Data-driven chance constrained stochastic program

Abstract

In this paper, we study data-driven chance constrained stochastic programs, or more specifically, stochastic programs with distributionally robust chance constraints (DCCs) in a data-driven setting to provide robust solutions for the classical chance constrained stochastic program facing ambiguous probability distributions of random parameters. We consider a family of density-based confidence sets based on a general $\phi$-divergence measure, and formulate DCC from the perspective of robust feasibility by allowing the ambiguous distribution to run adversely within its confidence set. We derive an equivalent reformulation for DCC and show that it is equivalent to a classical chance constraint with a perturbed risk level. We also show how to evaluate the perturbed risk level by using a bisection line search algorithm for general $\phi$-divergence measures. In several special cases, our results can be strengthened such that we can derive closed-form expressions for the perturbed risk levels. In addition, we show that the conservatism of DCC vanishes as the size of historical data goes to infinity. Furthermore, we analyze the relationship between the conservatism of DCC and the size of historical data, which can help indicate the value of data. Finally, we conduct extensive computational experiments to test the performance of the proposed DCC model and compare various $\phi$-divergence measures based on a capacitated lot-sizing problem with a quality-of-service requirement.

Appendices

Appendix 1: Proof of Lemma 3

Proof

We prove each property as follows:

• (i) By definition, $\phi ^*$ is a supremum of linear functions and hence convex.

• (ii) For any $x_1, x_2 \in {\mathbb {R}}$ such that $x_1 < x_2$, we have

  $$\begin{aligned} x_1t - g(t) \ \le \ x_2t - g(t), \ \ \ \forall t \ge 0. \end{aligned}$$

  Also, since $\phi (t) = +\infty$ for $t < 0$, we have

  $$\begin{aligned} \phi ^*(x) = \ \sup _{t \in {\mathbb {R}}} \left\{ xt-\phi (t)\right\} = \ \sup _{t \ge 0} \left\{ xt-\phi (t)\right\} , \end{aligned}$$

  and so

  $$\begin{aligned} \phi ^*(x_1) = \ \sup _{t \ge 0} \left\{ x_1t-\phi (t)\right\} \ \le \ \sup _{t \ge 0} \left\{ x_2t-\phi (t)\right\} = \ \phi ^*(x_2). \end{aligned}$$

• (iii) Since $\phi (1) = 0$, we have

  $$\begin{aligned} \phi ^*(x) = \ \sup _{t \ge 0} \left\{ xt-\phi (t)\right\} \ \ge \ x. \end{aligned}$$

• (iv) We prove by contradiction. Suppose that $\phi ^*(x) = m$ on the interval [a, b] and $\phi ^*(y) = m' \ne m$ for some $y < a$. First, we observe that $m' < m$ because $\phi ^*$ is nondecreasing. Second, there exists some $\lambda \in [0, 1]$ such that $a = \lambda y + (1-\lambda )b$. It follows that

  $$\begin{aligned} \phi ^*(a) \ \le \ \lambda \phi ^*(y) + (1-\lambda ) \phi ^*(b) = \ \lambda m' + (1-\lambda ) m < m, \end{aligned}$$

  which gives a desirable contradiction. $\square$

Appendix 2: Proof of Remark 2

Proof

If $\ell _{\phi } = +\infty$, then $\ell _{\phi } \ge \overline{m}(\phi ^*)$ and the claim holds. Hence, without loss of generality, we can assume that $\ell _{\phi } \,{<}\, +\infty$. In the remainder of the proof, we show that $\ell _{\phi } + \delta \ge \overline{m}(\phi ^*)$ for any $\delta \,{>}\, 0$, and the claim follows. Because $\ell _{\phi } = \lim _{x \rightarrow +\infty } \phi (x)/x$, there exists a $K \,{>}\, 0$ such that $|\ell _{\phi } - \phi (x)/x| \le \delta /2$, and accordingly $\ell _{\phi }x - \phi (x) \ge -\delta x /2$ for all $x \ge K$. If follows that, for any $M \in {\mathbb {R}}$, there exists an $N := \max \{K, 2M/\delta \}$ such that

$$\begin{aligned} x \ge N \ \Rightarrow&{\left\{ \begin{array}{ll} \ell _{\phi } x - \phi (x) \ge -\delta x /2, \\ \delta x / 2 \ge M \end{array}\right. } \\ \Rightarrow&(\ell _{\phi } + \delta ) x - \phi (x) = \left( \ell _{\phi } x - \phi (x)\right) + \delta x \ \ge \ \delta x / 2 \ \ge \ M. \end{aligned}$$

It follows that $\lim _{x \rightarrow +\infty }\{ (\ell _{\phi } + \delta ) x - \phi (x) \} = +\infty$. Hence,

$$\begin{aligned} \phi ^*(\ell _{\phi } + \delta ) = \ \sup _{x \in {\mathbb {R}}} \{ (\ell _{\phi } + \delta ) x - \phi (x) \} \ \ge \ \lim _{x \rightarrow +\infty }\{ (\ell _{\phi } + \delta ) x - \phi (x) \} = \ +\infty . \end{aligned}$$

Therefore, $\ell _{\phi } + \delta \ge \overline{m}(\phi ^*)$ for any $\delta > 0$, and the proof is completed. $\square$

Appendix 3: Proof of Proposition 2

Proof

First, since $\phi (x) = (x-1)^2$, we have

$$\begin{aligned} \phi ^*(x) = \left\{ \begin{array}{l@{\quad }l} -1, &{} \hbox {if} \, x \le -2,\\ \frac{1}{4}x^2 + x, &{} \hbox {if} \, x\ge -2. \end{array}\right. \end{aligned}$$

Hence, $\underline{m}(\phi ^*) = -2$ and $\overline{m}(\phi ^*) = +\infty$. Second, we solve the problem

$$\begin{aligned} \inf _{\begin{array}{c} z > 0,\\ z_0 + z \ge -2 \end{array}} \frac{\phi ^*(z_0+z) - z_0 - \alpha z + d}{\phi ^*(z_0+z) - \phi ^*(z_0)} = \ \inf _{\begin{array}{c} z > 0,\\ z_0 \ge -2 \end{array}} \frac{\phi ^*(z_0) - z_0 + (1-\alpha ) z + d}{\phi ^*(z_0) - \phi ^*(z_0-z)} \end{aligned}$$

to optimality, where we make a transform by replacing $z_0$ by $z_0-z$. We let $g(z_0, z)$ represent the objective function and discuss the following cases:

• 1. If $z_0 - z \le -2$, then $\phi ^*(z_0-z) = -1$ and $\phi ^*(z_0) = \frac{1}{4}z_0^2 + z_0$. It follows that

  $$\begin{aligned} g(z_0, z) = \ \frac{\left( \frac{1}{4}z_0^2 + z_0 \right) -z_0+(1-\alpha )z+d}{\left( \frac{1}{4}z_0^2 + z_0 \right) - (-1)} = \ \frac{\frac{1}{4}z_0^2 + (1-\alpha )z + d}{\left( \frac{1}{2}z_0 + 1 \right) ^2}, \end{aligned}$$

  and so

  $$\begin{aligned} \frac{\partial g(z_0, z)}{\partial z_0} = \frac{\frac{1}{2}z_0-(1-\alpha )z-d}{\left( \frac{1}{2}z_0 + 1 \right) ^3}. \end{aligned}$$

  Since $z_0 \le z -2,\,z_0 \ge -2$ and $\alpha < 1/2$ by assumption, we have $(1/2)z_0-(1-\alpha )z-d\le (\alpha -1/2)z-d-1 < 0$ and $\frac{1}{2}z_0 + 1 \ge 0$. Hence, $\partial g(z_0, z)/\partial z_0 < 0$ for any fixed z and it is optimal to choose $z^*_0 = z-2$. It follows that

  $$\begin{aligned} \inf _{\begin{array}{c} z > 0,\\ z_0 \ge -2 \end{array}} g(z_0, z) = \ \inf _{z > 0} g(z-2, z) = \inf _{z>0} \ 4(d+1)\left( \frac{1}{z}\right) ^2 - 4\alpha \left( \frac{1}{z}\right) + 1. \end{aligned}$$

  Therefore, it is optimal to choose $z^* = 2(d+1)/\alpha$ and

  $$\begin{aligned} \inf _{\begin{array}{c} z > 0,\\ z_0 \ge -2 \end{array}} g(z_0, z) = \ 1-\frac{\alpha ^2}{d+1}. \end{aligned}$$

• 2. If $z_0 -z \ge -2$, then $\phi ^*(z_0-z) = \frac{1}{4}(z_0-z)^2 + (z_0-z)$ and $\phi ^*(z_0) = \frac{1}{4}z_0^2 + z_0$. It follows that

  $$\begin{aligned} g(z_0, z) = \ \frac{\left( \frac{1}{4}z_0^2 + z_0 \right) -z_0+(1-\alpha )z+d}{\left( \frac{1}{4}z_0^2 + z_0 \right) - \left( \frac{1}{4}(z_0-z)^2 + (z_0-z) \right) } = \ \frac{\frac{1}{4}z_0^2 + (1-\alpha )z + d}{\frac{1}{2}zz_0 + z - \frac{1}{4}z^2}, \end{aligned}$$

  and so

  $$\begin{aligned} \frac{\partial g(z_0, z)}{\partial z_0} = \frac{z\left( z_0^2 + (4-z)z_0 - 4(1-\alpha )z-4d\right) }{8\left( \frac{1}{2}zz_0 + z - \frac{1}{4}z^2\right) ^2}. \end{aligned}$$

  For fixed z, we set $\partial g(z_0, z)/\partial z_0 = 0$ and obtain

  $$\begin{aligned} z_0 = \frac{(z-4)\pm \sqrt{z^2+8(1-2\alpha )z+16(d+1)}}{2}. \end{aligned}$$

  Since $z_0 \ge z-2$, we rule out the negative root and so

  $$\begin{aligned} z^*_0 = \frac{(z-4)+\sqrt{z^2+8(1-2\alpha )z+16(d+1)}}{2} \end{aligned}$$

  is a stationary point of $g(z_0, z)$ with z fixed and the corresponding objective value

  $$\begin{aligned} g(z_0^*, z) = \ \frac{1}{2}\sqrt{16(d+1)\left( \frac{1}{z}\right) ^2+8(1-2\alpha )\left( \frac{1}{z}\right) +1} \ + \frac{1}{2}\left( 1-4\left( \frac{1}{z}\right) \right) . \end{aligned}$$

  Now we show that $z_0^*$ is an optimal solution for $\inf _{z_0\ge z-2}g(z_0, z)$ with z fixed. We compare the value of $g(z_0^*, z)$ with $g(+\infty , z)$ and $g(z-2, z)$ because $+\infty$ and $z-2$ are the end points of the feasible region of $z_0$. We observe that $g(+\infty , z) = +\infty$. Also, we have

  $$\begin{aligned} g(z-2, z) = \ \frac{\frac{1}{4}(z-2)^2+(1-\alpha )z+d}{\frac{1}{2}z(z-2)+z-\frac{1}{4}z^2} = \ 4(d+1)\left( \frac{1}{z}\right) ^2 - 4\alpha \left( \frac{1}{z}\right) + 1, \end{aligned}$$

  and $g(z-2, z) \ge g(z_0^*, z)$. To see that, we compare the values of $g(z-2, z)$ and $g(z_0^*, z)$ by the following inequalities, where the inequalities below imply those above.

  $$\begin{aligned}&g(z-2, z) \ge g(z_0^*, z) \\&\quad \Leftarrow 8(d+1)\left( \frac{1}{z}\right) ^2 - 8\alpha \left( \frac{1}{z}\right) + 2 \\&\quad \ge \sqrt{16(d+1)\left( \frac{1}{z}\right) ^2+8(1-2\alpha )\left( \frac{1}{z}\right) +1} + \left( 1-4\left( \frac{1}{z}\right) \right) \\&\quad \Leftarrow \left[ 8(d+1)\left( \frac{1}{z}\right) ^2 + 4(1-2\alpha )\left( \frac{1}{z}\right) + 1\right] ^2 \\&\quad \ge 16(d+1)\left( \frac{1}{z}\right) ^2+8(1-2\alpha )\left( \frac{1}{z}\right) +1\\&\quad \Leftarrow 16\left( \frac{1}{z}\right) ^2\left[ 2(d+1)\left( \frac{1}{z}\right) + (1-2\alpha )\right] ^2 \ge 0. \end{aligned}$$

  Hence, $\inf _{z_0\ge z-2}g(z_0, z) = g(z_0^*, z)$ with z fixed. Therefore, we have

  $$\begin{aligned} \inf _{z > 0, z_0 \ge z-2} g(z_0, z) = \ \inf _{z > 0} \ \frac{1}{2}\sqrt{16(d+1)z^2+8(1-2\alpha )z+1} \ + \frac{1}{2}(1-4z), \end{aligned}$$

  where we have 1 / z replaced by z. Similarly, we set

  $$\begin{aligned} \frac{\partial g(z_0^*, z)}{\partial z} = \ \frac{8(d+1)z+2(1-2\alpha )}{\sqrt{16(d+1)z^2+8(1-2\alpha )z+1}}-2 = \ 0, \end{aligned}$$

  and obtain

  $$\begin{aligned} z^* = \frac{\sqrt{d^2 + 4d(\alpha -\alpha ^2)}-(1-2\alpha )d}{4d(d+1)}. \end{aligned}$$

  Therefore, we have

  $$\begin{aligned} g(z_0^*, z^*) = \ 1-\alpha +\frac{\sqrt{d^2 + 4d(\alpha -\alpha ^2)}-(1-2\alpha )d}{2(d+1)}. \end{aligned}$$

  Again, we shall compare the value of $g(z_0^*, z^*)$ with $g(z_0^*, +\infty )$ and $g(z_0^*, 0)$ since $+\infty$ and 0 are the end points of the feasible region of z. We observe that $g(z_0^*, +\infty ) = +\infty$ and $g(z_0^*, 0) = 1 \ge g(z_0^*, z^*)$, and hence

  $$\begin{aligned} \inf _{z > 0, z_0 \ge z-2} g(z_0, z) = \ 1-\alpha +\frac{\sqrt{d^2 + 4d(\alpha -\alpha ^2)}-(1-2\alpha )d}{2(d+1)}. \end{aligned}$$

Finally, we compare the optimal value of $g(z_0, z)$ in the two cases. We claim that the optimal value obtained in the latter case is smaller (and hence globally optimal). To see that, we compare the two values by the following inequalities, where the inequalities below imply those above.

$$\begin{aligned}&1 - \frac{\alpha ^2}{d+1} \ge 1-\alpha +\frac{\sqrt{d^2 + 4d(\alpha -\alpha ^2)}-(1-2\alpha )d}{2(d+1)} \\&\quad \Leftarrow d+2\alpha -2\alpha ^2 \ge \sqrt{d^2 + 4d(\alpha -\alpha ^2)} \\&\quad \Leftarrow (d+2\alpha -2\alpha ^2)^2 \ge d^2 + 4d(\alpha -\alpha ^2) \\&\quad \Leftarrow 4\alpha ^2(\alpha -1)^2 \ge 0. \end{aligned}$$

Therefore, the perturbed risk level is

$$\begin{aligned} \alpha ' = \ \alpha - \frac{\sqrt{d^2+4d(\alpha -\alpha ^2)}-(1-2\alpha )d}{2d+2}. \end{aligned}$$

$\square$

Appendix 4: Proof of Proposition 3

Proof

First, Since $\phi (x) = |x-1|$, we have

$$\begin{aligned} \phi ^*(x) = \left\{ \begin{array}{ll} -1, &{}\quad \hbox {if} \, x <-1, \\ x, &{}\quad \hbox {if} \, -1\le x\le 1, \\ +\infty , &{}\quad \hbox {if} \, x > 1. \end{array}\right. \end{aligned}$$

Hence, $\underline{m}(\phi ^*) = -1$ and $\overline{m}(\phi ^*) = 1$. Second, we solve the problem

$$\begin{aligned} \inf _{z > 0, -1 \le z_0 + z \le 1} g(z_0, z) := \frac{\phi ^*(z_0+z) - z_0 - \alpha z + d}{\phi ^*(z_0+z) - \phi ^*(z_0)} \end{aligned}$$

to optimality. We discuss the following cases:

• 1. If $z_0 \le -1$, then $\phi ^*(z_0) = -1$ and $\phi ^*(z_0+z) = z_0+z$. It follows that

  $$\begin{aligned} g(z_0, z) = \frac{(1-\alpha )z+d}{z_0+z+1}. \end{aligned}$$

  Note here that for any given $z,\,g(z_0, z)$ is a nonincreasing function of $z_0$, due to the fact that $z_0+z+1 \ge 0$. Meanwhile, $z_0+z \le 1$. Hence, it is optimal to choose $z_0^* = \min \{1-z, -1\}$ and so

  $$\begin{aligned} g(z_0^*, z) = \left\{ \begin{array}{ll} \frac{(1-\alpha )z+d}{2}, &{}\quad \hbox {if} \, z \ge 2, \\ \frac{(1-\alpha )z+d}{z}, &{}\quad \hbox {if} \, z \le 2. \end{array}\right. \end{aligned}$$

  Therefore, $g(z_0^*, z)$ is nonincreasing on z on the interval (0, 2] and nondecreasing on z on the interval $[2, +\infty )$, and so $g(z_0^*, z^*) = 1-\alpha +\frac{d}{2}$.

• 2. If $-1 \le z_0 \le 1$, then $\phi ^*(z_0) = z_0$. Also, we have $z \le 2$ and $\phi ^*(z_0+z) = z_0+z$ because $-1 \le z_0 + z\le 1$. Hence,

  $$\begin{aligned} g(z_0, z) = \frac{(1-\alpha )z+d}{z} = 1 - \alpha + \frac{d}{z} \ge 1-\alpha +\frac{d}{2}, \end{aligned}$$

  and the lower bound is attained at $z^* = 2$. Therefore, $g(z_0^*, z^*) = 1-\alpha +\frac{d}{2}$.

To sum up, we have $1 - \alpha ' = g(z_0^*, z^*) = 1-\alpha +\frac{d}{2}$, or equivalently $\alpha ' = \alpha -\frac{d}{2}$. $\square$

Appendix 5: Proof of Proposition 4

Proof

We divide the proof into two parts. In the first part, we show that the perturbed risk level

$$\begin{aligned} \alpha ' = 1 - \inf _{x\in (0, 1)} \left\{ \frac{e^{-d} x^{1-\alpha } - 1}{x - 1} \right\} . \end{aligned}$$

(15)

In the second part, we show how to compute $\alpha '$ by using bisection line search.

(Risk level) First, since $\phi (x) = x\log x -x + 1$, we have $\phi ^*(x) = e^x-1$. Hence, $\underline{m}(\phi ^*) = -\infty$ and $\overline{m}(\phi ^*) = +\infty$. Second, we solve the problem

$$\begin{aligned} \inf _{z > 0} g(z_0, z) := \frac{\phi ^*(z_0+z) - z_0 - \alpha z + d}{\phi ^*(z_0+z) - \phi ^*(z_0)} = \ \inf _{z > 0} \frac{e^z+(d-\alpha z-z_0-1)e^{-z_0}}{e^z-1} \end{aligned}$$

to optimality. Since

$$\begin{aligned} \frac{\partial g}{\partial z_0} = -\frac{d-\alpha z-z_0}{e^{z_0}(e^z-1)}, \end{aligned}$$

we have $z_0^* = d-\alpha z$ by setting $\partial g/\partial z_0 = 0$, and so

$$\begin{aligned} \inf _{z > 0} g(z_0, z) =&\ \inf _{z>0} \ g(z_0^*, z) \nonumber \\ =&\ \inf _{z > 0} \ \frac{e^z-e^{\alpha z-d}}{e^z-1} \nonumber \\ =&\ \inf _{z > 0} \ \frac{1-e^{-d}(1/e^z)^{1-\alpha }}{1-(1/e^z)} \nonumber \\ =&\ \inf _{x \in (0, 1)} \ \frac{e^{-d}x^{1-\alpha }-1}{x-1}, \end{aligned}$$

(16)

where Eq. (16) follows by replacing $(1/e^z)$ with x. Therefore, we have proved Eq. (15).

(Computation) We compute $\alpha '$ by searching the optimal solution of the minimization problem

$$\begin{aligned} \inf _{x\in (0, 1)} \Bigl \{ \frac{e^{-d} x^{1-\alpha } - 1}{x - 1} \Bigr \}. \end{aligned}$$

First, by denoting $1 - \alpha ' = \inf _{x\in (0, 1)} h(x)$, we have

$$\begin{aligned} h'(x) = \frac{1 - e^{-d} \alpha x^{1-\alpha } - e^{-d} (1-\alpha ) x^{-\alpha }}{(x-1)^2}, \ \ \forall x \in (0, 1). \end{aligned}$$

It is clear that $(x-1)^2$ decreases as x increases. Meanwhile, since $x < 1$ and $x^{-\alpha -1} > x^{-\alpha }$, we have

$$\begin{aligned} (1 - e^{-d} \alpha x^{1-\alpha } - e^{-d} (1-\alpha ) x^{-\alpha })'_x = e^{-d} \alpha (1-\alpha ) (x^{-\alpha -1} - x^{-\alpha }) > 0. \end{aligned}$$

Therefore, $h'(x)$ increase as x increases in (0, 1), and hence the function h(x) is convex over x in (0, 1). Because $\displaystyle \lim _{x \rightarrow 0^+} h'(x) = -\infty$ and $\displaystyle \lim _{x \rightarrow 1^-} h'(x) = +\infty$, we have:

$$\begin{aligned} \hbox {the infimum of } h(x) \hbox { is attained in the interval } (0, 1). \end{aligned}$$

(17)

We can compute the optimal $x^*$ by forcing

$$\begin{aligned} \frac{1 - e^{-d} \alpha (x^*)^{1-\alpha } - e^{-d} (1-\alpha ) (x^*)^{-\alpha }}{(x^*-1)^2} = 0, \end{aligned}$$

i.e., $(x^*)^{\alpha } = e^{-d}\alpha x^* + e^{-d}(1-\alpha )$. The intersection of functions $x^{\alpha }$ and $e^{-d}\alpha x + e^{-d}(1-\alpha )$ can be easily computed by a bisection line search. Finally, to achieve $\epsilon$ accuracy, i.e., $|\hat{x} - x^*| \le \epsilon$, of the incumbent probing value $\hat{x}$, we only have to conduct S steps of bisection, such that $2^{-S} \le \epsilon$. It follows that $S \ge \left\lceil \log _2(\frac{1}{\epsilon }) \right\rceil$. $\square$

Appendix 6: Proof of Proposition 6

Proof

First, the convergence claim follows from Proposition 5 because $x = 1$ is the unique minimizer of function $\phi _{\chi D2}(x) = (x-1)^2,\,x \ge 0$.

Second, we define

$$\begin{aligned} g(d) \ := \ \alpha - \alpha ' = \ \frac{\sqrt{d^2+4d(\alpha -\alpha ^2)}-(1-2\alpha )d}{2d+2} \end{aligned}$$

based on Proposition 2. Then, we have

$$\begin{aligned} g'(d) = \ \frac{1}{2(d+1)^2}\left[ \frac{(2\alpha ^2-2\alpha +1)d+2(\alpha -\alpha ^2)}{\sqrt{d^2+4d(\alpha -\alpha ^2)}} - (1-2\alpha ) \right] . \end{aligned}$$

(18)

Since $d = d(N)$ by assumption, we have

$$\begin{aligned} \hbox {VoD}_{\alpha }&= \frac{{\hbox {d}}\alpha '}{{\hbox {d}}N} = \left( \frac{{\hbox {d}}\alpha '}{{\hbox {d}}d}\right) \left( \frac{{\hbox {d}}d(N)}{{\hbox {d}}N}\right) \nonumber \\&= {-} g'(d) \ \left( \frac{{\hbox {d}}d(N)}{{\hbox {d}}N}\right) . \end{aligned}$$

(19)

The proof is completed by substituting the definition of $g'(d)$ in (18) into Eq. (19). $\square$

Appendix 7: Proof of Proposition 7

Proof

The convergence claim follows from Proposition 5 because $x = 1$ is the unique minimizer of function $\phi _{\hbox {KL}}(x) = x \log (x) - x + 1,\,x \ge 0$. We divide the remainder of the proof into two parts. We develop the relationship between $\alpha '$ and d in the first part, and compute the value of data in the second part.

(Relationship between $\alpha '$ and d) From Proposition 4, we have

$$\begin{aligned} 1 - \alpha ' = \inf _{x\in (0, 1)} \left\{ \frac{e^{-d} x^{1-\alpha } - 1}{x - 1} \right\} . \end{aligned}$$

(20)

Also, the optimal objective value of the embedded optimization problem in equality (20) can be attained by some $\bar{x} \in (0, 1)$ (based on claim (17) in the proof of Proposition 4), which is the stationary point of the objective function. It follows that

$$\begin{aligned} \left\{ \begin{array}{l} (e^{-d} \bar{x} ^{1-\alpha } - 1)/(\bar{x} - 1) = 1 - \alpha ' \\ \bar{x}^{\alpha } = e^{-d} \alpha \bar{x} + e^{-d} (1 - \alpha ). \end{array} \right. \end{aligned}$$

(21)

Solving this nonlinear equation system, we reformulate the first equation and then substitute the second equation into the first as follows:

$$\begin{aligned}&e^{-d} \bar{x} - (1-\alpha ') \bar{x} \bar{x}^{\alpha } = \alpha ' \bar{x}^{\alpha } \\&\quad \Rightarrow e^{-d} \bar{x} - (1-\alpha ') \bar{x} \Bigl ( e^{-d} \alpha \bar{x} + e^{-d} (1 - \alpha ) \Bigr ) = \alpha ' \Bigl ( e^{-d} \alpha \bar{x} + e^{-d} (1 - \alpha ) \Bigr ) \\&\quad \Rightarrow (\bar{x} - 1) \Bigl ( \alpha (1-\alpha ') \bar{x} - \alpha '(1 - \alpha ) \Bigr ) = 0. \end{aligned}$$

Ruling out the solution $\bar{x} = 1$, we have $\bar{x} = \frac{\alpha '(1 - \alpha )}{\alpha (1-\alpha ')} \in (0, 1)$. Finally, we substitute the solution of $\bar{x}$ back into the second equation in (21) and obtain

$$\begin{aligned} e^{-d} = {\bar{x}^{\alpha }}\big /{(\alpha \bar{x} + 1 - \alpha )} = \bigl ( {\alpha '}\big /{\alpha } \bigr )^{\alpha } \bigl ( {(1 - \alpha ')}\big /{(1 - \alpha )} \bigr )^{1 - \alpha }. \end{aligned}$$

(22)

Finally, by taking the natural logarithm on both sides of Eq. (22), we obtain that

$$\begin{aligned} d = \alpha \log \left( \frac{\alpha }{\alpha '} \right) + (1-\alpha ) \log \left( \frac{1-\alpha }{1-\alpha '} \right) . \end{aligned}$$

(23)

(Value of data) From Eq. (23) we have

$$\begin{aligned} \frac{{\hbox {d}}d}{{\hbox {d}}\alpha '} = -\frac{\alpha }{\alpha '} + \frac{1-\alpha }{1-\alpha '} = \frac{\alpha '-\alpha }{\alpha '(1-\alpha ')}. \end{aligned}$$

It is easy to observe that ${{\hbox {d}}d}\big /{{\hbox {d}}\alpha '}$ is a monotone function of $\alpha '$ and ${{\hbox {d}}d}\big /{{\hbox {d}}\alpha '} \ne 0$. Hence, we have

$$\begin{aligned} \frac{{\hbox {d}}\alpha '}{{\hbox {d}}d} = 1 \Bigl / \left( \frac{{\hbox {d}}d}{{\hbox {d}}\alpha '}\right) = \frac{\alpha '(1-\alpha ')}{\alpha '-\alpha }. \end{aligned}$$

Therefore,

$$\begin{aligned} \hbox {VoD}_{\alpha } = \left( \frac{{\hbox {d}}\alpha '}{{\hbox {d}}d}\right) \left( \frac{{\hbox {d}}d(N)}{{\hbox {d}}N} \right) = \left[ \frac{\alpha '(1-\alpha ')}{\alpha '-\alpha }\right] d'(N). \end{aligned}$$

$\square$