Portfolio optimization in a defaultable Lévy-driven market model

Abstract

In this paper, we analyse a market where the risky assets follow defaultable exponential additive processes, with coefficients depending on the default state of the assets. In this market we show that when an investor wants to maximize a utility function which is logarithmic on both his/her consumption and terminal wealth, his/her optimal portfolio strategy consists in keeping proportions of wealth in the risky assets which only depend on time and on the default state of the risky assets, but not on their price or on current wealth level; this generalizes analogous results of Pasin and Vargiolu (Econ Notes 39:65–90, 2010) in non-defaultable markets without intermediate consumption. We then present several examples of market where one, two or several assets can default, with the possibility of both direct and information-induced contagion, obtaining explicit optimal investment strategies in several cases. Finally, we study the growth-optimal portfolio in our framework and show an example with necessary and sufficient conditions for it to be a proper martingale or a strict local martingale.

Appendix: Proof of Theorem 5.4

We first need to introduce the following notation.

Definition 8.1

Given $d\in \{0,1\}^n$, we call the length of $d$ the positive integer defined as

$$\begin{aligned} l(d):=n-\sum _{i=0}^n d_i. \end{aligned}$$

Moreover we establish on $\{0,1\}^n$ the following (partial) order relation:

$$\begin{aligned} d\le d'\quad \hbox {if}\ d_i\ge d'_i\quad \forall i=1,\ldots ,n. \end{aligned}$$

Note that given $D_t=d$ for a certain $t\le 0$, the states $d'<d$ are the only states accessible for $D$ after the time $t$.

Roughly speaking, the length of $d$ is equal to the number of risky asset that are still alive. In particular, when every firm has already defaulted we have $l(d)=0$, whereas when every firm did not we have $l(d)=n$. We also explicitly observe that

$$\begin{aligned} d+\chi (d,x)\le d\quad \forall d\in \{0,1\}^n,x\in X^n. \end{aligned}$$

Hence by (5.4), given a state $d\in \{0,1\}^n$, $A^h J(t,x,d)$ depends only on the states $d'\le d$, i.e. the states whose alive assets are a subset of the alive ones in $d$; in other words, $A^{h,c} J(t,x,d)$ does not depend on the assets whose entities have already defaulted.

We also need the following

Lemma 8.2

Consider the function

$$\begin{aligned} \psi (t,v,c)= \left\{ \begin{array}{ll} B \mathrm{e}^{-\delta (T-t)} \log c - c\frac{A +\frac{B}{\delta }\left( 1- \mathrm{e}^{-\delta (T-t)} \right) }{v},&{} \quad \delta >0 \\ B \log c - c\frac{A +B(T-t)}{v},&{} \quad \delta =0 \end{array} \right. \end{aligned}$$

(8.1)

with $A,B\ge 0,A+B>0$. Then, for any $t\in [0,T]$ and $v>0$ we have

$$\begin{aligned} {\bar{c}}(t)v=\arg \max _{c>0}\psi (t,v,c), \end{aligned}$$

(8.2)

where ${\bar{c}}$ is defined as in (5.16). Moreover,

$$\begin{aligned} \max \limits _{c>0} \psi (t,v,c) =\psi (t,v,{\bar{c}}(t)v)&= B \mathrm{e}^{-\delta (T-t)}\left( \log \left( \frac{B \mathrm{e}^{-\delta (T-t)}}{A +\frac{B}{\delta }\left( 1- \mathrm{e}^{-\delta (T-t)} \right) }\right) -1 \right) \nonumber \\&+ B \mathrm{e}^{-\delta (T-t)} \log v \end{aligned}$$

(8.3)

if $\delta >0,$ whereas

$$\begin{aligned} \max _{c>0} \psi (t,v,c) =\psi (t,v,{\bar{c}}(t)v)=B \left( \log \left( \frac{B}{A +B(T-t)}\right) -1 \right) + B \log v \end{aligned}$$

if $\delta =0$.

Proof

We only prove the case $\delta >0$. For any $t\in [0,T],v>0$ we have

$$\begin{aligned} \psi _c(t,v,c)=\frac{B \mathrm{e}^{-\delta (T-t)}}{c}-\frac{A+\frac{B}{\delta }\left( 1-\mathrm{e}^{-\delta (T-t)} \right) }{v}=0 \end{aligned}$$

if and only if $c={\bar{c}}(t)v$. Thus, ${\bar{c}}(t)v$ is the only stationary point for $\psi (t,v,\cdot )$, and since $\lim _{c\rightarrow 0}\psi (t,v,c)=\lim _{c\rightarrow \infty }\psi (t,v,c)=-\infty$, we obtain (8.2). Eventually, (8.3) follows from a direct computation. $\square$

We now prove Theorem 5.4.

Proof of Theorem 5.4

We only prove the theorem for $\delta >0$, as the case $\delta =0$ is totally analogous.

Part (a). By induction on $k=l(d)$. We start proving the statement when $k=0$. In this case we clearly have $d={\mathbf {1}}:=(1,\ldots ,1)$, i.e. all the entities have defaulted. If we search for a solution of the kind $K(t,v,d)$ as in (5.13), we clearly obtain

$$\begin{aligned} A^{h,c} K(t,v,{\mathbf {1}})=A^h K(t,v,1,\ldots ,1)=-c\frac{A +\frac{B}{\delta }\left( 1- \mathrm{e}^{-\delta (T-t)} \right) }{v}, \end{aligned}$$

(8.4)

so that the HJB equation becomes

$$\begin{aligned} \frac{{\mathrm {d}}}{{\mathrm {d}}t}\Phi ^{{\mathbf {1}}}(t)&= B \mathrm{e}^{-\delta (T-t)}\log v -\sup _{c>0}\left( B \mathrm{e}^{-\delta (T-t)} \log c - c\frac{A +\frac{B}{\delta }\left( 1- \mathrm{e}^{-\delta (T-t)} \right) }{v} \right) \nonumber \\&= B \mathrm{e}^{-\delta (T-t)}\log v -\sup _{c>0} \psi (t,v,c), \end{aligned}$$

(8.5)

with $\psi (t,v,c)$ as in (8.1). Thus by Lemma 8.2 we have

$$\begin{aligned} \frac{{\mathrm {d}}}{{\mathrm {d}}t}\Phi ^{{\mathbf {1}}}(t)&= B \mathrm{e}^{-\delta (T-t)}\log v - \psi (t,v,{\bar{c}}(t)v) \nonumber \\&= - B \mathrm{e}^{-\delta (T-t)}\left( \log \left( \frac{B \mathrm{e}^{-\delta (T-t)}}{A +\frac{B}{\delta }\left( 1- \mathrm{e}^{-\delta (T-t)} \right) }\right) -1 \right) \end{aligned}$$

(8.6)

Therefore, we define $\Phi ^{{\mathbf {1}}}$ as the unique solution of (8.6) provided with the terminal condition $\Phi ^{{\mathbf {1}}}(T)=0$, so that $K(t,v,{\mathbf {1}})$ solves Eq. (5.3) with the terminal condition (5.7).

We now assume the statement to be true for any $d'\in \{0,1\}^n$ such that $l(d')\le k-1$ and prove it to be true for any $d$ such that $l(d)=k$. We set

$$\begin{aligned} K(t,v,d)=\left( A +\frac{B}{\delta }\left( 1- \mathrm{e}^{-\delta (T-t)} \right) \right) \log v + \Phi ^{d}(t), \end{aligned}$$

(8.7)

where $\Phi ^{d}$ is a $C^1$ deterministic function such that $\Phi ^{d}(T)=0$. Then we have

$$\begin{aligned} \frac{\partial K}{\partial t}(t,v,d)&= \frac{{\mathrm {d}}}{{\mathrm {d}}t}\Phi ^{d}(t)- B \mathrm{e}^{-\delta (T-t)}, \\ v \frac{\partial K}{\partial v}(t,v,d)&= -v^2 \frac{\partial ^2 K}{\partial v^2}(t,v,d)=A +\frac{B}{\delta }\left( 1- \mathrm{e}^{-\delta (T-t)} \right) . \end{aligned}$$

Therefore we obtain

$$\begin{aligned}&\frac{A^{h,c} K(t,v,d)}{A +\frac{B}{\delta }\left( 1- \mathrm{e}^{-\delta (T-t)} \right) }=\langle \mu (t,d),h^{d} \rangle - \frac{c}{v}-\frac{1}{2}\langle h^{d}\Sigma (t,d),h^{d} \rangle \\&\quad +\int _{X^n}\left( \frac{ K\big (t,v(1+\langle x,h^{d} \rangle ),d+\chi (d,x)\big )-\Phi ^{d}(t)}{A +\frac{B}{\delta }\left( 1- \mathrm{e}^{-\delta (T-t)} \right) } -\log {v}-\langle x,h^{d} \rangle \right) \nu ^d_t({\mathrm {d}}x) \end{aligned}$$

[by (5.6)]

$$\begin{aligned} =\langle \mu (t),h^{d} \rangle - \frac{c}{v}-\frac{1}{2}\langle h^{d}\Sigma (t),h^{d} \rangle + I_1 +I_2, \end{aligned}$$

where

$$\begin{aligned} I_1=\int _{X^n\backslash \Theta ^{d}}\left( \frac{ K\big (t,v(1+\langle x,h^{d} \rangle ),d\big )-\Phi ^{d}(t)}{A +\frac{B}{\delta }\left( 1- \mathrm{e}^{-\delta (T-t)} \right) }-\log {v}-\langle x,h^{d} \rangle \right) \nu ^d_t({\mathrm {d}}x), \end{aligned}$$

[by (8.7)]

$$\begin{aligned} =\int _{X^n\backslash \Theta ^{d}}\Big ( \log {(1+\langle x,h^{d} \rangle )}-\langle x,h^{d} \rangle \Big )\nu ^d_t({\mathrm {d}}x), \end{aligned}$$

with

$$\begin{aligned} \Theta ^{d}=X^n\bigcap \left( \mathop {\bigcup }\limits _{\begin{array}{c} 1\le i\le n,\\ d_i=0 \end{array}}\{x_{i}= -1\}\right) , \end{aligned}$$

(8.8)

and where

$$\begin{aligned} I_2=\int _{\Theta ^{d}}\left( \frac{ K\big (t,v(1+\langle x^{d},h^{d} \rangle ),d+\chi (d,x)\big )-\Phi ^{d}(t)}{A +\frac{B}{\delta }\left( 1- \mathrm{e}^{-\delta (T-t)} \right) }-\log {v}-\langle x^{d},h^{d} \rangle \right) \nu ^d_t({\mathrm {d}}x) \end{aligned}$$

(by induction hypothesis)

$$\begin{aligned} =\int _{\Theta ^{d}}\left( \frac{ \Phi ^{d+\chi (d,x)}(t)-\Phi ^{d}(t)}{A +\frac{B}{\delta }\left( 1- \mathrm{e}^{-\delta (T-t)} \right) }+\log {(1+\langle x^{d},h^{d} \rangle )}-\langle x^{d},h^{d} \rangle \right) \nu ^d_t({\mathrm {d}}x) \end{aligned}$$

[by (5.6)]

$$\begin{aligned} =\frac{ \varphi ^{d}(t)-\nu _t\left( \Theta ^{d}\right) \Phi ^{d}(t)}{A +\frac{B}{\delta }\left( 1- \mathrm{e}^{-\delta (T-t)} \right) }+\int _{\Theta ^{d}}\Big ( \log {(1+\langle x^{d},h^{d} \rangle )}-\langle x^{d},h^{d} \rangle \Big )\nu ^d_t({\mathrm {d}}x), \end{aligned}$$

where $\varphi ^{d}$ is the continuous deterministic function

$$\begin{aligned} \varphi ^{d}(t)=\sum _{d'<d}\nu ^d_t\left( \Lambda ^{d'}\right) \Phi ^{d'}(t), \end{aligned}$$

with

$$\begin{aligned} \Lambda ^{d'}=X^n\bigcap \left( \mathop {\bigcap }\limits _{\begin{array}{c} 1\le i\le n,\\ d'_i=0 \end{array}}\{x_{i}\ne -1\}\right) \bigcap \left( \mathop {\bigcap }\limits _{\begin{array}{c} 1\le i\le n,\\ d'_i=1 \end{array}}\{x_{i}= -1\}\right) . \end{aligned}$$

Thus we obtain

$$\begin{aligned} A^{h,c} K(t,v,d)&= \varphi ^{d}(t)-\nu ^d_t\left( \Theta ^{d}\right) \Phi ^{d}(t) \nonumber \\&+\left( A +\frac{B}{\delta }\left( 1- \mathrm{e}^{-\delta (T-t)} \right) \right) \left( F^{d}(t,h)- \frac{c}{v}\right) \end{aligned}$$

(8.9)

with $F^d$ as in (5.12), and the HJB equation becomes

$$\begin{aligned} \frac{\mathrm{d}}{{\mathrm {d}}t}\Phi ^{d}(t)&= B \mathrm{e}^{-\delta (T-t)}\log v -\sup _{h\in H,\, c>0}\left( A^{h,c} K(t,v,d)+ B \mathrm{e}^{-\delta (T-t)} \log c \right) \nonumber \\&= B \mathrm{e}^{-\delta (T-t)}\log v + \nu ^d_t\left( \Theta ^{d}\right) \Phi ^{d}(t)-\varphi ^{d}(t)\nonumber \\&- \left( A +\frac{B}{\delta }\left( 1- \mathrm{e}^{-\delta (T-t)} \right) \right) \sup _{h\in H} F^{d}(t,h) \nonumber \\&- \sup _{c>0}\left( B \mathrm{e}^{-\delta (T-t)} \log c - c \frac{A +\frac{B}{\delta }\left( 1- \mathrm{e}^{-\delta (T-t)} \right) }{v} \right) \nonumber \\&= \nu ^d_t\left( \Theta ^{d}\right) \Phi ^{d}(t)-\varphi ^{d}(t)- \left( A +\frac{B}{\delta }\left( 1- \mathrm{e}^{-\delta (T-t)} \right) \right) \sup _{h\in H} F^{d}(t,h) \nonumber \\&+B \mathrm{e}^{-\delta (T-t)}\log v - \sup _{c>0}\psi (t,v,c), \end{aligned}$$

(8.10)

with $\psi (t,v,c)$ as in (8.1). Let us observe that $\arg \max _{h\in H}F^{d}(t,h)$ is not empty for any $t\in [0,T]$ because $H$ is a compact subset of ${\mathbb {R}}^n$ and $F^{d}$ is continuous, and thus

$$\begin{aligned} \sup _{h\in H} F^{d}(t,h)=F^{d}(t,{\bar{h}}(t,d)), \end{aligned}$$

(8.11)

with $F^d$ as in (8.1). Therefore, plugging (8.11)–(8.2) into (8.10) we get

$$\begin{aligned} \frac{\mathrm{d}}{{\mathrm {d}}t}\Phi ^{d}(t)&= \nu ^d_t\left( \Theta ^{d}\right) \Phi ^{d}(t)-\varphi ^{d}(t)- \left( A +\frac{B}{\delta }\left( 1- \mathrm{e}^{-\delta (T-t)} \right) \right) F^{d}(t,{\bar{h}}(t,d)) \\&+B \mathrm{e}^{-\delta (T-t)}\log v - \psi (t,v,{\bar{c}}(t)v) \end{aligned}$$

[by (8.3)]

$$\begin{aligned}&=\nu ^d_t\left( \Theta ^{d}\right) \Phi ^{d}(t)-\varphi ^{d}(t)- \left( A +\frac{B}{\delta }\left( 1- \mathrm{e}^{-\delta (T-t)} \right) \right) F^{d}(t,{\bar{h}}(t,d)) \nonumber \\&\qquad - B \mathrm{e}^{-\delta (T-t)}\left( \log \left( \frac{B \mathrm{e}^{-\delta (T-t)}}{A +\frac{B}{\delta }\left( 1- \mathrm{e}^{-\delta (T-t)} \right) }\right) -1 \right) . \end{aligned}$$

(8.12)

Furthermore, note that $\varphi ^{d}(t)$ and $\nu ^d_t\left( \Theta ^{d}\right)$ are continuous in $t$ by Assumption 3.5. Thus, setting $\Phi ^{d}(\cdot )$ as the unique solution of the ODE (8.12) with terminal condition $\Phi ^{d}(T)=0$, we have that $K(t,v,d)$ solves Eq. (5.3) with terminal condition (5.7), and Part (a) is proved.

Part (b). To prove $K\in {\mathcal D}$ it is sufficient to prove that for any $\bar{t}\in [0,T]$ and $({\mathfrak {h}},{\mathfrak {c}})\in {\mathcal A}[\bar{t},T]$,

$$\begin{aligned} K(t,(V,D)^{{\mathfrak {h}},{\mathfrak {c}};\bar{t},v,d}_{t})-\int _{\bar{t}}^{t}A^{{\mathfrak {h}}_{u},{\mathfrak {c}}_u}K(u,(V,D)^{{\mathfrak {h}},{\mathfrak {c}};\bar{t},v,d}_{u}){\mathrm {d}}u \end{aligned}$$

(8.13)

is a martingale. Now, by applying the Itô formula, we obtain

$$\begin{aligned} {\mathrm {d}}K(t,(V,D)^{{\mathfrak {h}},{\mathfrak {c}}}_t)=A^{{\mathfrak {h}}_{t},{\mathfrak {c}}_{t}}K(t,(V,D)^{{\mathfrak {h}},{\mathfrak {c}}}_{t}){\mathrm {d}}t+{\mathrm {d}}M_t, \end{aligned}$$

where

$$\begin{aligned} {\mathrm {d}}M_t&= a(t) {\mathfrak {h}}^{D_t}_t\sigma (t){\mathrm {d}}W_t+\int _{{\mathbb {R}}^n}\Big ( a(t) \log {\big ( 1+\langle x,{\mathfrak {h}}^{D_{t-}}_{t-} \rangle \big )}\\&+\Phi ^{D_{t-}+\chi (D_{t-},x)}(t)-\Phi ^{D_{t-}(t)}\Big )\big ( N(D_{t-},{\mathrm {d}}x,{\mathrm {d}}t)-\nu _t(D_{t-},{\mathrm {d}}x){\mathrm {d}}t \big ), \end{aligned}$$

and where we have set $a(t):=A +\frac{B}{\delta }\left( 1- \mathrm{e}^{-\delta (T-t)} \right)$. Therefore, to prove the theorem it is sufficient to check that $M_t$ is a martingale. Since

$$\begin{aligned} {\mathbb {E}}\left[ \int _{\bar{t}}^T | a(t) {\mathfrak {h}}^{D_t}_t\sigma (u)|^2 {\mathrm {d}}t \right]&\le a^2(0) {\mathbb {E}}\left[ \int _{\bar{t}}^T \left\| {\mathfrak {h}}_t\right\| ^2\left\| \sigma (t)\right\| ^2 {\mathrm {d}}t \right] \\&\le a^2(0) \sup _{h\in H}\left\| h\right\| ^2\int _{\bar{t}}^{T}\left\| \sigma (t)\right\| ^2 {\mathrm {d}}u<+\infty , \end{aligned}$$

the continuous part is a martingale. We observe now that since ${\mathfrak {h}}$ takes values in the compact set $H\subset \hbox {int}(\cap _{t\in [\bar{t},T]}H_t)$, there exists a constant $\delta >0$ such that $1+\langle {\mathfrak {h}}_t,x \rangle \ge \delta$ $\nu ^d({\mathrm {d}}x)$-a.s. for any $d\in \{0,1\}^n$, $t\in [\bar{t},T]$. Thus, the function $x\rightarrow \log {(1+\langle {\mathfrak {h}}_t,x \rangle )}$ is bounded from below and with linear growth, and so there is a constant $C>0$ such that

$$\begin{aligned} |\log {(1+\langle {\mathfrak {h}}_t,x \rangle )}|\le C \sup _{h\in H}\left\| h\right\| \left\| x\right\| \quad \nu ^d({\mathrm {d}}x)\text {-a.s.} \end{aligned}$$

(8.14)

for any $d\in \{0,1\}^n$, $t\in [\bar{t},T]$. According now to the notation used in the proof of part (a), we define

$$\begin{aligned} \Theta :=\Theta ^{(0,\ldots ,0)}=\bigcup _{i=1}^{n}\{x_{i}= -1\}, \end{aligned}$$

(8.15)

and finally, to verify the pure jump stochastic integral to be a martingale, we only need to check that for any $d\in \{0,1\}^n$

$$\begin{aligned}&{\mathbb {E}}\left[ \int _{\bar{t}}^{T}\int _{{\mathbb {R}}^n}\left| a(t) \log {\big ( 1+\langle x,{\mathfrak {h}}^{D_{t-}}_{t-} \rangle \big )} +\Phi ^{D_{t-}+\chi (D_{t-},x)}(t)-\Phi ^{D_{t-}(t)}\right| ^2 \nu ^d_t({\mathrm {d}}x){\mathrm {d}}t \right] \end{aligned}$$

[by (5.6) and (8.15)]

$$\begin{aligned}&=\!{\mathbb {E}}\left[ \!\int _{\bar{t}}^{T}\!\int _{{\mathbb {R}}^n}\left| a(t) \log {\big ( 1\!+\!\langle x,{\mathfrak {h}}^{D_{t-}}_{t-} \rangle \big )} \!+\!{1\!\!1}_{\Theta }(x)\left( \Phi ^{D_{t-}\!+\!\chi (D_{t-},x)}(t)\!-\!\Phi ^{D_{t-}(t)}\right) \right| ^2 \nu ^d_t({\mathrm {d}}x){\mathrm {d}}t \right] \end{aligned}$$

(by the triangular inequality)

$$\begin{aligned}&\le 2 a^2(0)\, {\mathbb {E}}\left[ \int _{\bar{t}}^{T}\int _{{\mathbb {R}}^n}\left| \log {\big ( 1+\langle x,{\mathfrak {h}}^{D_{t-}}_{t-} \rangle \big )}\right| ^2 \nu ^d_t({\mathrm {d}}x){\mathrm {d}}t \right] \\&\quad +2{\mathbb {E}}\left[ \int _{\bar{t}}^{T}\int _{\Theta }\left| \Phi ^{D_{t-}+\chi (D_{t-},x)}(t)-\Phi ^{D_{t-}}(t)\right| ^2 \nu ^d_t({\mathrm {d}}x){\mathrm {d}}t \right] \end{aligned}$$

[by (8.14) and Assumption 3.3]

$$\begin{aligned} \le 2 a^2(0) \int _{\bar{t}}^{T} C^2 \sup _{h\in H}\left\| h\right\| ^2 \int _{{\mathbb {R}}^n}\left\| x\right\| ^2\nu ^d_t({\mathrm {d}}x)\!+\!2 \left( \max _{d'\in \{0,1\}^n}\Phi ^{d'}(t)\right) ^2\nu ^d_t(\Theta ){\mathrm {d}}t\!<\!+\infty . \end{aligned}$$

Part (c). By (8.9), (8.11) and Lemma 8.2 we have

$$\begin{aligned} ({\bar{h}}(t,d),{\bar{c}}(t)v)\in \arg \max _{h\in H,\, c>0}A^{h,c} K(t,v,d)\quad \forall v\in {\mathbb {R}}^{+} \end{aligned}$$

(8.16)

for any $t\in [0,T]$ and $d\in \{0,1\}^n$. Therefore, the process $(\bar{{\mathfrak {h}}}_t,\bar{{\mathfrak {c}}}_t)$ defined in (5.14) satisfies (5.9) and the statement follows by Theorem 5.2. $\square$