Chance-constrained problems and rare events: an importance sampling approach

Abstract

We study chance-constrained problems in which the constraints involve the probability of a rare event. We discuss the relevance of such problems and show that the existing sampling-based algorithms cannot be applied directly in this case, since they require an impractical number of samples to yield reasonable solutions. We argue that importance sampling (IS) techniques, combined with a Sample Average Approximation (SAA) approach, can be effectively used in such situations, provided that variance can be reduced uniformly with respect to the decision variables. We give sufficient conditions to obtain such uniform variance reduction, and prove asymptotic convergence of the combined SAA-IS approach. As it often happens with IS techniques, the practical performance of the proposed approach relies on exploiting the structure of the problem under study; in our case, we work with a telecommunications problem with Bernoulli input distributions, and show how variance can be reduced uniformly over a suitable approximation of the feasibility set by choosing proper parameters for the IS distributions. Although some of the results are specific to this problem, we are able to draw general insights that can be useful for other classes of problems. We present numerical results to illustrate our findings.

Appendices

Appendix 1: MIP formulation for $\hat{p}^{\text {IS}_0}_a$ estimator under heterogeneous demand

We can formulate an integer linear programming model for this problem

$$\begin{aligned} \min \sum \limits _{a\in A} w_a&\ \end{aligned}$$

(52)

$$\begin{aligned} \mathcal {N}y^c = d^c&\quad \quad \forall c=1,\ldots ,C \end{aligned}$$

(53)

$$\begin{aligned} \sum \limits _{c=1}^{C} \hat{\xi }^s_c y_{c,a} \le w_a + \sum \limits _{k=1}^C k\cdot u_{a,s,k}&\ \quad \quad \forall a\in A, \quad \forall s= 1,\ldots ,N \end{aligned}$$

(54)

$$\begin{aligned} \sum \limits _{c=1}^{C} \hat{\xi }^s_c y_{c,a} \ge \sum \limits _{k=1}^C k\cdot u_{a,s,k}&\ \quad \quad \forall a\in A, \quad \forall s= 1,\ldots ,N \end{aligned}$$

(55)

$$\begin{aligned} \sum \limits _{s=1}^N \sum \limits _{k=1}^C e^{-k\lambda } u_{a,s,k} \le \alpha N \sum \limits _{k=0}^C G_a(k) v_{a,k}&\ \quad \quad \forall a\in A \end{aligned}$$

(56)

$$\begin{aligned} \sum \limits _{c=1}^{C} y_{c,a} = \sum \limits _{k=0}^C k v_{a,k}&\ \quad \quad \forall a\in A \end{aligned}$$

(57)

$$\begin{aligned} \sum \limits _{k=0}^C v_{a,k} = 1&\ \quad \quad \forall a\in A \end{aligned}$$

(58)

$$\begin{aligned} \sum \limits _{k=1}^C u_{a,s,k} \le 1&\ \quad \quad \forall a\in A, \quad \forall s= 1,\ldots ,N \end{aligned}$$

(59)

$$\begin{aligned} w_a\in \mathbb {N},\ y_{c,a} \in \{0,1\}, u_{a,s,k}\in \{0,1\},\nonumber \\ v_{a,k}\in \{0,1\}&\quad \quad \forall a\in A, \quad \forall c=1,\ldots ,C,\nonumber \\&\qquad \forall s= 1,\ldots ,N \end{aligned}$$

(60)

Binary variables $v_{a,k}$, together with Eqs. (58) and (57), satisfy that $v_{a,k}=1$ if and only if $\sum \nolimits _{c=1}^C y_{c,a}=k$. The role of binary variables u is explained in the following lemma

Lemma 2

Let (x, w, u, v) be an optimal solution of previous formulation, then there exist an optimal solution $(x,w,\hat{u},v)$ such that

• 1. $\sum _{c=1}^C \hat{\xi }^s_c y_{c,a} \le w_a$ if and only if $\hat{u}_{a,s,k}=0$ for all $k=1,\ldots ,C$.

• 2. if $\hat{u}_{a,s,k}=1$ then $\sum _{c=1}^C \hat{\xi }^s_c y_{c,a} =k$,

Hence,

$$\begin{aligned} \sum _{k=1}^C e^{-k\lambda } \hat{u}_{a,s,k} = e^{\sum _{c=1}^C \hat{\xi }^s_c y_{c,a}} {\mathbbm {1}}_{\left\{ \sum \limits _{c=1}^C \hat{\xi }^s_c y_{c,a} > w_a\right\} } \end{aligned}$$

Proof

First, note that constraint (54) impose that if $u_{a,s,k}=0$ for all k then $\sum _{c=1}^C \hat{\xi }^s_c y_{c,a} \le w_a$. Suppose that $\sum _{c=1}^C \hat{\xi }^s_c y_{c,a} \le w_a$ but $u_{a,s,k'}=1$ for some $k'$. It is easy to see that defining $\hat{u}_{a,s,k}=0$ for all k and $\hat{u}=u$ for the other variables, then $\hat{u}$ also satisfy Eqs. (54) and (55), and since $\hat{u}\le u$ then it also satisfy Eqs. (59) and (56), hence $(x,w,\hat{u},v)$ is also optimal. Repeating this procedure is easy to see that we obtain a solution that satisfies condition (1). For the second condition, suppose that $u_{a,s,k}=1$ for some k but $\sum _{c=1}^C \hat{\xi }^s_c y_{c,a} > k$. Let $\hat{k}=\sum _{c=1}^C \hat{\xi }^s_c y_{c,a}$ and define $\hat{u}_{a,s,\hat{k}}=1$, $\hat{u}_{a,s,k}=0$ $\forall k\ne \hat{k}$ and $\hat{u}=u$ for the other variables. By definition, $(w,x,\hat{u},v)$ satisfies (54) and (59), and since $\hat{k}>k$ then it also satisfies (54). On the other hand, since $\lambda >0$ then $e^{-\lambda k}>e^{-\lambda \hat{k}}$ so it also satisfy (56) and then $(x,w,\hat{u},v)$ is also optimal. Repeating this procedure is easy to see that we obtain a solution that satisfies condition (2). $\square$

Lemma 2 shows that the optimal solution (y, w) of this MIP formulation satisfies

$$\begin{aligned} \frac{1}{N} \sum _{s= 1}^N e^{\sum _{c=1}^C \hat{\xi }^s_c y_{c,a}} {\mathbbm {1}}_{\left\{ \sum \limits _{c= 1}^C\hat{\xi }^s_c y_{c,a} > w_a\right\} } \le \alpha \cdot G_a\left( \sum _{c=1}^C y_{c,a} \right) \quad \forall a\in A, \end{aligned}$$

which is the desired approximation of equation $\hat{p}^{\text {IS}_0}_a \le \alpha$.

Appendix 2: Proofs of results

1.1 Proof of Lemma 1

Lemma 1

Suppose that the set-function $I_x$ is such that $G(x,\cdot )$ is $I_x$-determined for each $x \in X$. Given an i.i.d. sample $(\hat{\xi }^1,\ldots ,\hat{\xi }^N)$ from the distribution of $\hat{\xi }$, let

$$\begin{aligned} \hat{p}^{\text {IS}_0}(x)\ :=\ \frac{1}{N} \sum _{j=1}^{N} {\mathbbm {1}}_{\{G\big (x,\hat{\xi }^j\big )>0\}}L_x(\hat{\xi }^j). \end{aligned}$$

(61)

Then $\hat{p}^{\text {IS}_0}(x)$ is also an unbiased estimator of p(x). Moreover,

$$\begin{aligned} \text {Var}(\hat{p}^{\text {IS}_0}(x))\ = \ \text {Var}(\hat{p}^{\text {IS}}(x)) - \frac{1}{N}\mathbb {E}_{\hat{\xi }}\left[ \text {Var}\left( {\mathbbm {1}}_{\{G(x,\hat{\xi })>0\}}\,|\,(\hat{\xi }_i)_{i\in I_x}\right) \right] \end{aligned}$$

(62)

Proof

First let us prove that the estimator $\hat{p}^{\text {IS}_0}(x)$ is unbiased, for which it suffices to show that $\mathbb {E}_{\hat{\xi }}\left[ {\mathbbm {1}}_{\{G(x,\hat{\xi })>0\}}L_x(\hat{\xi })\right] =p(x)$. Indeed, we have

$$\begin{aligned} \mathbb {E}_{\hat{\xi }}\left[ {\mathbbm {1}}_{\{G(x,\hat{\xi })>0\}}L_x(\hat{\xi })\right]= & {} \mathbb {E}_{\hat{\xi }}\left[ {\mathbbm {1}}_{\{G(x,\hat{\xi })>0\}}\mathbb {E}_{\hat{\xi }}\left[ L(\hat{\xi })~|(\hat{\xi }_i)_{i \in I_x}\right] \right] \nonumber \\= & {} \mathbb {E}_{\hat{\xi }}\left[ \mathbb {E}_{\hat{\xi }}\left[ {\mathbbm {1}}_{\{G(x,\hat{\xi })>0\}}L(\hat{\xi })~|~(\hat{\xi }_i)_{i \in I_x}\right] \right] \end{aligned}$$

(63)

$$\begin{aligned}= & {} \mathbb {E}_{\hat{\xi }}\left[ {\mathbbm {1}}_{\{G(x,\hat{\xi })>0\}}L(\hat{\xi })\right] = p(x), \end{aligned}$$

(64)

where the second equality follows from the assumption that $G(x,\cdot )$ is $I_x$-determined, which implies that $G(x,\hat{\xi })$ is measurable with respect to the sigma-algebra generated by $(\hat{\xi }_i)_{i \in I_x}$.

For the second assertion of the theorem, note that

$$\begin{aligned} \mathbb {E}_{\hat{\xi }}\left[ {\mathbbm {1}}_{\{G(x,\hat{\xi })>0\}}^2 L_x(\hat{\xi })^2\right]&= \mathbb {E}_{\hat{\xi }}\left[ {\mathbbm {1}}_{\{G(x,\hat{\xi })>0\}} \left( \mathbb {E}_{\hat{\xi }}\left[ L(\hat{\xi })~|~(\hat{\xi }_i)_{i \in I_x}\right] \right) ^2\right] \\&= \mathbb {E}_{\hat{\xi }}\left[ \left( \mathbb {E}_{\hat{\xi }}\left[ {\mathbbm {1}}_{\{G(x,\hat{\xi })>0\}}L(\hat{\xi })~|~(\hat{\xi }_i)_{i \in I_x}\right] \right) ^2\right] \\&= \ \mathbb {E}_{\hat{\xi }}\left[ \mathbb {E}_{\hat{\xi }}\left[ {\mathbbm {1}}_{\{G(x,\hat{\xi })>0\}}L(\hat{\xi })^2~|~(\hat{\xi }_i)_{i \in I_x}\right] \right. \\&\qquad - \left. \text {Var}\left( {\mathbbm {1}}_{\{G(x,\hat{\xi })>0\}}L(\hat{\xi })~|~(\hat{\xi }_i)_{i \in I_x}\right) \right] \\&= \ \mathbb {E}_{\hat{\xi }}\left[ {\mathbbm {1}}_{\{G(x,\hat{\xi })>0\}}L(\hat{\xi })^2\right] \\&\qquad - \mathbb {E}_{\hat{\xi }}\left[ \text {Var}\left( {\mathbbm {1}}_{\{G(x,\hat{\xi })>0\}}L(\hat{\xi })~|~(\hat{\xi }_i)_{i \in I_x}\right) \right] \end{aligned}$$

and therefore

$$\begin{aligned} N \text {Var}(\hat{p}^{\text {IS}_0}(x))&= \mathbb {E}_{\hat{\xi }}\left[ {\mathbbm {1}}_{\{G(x,\hat{\xi })>0\}}^2 L_x(\hat{\xi })^2\right] - p(x)^2 = \ \mathbb {E}_{\hat{\xi }}\left[ {\mathbbm {1}}_{\{G(x,\hat{\xi })>0\}}L(\hat{\xi })^2\right] \\&\quad - \mathbb {E}_{\hat{\xi }}\left[ \text {Var}\left( {\mathbbm {1}}_{\{G(x,\hat{\xi })>0\}}L(\hat{\xi })~|~(\hat{\xi }_i)_{i \in I_x}\right) \right] - p(x)^2 \\&= N \text {Var}(\hat{p}^{\text {IS}}(x)) - \mathbb {E}_{\hat{\xi }}\left[ \text {Var}\left( {\mathbbm {1}}_{\{G(x,\hat{\xi })>0\}}\,|\,(\hat{\xi }_i)_{i\in I_x}\right) \right] . \end{aligned}$$

$\square$

1.2 Proof of Proposition 3

Proposition 3

Let $\zeta _1,\ldots ,\zeta _m$ be $m\ge 1$ independent Bernoulli random variables with $\mathbb {P}\{\zeta _i=1\}=p_i$, and suppose that $0<p_i<1$ for all i. Let $Z:= \sum _{i=1}^m \zeta _i$, and define $\delta := \min _i\, p_i(1-p_i) > 0$. Then, we have

$$\begin{aligned} \mathbb {P}\left\{ Z > \mathbb {E}[Z]\right\} \ > \ \frac{\delta }{2m}. \end{aligned}$$

(65)

Proof

Let $u:[0,m]\mapsto \mathbb {R}$ be the function defined as $u(t):= m^2 - t^2$. Since $u(\cdot )$ is nonnegative and decreasing on [0, m], we have that

$$\begin{aligned} \mathbb {P}\left\{ Z \le \mathbb {E}[Z]\right\}&= \mathbb {P}\left\{ u(Z) \ge u(\mathbb {E}[Z])\right\} \\&= \ \mathbb {P}\left\{ m^2 -Z^2 \ge m^2 - (\mathbb {E}[Z])^2\right\} \\&\le \ \frac{\mathbb {E}\left[ m^2 - Z^2\right] }{m^2 - (\mathbb {E}[Z])^2}, \end{aligned}$$

where the last inequality follows from Markov’s inequality. Thus, we have

$$\begin{aligned} \mathbb {P}\left\{ Z > \mathbb {E}[Z]\right\}&\ge \ 1 - \frac{\mathbb {E}\left[ m^2 - Z^2\right] }{m^2 - (\mathbb {E}[Z])^2} \ =\ \frac{\mathbb {E}\left[ Z^2\right] -(\mathbb {E}[Z])^2}{m^2 - (\mathbb {E}[Z])^2} \nonumber \\&=\ \frac{\text {Var}(Z)}{(m + \mathbb {E}[Z])(m - \mathbb {E}[Z])} . \end{aligned}$$

(66)

Next, notice that independence of $\{\zeta _i\}$ implies that $\text {Var}(Z)=\sum _{i=1}^m p_i(1-p_i)$. Moreover, since $0<\mathbb {E}[Z]<m$ we have that $m + \mathbb {E}[Z] < 2m$, $m - \mathbb {E}[Z]< m$ and thus from (66) we have that

$$\begin{aligned} \mathbb {P}\left\{ Z > \mathbb {E}[Z]\right\}&> \ \frac{\sum _{i=1}^m p_i(1-p_i)}{2m^2} \ \ge \ \frac{\delta m}{2m^2} \ = \ \frac{\delta }{2m}. \end{aligned}$$

$\square$

1.3 Proof of Theorem 1

Theorem 1

Suppose that $0<\rho _c<1$ for all $c=1,\dots , C$. Let $x=(y,w)$ be such that $w \in \mathbb {N}$ satisfies $\sum _{c=1}^C \rho _c y_c < w \le \sum _{c=1}^C y_c-1$. Then the function $B_{x}(\varvec{\lambda })$ is convex and there exists $\lambda ^*_{x}\in \mathbb {R}_+\cup \{\infty \}$ such that the vector $\varvec{\lambda }$ defined as $\lambda _c=\lambda ^*_{x}$ $\forall c \in C$ minimizes $B_{x}(\varvec{\lambda })$. If $w=\sum _{c=1}^C y_c-1$, then the optimal $\lambda ^*_{x}$ is $\lambda ^*_{x}=\infty$ and $\hat{\rho }_c(\lambda ^*_{x})=1$; otherwise, $\lambda ^*_{x}$ and $\hat{\rho }_c(\lambda ^*_{x})$ satisfy

$$\begin{aligned} \sum _{c=1}^C \hat{\rho }_c(\lambda ^*_{x}) y_c = w+1 \qquad \text{ and } \qquad \hat{\rho }_c(\lambda ^*_{x}) =\frac{e^{\lambda ^*_{x}} \rho _c}{e^{\lambda ^*_{x}} \rho _c +(1-\rho _c)}. \end{aligned}$$

To prove the theorem, we need the following lemma, the proof of which is shown after the proof of the theorem.

Lemma 3

For $n\ge 1$, let $\rho _i$, $i=1,\ldots ,n$ be numbers such that $\rho _i \in (0,1)$ and $\rho _1 \ge \rho _2 \ge \ldots \ge \rho _n$. Given an integer w such that $0 \le w \le n-1$, consider problem (P) defined as follows:

$$\begin{aligned} \min _{\lambda \in \mathbb {R}^n_+} \max _{\begin{array}{c} z_i\in \{0,1\}^n\\ \sum _i z_i =w+1 \end{array}} -\sum _{i=1}^n z_i \lambda _i + \sum _{i=1}^n \log (e^{\lambda _i} \rho _i + (1-\rho _i)) . \end{aligned}$$

(P)

Then, there exists an optimal solution to (P) that satisfies $\lambda _1 \le \lambda _2 \le \ldots \le \lambda _n$.

Proof

(of Theorem 1) Let $n=\sum _{c=1}^C y_c$. Without loss of generality, let us assume for the sake of simplifying notation that the set $\{c: y_c=1\}$ corresponds to $\{1,\ldots ,n\}$. Since the $\log$ function is increasing, we have that

$$\begin{aligned} \log (B_{x}(\varvec{\lambda }))\ =\ \max _{\begin{array}{c} z_i\in \{0,1\}^n\\ \sum _i z_i =w+1 \end{array}} -\sum _{i=1}^n z_i \lambda _i + \sum _{i=1}^n \log (e^{\lambda _i} \rho _i + (1-\rho _i)) \end{aligned}$$

By Lemma 3, minimizing $\log (B_{x}(\varvec{\lambda }))$ amounts to solving the following problem:

$$\begin{aligned} \min _{\varvec{\lambda }\in \mathbb {R}^n}\, \psi (\varvec{\lambda }):= -\sum _{i=1}^{w+1} \lambda _i +&\sum _{i=1}^n \log (e^{\lambda _i} \rho _i + (1-\rho _i)) \end{aligned}$$

(67)

$$\begin{aligned} \lambda _i \le \lambda _{i+1}&\quad i=1\ldots n-1 \end{aligned}$$

(68)

$$\begin{aligned} \lambda _1 \ge 0&\end{aligned}$$

(69)

Note that the objective function of the above problem is strictly convex in $\varvec{\lambda }$. In fact, its second derivatives are

$$\begin{aligned} \frac{\partial ^2 \psi }{\partial \lambda _i^2} = \frac{e^{\lambda _i} \rho _i (1- \rho _i)}{(e^{\lambda _i}\rho _i + (1-\rho _i))^2}>0, \qquad \frac{\partial ^2 \psi }{\partial \lambda _i\partial \lambda _j}=0. \end{aligned}$$

Since $B_x(\varvec{\lambda })=\exp (\log (B_x(\varvec{\lambda }))$ and $\log (B_x(\varvec{\lambda }))$ is convex—though not strictly convex due to the components $\lambda _c$ such that $y_c=0$—it follows that $B_x$ is convex in $\varvec{\lambda }$. Of course, the components $\lambda _c$ such that $y_c=0$ do not affect the value of $B_x(\varvec{\lambda })$.

Suppose first that $w=n-1$. Then, the first derivative of the objective function in (67) is given by

$$\begin{aligned} \frac{\partial \psi }{\partial \lambda _i} \ = \ -1 + \frac{e^{\lambda _i} \rho _i}{e^{\lambda _i} + (1- \rho _i)}, \quad i=1,\ldots ,n, \end{aligned}$$

so we see that $\lim _{\varvec{\lambda }\rightarrow \infty } \nabla \psi (\varvec{\lambda }) =0$. Notice that we can in particular interpret $\lim _{\varvec{\lambda }\rightarrow \infty }$ as $\lim _{{\lambda }\rightarrow \infty }$ with $\lambda _i=\lambda$. That is, in that case the optimal solution of (67)–(69) is $\lambda _i=\infty$, $i=1,\ldots ,n$.

Consider now the case $w<n-1$. We will show that problem (67)–(69) has a unique optimal solution, which can be found by writing the Karush-Kuhn-Tucker conditions as follows:

$$\begin{aligned} -1_{(i\le w+1)} + \frac{e^{\lambda _i} \rho _i}{e^{\lambda _i} \rho _i +(1-\rho _i)} +\mu _{i}-\mu _{i-1}&= 0&i=1\ldots n-1 \end{aligned}$$

(70)

$$\begin{aligned} \frac{e^{\lambda _n} \rho _n}{e^{\lambda _n} \rho _n +(1-\rho _n)} -\mu _{n-1}&= 0&\end{aligned}$$

(71)

$$\begin{aligned} \mu _i (\lambda _{i+1}-\lambda _i)&= 0&i=1\ldots n-1 \end{aligned}$$

(72)

$$\begin{aligned} \mu _0 \lambda _1&= 0&\end{aligned}$$

(73)

$$\begin{aligned} \mu _i&\ge 0&i=0\ldots n-1 \end{aligned}$$

(74)

where $\varvec{\mu }=(\mu _i)$ is the vector of Lagrangean multipliers of constraints (68) and $\mu _0$ is the Lagrangean multiplier of constraint (69).

Consider now a particular choice of vectors $\varvec{\mu }$ and $\varvec{\lambda }$ defined as follows. All components of $\varvec{\lambda }$ are identical, with $\lambda _i=\lambda ^*$, where $\lambda ^*\in \mathbb {R}_{+}$ solves the equation

$$\begin{aligned} \varphi (\lambda ^*):= \sum _{i=1}^{n} \frac{e^{\lambda ^*} \rho _i}{e^{\lambda ^*} \rho _i +(1-\rho _i)}\ = \ w+1. \end{aligned}$$

(75)

Note that we can always find such $\lambda ^*$, since the function $\varphi (\lambda )$ is continuous and increasing, and

$$\begin{aligned} \varphi (0)&= \ \sum _{i=1}^n \rho _i\ < w \ < \ w+1 \end{aligned}$$

(76)

$$\begin{aligned} \lim _{\lambda \rightarrow \infty } \varphi (\lambda )&= \ n\ > \ w+1. \end{aligned}$$

(77)

The inequalities in (76) follow from the assumptions of the theorem on w and the fact that we are analyzing the case $w<n-1$. The components of $\varvec{\mu }$ are defined as

$$\begin{aligned} \mu _0&:= 0 \end{aligned}$$

(78)

$$\begin{aligned} \mu _i&:= \min \{i,w+1\} - \sum _{j=1}^i \frac{e^{\lambda ^*} \rho _j}{e^{\lambda ^*} \rho _j +(1-\rho _j)} \quad i=1,\ldots ,n-1. \end{aligned}$$

(79)

We claim that $\varvec{\mu }$ and $\varvec{\lambda }$ satisfy the KKT conditions (70)–(74) laid out above. To see that, observe that Eqs. (78)–(79) imply (70). Equation (71) follows from (75), since we have

$$\begin{aligned} \frac{e^{\lambda ^*} \rho _n}{e^{\lambda ^*} \rho _n +(1-\rho _n)}\ = \ w+1- \sum _{i=1}^{n-1} \frac{e^{\lambda ^*} \rho _i}{e^{\lambda ^*} \rho _i +(1-\rho _i)} \end{aligned}$$

and the latter term coincides with $\mu _{n-1}$ defined in (79). Equations (72) and (73) are trivially satisfied. Finally, we show that (74) holds, with strict inequality if $i\ge 1$. Indeed, (75) implies that

$$\begin{aligned} \sum _{j=1}^{i} \frac{e^{\lambda ^*} \rho _j}{e^{\lambda ^*} \rho _j +(1-\rho _j)}\ < \ w+1 \quad i=1,\ldots ,n-1 \end{aligned}$$

and clearly have

$$\begin{aligned} \sum _{j=1}^{i} \frac{e^{\lambda ^*} \rho _j}{e^{\lambda ^*} \rho _j +(1-\rho _j)}\ < \ i \quad i=1,\ldots ,n \end{aligned}$$

as each term in the summand is less than 1. $\square$

Proof

(of Lemma 3) Suppose that $\varvec{\lambda }=(\lambda _1, \ldots , \lambda _n)$ is an optimal solution and there exists some $j<n$ such that $\lambda _j>\lambda _{j+1}$. We will show that $\bar{\varvec{\lambda }}$ defined as $\bar{\lambda }_j=\lambda _{j+1}$, $\bar{\lambda }_{j+1}=\lambda _j$ and $\bar{\lambda }_i=\lambda _i$ for $i\ne \{j,j+1\}$ has no worse objective function than $\varvec{\lambda }$. Let $\varDelta$ be defined as the difference in objective function between $\varvec{\lambda }$ and $\bar{\varvec{\lambda }}$, i.e.,

$$\begin{aligned} \varDelta =&\max _{\begin{array}{c} z_i\in \{0,1\}^n\\ \sum _i z_i =w+1 \end{array}} -\sum _{i=1}^n z_i \lambda _i + \sum _{i=1}^n \log (e^{\lambda _i} \rho _i + (1-\rho _i)) \end{aligned}$$

(80)

$$\begin{aligned}&\ \ -\left( \max _{\begin{array}{c} z_i\in \{0,1\}^n\\ \sum _i z_i =w+1 \end{array}} -\sum _{i=1}^n z_i \bar{\lambda }_i + \sum _{i=1}^n \log (e^{\bar{\lambda }_i} \rho _i + (1-\rho _i)) \right) . \end{aligned}$$

(81)

We will prove that $\varDelta \ge 0$, showing that ${\bar{\varvec{\lambda }}}$ is no worse than $\varvec{\lambda }$. Note initially that

$$\begin{aligned} \max _{\begin{array}{c} z_i\in \{0,1\}^n\\ \sum _i z_i =w+1 \end{array}} -\sum _{i=1}^n z_i \lambda _i \ = \ \max _{\begin{array}{c} z_i\in \{0,1\}^n\\ \sum _i z_i =w+1 \end{array}} -\sum _{i=1}^n z_i \bar{\lambda }_i, \end{aligned}$$

since the maximum value on both sides is equal to the sum of the smallest $w+1$ components of the vector $\varvec{\lambda }$. Thus, we only need to compare remaining part of the objective function, i.e., we have

$$\begin{aligned} \varDelta&=\ \sum _{i=1}^n \log (e^{\lambda _i} \rho _i + (1-\rho _i)) -\sum _{i=1}^n \log (e^{\bar{\lambda }_i} \rho _i + (1-\rho _i)) \\&=\ \log (e^{\lambda _j} \rho _j + (1-\rho _j))+\log (e^{\lambda _{j+1}} \rho _{j+1} + (1-\rho _{j+1})) \\&\qquad - \log (e^{\bar{\lambda }_j} \rho _j + (1-\rho _j))-\log (e^{\bar{\lambda }_{j+1}} \rho _{j+1} + (1-\rho _{j+1})). \end{aligned}$$

Since $\bar{\lambda }_{j}=\lambda _{j+1}$ and $\bar{\lambda }_{j+1}=\lambda _{j}$, it follows that

$$\begin{aligned} \varDelta&=\ \log \left( \frac{e^{\lambda _j} \rho _j + (1-\rho _j)}{e^{\lambda _{j+1}} \rho _j + (1-\rho _j)}\right) -\log \left( \frac{e^{\lambda _{j}} \rho _{j+1} + (1-\rho _{j+1})}{e^{\lambda _{j+1}} \rho _{j+1} + (1-\rho _{j+1})}\right) \\&=\ \log \left( \frac{e^{\lambda _{j}}-e^{\lambda _{j+1}}}{e^{\lambda _{j+1}} + \frac{1}{\rho _{j}}-1}+1\right) - \log \left( \frac{e^{\lambda _{j}}-e^{\lambda _{j+1}}}{e^{\lambda _{j+1}} + \frac{1}{\rho _{j+1}}-1}+1\right) . \end{aligned}$$

Note that the argument inside the $\log$ is positive, since $\lambda _j>\lambda _{j+1}$. Moreover, since $\rho _j \ge \rho _{j+1}$, we see that $1/\rho _{j}-1\le 1/\rho _{j+1}-1$ and hence we conclude that $\varDelta \ge 0$. $\quad \square$