Optimal Stopping and Maximal Inequalities for Poisson Processes

D.O. Kramkov ^{1}  E. Mordecki ^{2}

August 14, 2002

1Steklov Mathematical Institute, Moscow, Russia 2Universidad de la República, Montevideo, Uruguay

Abstract

Closed form solutions for some optimal stopping problems for stochastic processes driven by a Poisson processes N = (N_{t})_{t≥0} are given.

First, cost functions and optimal stopping rules are described for the problems

s(x) = sup_{τ} E(max[x, sup_{0≤t≤τ}(N_{t} − at)] − cτ),

v(x) = sup_{τ} E(max[x, sup_{0≤t≤τ}(bt − N_{t})] − cτ),

with a, b, c positive constants and τ a stopping time.

Based on the obtained results, maximal inequalities in the spirit of [1] are obtained.

To complete the picture, solutions to the problems

c(x) = sup_{τ} E(x + N_{τ} − aτ)^{+}, p(x) = sup_{τ} E(x + bτ − N_{τ})^{+}

are given.

1 Introduction and main results

1.1. Let be given a Poisson process N = (N_{t})_{t≥0} with intensity λ > 0 defined on a filtered probability space (Ω, F, (F_{t})_{t≥0}, P). Denote by \overline{M} the class of all stopping times (that can take the value +∞), by M the class of finite valued stopping times, and by M_{0} the class of stopping times with finite expectation. All stopping times are considered with respect to the filtration (F_{t})_{t≥0}.

In this paper we face the problem of giving the cost functions and optimal stopping times in the following optimal stopping problems:

(1) s(x) = sup_{τ} E(max[x, sup_{0≤t≤τ}(N_{t} − at)] − cτ),

(2) v(x) = sup_{τ} E(max[x, sup_{0≤t≤τ}(bt − N_{t})] − cτ),

where a, b and c are positive constants, (c is the “price” of one unity of observation), and x ∈ R.

Problems (1) and (2) are related to the pricing of “Russian Options” introduced by L. Shepp and A.N. Shiryaev in [2]. In our case, the price process for the risky asset is driven by a Poisson process instead of a geometric Brownian motion.

The solutions to these problems are then used to obtain maximal inequalities of the form

sup_{τ∈M_{0}} E[ sup_{0≤t≤τ}(N_{t} − at))] ≤ ϕ(Eτ) (3)

sup_{τ∈M_{0}} E[ sup_{0≤t≤τ}(bt − N_{t}))] ≤ ψ(Eτ) (4)

where ϕ = ϕ(x), ψ = ψ(x), x > 0 are the minimal possible functions, that satisfy (3) and (4).

In relation with this, we refer to the paper of L. Dubbins, L.A. Shepp and A.N. Shiryaev [1] devoted to the study of optimal stopping rules and maximal inequalities for Bessel Processes, the work of L. Dubbins and G. Schwartz [2], and to some recent results for linear diffusions of S.E. Graversen and G. Peškir [4].

Finally, to complete the picture, we give the cost function and optimal stopping time in the following problems

c(x) = sup_{τ∈\overline{M}} E(x + N_{τ} − aτ)^{+} (5)

p(x) = sup_{τ∈\overline{M}} E(x + bτ − N_{τ})^{+} (6)

where x ∈ R, z^{+} denotes max(z, 0).

1.2. In order to formulate our results, let us introduce the function u = u(x) = u(x; d), x ≥ 0 with d a positive constant, defined by

u(x; d) = ∑_{k=0}^{+∞}[e^{\frac{x−k}{d}} P_{k}(\frac{x − k}{d}) − (k + 1)]I_{[k,k+1)}(x) (7)

where P_{k} = P_{k}(x) is a polynomial of order k:

P_{k}(x) = ∑_{l=0}^{k} d_{k−l} \frac{(−1)^{l}x^{l}}{l!} , (8)

and the coefficients d_{k}, k ≥ 0 are defined by the following recurrence relation:

d_{0} = 1, d_{k+1} = 1 + e^{1/d} ∑_{l=0}^{k} d_{k−l} \frac{(−1)^{l}}{d^{l}l!} = 1 + e^{1/d}P_{k}(1/d). (9)

Some properties of the function u = u(x) are revisted in section 2.

Denote now by τ_{s}^{∗} the optimal stopping time in the problem (1), that is, the stopping time for which the supremum is realized. The price function s = s(x) and the optimal stopping time τ_{s}^{∗} are given in the following

Theorem 1 (i) If a + c ≤ λ, then s(x) = +∞, for all x ≥ 0. (ii) If a > λ and c = 0, then

s(x) = E(max[x, sup_{0≤t≤+∞}(N_{t} − at)]) = \frac{λ}{2(a − λ)} + \frac{a − λ}{λ}u(x; \frac{a}{λ}) (10)

(iii) If a > 0 and c ≥ λ, then s(x) = x and τ_{s}^{∗} = 0.

(iv) If c < λ < a + c, then

s(x) = \{ \begin{matrix}x, & if x ≥ x^{∗} \\ x_{s}^{∗} + \frac{a+c−λ}{λ}[u(x, \frac{a}{λ}) − u(x_{s}^{∗}; \frac{a}{λ})], & if 0 ≤ x < x^{∗} \end{matrix} (11)

and τ_{s}^{∗} = inf{t ≥ 0: X_{t} ≥ x_{s}^{∗}}, (12) where X = (X_{t})_{t≥0} is a stochastic process defined by

X_{t} = max[x, sup_{0≤r≤t}(N_{r} − ar)] − (N_{t} − at), (13)

and the positive constant x_{s}^{∗} is the solution of the equation

\frac{a + c − λ}{λ}u′(x; \frac{a}{λ}) = 1. (14)

We also have, when x < x_{s}^{∗}

E(τ_{s}^{∗}) = \frac{1}{λ}[u(x_{s}^{∗}, \frac{a}{λ}) − u(x; \frac{a}{λ})]. (15)

Denote now by τ_{v}^{∗} the optimal stopping time for the problem (2).The price function v = v(x) and the optimal stopping time are given in the following

Theorem 2 (i) If c + λ ≤ b, then v(x) = +∞, for all x ≥ 0. (ii) If λ > b and c = 0, then

v(x) = E(max[x, sup_{0≤t<+∞}(bt − N_{t})]) = x + \frac{1}{α^{∗}} e^{−α^{∗}x}, (16)

where the constant α^{∗} is the unique positive root of the equation

\frac{b}{λ}α + e^{−α} − 1 = 0. (17)

(iii) If c ≥ b, then v(x) = x and τ_{v}^{∗} = 0.

(iv) If c < b < c + λ, then

v(x) = \{ \begin{matrix}x, & if x ≥ x_{v}^{∗}, \\ x + \frac{c}{λ}u(x_{v}^{∗} − x, \frac{b}{λ}), & if 0 ≤ x < x_{v}^{∗}, \end{matrix} (18)

and

τ_{v}^{∗} = inf{t ≥ 0: Y_{t} ≥ x_{v}^{∗}},

where Y = (Y_{t})_{t≥0} is a stochastic process defined by

Y_{t} = max[x, sup_{0≤r≤t}(br − N_{r})] − (bt − N_{t}), (19)

and the positive constant x_{v}^{∗} is the solution of the equation

\frac{c}{λ}u′(x; \frac{b}{λ}) = 1. (20)

We also have, when 0 ≤ x < x_{v}^{∗}

E(τ_{v}^{∗}) = \frac{1}{c + λ − b}[x_{v}^{∗} − x − \frac{c}{λ}u(x_{v}^{∗} − x, \frac{b}{λ})]. (21)

1.3. Let M_{T} = {τ ∈ M_{0}: E(τ) ≤ T } denote the set of all the stopping times τ with expected value less or equal than T > 0. Functions ϕ and ψ in (3) and (4) can be defined in the following way

ϕ(T) = sup_{τ∈M_{T}} E( sup_{0≤t≤τ}(N_{t} − at)), (22)

ψ(T) = sup_{τ∈M_{T}} E( sup_{0≤t≤τ}(bt − N_{t})). (23)

In what follows, the stopping times such that the supremum in (22) and (23) is realized will be called M_{T} -optimal.

Theorem 3 Let 0 < T < +∞.

(i) Denote by x_{s}^{∗}(T) the root of the equation

u(x; \frac{a}{λ}) = λT. (24)

Then ϕ(T) = x_{s}^{∗}(T) + (λ − a)T (25)

and the stopping time

τ_{s}^{∗}(T) = inf{t ≥ 0: X_{t} ≥ x_{s}^{∗}(T)},

is M_{T} -optimal for the problem (22), where in the process X = (X_{t})_{t≥0} defined in (13) we take x = 0.

Furthermore, when λ = a we have

\frac{\sqrt{4T + 1} − 1}{2} ≤ ϕ(T) ≤ \sqrt{T} (26)

and in consequence

lim_{T →+∞} \frac{ϕ(T)}{\sqrt{T}} = 1. (27)

(ii) Denote by x_{v}^{∗}(T) the positive root of the equation

\frac{xu′(x; \frac{b}{λ}) − u(x; \frac{b}{λ})}{1 + (1 − \frac{b}{λ})u′(x; \frac{b}{λ})} = λT, (28)

Then ψ(T) = x_{v}^{∗}(T) + (b − λ)T, (29)

and the stopping time

τ_{v}^{∗}(T) = inf{t ≥ 0: Y_{t} ≥ x_{v}^{∗}(T)},

is M_{T} -optimal for the problem (23), where in the process Y = (Y_{t})_{t≥0} defined in (19) we take x = 0.

Furthermore, when λ = b we have

lim_{T →+∞} \frac{ψ(T)}{\sqrt{T}} = 1. (30)

1.4. Denote by τ_{c}^{∗} the optimal stopping time in the problem (5). The price function and the optimal stopping time for this problem are given in the following

Theorem 4 (i) If a ≤ λ, then c(x) = +∞ for all x ∈ R. (ii) If λ < a, then

c(x) = \{ \begin{matrix}x, & if x ≥ x_{c}^{∗}, \\ x + (\frac{a}{λ} − 1)u(x_{c}^{∗} − x, \frac{a}{λ}), & if x < x_{c}^{∗}, \end{matrix} (31)

with x_{c}^{∗} = \frac{λ}{2(a−λ)}. The optimal stopping time is

τ_{c}^{∗} = inf{t ≥ 0: x + N_{t} − at ≥ x_{c}^{∗}}. (32)

Denote finally by τ_{p}^{∗} the optimal stopping time in the problem (6). The cost function and the optimal stopping time for this problem are given in the following Theorem 5

(i) If b ≥ λ, then p(x) = +∞ for all x ∈ R.

(ii) Assume b < λ. Denote by α^{∗} the positive root of the equation (17). Then, the cost function is

p(x) = \{ \begin{matrix}x, & if x ≥ x_{p}^{∗}, \\ x_{p}^{∗}e^{α^{∗}(x−x_{p}^{∗})}, & if x < x_{p}^{∗}, \end{matrix} (33)

with x_{p}^{∗} = \frac{1}{α^{∗}}, and the optimal stopping time is

τ_{p}^{∗} = inf{t ≥ 0: x + bt − N_{t} ≥ x_{p}^{∗}}. (34)

2 Some auxiliar results

2.1. In this section we formulate some technical results concerning the function u = u(x), x ≥ 0 defined in (7). Let us introduce the operator K = K(d) defined on the set of continuously differentable functions by the following relation Kw(x) = dw′(x) + w(x − x ∧ 1) − w(x), x ≥ 0. (35)

As will be shown in section 3, the operator λK is the infinitesimal operator associated to the process X = (X_{t})_{t≥0} defined in (13) with d = \frac{a}{λ}.

Lemma 1 The function u = u(x) defined in (7) is continously differentable, and satisfies the following relation

Ku(x) = du′(x) − u(x) + u(x − x ∧ 1) = 1, x ≥ 0. (36)

Proof. From the definition of the function u = u(x) (see (7), (8) and (9)) follows, that this function is smooth on the intervals (k, k + 1) for any nonnegative integer k.

If x ∈ [0, 1), we have u(x) = e^{x/d} − 1, giving that u(0) = 0 and

Ku(x) = du′(x) − u(x) = 1.

If k ≥ 1 and x ∈ (k, k + 1), equation (36) is

du′(x) − u(x) = 1 − u(x − 1).

In view of (7) for x ∈ (k, k + 1)

u(x) = e^{\frac{x−k}{d}} P_{k}(\frac{x − k}{d}) − (k + 1),

and from (8) follows that P′_{k}(x) = −P_{k−1}(x). Then

du′(x) − u(x) = e^{\frac{x−k}{d}} P′_{k}(\frac{x − k}{d}) + k + 1

= 1 − e^{\frac{x−k}{d}} P_{k−1}(\frac{x − k}{d}) + k = 1 − u(x − 1) (37)

In this way (36) is proved for x ∈ (k, k + 1). It is clear from (35), that in order to verify (36) when x = 1, 2, … it is enough to see that the functions u = u(x) and u′ = u′(x) are continuous in these points.

Let us examine the continity of u. In view of (7)

u(k) = P_{k}(0) − (k + 1) = d_{k} − (k + 1)

and by (9)

lim_{x↑k} u(x) = e^{1/d}P_{k−1}(1/d) − k = d_{k} − (k + 1),

and the continuity of u follows. For the function u′ in view of (37)

lim_{x↓k} u′(x) = \frac{1}{d} lim_{x↓k}[1 + u(x − 1) − u(x)]

= \frac{1}{d}[1 + u(k − 1) − u(k)],

and

lim_{x↑k} u′(x) = \frac{1}{d} lim_{x↑k}[1 + u(x − x ∧ 1) − u(x)]

= \frac{1}{d}[1 + u(k − 1) − u(k)],

concluding the proof. □

2.2. In the following Lemma we study the behaviour of u = u(x) and its derivative.

Lemma 2 The functions u = u(x) and u′ = u′(x) satisfy a) u(0) = 0, u′(0) = \frac{1}{d}.

b) u and u′ are strictly increasing.

c) lim_{x→+∞} u(x) = +∞. lim_{x→+∞} u′(x) = \{ \begin{matrix}\frac{1}{d−1}, & d > 1 \\ +∞, & 0 < d ≤ 1. \end{matrix}

d) If d = 1, for all x ≥ 0 we have

x^{2} ≤ u(x) ≤ x^{2} + x.

e) If d = 1, then

lim_{x→+∞} \frac{xu′(x) − u(x)}{x^{2}} = 1.

Proof. a) As u(x) = e^{x/d} − 1 when x ∈ [0, 1) we obtain

u(0) = 0 and u′(0) = \frac{1}{d}.

b) In order to prove that u and u′ are strictly increasing, as they are both absolutely continuous, we will see that

(38) u′′(x) is positive and continuous for all x ≠ 1, x ≥ 0,

As u(x) = e^{x/d} − 1 if x ∈ (0, 1), by differentiation u′′(x) = \frac{1}{d^{2}} e^{x/d} is continuous and positive if x ∈ [0, 1). Take now x > 1. From (36), for x > 1 we obtain du′′(x) = ∫_{x−1}^{x} u′′(y)dy. (39)

As u′′ is bounded on compact intervals, the continuity if x ≠ 1 follows. In view of (39) follows that

u′′(x) > 0, x > 1.

In fact, let x_{0} = inf{x ≥ 0: u′′(x) = 0}. We have u′′(x_{0}) = 0 and u′′(x) > 0 when x < x_{0} contradicting (39).

So u = u(x) and u′ = u′(x) are both strictly increasing functions.

c) Let us now compute the limits at infinite. If x > 1, then from (36) follows that

du′(x) = u(x) − u(x − 1) + 1 = 1 + ∫_{x−1}^{x} u′(y)dy.

From this, taking into account the monotonicity of the function u′ = u′(x) we obtain

1 + u′(x − 1) < du′(x) < 1 + u′(x)

and taking limits we obtain the second limit in c). Finally, this ensures the convergence of u(x) → +∞ when x → +∞, proving c).

d) Let d = 1. Denote Δ(x) = u(x) − x^{2}. We want to establish

Δ(x) ≥ 0 for all x ≥ 0. (40)

For this function

KΔ(x) = Ku − K(x^{2}) = 0.

This means

Δ′(x) = Δ(x) − Δ(x − 1) = ∫_{x−1}^{x} Δ′(t)dt.

If x ∈ (0, 1), we know Δ(x) = e^{x} − 1 − x^{2}, and in consequence Δ′(x) > 0 on this interval. So, with the same argument as in the proof of (38) we obtain that Δ′(x) > 0 for all x > 0. As Δ(0) = 0, (40) is proved.

Denote now

δ(x) = x^{2} + x − u(x).

In a similar way, as was done for Δ, we obtain that

δ′(x) = 2x + 1 − e^{x} > 0 for 0 < x < 1,

and for x > 1

δ′(x) = ∫_{x−1}^{x} δ′(t)dt.

concluding that δ(x) ≥ 0 for all x ≥ 0 and the proof of d).

e) From d) we obtain

lim_{x→+∞} \frac{u(x)}{x^{2}} = 1 (41)

Now, denoting w(x) = xu′(x) − u(x), we have

w′(x) = xu′′(x).

So, by L’Hôpital rule

lim_{x→+∞} \frac{w(x)}{x^{2}} = lim_{x→+∞} \frac{xu′′(x)}{2x} = lim_{x→+∞} \frac{u′′(x)}{2}.

But, in view of (41), and L’Hôpital rule again

lim_{x→+∞} \frac{u′′(x)}{2} = lim_{x→+∞} \frac{u′(x)}{2x} = lim_{x→+∞} \frac{u(x)}{x^{2}} = 1.

proving d). □

3. We will also need the following

Lemma 3 Let d > 1. Then the function w = w(x) defined by

w(x) = x − (d − 1)u(x), x ≥ 0,

is strictly increasing and

lim_{x→+∞} w(x) = \frac{1}{2(d − 1)}. (42)

Proof. By part c) of Lemma 2

w′(x) = 1 − (d − 1)u′(x) > 0 for all x ≥ 0

so the funtion w is stricltly incresing. In consequence the limit in (42) exists. Now, by (36)

Kw(x) = dw′(x) − w(x) = 1 − x, x < 1,

and Kw(x) = dw′(x) − w(x) + w(x − 1) = 0, x ≥ 1. (43)

Integrating (43) over [1, x], follows

d[w(x) − w(1)] = ∫_{1}^{x}[w(y) − w(y − 1)]dy
= ∫_{x−1}^{x} w(y)dy + ∫_{0}^{1} w(y)dy

Taking into account (36)

dw(x) = ∫_{x−1}^{x} w(y)dy + \frac{1}{2}.

Now, the monotonicity of w(x) gives

w(x − 1) + \frac{1}{2} < dw(x) < w(x) + \frac{1}{2},

and taking limits as x → +∞ we conclude the proof. □

3 Proofs of the Theorems

3.1. The process X = (X_{t})_{t≥0}, defined in (13), is an homogeneous Markov process with stochastic differential

dX_{t} = adt − (X_{t−} ∧ 1)dN_{t}.

If w = w(x), x ≥ 0, is a continously differentiable function, then, Itô’s formula for pure jump processes gives (see [6] ch. 3 §6)

w(X_{t}) = w(x) + λ ∫_{0}^{t} Kw(X_{r−})dr + M_{t}, t ≥ 0, (44)

with K = K(\frac{a}{λ}) in (35) and the process M = (M_{t})_{t≥0} given by

M_{t} = ∫_{0}^{t}[w(X_{r−} − X_{r−} ∧ 1) − w(X_{r−})]d(N_{r} − λr)

is a local martingale. From (44) and Lemma 1 follows, that the process (u(X_{t}) − λt)_{t≥0} is a local martingale, with the function u = u(x; \frac{a}{λ}) defined in (7) and d = \frac{a}{λ}.

Proof of Theorem 1.

(i) Assume a + c ≤ λ. As E(N_{t} − (a + c)t) = (λ − a − c)t ≥ 0 and the process {N_{t} − (a + c)t} has stationary independent increments, it follows that P (sup_{t}(N_{t} − (a + c)t) = +∞) = 1 (see [7]). Thi s means, that for the stopping time

τ_{H} = inf{t ≥ 0: N_{t} − (a + c)t ≥ H}.

we have P (τ_{H} < +∞) = 1. Then

s(x) ≥ s(0) ≥ E( sup_{0≤r≤τ_{H}}(N_{r} − ar) − cτ_{H})

≥ E((N_{τ_{H}} − (a + c)τ_{H}) ≥ H.

As H is arbitrary, the proof of (i) is concluded.

(ii) Let c = 0 and a > λ. For t > 0 denote

s_{t}(x) = E(max[x, sup_{0≤r≤t}(N_{r} − ar)]).

By monotnonous convergence

s_{t}(x) ↑ s(x), t → +∞.

Taking into account, that E(N_{t}) = λt, t ≥ 0, we obtain that

s_{t}(x) = E(X_{t} − (a − λ)t), X_{0} = x,

with the process X = (X_{t})_{t≥0} defined in (13).

Denote w(x) = x − \frac{a−λ}{λ}u(x). By Lemma 3, we have

lim_{x→+∞} w(x) = \frac{λ}{2(a − λ)}.

Furthermore, as X_{t} ≥ at − N_{t}, and a > λ, then, X_{t} → +∞ with probability one as t → +∞. As the process \{\frac{a−λ}{λ}u(X_{t}) − (a − λ)t\} is a martingale, we obtain

s(x) − \frac{a − λ}{λ}u(x) = lim_{t→+∞} E(s_{t}(x)) − E(\frac{a − λ}{λ}u(X_{t}) − (a − λ)t)

= lim_{t→+∞} E(X_{t} − \frac{a − λ}{λ}u(X_{t})) = lim_{t→+∞} E(w(X_{t})) = \frac{λ}{2(a − λ)}

by bounded convergence, concluding the proof of (ii).

(iii) First we observe, that when c > 0 and a+c > λ the supremum in (1) can be taken over M_{0}. To see this, take an arbitrary stopping time τ ∈ M with E(τ) = +∞. Consider δ such that λ − a < δ < c.

E(max[x, sup_{0≤t≤τ}(N_{t} − at)] − cτ) ≤ x + E[ sup_{0≤t<+τ}(N_{t} − at) − cτ]

≤ x + E[ sup_{0≤t<+τ}(N_{t} − (a + δ)t) − (c − δ)τ] = −∞

because, in accordance with (10)

E( sup_{0≤t<+∞}(N_{t} − (a + δ)t)) = \frac{λ}{a + δ − λ} < +∞.

Now, if c ≥ λ, a > 0 and E(τ) < +∞, then E(N_{τ}) = λE(τ) and

E(max[x, sup_{0≤t≤τ}(N_{t} − at)] − cτ)) ≤ x + E( sup_{0≤t≤τ}(N_{t} − at) − cτ)

≤ x + E(N_{τ} − cτ) ≤ x − (c − λ)E(τ) ≤ x.

On the other hand, taking τ_{s}^{∗} = 0

E(max[x, sup_{0≤t≤τ_{s}^{∗}}(N_{t} − at)] − cτ_{s}^{∗}) = x,

and follows that τ_{s}^{∗} = 0 is the optimal stopping time, and s(x) = x, x ≥ 0.

(iv) Let c < λ < a + c. In view of Lemma 2 (14) has an unique solution x_{s}^{∗} such that 0 < x_{s}^{∗} < +∞.

Let us first prove, that for τ_{s}^{∗} defined by (12), identity (15) holds.

As u(x) ≤ u(x_{s}^{∗}) if x ≤ x_{s}^{∗}, then

−λt ≤ u(X_{τ_{s}^{∗}∧t}) − λ(τ_{s}^{∗} ∧ t) ≤ u(x_{s}^{∗})

and, as a consequence, the local martingale {u(X_{τ_{s}^{∗}∧t}) − λ(τ_{s}^{∗} ∧ t)}_{t≥0} is in fact a martingale. Therefore

λE(τ_{s}^{∗} ∧ t) = E(u(X_{τ_{s}^{∗}∧t})) − u(x) (45)

From this, we deduce that E(τ_{s}^{∗}) < +∞, so P (τ_{s}^{∗} < +∞) = 1, concluding that

u(X_{τ_{s}^{∗}∧t}) → u(x_{s}^{∗}) as t → +∞.

Now, taking limits in (45) as t → +∞ we obtain (15).

We know by (iii) that we can take the supremum over M_{0}. If τ ∈ M_{0}, E(τ) < +∞, then E(N_{τ}) = λE(τ). In consequence

E(max[x, sup_{0≤t≤τ}(N_{t} − at)] − cτ) = E(X_{τ} − (a + c − λ)τ).

Denote by s̃ = s̃(x) by

s(x) = \{ \begin{matrix}x, & if x ≥ x^{∗} \\ x_{s}^{∗} + \frac{a+c−λ}{λ}[u(x, \frac{a}{λ}) − u(x_{s}^{∗}; \frac{a}{λ})], & if 0 ≤ x < x^{∗} \end{matrix}

We want to prove s̃ = s with s in (1). The following relation takes place (A)s(x) = max(x, x_{s}^{∗}) − (a + c − λ)E(τ_{s}^{∗}) = E(X_{τ_{s}^{∗}} − (a + c − λ)τ_{s}^{∗}), that is direct if x ≥ x_{s}^{∗} and follows from (15) when x < x_{s}^{∗}. In consequence,

in order to complete the proof of the Theorem, it remains to see that for any stopping time τ ∈ M_{0}

(B)s̃(x) ≥ E(X_{τ} − (a + c − λ)τ).

In view of Lemma 2 the function u = u(x) is convex, so s(x) ≥ x. To prove (B) it is enough to see that the process Z = (Z_{t})_{t≥0} defined by

Z_{t} = s̃(X_{τ∧t}) − (a + c − λ)(τ ∧ t)

is a supermartingale. From (44) follows that the process Z is a local supermartingale, if λKs̃(x) − (a + c − λ) ≤ 0, x ≥ 0, (46)

with K = K(\frac{a}{λ}) defined in (35).

For x ≤ x_{s}^{∗} (46) takes places in view of Lemma 1 (in fact we have an identity). If x > x_{s}^{∗}, then s̃(x) = x and

λKs̃(x) − (a + c − λ) = a + λs̃(x − x ∧ 1) − λx − (a + c − λ)

≤ a + λs̃(x_{s}^{∗} − x_{s}^{∗} ∧ 1) − λx_{s}^{∗} − (a + c − λ) = λKs̃(x_{s}^{∗}) − (a + c − λ) = 0,

where we used the convexity of the function s̃. In this way the process Z = (Z_{t})_{t≥0} is a local supermartingale, and as

Z_{t} ≥ −(a + c − λ)t, t ≥ 0,

and E(τ) < +∞, we deduce that Z is a supermartingale and the proof is concluded. □

3.2. The process Y = (Y_{t})_{t≥0} defined in (19) is an homogeneous Markov process with stochastic differential

dY_{t} = dN_{t} − I_{{Y_{t−}>0}}bdt. (47)

If w = w(x), x ≥ 0, is a continously differentiable function, Itô’s formula gives (see [6])

w(Y_{t}) = w(x) + λ ∫_{0}^{t} Lw(Y_{r−})dr + M_{t}, (48)

where

Lw(x) = −\frac{b}{λ}I_{\{x>0\}}w′(x) + w(x + 1) − w(x), (49)

and the process M = (M_{t})_{t≥0}, given by

M_{t} = ∫_{0}^{t}[w(Y_{r−} + 1) − w(Y_{r−})]d(N_{r} − λr)

is a local martingale.

Proof of Theorem 2.

(i) Let λ + c ≤ b. Take H > 0 and define

τ_{H} = inf{t ≥ 0: (b − c)t − N_{t} ≥ H}.

As on the proof of (i) in Theorem 1 we conclude that P (τ_{H} < +∞) = 1. We have

v(x) ≥ E(max[x, sup_{0≤t≤τ_{H}}(bt − N_{t})] − cτ_{H})

≥ E((b − c)τ_{H} − N_{τ_{H}}) ≥ H.

As H is arbitrary, the proof of (i) is concluded.

(ii) Let c = 0 and b < λ. For t > 0 denote

v_{t}(y) = E(max[x, sup_{0≤r≤t}(br − N_{r})]).

By monotonous convergence

v_{t}(y) ↑ v(y) as t → +∞.

Taking into account the definition of Y in (19) and the fact that E(N_{t}) = λt, we obtain that

v_{t}(x) = E[Y_{t} − (λ − b)t], Y_{0} = x.

Consider the function ṽ = ṽ(x) defined by

ṽ(x) = x + \frac{1}{α^{∗}} e^{−α^{∗}x},

with α^{∗} the positive root of the equation (17) that, as b < λ has an unique positive solution. A direct computation shows that the function ṽ = ṽ(x) satisfies the equation

λLṽ(x) − (λ − b)t = 0.

In view of (46) the process K = (K_{t})_{t≥0} defined by

K_{t} = ṽ(Y_{t}) − (λ − b)t

is a local martingale.

From the definition of ṽ = ṽ(x) follows that

x ≤ ṽ(x) ≤ x + \frac{1}{α^{∗}} , x ≥ 0, (50)

and

lim_{x→+∞} |ṽ(x) − x| = 0.

Notice now, that max(0, N_{t} − bt) ≤ Y_{t} ≤ x + N_{t} (51)

so the local martingale K = (K_{t})_{t≥0} is uniformly integrable and in consequence a martingale. Finally, from (51) and λ > b follows that

Y_{t} → +∞, t → +∞ P -a.s. (52)

From this fact, and (50), we obtain

lim_{t→+∞} |ṽ(Y_{t}) − Y_{t}| = 0.

So, we have

v(x) = lim_{t→+∞} v_{t}(x) = lim_{t→+∞} E(Y_{t} − (b − λ)t)

= lim_{t→+∞} E(ṽ(Y_{t}) − (b − λ)t) = ṽ(x),

and the proof of (ii) is concluded.

(iii) Let c ≥ b. Then, for any τ ∈ M

E(max[x, sup_{0≤t≤τ}(bt − N_{t})] − cτ) ≤ y + (b − c)E(τ) ≤ y,

and, if we take τ_{s}^{∗} = 0

E(max[x, sup_{0≤t≤τ_{s}^{∗}}(bt − N_{t}) − cτ_{s}^{∗})] = x,

proving (iii).

(iv) Assume c < b < c + λ. Let us see, that in (2) it is enough to take the supremum over M_{0}. Take δ positive, such that b − λ < δ < c. Consider an arbitrary stopping time τ with E(τ) = +∞. As by (16)

E( sup_{0≤t<+∞}((b − δ)t − N_{t}) < ∞,

we obtain

E(max[x, sup_{0≤t≤τ}(bt − N_{t})] − cτ)) ≤ x+

+E( sup_{0≤t<+∞}((b − δ)t − N_{t}) − (c − δ)E(τ) = −∞,

See now, that in view of Lemma 2 equation (20) has a unique positive solution x_{v}^{∗}. Define now the function ṽ = ṽ(x) by

ṽ(x) = \{ \begin{matrix}x, & x ≥ x_{v}^{∗} \\ x + \frac{c}{λ}u(x_{v}^{∗} − x; \frac{b}{λ}), & 0 ≤ x < x_{v}^{∗}. \end{matrix}

In order to prove (iv) we verify ṽ = v, with v in (2). For this, it is enough to verify the following two assertions:

(A) If τ_{v}^{∗} = inf{t ≥ 0 : Y_{t} ≥ x_{v}^{∗}}, then

ṽ(y) = E(max(y, sup_{0≤t≤τ_{v}^{∗}}(bt − N_{t})) − cτ_{v}^{∗}).

(B) For all τ ∈ M_{0}

ṽ(x) ≥ E(max[x, sup_{0≤t≤τ}(bt − N_{t})] − cτ),

We begin by (B). If τ ∈ M_{0} then E(N_{τ}) = λE(τ) and in consequence

E(max[x, sup_{0≤t≤τ}(bt − N_{t})] − cτ) = E(Y_{τ} − (c + λ − b)τ)

≤ E(ṽ(Y_{τ}) − (c + λ − b)τ),

because ṽ(x) ≥ x, as the function u is convex (Lemma 2).

Then, in order to conclude the proof of (B) it is enough to see that the process V = (V_{t})_{t≥0} given by

V_{t} = ṽ(Y_{t∧τ}) − (c + λ − b)(t ∧ τ)

is a supermartingale. As V_{t} ≥ −(c + λ − b)t, by (46) it is enuogh to verify

λLṽ(x) − (c + λ − b) ≤ 0 for x ≥ 0. (53) with L defined in (48). Extend the definition of the function u in a continuous way, as u(x) = 0 if x < 0. Now, using Lemma 1 we obtain, for 0 ≤ x < x_{v}^{∗}

λLṽ(x) = −bI_{\{y>0\}}[1 − \frac{c}{λ}u′(x_{v}^{∗} − x)] + λ + c[u(x_{v}^{∗} − x − 1) − u(x_{v}^{∗} − x)]

= −b + λ + cKu(y)|_{y=x_{v}^{∗}−x} = c + λ − b ≤ 0.

When x ≥ x_{v}^{∗}, then ṽ(x) = x and

λLṽ(x) − (c + λ − b) = −c < 0,

and in this way the inequality (51) holds and (B) is proved.

Let us see (A). Define V^{∗} = (V_{t}^{∗})_{t≥0} by

V_{t}^{∗} = ṽ(Y_{t∧τ_{v}^{∗}}) − (c + λ − b)(t ∧ τ_{v}^{∗}).

In view of (46) and (52) V^{∗} is a local martingale. Furthermore, this process is bounded on each interval of the form [0, T], T < +∞, and in consequence is a martingale. So

ṽ(x) = E(V_{t}^{∗}) = E(ṽ(Y_{t∧τ_{v}^{∗}})) − (c + λ − b)E(t ∧ τ_{v}^{∗}). (54)

As 0 ≤ ṽ(Y_{t∧τ_{v}^{∗}}) ≤ max(x, x_{v}^{∗}), we can take limits as t → +∞ in (54) obtaining

ṽ(x) = E(ṽ(Y_{τ_{v}^{∗}} − (c + λ − b)τ_{v}^{∗}) = E(Y_{τ_{v}^{∗}} − (c + λ − b)τ_{v}^{∗})

= E(max[x, sup_{0≤tτ_{v}^{∗}}(bt − N_{t}) − cτ_{v}^{∗}).

concluding (A). From (54), we also deduce (21)

E(τ_{v}^{∗}) = \frac{1}{c + λ − b}(ṽ(x_{v}^{∗}) − ṽ(x)) = \frac{1}{c + λ − b}(x_{v}^{∗} − x − \frac{c}{λ}u(x_{v}^{∗} − x)).

concluding the proof of Theorem 2. □

3.3. Proof of Theorem 3.

(i) For T > 0 equation (24) has an unique positive root x_{s}^{∗}(T) by Lemma 2. The constant c(T) defined by

\frac{a + c − λ}{λ}u′(x_{s}^{∗}(T); \frac{a}{λ}) = 1,

satisfies

λ − a < c(T) < λ.

So, by (iv) in Theorem 1 the stopping time

τ_{s}^{∗}(T) = inf{t ≥ 0: X_{t} ≥ x_{s}^{∗}(T)},

with X = (X_{t})_{t≥0} defined by (11) is optimal for the problem (1). By the election of c = c(T), and (15)

E(τ_{s}^{∗}(T)) = T.

Also, in view of (1), (11) and (24)

E( sup_{0≤t≤τ_{s}^{∗}(T)}(N_{t} − at)) = s(0) + c(T)T = x_{s}^{∗}(T) − (λ − a)T,

and for any τ ∈ M_{T}

E( sup_{0≤t≤τ}(N_{t} − at)) ≤ s(0) + c(T)E(τ) ≤ x_{s}^{∗}(T) − (λ − a)T.

proving (25). Finally, (26) and (27) are direct consequences of c) in Lemma 2, concluding (i).

(ii) For a positive T, define x_{v}^{∗}(T) as the root of the equation

\frac{xu′(x) − u(x)}{1 + (1 − \frac{b}{λ})u′(x)} = λT, (55)

where u = u(x) = u(x; \frac{b}{λ}). In order to see that the equation (55) has an unique root, we note, that for b > λ the function

f(x) = \frac{xu′(x) − u(x)}{1 + (1 − \frac{b}{λ})u′(x)}

is increasing (because is the quotient of an increasing function over a decreasing one), f(0) = 0 and f(x) → +∞ if u′(x) → (\frac{b}{λ} − 1)^{−1}.

If b ≤ λ, L’Hôpital rule gives lim_{x→+∞} f(x) = +∞, and, as f(0) = 0, we confirm the existence of only one root, because

f′(x) = \frac{u′′(x)[x + u(x)(1 − \frac{b}{λ})]}{[1 + (1 − \frac{b}{λ})u′(x)]^{2}} ≥ 0.

Let us see that the constant c(T) defined by the relation

\frac{c(T)}{λ}u′(x_{v}^{∗}(T); \frac{b}{λ}) = 1 (56)

satisfies b − λ < c(T) < b. As, by Lemma 2 u′(x_{v}^{∗}(T)) > u′(0) = \frac{λ}{b},

c = \frac{λ}{u′(x_{v}^{∗}(T))} < b.

On the other side, if b ≤ λ the second inequality is inmediate. If λ < b, then by b) in Lemma 2,

u′(x_{v}^{∗}(T)) < lim_{x→+∞} u′(x) = \frac{λ}{b − λ}

and the second inequality follows.

Then, we are in case (iv) of Theorem 2. The stopping time

τ_{v}^{∗}(T) = inf{t ≥ 0: Y_{t} ≥ x_{v}^{∗}(T)}

is optimal for the problem (2) with c = c(T), x = 0. Then,

E(τ_{v}^{∗}(T)) = \frac{1}{c + λ − b}(x_{v}^{∗}(T) − \frac{c}{λ}u(x_{v}^{∗})) = T.

Furthermore, (2), (18) and (56) gives

E( sup_{0≤t≤τ_{v}^{∗}(T)}(bt − N_{t})) = v(0) + c(T)E(τ_{v}^{∗}(T)) = \frac{c(T)}{λ}u(x_{v}^{∗}(T)) + c(T)T

= \frac{u(x_{v}^{∗})(T) + λT}{u′(x_{v}^{∗}(T))} = x_{v}^{∗}(T) − (λ − b)T,

concluding the proof of (31).

To see (30), take b = λ, and denote w(x) = xu′(x) − u(x). We have x_{v}^{∗}(T) = w^{−1}(T) = ψ(T). We know,

lim_{x→+∞} \frac{w(x)}{x^{2}} = 1,

this means

\frac{T}{(w^{−1}(T))^{2}} → 1,

and taking square roots, we obtain (30) concluding the proof of Theorem 3. □

3.4. Proof of Theorem 4.

(i) Assume λ > a. For H > 0, denote

τ_{H} = {t ≥ 0 : x + N_{t} − at ≥ H}.

As in the proof of (i) in Theorem 1, we obtain P (τ_{H} < +∞ = 1). Then

c(x) = sup_{τ≥0} E(x + N_{τ} − aτ)^{+} ≥ E(x + N_{τ_{H}} − aτ_{H})^{+} ≥ H.

As H is arbitrary, the proof of (i) is concluded.

(ii) Take a > λ. Denote U_{t} = x + N_{t} − at. The process U = (U_{t})_{t≥0} has an infinitesimal operator of the form

Uf(x) = λ(f(x + 1) − f(x)) − af′(x).

Itô’s formula in this case reads, with c in (31)

c(U_{t}) = c(x) + ∫_{0}^{t} Uc(U_{r−})dr + M_{t}, t ≥ 0, (57)

with the process M = (M_{t})_{t≥0} given by

M_{t} = ∫_{0}^{t}[c(U_{r−} + 1) − c(U_{r−}))]d(N_{r} − λr)

a local martingale. Let us now verify that for the function c = c(x) defined in (31) Uc(x) = 0 if x < x_{c}^{∗}. (58)

Uc(x) < 0 if x ≥ x_{c}^{∗}. (59)

In fact, (59) is inmediate. In order to see (58), taking into account (36)

Uc(x) = λ(c(x + 1) − c(x)) − ac′(x)

= λ − a + (a − λ)(u(x_{c}^{∗} − 1) − u(x_{c}^{∗} − x) + \frac{a}{λ}u′(x_{c}^{∗} − x)) = 0.

Now, from (57) we obtain that the stopped local martingale M^{∗} = {M_{t∧τ_{c}^{∗}}}_{t≥0} with τ_{c}^{∗} in (32) is uniformly bounded:

−c(x) ≤ M_{t∧τ_{c}^{∗}} ≤ c(x_{c}^{∗}) + 1,

so, taking expected values and limits, we obtain E(M_{τ_{c}^{∗}}) = 0 that means, by (31). As on the set {τ_{c}^{∗} = +∞} we have

c(x + N_{τ_{c}^{∗}} − aτ_{c}^{∗}) = (x + N_{τ_{c}^{∗}} − aτ_{c}^{∗})^{+} = 0.

we deduce (A) c(x) = E(c(x + N_{τ_{c}^{∗}} − aτ_{c}^{∗})) = E(x + N_{τ_{c}^{∗}} − aτ_{c}^{∗})^{+}. To complete the proof, we will see (B) For any τ ∈ M̄

c(x) ≥ E(x + N_{τ} − aτ)^{+}.

For the process M defined by (57), as −Uc(x) ≥ 0 we have

M_{t} ≥ −c(x).

This fact, and Fatou’s Lemma gives, that the process M is a supermartingale. Now, using this fact, and c(x) ≥ x^{+} we obtain

c(x) ≥ E(c(x + N_{τ} − aτ)) ≥ E(x + N_{τ} − aτ)^{+},

concluding the proof of Theorem 4. □

3.5. Proof of Theorem 5.

(i) Is analogous to the proof of (i) in Theorem 4.

(ii) Take b < λ. Denote D_{t} = x + bt − N_{t}. The process D = (D_{t})_{t≥0} has in infinitesimal operator of the form

Df(x) = bf′(x) − λ(f(x − 1) − f(x)).

It is direct to see that

\{ \begin{matrix}Dp(x) = 0, & if x < x_{p}^{∗}. \\ Dp(x) = b − λ, & if x > x_{p}^{∗}. \end{matrix} (60)

Itô’s formula in this case is

p(D_{t}) = p(x) + ∫_{0}^{t} D(D_{r−})dr + M_{t}, t ≥ 0, (61)

with the process M = (M_{t})_{t≥0} given by

M_{t} = ∫_{0}^{t}[p(D_{r−} − 1) − p(D_{r−}))]d(N_{r} − λr)

a local martingale. Now, from (60) we obtain that the stopped local martingale M^{∗} = {M_{t∧τ_{p}^{∗}}}_{t≥0} with τ_{p}^{∗} defined in (34) is uniformly bounded:

−p(x) ≤ M_{t∧τ_{p}^{∗}} ≤ p(x_{p}^{∗}) = x_{p}^{∗},

so, taking expected values and limits, we obtain E(M_{τ_{p}^{∗}}) = 0. As on the set {τ_{p}^{∗} = +∞} we have = p(D_{τ_{p}^{∗}}) = (D_{τ_{c}^{∗}})^{+} = 0, (61) gives

(A) p(x) = E(p(x + bτ_{p}^{∗} − N_{τ_{p}^{∗}}) = E(x + bτ_{p}^{∗} − N_{τ_{p}^{∗}})^{+}.

To complete the proof, we will see

(B) For any τ ∈ M̄

p(x) ≥ E(x + bτ − N_{τ})^{+}.

For the process M defined by (61), as −Dp(x) ≥ 0 we have

M_{t} ≥ −p(x).

This fact, and Fatou’s Lemma gives, that the process M is a supermartingale. Now, using this fact, and p(x) ≥ x^{+} we obtain

p(x) ≥ E(p(x + bτ − N_{τ})) ≥ E(x + bτ − N_{τ})^{+},

concluding the proof of the Theorem □

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