- x} - 2 x - 1)} .$

For other Propositions, they may be treated in a similar fashion.

4. Conclusions

The question of the characterization of a distribution is important in many fields and has recently aroused the interest of many researchers. As a result, several characterization results have been published in the literature. The purpose of this study is to present several characterizations of the distribution in their generality in hoping they would be beneficial to researchers wishing to know whether their model meets the requirements of a certain underlying distribution.

Disclosure statement

No potential conflict of interest was reported by the author.

References

Koudou AE, Ley C. Characterizations of GIG laws: A survey. Probab Surv. 2014;11:161–176. DOI:10.1214/13-PS227
Google Scholar
Nagaraja H, Characterizations of probability distributions. In: Springer handbook of engineering statistics. 2006. p. 79–95. London: Springer. DOI:10.1007/978-1-84628-288-1_4
Google Scholar
Dimaki C, Xekalaki E. Towards a unification of certain characterizations by conditional expectations. Ann Inst Stat Math. 1996;48:157–168. DOI:10.1007/BF00049296
Web of Science ®Google Scholar
Ghitany M, Gupta R, Wang S. Some characterization results by conditional expectations and their applications to lindley-type distributions. Int J Probab Stat. 2017;7:86. DOI:10.5539/ijsp.v7n1p86
Google Scholar
Khan A, Athar H, Yaqub M. Characterization of probability distributions through conditional expectation of function of two order statistics. Calcutta Stat Assoc Bull. 2001;51:259–266. DOI:10.1177/0008068320010309
Google Scholar
Ruiz JM, Navarro J. Characterizations based on conditional expectations of the doubled truncated distribution. Ann Inst Stat Math. 1996;48:563–572. DOI:10.1007/BF00050855
Web of Science ®Google Scholar
Su JC, Huang WJ. Characterizations based on conditional expectations. Stat Pap. 2000;41:423. DOI:10.1007/BF02925761
Web of Science ®Google Scholar
Zoroa P, Ruiz J, Marín J. A characterization based on conditional expectations. Commun Stat Theory Methods. 1990;19:3127–3135. DOI:10.1080/03610929008830368
Web of Science ®Google Scholar
Ahsanullah M, Shakil M, Kibria BMG. Characterizations of folded student's t distribution. J Stat Distrib Appl. 2015;2:15. DOI:10.1186/s40488-015-0037-5
Google Scholar
Ahsanullah M, Shakil M, Kibria BG. On a generalized raised cosine distribution: some properties, characterizations and applications. Moroc J Pure Appl Anal. 2019;5:63–85. DOI:10.2478/mjpaa-2019-0006
Google Scholar
Ahsanullah M. Characterizations of univariate continuous distributions. Paris: Springer; 2017. DOI:10.2991/978-94-6239-139-0
Google Scholar
Ahsanullah M, Shakil M. A note on the characterizations of pareto distribution by upper record values. Commun Korean Math Soc. 2012;27:835–842. DOI:10.4134/CKMS.2012.27.4.835
Google Scholar
Ahsanullah M, Shakil M. Characterizations of continuous probability distributions occurring in physics and allied sciences by truncated moment. Int J Adv Stat Probab. 2015;3:100–114.
Google Scholar
Ahsanullah M, Shakil M, Kibria BMG. Characterizations of continuous distributions by truncated moment. J Mod Appl Stat Methods. 2016;15:17. DOI:10.22237/jmasm/1462076160
Google Scholar
Galambos J, Kotz S. Characterizations of probability distributions: a unified approach with an emphasis on exponential and related models. Vol. 675. Berlin: Springer; 2006. DOI:10.1007/BFb0069530
Google Scholar
Hamedani G. Characterizations of recently introduced univariate continuous distributions II. Nova Science Publishers; 2019. (Mathematics research developments). Available from: https://books.google.com.sa/books?id=rtAywAEACAAJ.
Google Scholar
Khan MI. Characterization of some continuous distributions by truncated moment. J Stat Manag Syst. 2021;0:1–10. DOI:10.1080/09720510.2021.1933709
Google Scholar
Kotz S, Shanbhag DN. Some new approaches to probability distributions. Adv Appl Probab. 1980;12:903–921. DOI:10.2307/1426748
Web of Science ®Google Scholar
Shakil M, Ahsanullah M, MG Kibria B. Some characterizations and applications of a size-biased weighted distribution useful in lifetime modelling. J Stat Appl Probab. 2021;10:607–624. DOI:10.18576/jsap/100301
Google Scholar
Hamedani G. A few characterizations of the univariate continuous distributions. JIRSS. 2016;15:63–71. Available from: https://iranjournals.nlai.ir/bitstream/handle/123456789/600330/C1AEA206AD3A27B2CEAF61EFEA8CB376.pdf?sequence=-1
Google Scholar
Hamedani G, Mameli V. Characterizations of the generalized beta-generated family of distributions. J Stat Theory Appl. 2017;16:18–25. DOI:10.2991/jsta.2017.16.1.2
Google Scholar
Merovci F, Alizadeh M, Hamedani GG. Another generalized transmuted family of distributions:properties and applications. Austrian J Stat. 2016;45:71–93. Available from: https://www.ajs.or.at/index.php/ajs/article/view/doi
Web of Science ®Google Scholar
Abramowitz M. Handbook of mathematical functions: with formulas, graphs, and mathematical tables. New York: Dover Publications; 1970. Available from: https://books.google.com.sa/books?id=4j2hwwEACAAJ.
Google Scholar
Gradshteyn IS, Ryzhik IM. Table of integrals, series, and products. New York: Academic Press; 2007.
Google Scholar
Al-Shomrani A, Arif O, Shawky A, et al. Topp–Leone family of distributions: some properties and application. Pakistan J Stat Oper Res. 2016;12:443–451. DOI:10.18187/pjsor.v12i3.1458
Web of Science ®Google Scholar

Further characterization based on conditional expectations: new and extended findings

ABSTRACT

1. Introduction

2. Main results

3. Applications

4. Conclusions

Disclosure statement

References

Information for

Open access

Opportunities

Help and information

Further characterization based on conditional expectations: new and extended findings

ABSTRACT

1. Introduction

2. Main results

3. Applications

4. Conclusions

Disclosure statement

References

Related research

To cite this article:

Download citation

Your download is now in progress and you may close this window

Login or register to access this feature

Information for

Open access

Opportunities

Help and information

Keep up to date