Lecture 4 Two-stage stochastic linear programs · Uday V. Shanbhag Lecture 4 Introduction Consider the two-stage stochastic linear program de ned as SLP minimize x cTx+ Q(x) subject

Lecture 4

Two-stage stochastic linear programs

January 24, 2018

Uday V. Shanbhag Lecture 4

Introduction

Consider the two-stage stochastic linear program defined as

SLP minimizex

cTx+Q(x)

subject toAx = b,

x ≥ 0,

where Q(x) is defined as follows:

Q(x) , E[Q(x, ξ)].

If ξ(ω) , (T (ω),W (ω), q(ω), b(ω)), then the integrand of this expecta-

tion, denoted by Q(x, ξ), is the optimal value of the following second-stage

Stochastic Optimization 1


problem:

SecLP(ξ) minimizey(ω)

q(ω)Ty(ω)

subject toT (ω)x+W (ω)y(ω) = b(ω)

y(ω) ≥ 0.

• T (ω) is referred to as the tender

•W (ω) is called the recourse matrix



Polyhedral functions

Definition 1 (Polyhedra) A polyhedron (or a polyhedral convex set) is the

intersection of a finite number of linear inequalities.

Definition 2 (Polyhedral function) A function f : Rn → (−∞,+∞] is

said to be polyhedral if its epigraph is a polyehdral set in Rn+1 or epi(f) is

a polyhedral set where

epi(f) , (x,w) : x ∈ Rn, w ∈ R, f(x) ≤ w .

A polyhedral function f is closed, convex, and proper since epi(f) is a

closed, convex, and nonempty set.



Discrete random variables

• Suppose ξ is a discrete random variable taking on values in Ξ.

• Suppose K1 and K2(ξ) (elementary feasibility set) are defined as

K1 , x : Ax = b, x ≥ 0

and

K2(ξ) , x : y ≥ 0 subject to W (ω)y = h(ω)− T (ω)x .

• Moreover, K2 is defined as

K2 ,⋂ξ∈Ξ

K2(ξ).



Next, we prove some useful properties of K2(ξ) but need to define the

positive hull:

Definition 3 (Positive hull) The positive hull of W is defined as

posW , t : Wy = t, y ≥ 0 .

In fact, pos W is a finitely generated cone which is the set of nonnegative

linear combinations of finitely many vectors. Note that it is convex and

polyhedral.

Proposition 1 (Polyhedrality of K2(ξ) and K2) The following hold:

1. For a given ξ, K2(ξ) is a convex polyhedron;

2. When ξ is a finite discrete random variable, K2 is a convex polyhedron.

Proof:



(i) Consider an x 6∈ K2(ξ) and ξ = (T (ω),W (ω), q(ω), b(ω)). Con-

sequently there exists no nonnegative y such that W (ω)y =

h(ω) − T (ω)x; equivalently h(ω) − T (ω)x 6∈ posW (ω), where

posW (ω) is a finitely generated convex cone. Therefore, by the sep-

arating hyperplane theorem, there exists a hyperplane, defined as H,

such that H = v : uTv = 0 such that uT(h(ω) − T (ω)x)) < 0

and uTv > 0 where v ∈ posW (ω). Since posW (ω) is a finitely

generated cone, there can only be finitely many such hyperplanes.

It follows that K2(ξ) can be viewed as an intersection of a finite

number of halfspaces and is therefore a convex polyhedron.

(ii) Since K2 is an intersection of a finite number of polyhedral sets, K2

is a convex polyhedron.

• Next, we examine the properties of Q(x, ξ)



Proposition 2 (Properties of Q(x, ξ)) For a given ξ, Q(x, ξ) satisfies

the following:

(i) Q(x, ξ) is a piecewise linear convex function in (h, T )

(ii) Q(x, ξ) is a piecewise linear concave function in q

(iii) Q(x, ξ) is a piecewise linear convex function in x for x ∈ K2

Proof:

(i) Suppose f(b) is defined as f(b) = minqTy : Wy = b, y ≥ 0

.

We proceed to show that f(b) is convex in b. Consider b1, b2

such that bλ = λb1 + (1 − λ)b2 where λ ∈ (0,1). Let y1 and

y2 denote solutions ofqTy : Wy = b, y ≥ 0

for b = b1 and

b2 respectively. Furthermore, suppose yλ denotes the solution of

minqTy : Wy = bλ, y ≥ 0

and it follows that

f(bλ) = qTyλ ≤ qT(λy1 + (1− λ)y2),



where λy1 + (1− λ)y2 is feasible i.e.

W (λy1 + (1− λ)y2) = λWy1 + (1− λ)Wy2 = b

(λy1 + (1− λ)y2) ≥ 0.

Consequently,

f(bλ) = qTyλ ≤ qT(λy1 + (1− λ)y2) = qT y (By optimality of yλ)

= λ qTy1︸︷︷︸=f(b1)

+(1− λ) qTy2︸︷︷︸=f(b2)

= λf(b1) + (1− λ)f(b2),

where λy1 + (1 − λ)y2 is feasible with respect to the constraint

y : Wy = bλ. The convexity of f in (h, T ) follows.



(ii) To show the concavity of Q(x, ξ) in q, we may write f(q) as

f(q) = maxπTb : W Tπ ≤ q

.

Proceeding as earlier, it is relatively easy to show that f(q) is concave

in q.

(iii) The convexity of Q(x, ξ) in x follows from (i). Next, we show the

piecewise linearity of Q(x, ξ) in (h, T ). Suppose y is a solution of

minqTy : Wy = b, y ≥ 0

. Then if yB and yN(≡ 0) denote the

basic and non-basic variables and B(ω) denotes the basis and N(ω)



represents the remaining columns, we have

B(ω)yB +N(ω)yN = h(ω)− T (ω)x

B(ω)yB = h(ω)− T (ω)x

yB = (B(ω))−1(h(ω)− T (ω)x).

Recall that when a feasible basis is optimal, we have that

q(ω)T y ≥ qB(ω)TyB

= qB(ω)T(B(ω))−1(h(ω)− T (ω)x)

= Q(x, ξ),

implying that Q(x, ξ) is linear in (h, T, q, x) on a domain prescribed

by feasibility and optimality conditions. Piecewise linearity follows



from noting the existence of a finite number of different optimal bases

for the second-stage program.



Support functions

The support function sC(•) of a nonempty closed and convex set C is

defined as

sC(h) , supz∈C

zTh.

Given a convex set C, its support function, denoted by sC(•), is convex,

positively homogeneous,∗ and lower semicontinuous.†

If s1(•) and s2(•) are support functions of C1 and C2 respectively, then

s1(•) ≤ s2(•)⇔ C1 ⊂ C2.

Furthermore,

s1(•) = s2(•)⇔ C1 = C2.∗A function f(x) is positively homogeneous if f(λx) = λf(x) where λ ≥ 0 and x ∈ Rn.†A function f(x) is lower semicontinuous at x0 if lim inf

x→x0f(x) ≥ f(x0).



Lemma 1 (Support function of unit ball) Consider the unit ball B de-

fined with the Euclidean norm as follows

B , π : ‖π‖2 ≤ 1.

Then sB(.) = ‖.‖2.

Proof: By definition, we have that

sB(χ) = supπ∈B

πTχ.

Since B is a compact set, by the continuity of the objective, the supremum

is achieved. Suppose the maximizer is denoted by π∗. Then, ‖π∗‖ ≤ 1 and



it follows that π∗ = χ‖χ‖. Consequently,

(π∗)Tχ =χT

‖χ‖χ =

‖χ‖2

‖χ‖= ‖χ‖.

Lemma 2 (Support function of Minkowski addition of sets) Consider a

set X = X1 +X2 + . . .+Xn where X1, X2, . . . , Xn are nonempty closed

and convex sets. Then for any y ∈ Rn, the support function sX(y) is

given by the following:

sX(y) = sX1(y) + sX2(y) + . . .+ sXn(y).

Proof: Recall that by the definition of a support function, we have the



following sequence of equalities:

sX(y) = supz∈X

zTy

= supz∈X1+...+Xn

zTy

= supz1∈X1,...,zn∈Xn

zTy

=

(supz1∈X1

zT1 y

)+ . . .+

(supzn∈Xn

zTny

)= sX1(y) + . . .+ sXn(y).

Lemma 3 (Support function is positively homogeneous) Consider the sup-

port function of a closed and convex set X. Then given any y ∈ Rn, we

have that sX(λy) = λsX(y) when λ ≥ 0.



Proof: This result follows by noting that for λ ≥ 0, we have that

sX(λy) = supz∈X

λyTz

= λ supz∈X

yTz

= λsX(y).

Other examples of support functions:

•C =

x :

n∑i=1

xi = 1, xi ≥ 0

, sC(χ) = max

j∈1,...,nχj.



Subgradients of Q(x, ξ)

Consider the second-stage linear program given by

SecLP(ξ) minimizey(ω)

q(ω)Ty(ω)

subject toT (ω)x+W (ω)y(ω) = b(ω)

y(ω) ≥ 0.

Then the associated dual problem is given by

D-SecLP(ξ) maximizeπ(ω)

b(ω)Tπ(ω)

subject to W T(ω)π(ω) ≤ q(ω).



The recourse function Q(x, ξ) is defined as follows:

Q(x, ξ) , infq(ω)Ty : W (ω)y = h(ω)− T (ω)x, y ≥ 0

≡ inf

q(ω)Ty : W (ω)y = χ, y ≥ 0

, sq(χ).

(Suppressing ω) Suppose Π(q) is defined as

Π(q) ,π : W Tπ ≤ q

.

It follows that from LP duality, we have that

infq(ω)Ty : W (ω)y = χ, y ≥ 0

≡ sup

χTπ : W (ω)Tπ ≤ q(ω)

.



As a consequence,

sq(χ) = supπ∈Π(q)

πTχ.

This implies that sq(χ) is the support function of Π(q), a closed and convex

polyhedral set. Next we examine the subdifferential of Q(x, ξ) and need

some definitions.

Definition 4 (Differentiability) Given a mapping g : Rn → Rm. It is

said that g is directionally differentiable at a point x0 ∈ Rn in a direction

h ∈ Rn if the following limit exists:

g′(x0;h) := limt→0

(g(x0 + th)− g(x0)

t

).

If g is directionally differentiable at x0 for every h ∈ Rn, then it is said

to be directionally differentiable at x0. Note that whenever this limit

exists, g′(x0;h) is said to be positively homogeneous in h or g′(x0; th) =



tg′(x0;h) for any t ≥ 0.

If g(x) is directionally differentiable at x0 and g′(x0;h) is linear in h,

then g(x) is said to be Gateaux differentiable at x0. This limit can also

be written as follows:

g(x0 + h) = g(x0) + g′(x0;h) + r(h),

where r(h) is such that r(th)t→ 0 as t→ 0 for any fixed h ∈ Rn.

If g′(x0;h) is linear in h and r(h)‖h‖ → 0 as h→ 0 (or r(h) = o(h)), then

g(x) is said to be Frechet differentiable at x0 or merely differentiable at

x0

Note that Frechet implies Gateaux differentiability and when the func-

tions are locally Lipschitz, both notions coincide. Recall that a mapping

g is said to be locally Lipschitz on a set X ⊂ Rn, if it is Lipschitz con-

tinuous on a neighborhood of every point of X (with possibly different



constants).



Summary of differentiability concepts

• Directionally differentiable at x0 =⇒ Directionally differentiable for all

h at x0

• g(x) is Gauteaux differentiable at x0 =⇒ Directionally differentiable

at x0 and g′(x0;h) is linear in h

• g(x) is Frechet differentiable at x0 =⇒ g′(x0;h) is linear in h andr(h)‖h‖ → 0 as h→ 0

• In fact Frechet differentiable =⇒ Gauteaux differentiable; if g is locally

Lipschitz continuous, then both notions coincide.



Subgradients and subdifferentials

A vector z ∈ Rn is said to be a subgradient of f(x) at x0 if

f(x)− f(x0) ≥ zT(x− x0), ∀x ∈ Rn.

The set of all subgradients of f(x) at x0 is referred to as a subdifferential

and is denoted by ∂f(x0). The subdifferential ∂f(x0) is a closed and

convex subset of Rn and f is subdifferentiable at x0 if ∂f(x0) 6= ∅.

• Consider the function f(x) = |x|. Then ∂f(x) is defined as follows:

∂f(x) =

−1, x < 0

[−1,1], x = 0

1, x > 0.



• Consider the function f(x) = maxx,0

∂f(x) =

0, x < 0

[0,1], x = 0

1, x > 0.



Convexity and polyhedrality of Q(x, ξ)

Proposition 3 (Convexity and polyhedrality of Q(x, ξ)) For any given

ξ, the function Q(., ξ) is convex. Moreover, if the set Π(q) is nonempty

and D-SecLP(ξ) is feasible for at least one x, then the function Q(., ξ) is

polyhedral.

Proof: Recall that the set Π(q) is closed, convex, and polyhedral. If Π(q)

is nonempty, then sq(χ), defined as


χTπ,

is a polyhedral function,‡ where χ = h−Tx. Finally, sinceQ(x, ξ) = sq(χ),

the result follows.‡Recall that an extended real-valued function is called polyhedral, if it is proper, convex, and lower semicontinuous, its domain

is a closed and convex polyhedron, and it is piecewise linear on its domain. A function f is said to be proper if f(x) > −∞ forall x ∈ Rn and its domain dom(f) is nonempty where dom(f) , x : f(x) < +∞.



Next we characterize the subdifferential of Q(x, ξ) but require the Fenchel-

Moreau theorem:

Theorem 4 (Fenchel-Moreau) Suppose that f is a proper, convex, and

lower semicontinuous function. Then f∗∗ = f , where f∗, the conjugate

function of f , is defined as

f∗(z) , supx∈RnzTx− f(x).

Note that the conjugate function f∗ : Rn → R is always convex and lsc.

Furthermore, we have that

∂f∗(x) = argmaxz∈Rn

zTx− f(z)

.

Proposition 5 (Subdifferential of Q(x, ξ)) Suppose that for a given x =



x0 and for ξ ∈ Ξ, the function Q(x0, ξ) is finite. Then Q(., ξ) is

subdifferentiable at x0 and

∂Q(x0, ξ) = −T TD(x0, ξ),

where D(x, ξ) the set of dual solutions is defined as

D(x, ξ) := arg maxπ∈Π(q)

πT(h− Tx).

Proof: Since Q(x0, ξ) is finite, then Π(q) is nonempty and sq(χ) is its

support function, defined as

sq(χ) , supπ∈Π(q)

χTπ.



By the definition of a conjugate function, it can be seen that sq(χ) is the

conjugate of the indicator function 1lq(π), defined as

1lq(π) :=

0, if π ∈ Π(q)

+∞, otherwise.

Since Π(q) is closed and convex, the function 1lq(π) is convex and lower

semicontinuous. By the Fenchel-Moreau theorem, the conjugate of sq(.) is

1lq(.). This can be easily seen by noting the following:

sq(z) = supπ∈Π(q)

zTπ = supπzTπ − 1lq(π),

implying that sq(·) is the conjugate of 1lq(·). Since 1lq(·) is convex and

lower semicontinuous, by Fenchel-Moreau, we have that 1lq(·) = 1l∗∗q (·)



(the biconjugate) where sq(·) = 1l∗q(.) implying that 1lq(·) = (sq(·))∗.

Then for χ0 = h− Tx0,

∂sq(x0) = arg maxπ

πTχ0 − 1lq(π)

= arg max

π∈Π(q)πTχ0.

Since Π(q) is polyhedral and sq(χ0) is finite, it follows that ∂sq(χ0) is

nonempty. Moreover, sq(.) is piecewise linear (earlier in this lecture) and

by the chain rule

∂xQ(x, ξ) = (∂xχ0)T∂χ0sq(χ0) = −T TD(x0, ξ).



Expected recourse for discrete distributions

In this section, we consider the expected cost of recourse, denoted by Q(x),

where

Q(x) , E[Q(x, ξ)].

Suppose, the distribution of ξ has finite support, implying that ξ can take

on a finite number of realizations given by ξ1, . . . , ξK with a mass function

denoted by p1, . . . , pK. For a given x, yj is given by the solution to

SecLP(ξj) minimizeyj

qTj yj

subject toTjx+Wjyj = bj

yj ≥ 0,

where (T (ξj),W (ξj), q(ξj), b(ξj)) = (Tj,Wj, qj, bj). Given that the



problems in y are separable, we may specify the expected cost of recourse

as the following problem in (y1, . . . , yK) as follows:

SecLP(ξ1, . . . , ξK) minimizey1,...,yK

K∑j=1

pjqTj yj

subject toTjx+Wjyj = bj, j = 1, . . . ,K

yj ≥ 0, j = 1, . . . ,K



Note that the entire two-stage problem can then be specified as follows:

SLP minimizex,y1,...,yK

cTx+K∑j=1

pjqTj yj

subject to

Ax = b

Tjx+Wjyj = bj, j = 1, . . . ,K

x, yj ≥ 0, j = 1, . . . ,K

Theorem 6 (Moreau-Rockafellar) Let fi : Rn → R be proper convex

functions for i = 1, . . . , N . Let f(.) =∑Ni=1 fi(.) and x0 be a point such

that fi(x0) are finite or x0 ∈ ∩Ni=1dom fi. Then

∂f1 + . . .+ ∂fN ⊂ ∂f.



Furthermore, we have equality in this relationship or

∂f1 + . . .+ ∂fN = ∂f,

if one of the following hold:

1. The set ∩mi=1ri(domfi) is nonempty;

2. the functions f1, . . . , fk for k ≤ m are polyhedral and the intersection of

the sets ∩ki=1domfi and ∩mi=k+1ri(domfi) is nonempty;

3. there exists a point x ∈ dom(fm) such that x ∈ int(domfi) for i =

1, . . . ,m− 1.

The subdifferential of Q(x) is specified by the following proposition.



Proposition 7 (Subdifferential of ∂Q(x)) Suppose that the probability

distribution function for ξ has finite support or Ξ , ξ1, . . . , ξK and the

expected recourse cost has a finite value for at least some x ∈ Rn. Then

Q(x) is polyhedral and

∂Q(x0) =K∑j=1

∂Q(x, ξj).

Proof: This follows as a consequence of the Moreau-Rockafellar theorem.



Expected recourse for general distributions

Consider Q(x, ξ) where ξ : Ω→ Rd. The cost of taking recourse given ξ is

the minimum value of q(ω)Ty(ω).

Lemma 4 Q(x, ξ) is a random lower semicontinuous function.

Proof: This follows from noting that Q(x, .) is measurable§ with respect to

(.). Furthermore, Q(., ξ) is lower semicontinuous. It follows that Q(x, ξ)

is a random lower semicontinuous function.

Proposition 8 Suppose either E[Q(x, ξ)]+ or E[Q(x, ξ)]− is finite. Then,

Q(x) is well defined.

Proof: This requires verifying whether Q(x, .) is measurable with respect

to the Borel sigma algebra on Rd. This follows by directly employing

Theorem 7.37 [SDR09].§Given a probability space (Ω,F ,P), a function ξ : Ω → [−∞,+∞] is said to be measurable if the set ξ−1((a,∞]) =

ω ∈ Ω : ξ(ω) > a is a measurable set. Any element of F is said to be a measurable set.



Theorem 9 Let F : Rn × Ω → R is a random lower semicontinuous

function. Then the optimal value function ϑ(ω) and the optimal solution

multifunction X∗(ω) are both measurable where

ϑ(ω) , infy∈Rn

F (y, ω) and X∗(ω) , arg minx∈Rn

F (x, ω).

Next, we consider several settings where either E[Q(x, ξ)]+ or E[Q(x, ξ)]−

is finite(ii). However, this requires taking a detour in discussing various

notions of recourse.



Notions of recourse

• The two-stage problem is said to have fixed recourse if W (ω) = W for

every ω ∈ Ω.

• The problem is said to have complete recourse if there exists a solution

y satisfying Wy = χ and y ≥ 0 for every χ. In effect, for any first-stage

decision, the primal second-stage problem is feasible.

More formally, this can be stated as follows:

posW = Rm,

where h− Tx ∈ Rm.

• By employing duality, the fixed recourse is said to be complete if and

only if the feasible set Π(q) of the second-stage dual problem is bounded



(could also be empty).

Complete recourse⇔ Primal second-stage is feasible

⇔ Dual second-stage is bounded (could be empty).

• Given a set X, then X∞ refers to the recession cone of X and is defined

as follows:

X∞ , y : ∀x ∈ X, ∀λ ≥ 0, x+ λy ∈ X, .

• Boundedness of Π(q) implies that its recession cone denoted by

Π0 = Π(0) contains only the zero element; Π0 = 0, provided

that Π(q) is nonempty.



• Claim: Π(q) = π : W Tπ ≤ q is nonempty and bounded implies that

Π0 contains only the zero element.

Proof sketch: Suppose this claim is false. Then Π0 6= ∅ and suppose

u ∈ Π0. This implies that W Tu ≤ 0 where u 6≡ 0. Since Π(q) is

nonempty, then it contains at least one element, say π. It follows that

W T(π + λu) ≤W T π ≤ q,

for all λ ≥ 0. This implies that Π(q) is unbounded since it contains a

ray given by π + λu where λ ≥ 0 and π ∈ Π(q).

• A subclass of problems with fixed and complete recourse are simple

recourse problems where the recourse decisions are either surplus or

shortage decisions determined by the first-stage decisions. Furthermore,

T and q are deterministic and q > 0.



Specifically, y(ω) = y+(ω) + y−(ω). If [h − Tx]i ≥ 0, y+i (ω) =

[h−Tx]i and y−i (ω) = 0. Similarly, if [h−Tx]i ≤ 0, y−i (ω) = [h−Tx]iand y+

i (ω) = 0. This is compactly captured by defining W =(I −I

),

leading to the following system:

y+(ω)− y−(ω) = h(ω)− Txy+(ω), y−(ω) ≥ 0.

• The recourse is said to be relatively complete if for every x in the set

X , x : Ax = b, x ≥ 0, the feasible set of the primal second-stage

problem is nonempty for a.e. ω ∈ Ω. Note that there may be events that

occur with zero probability (measure) that may have Q(x, ξ) = +∞.

However, these do not affect the overall expectation.• A sufficient condition for relatively complete recourse is the following:

for every x ∈ X,Q(x, ξ) < +∞ for all ξ ∈ Ξ.



• Note that this condition becomes necessary and sufficient in two cases:

• The vector ξ has finite support

• Fixed recourse

• Next we consider an instance of a problem with random recourse and

discuss some concerns.

Example 1 (Random recourse) Suppose Q(x, ξ) := infy : ξy =

x, y ≥ 0, with x ∈ [0,1] and ξ having a density function given by

p(z) = 2z,0 ≤ z ≤ 1. Then for ξ > 0, x ∈ [0,1], Q(x, ξ) = x/ξ

and E[Q(x, ξ)] = 2x. For ξ = 0 and x > 0, Q(x, ξ) = +∞.

It can be seen that for almost every ξ ∈ [0,1], Q(x, ξ) < +∞. –

relatively complete recourse. Furthermore, P(ξ : ξ = 0) = 0 and

E[Q(x, ξ)] < +∞ for all x ∈ [0,1].

However, a small perturbation in the distribution may change that.

For instance, a discretization of the distribution with the first point at



ξ = 0, immediately leads to E[Q(x, ξ)] = +∞ for x > 0. In effect,

this problem is unstable and this issue cannot occur if the recourse is

fixed.

•We now consider the support function sq associated with Π(q). Recall

that this is defined as


πTχ.

Our goal lies in finding sufficiency conditions for the finiteness of

the expectation E[sq(χ)]. We will employ Hoffman’s Lemma [SDR09,

Theorem 7.11] which is defined as follows:

Lemma 5 (Hoffman’s lemma) Consider the multifunction M(b) :=

x : Ax ≤ b, where A ∈ Rm×n. Then there exists a positive constant



κ, depending on A, such that for any given x ∈ Rn and any b ∈ domM,

dist(x,M(b)) ≤ κ‖(Ax− b)+‖.

Lemma 6 Suppose there exists a constant κ depending on W such that

if for some q0, the set Π(q0) is nonempty, then for every q, the following

inclusion holds:

Π(q) ⊂ Π(q0) + κ‖q − q0‖B,

where B := π : ‖π‖ ≤ 1 and ‖.‖ denotes the Euclidean norm. Then

the following holds:

sq(•) ≤ sq0(•) + κ‖q − q0‖‖ • ‖.



If q0 = 0, then

| sq(χ1)− sq(χ2) |≤ κ‖q‖‖χ1 − χ2‖, ∀χ1, χ2 ∈ dom(sq).

Proof. Since the support function of the unit ball is the Euclidean

norm, when Π(q0) is nonempty, and that

s1(•) ≤ s2(•)⇔ C1 ⊂ C2.



If C1 = Π(q) and C2 = Π(q0) + κ‖q − q0‖B, then we have that

sq(•) ≤ sΠ(q0)+κ‖q−q0‖B(•)= sΠ(q0)(•) + sκ‖q−q0‖B(•) (Support function of Minkowski sum)

= sΠ(q0)(•) + κ‖q − q0‖sB(•) (Positive homogeneity of support function)

= sq0(•) + κ‖q − q0‖‖ • ‖. (Support function of unit ball is Euclidean norm) (1)

• Consider q0 = 0 and the set Π(0) is defined as

Π(0) ,π : W Tπ ≤ 0

.

This set is a cone and suppose its support function is denoted by

s0 = sq0.



• The support function is defined as follows (proof omitted):

s0(χ) =

0, if χ ∈ pos W

+∞, otherwise

• Therefore, if q0 = 0 in (1) allows one to claim that if Π(q) is nonempty,

then sq(χ) ≤ κ‖q‖‖χ‖ for all χ ∈ pos W , and sq(χ) = +∞ if

χ 6∈ pos W .

• By the polyhedrality of Π(q), if Π(q) is nonempty, we have that sq(·)is piecewise linear on its dom(sq(·)) which is given by posW . In fact,

sq(.) is Lipschitz continuous on its domain or

| sq(χ1)− sq(χ2) |≤ κ‖q‖‖χ1 − χ2‖, ∀χ1, χ2 ∈ dom(sq).



•We now provide a necessary and sufficient condition for a fixed recourse

problem having a finite E[Q(x, ξ)+].

Proposition 10 (Nec. and suff. condition for finiteness of E(Q(x, ξ)+))

Suppose that recourse is fixed and E[‖q‖‖h‖] < +∞ and E[‖q‖‖T‖] <+∞. Consider a point x ∈ Rn. Then E[Q(x, ξ)+] is finite if and only

if h(ξ)− T (ξ)x ∈ pos W for almost every ξ.

Proof:

(⇒) Suppose h(ξ) − T (ξ)x 6∈ pos W with probability one. This implies

that for some ξ ∈ U ⊆ Ξ, h(ξ) − T (ξ)x 6∈ pos W where µ(U) =

P[ξ : ξ ∈ U ] > 0. But for any such ξ, Q(x, ξ) = +∞. But

(Q(x; ξ))+ , max(Q(x; ξ),0)) ≥ Q(x; ξ) = +∞ with positive

probability. Consequently, it follows that E[Q(x, ξ)+] = +∞.(⇐) Suppose h(ξ) − T (ξ)x ∈ pos W with probability one. Then



Q(x, ξ) = sq(h− Tx). By (1), we have that

sq(χ) ≤ s0(χ) + κ‖q‖‖χ‖.

Furthermore, we recall that s0(χ) = 0 when χ ∈ pos W implying

that

sq(χ) ≤ κ‖q‖‖χ‖.

By using the triangle inequality, we have that

Q(x; ξ) = sq(h(ξ)− T (ξ)x)

≤ κ‖q(ξ)(‖h(ξ)‖+ ‖T (ξ)‖‖x‖) (with probability one) (wp1)

Since

0 ≤ κ‖q(ξ)(‖h(ξ)‖+ ‖T (ξ)‖‖x‖) wp1,



it follows that

max(Q(x; ξ),0) ≤ κ‖q(ξ)(‖h(ξ)‖+ ‖T (ξ)‖‖x‖) wp1.

Taking expectations, we have the following:

E[Q(x; ξ)+] ≤ κE[‖q(ξ)‖h(ξ)‖] + κE[‖q(ξ)‖‖T (ξ)‖]‖x‖.

By assumption, we have that E[‖q‖‖h‖] < +∞ and E[‖q‖‖T‖] <+∞. As a result, E[Q(x, ξ)+] < +∞.

• Next, instead of requiring that (h(ξ) − T (ξ)x) ∈ pos(W ) wp1, we

assume that Π(q(ξ)) is nonempty wp1.

Proposition 11 Suppose that (i) the recourse is fixed, (ii) for a.e. q the



set Π(q) is nonempty, and (iii) the following holds:

E[‖q‖‖h‖] < +∞ and E[‖q‖‖T‖] < +∞.

Then the following hold: (a) the expectation function Q(x) is well-

defined and Q(x) > −∞ for all x ∈ Rn; (b) Moreover, Q(x) is

convex, lower semicontinuous, and Lipschitz continuous on dom φ, and

its domain is a convex closed subset of Rn is given by

domQ = x : h− Tx ∈ pos W w.p.1.

Proof:

(a): By (ii), Π(q) is nonempty with probability one. Consequently, for

almost every ξ ∈ Ξ, Q(x, ξ) = sq(h(ξ) − T (ξ)x) for every x and for

almost every ξ.



Suppose π(q) denotes the element of Π(q) closest to zero; by the

closedness of Π(q), it exists and by Hoffman’s lemma, there exists a

constant κ such that

‖π(q)‖ ≤ κ‖q‖.

By the definition of the support function,

sq(h− Tx) = supπ(q)∈Π(q)

π(q)T(h− Tx) ≥ π(q)T(h− Tx)

and for every x, the following holds with probability one:

−π(q)T(h− Tx) ≤ ‖π(q)‖ (‖h‖+ ‖T‖‖x‖) (Cauchy-Schwartz and triangle inequality)

≤ κ‖q‖ (‖h‖+ ‖T‖‖x‖)=⇒ π(q)T(h− Tx) ≥ −κ‖q‖ (‖h‖+ ‖T‖‖x‖)



By (iii), we have that

E[‖q‖‖h‖] < +∞ and E[‖q‖‖T‖] < +∞

and it can be concluded that Q(•) is well defined and Q(x) > −∞ for

all x ∈ Rn.(b): Since sq(•) is lower semicontinuous in (•) (since it is a support

function), it follows that Q(•) is also lower semicontinuous by Fatou’s

Lemma¶

Lemma 7 (Fatou’s Lemma) Suppose that there exists a P-integrable‖

function g(ω) such that fn(•) ≥ g(•). Then

lim infn→∞

E[fn] ≥ E[lim infn→∞

fn].

¶Just to jog your memory, Fatou’s Lemma allows for interchanging integrals and limits under assumptions on the integrand.‖A random variable Z : Ω → Rd is P-integrable if E[Z] is well-defined and finite. Recall that E[Z(ω)] is well-defined if it

does not happen that E[Z(ω)+] and E[(−Z(ω))+] are +∞ in which case E[Z] = E[Z+]− E[(−Z)+]



• Convexity of Q(x) can be claimed as follows. From convexity of Q(x, ξ)

in x, we have that for any x, y

Q(x, ξ) ≥ Q(y, ξ) + u(ξ)T(x− y), where u(ξ) ∈ ∂Q(y, ξ). (2)

Taking expectations on both sides and by the well-defined nature of

Q(x) for all x, we obtain the following:

E[Q(x, ξ)] ≥ E[Q(y, ξ)] + E[u(ξ)]T(x− y), where u(ξ) ∈ ∂Q(y, ξ).

(3)

But under suitable assumptions, we may interchange subdifferential and

expectation, implying that

u = E[u(ξ)] ∈ E[∂Q(x, ξ)] = ∂E[Q(x, ξ)] = ∂Q(x).



It follows that for any x, y, we have that

Q(x) ≥ Q(y) + uT(x− y), where u ∈ ∂Q(y). (4)

In other words, Q(x) is a convex function in x.

• Convexity and closedness of domQ is a consequence of the convexity

and lower semicontinuity of Q. By Prop. 10, Q(x) < +∞ if and only

if h(ξ)− T (ξ)x ∈ pos W with probability one. But this implies that

domQ = x ∈ Rn : Q(x) < +∞= x ∈ Rn : h(ξ)− T (ξ)x ∈ pos W, with probability one. .

(c): In the remainder of the proof, we show the Lipschitz continuity of



Q(x). Consider x1, x2 ∈ domQ. Then, we have

h(ξ)− T (ξ)x1 ∈ pos W, w.p.1 and h(ξ)− T (ξ)x2 ∈ pos W, w.p.1 .

By leveraging this claim, if Π(q) is nonempty, we have that

|sq(h− Tx1)− sq(h− Tx2)| ≤ κ‖q‖‖T‖‖x1 − x2‖.

Taking Expectations on both sides, we have that

|sq(h− Tx1)− sq(h− Tx2)| ≤ κ‖q‖‖T‖‖x1 − x2‖.



|Q(x1)−Q(x2)| ≤ E[|sq(h− Tx1)− sq(h− Tx2)|]≤ κE[‖q‖‖T‖]‖x1 − x2‖,

where the first inequality follows from the application of Jensen’s in-

equality∗∗ and the convexity of the norm. The Lipschitz continuity of

Q(x) on its domain follows.

∗∗Recall that Jensen’s inequality implies that f(E[X]) ≤ E[f(X)], when f is a convex function and E[X] denotes theexpectation of a random variable X.



References

[SDR09] A. Shapiro, D. Dentcheva, and A. Ruszczynski. Lectures on

stochastic programming, volume 9 of MPS/SIAM Series on Opti-

mization. Society for Industrial and Applied Mathematics (SIAM),

Philadelphia, PA, 2009. Modeling and theory.


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