A note on “Optimal stopping and perpetual options for Lévy processes”

Ernesto Mordecki

Facultad de Ciencias, Centro de Matemática Iguá 4225. CP 11400. Montevideo. Uruguay

May 10, 2004

Abstract In this note some details of the proof of Theorem 1 in the article [3]

are given^{1}.

For convenience of the reader we state the Theorem. For all unexplained notation we refer to [3].

Theorem 1 (Perpetual Call options) Consider a Lévy market and M given in (3 of [3]).

(a) If E(e^{X_{1}}) < e^{r} then E(e^{M}) < ∞, the price of a perpetual call option is given by

V_{c}(S_{0}) = \frac{E[S_{0}e^{M} − KE(e^{M})]^{+}}{E(e^{M})} ,

and is optimally exercised at the stopping time

τ_{c}^{∗} = inf{t ≥ 0: S_{t} ≥ S_{c}^{∗}},

with S_{c}^{∗} = KE(e^{M}).

(b) If E(e^{X_{1}}) = e^{r} , then E(e^{M}) = ∞, no optimal stopping time exists, and the cost function takes the form V_{c}(S_{0}) = S_{0}. For each 0 < ε < K/S_{0}, the deterministic time

τ_{c}^{ε} = \frac{1}{r} log (\frac{K}{S_{0}ε})

is ε-optimal in the sense that

S_{0}(1 − ε) ≤ E(e^{−rτ_{c}^{ε}}(S_{τ_{c}^{ε}} − K)^{+}) ≤ S_{0}.

(c) If e^{r} < E(e^{X_{1}}) < ∞ then E(e^{M}) = ∞ and V_{c}(S_{0}) = ∞. Given H > 0, the deterministic time

τ_{c}^{H} = \frac{1}{log E(e^{X_{1}−r})} log (\frac{H + K}{S_{0}} )

1The author is indebted to Serguei Levendorskii for suggesting writing this note.

satisfies

E(e^{−rτ_{c}^{H}} (S_{τ_{c}^{H}} − K)^{+}) ≥ H.

(d) If E(e^{X_{1}}) = ∞ then V_{c}(S_{0}) = ∞ and E(e^{−rt}(S_{t} −K)^{+}) = ∞ for each t > 0.

Proof of Theorem 1. In all cases, the finiteness or not of E(e^{M}) follows from Lemma 1 in [3].

To prove that (a) in Theorem 1 provide the solution to the optimal stopping problem we verify the following two statements:

sup_{τ∈M} E(e^{−rτ} (e^{x+X_{τ}} − K)^{+}) = \frac{E[e^{x+M} − KE(e^{M})]^{+}}{E(e^{M})}  (1)

\frac{E[e^{x+M} − KE(e^{M})]^{+}}{E(e^{M})} = E(e^{−rτ_{c}^{∗}}(e^{x+X_{τ^{∗}_{c}}} − K)^{+}).  (2)

The proof of (1) is done by approximation and given in [3].

Let us now see (2). Denote A = log S_{c}^{∗}, and observe that

{x + M ≥ A} = {τ_{c}^{∗} < τ(r)}.  (3)

Assume first that:

E(e^{x+M}1_{{x+M≥A}}) = E(e^{M}) × E(e^{x+X_{τ^{∗}_{c}}} 1_{{x+M≥A}}).  (4)

Then, based on (4) we obtain

E[e^{x+M} − KE(e^{M})]^{+} = E[e^{x+M}1_{{x+M≥A}}] − KE(e^{M})E[1_{{x+M≥A}}]

= E(e^{M})[E(e^{x+X_{τ^{∗}_{c}}} 1_{{x+M≥A}}) − KE[1_{{x+M≥A}}]]

= E(e^{M})E[(e^{x+X_{τ^{∗}_{c}}} − K)1_{{x+M≥A}}].

To obtain (2) from this, based on (3) and on the independence of the exponential random time τ(r), we verify:

E[(e^{x+X_{τ^{∗}_{c}}} − K)1_{{x+M≥A}}]

= E[(e^{x+X_{τ^{∗}_{c}}} − K)1_{{τ_{c}^{∗}<τ(r)}}] = ∫_{Ω} ∫_{τ_{c}^{∗}}^{∞}(e^{x+X_{τ^{∗}_{c}}} − K)^{+}re^{−rt}dtdP

= ∫_{Ω} e^{−rτ_{c}^{∗}}(e^{x+X_{τ^{∗}_{c}}} − K)^{+}dP = E[e^{−rτ_{c}^{∗}}(e^{x+X_{τ^{∗}_{c}}} − K)^{+}].

This is essentially given in Darling et al. (1972) page 1368.

It remains to verify (4). In order to do this, we state the strong Markov property from Bertoin (Theorem 6, Chapter 1, 1996).

Proposition 1 Let τ be a stopping time with P (τ < ∞) > 0. Then, conditionally on {τ < ∞} the process {X_{τ+t} − X_{τ} }_{t≥0} is independent of F_{τ} and has the law P.

More precisely, if A ∈ F_{τ}, t_{1}, …, t_{k} are positive real numbers and I_{1}, …, I_{k} are real intervals, the strong Markov property is the statement (see (4.24) in Skorokhod (1986)):

P (A, τ < ∞, X_{τ+t_{1}} − X_{τ} ∈ I_{1}, …, X_{τ+t_{k}} − X_{τ} ∈ I_{k}) = P (A, τ < ∞) × P (X_{τ+t_{1}} − X_{τ} ∈ I_{1}, …, X_{τ+t_{k}} − X_{τ} ∈ I_{k}).

The verification of (4), for r > 0, is the following computation:

E(e^{x+M}1_{{τ_{c}^{∗}<τ(r)}}) = E( exp ( sup_{0≤t<τ(r)}(X_{t} − X_{τ_{c}^{∗}})) exp(x + X_{τ_{c}^{∗}})1_{{τ_{c}^{∗}<τ(r)}}) (5)

= E( exp ( sup_{0≤s<τ(r)−τ_{c}^{∗}}(X_{τ_{c}^{∗}+s} − X_{τ_{c}^{∗}})) exp(x + X_{τ_{c}^{∗}})1_{{τ_{c}^{∗}<τ(r)}}) (6)

= ∫_{Ω} ∫_{τ_{c}^{∗}}^{∞} e^{sup_{0≤s<t−τ^{∗}_{c}} (X_{τ^{∗}_{c}+s}−X_{τ^{∗}_{c}})}e^{x+X_{τ^{∗}_{c}}} re^{−rt}dtdP  (7)
= ∫_{Ω} e^{−rτ_{c}^{∗}} ∫_{0}^{∞} e^{sup_{0≤s<v}(X_{τ^{∗}_{c}+s}−X_{τ^{∗}_{c}})}e^{x+X_{τ^{∗}_{c}}} re^{−rv}dvdP  (8)
= E(e^{−rτ_{c}^{∗}}e^{x+X_{τ^{∗}_{c}}} [ ∫_{0}^{∞} e^{sup_{0≤s<v}(X_{τ^{∗}_{c}+s}−X_{τ^{∗}_{c}})}re^{−rv}dv])
= E( ∫_{0}^{∞} e^{sup_{0≤s<v}(X_{τ^{∗}_{c}+s}−X_{τ^{∗}_{c}})}re^{−rv}dv) × E(e^{x+X_{τ^{∗}_{c}}} 1_{{τ_{c}^{∗}<τ(r)}}) (9)
= E( ∫_{0}^{∞} e^{sup_{0≤s<v} X_{s}}re^{−rv}dv) × E(e^{x+X_{τ^{∗}_{c}}} 1_{{x+M≥A}})  (10)
= E(e^{M}) × E(e^{x+X_{τ^{∗}_{c}}} 1_{{x+M≥A}}).

Here, in line (5) we add and substract X_{τ_{c}^{∗}}, in line (6) we observe that the supremum is realized when t ≥ τ_{c}^{∗}, in line (7) we relay on the independence of τ(r), in line (8) we change variables according to v = t − τ_{c}^{∗}. Finally, to obtain (9) and (10), we apply the strong Markov property: first independence and second identity of distributions, as we integrate on the set {τ_{c}^{∗} < ∞} (the integrand vanishes on the set {τ_{c}^{∗} = ∞}): the integral in square brackets is independent of τ_{c}^{∗} and X_{τ_{c}^{∗}}, that are F_{τ_{c}^{∗}} measurable.

Case r = 0 of equation (4) is simpler and can be obtained taking limits as r → 0; or directly, under similar arguments as the case r > 0, in the following way:

E(e^{x+M}1_{{τ_{c}^{∗}<τ(r)}}) = E( exp ( sup_{t≥0}(X_{t} − X_{τ_{c}^{∗}})) exp(x + X_{τ_{c}^{∗}})1_{{τ_{c}^{∗}<∞}})

= E( exp ( sup_{s≥0}(X_{τ_{c}^{∗}+s} − X_{τ_{c}^{∗}})) exp(x + X_{τ_{c}^{∗}})1_{{τ_{c}^{∗}<∞}})

= E(e^{M}) × E(e^{x+X_{τ^{∗}_{c}}} 1_{{x+M≥A}}).

For the rest of the proof of Theorem 1, that is, parts (b), (c) and (d) see [3].

References

[1] Bertoin, J.: Lévy Processes. Cambridge: Cambridge University Press 1996 [2] Darling, D.A., Ligget, T. Taylor, H.M.: Optimal stopping for partial sums.

The Annals of Mathematical Statistics 43, 1363–1368 (1972)

[3] Mordecki, E. Optimal stopping and perpetual options for Lévy processes, Finance and Stochastics. Volume VI (2002) 4, 473-493.

[4] Skorokhod, A. V.: Random processes with independent increments Dordrecht: Kluwer Academic Publishers 1991