Pricing vulnerable claims in a Lévy-driven model

Abstract

We obtain an explicit expression for the price of a vulnerable claim written on a stock whose predefault dynamics follows a Lévy-driven SDE. The stock jumps to zero at default with a hazard rate given by a negative power of the stock price. We recover the characteristic function of the terminal log price as the solution of an infinite-dimensional system of complex-valued first-order ordinary differential equations. We provide an explicit eigenfunction expansion representation of the characteristic function in a suitably chosen Banach space and use it to price defaultable bonds and stock options. We present numerical results to demonstrate the accuracy and efficiency of the method.

Appendix

This section contains the proofs of Lemmas 3.7, 3.12, 3.14, and 3.15.

Proof of Lemma 3.7

We only prove (3.24) and (3.25). Relations (3.26)–(3.28) can be proved analogously by using the definition of f _n in (3.17).

We first prove (3.24). By (3.12) and (3.13) we have

$$ f(n i)=-p\Big(r-q+\frac{{\sigma}^2 }{2}\Big)n+\frac {1}{2}p^2{\sigma}^2n^2+\psi(n i) $$

(A.1)

with

$$ \psi(n i)=\int_{{\mathbb{R}}}\big( -n p( 1-e^z ) + e^z( e^{-n p z} -1) \big)\nu (dz) . $$

(A.2)

Clearly, the first part grows as n ² as n goes to ∞. Now, for n large enough, we have

$$\frac{d}{dz} \big( -n p( 1-e^z ) + e^z( e^{-n p z} -1) \big)=(n p - 1) e^{z}(1-e^{-n p z})\geq0 \quad\Longleftrightarrow\quad z \geq0, $$

and thus z↦−np(1−e ^z)+e ^z(e ^−npz−1) has a minimum at z=0. Therefore, for n large enough,

$$\psi(n i) \geq\big( -n p( 1 - e^{- \bar{z} } ) + e^{- \bar{z} }( e^{n p \bar{z} } -1) \big) \nu\big((-\infty,- \bar{z} ]\big), $$

and this proves (3.24). We now prove (3.25). By (A.1) and (A.2) we have

$$f\big((n+1) i\big)-f(n i)=-p\Big(r-q+\frac{{\sigma}^2 }{2}\Big)+\frac {1}{2}p^2{\sigma}^2(2n +1)+\psi\big((n+1) i\big)-\psi(n i) $$

with

$$\psi\big((n+1) i\big)-\psi(n i)=\int_{{\mathbb{R}}}\big( -p( 1-e^z ) + e^{z(1-np)}( e^{- p z} - 1) \big)\nu(dz) . $$

Also in this case, one can see that for n large enough,

$$\frac{d}{dz}\big( -p( 1-e^z ) + e^{z(1-np)}( e^{- p z} - 1 )\big) \geq 0 \quad\Longleftrightarrow\quad z \geq0, $$

and thus z↦−p(1−e ^z)+e ^z(1−np)(e ^−pz−1)≥0 has a minimum at z=0. Again,

$$\psi\big((n+1) i\big)-\psi(n i) \geq\big( -p( 1-e^{- \bar{z} }) + e^{{- \bar{z} }(1-np)}( e^{p \bar{z} } - 1) \big) \nu\big((-\infty,- \bar{z} ]\big) $$

for any n suitably large. □

Proof of Lemma 3.12

Clearly, apart from rescaling the summation indices, we can assume without loss of generality that k=0. Let us fix m>0 and note that (3.33) is true if and only if

$$ p(z_m)=0, $$

(A.3)

where

$$\begin{aligned} p(y)=1-\sum_{n=0}^{m-1} \prod_{\scriptstyle\ell=0 \atop \scriptstyle\ell\neq n}^{m-1}\frac{y-z_\ell}{z_n-z_\ell}. \end{aligned}$$

Indeed,

$$\begin{aligned} p(z_m)&= 1-\sum_{n=0}^{m-1} \prod_{\scriptstyle\ell=0 \atop \scriptstyle\ell\neq n}^{m-1}\frac{z_m-z_\ell}{z_n-z_\ell}= \left ( \prod _{\ell=0}^{m-1}\frac{z_m-z_\ell}{z_m-z_\ell} \right)-\sum_{n=0}^{m-1} \prod_{\scriptstyle\ell=0 \atop\scriptstyle\ell\neq n}^{m-1}\frac {z_m-z_\ell}{z_n-z_\ell} \\ &= \bigg( \prod_{\ell=0}^{m-1}\frac{z_m-z_\ell}{z_m-z_\ell} \bigg)+\sum _{n=0}^{m-1}\bigg(\frac{z_m-z_n}{z_n-z_m} \prod_{\scriptstyle\ell=0 \atop\scriptstyle\ell\neq n}^{m-1}\frac{z_m-z_\ell}{z_n-z_\ell }\bigg) \\ &=\sum_{n=0}^m\bigg( \prod_{\ell=0}^{m-1} ( z_m-z_\ell) \prod _{\scriptstyle\ell=0 \atop\scriptstyle\ell\neq n}^{m}\frac {1}{z_n-z_\ell} \bigg)=\bigg(\prod_{\ell=0}^{m-1} (z_m-z_\ell) \bigg)\sum_{n=0}^m \prod_{\scriptstyle\ell=0 \atop\scriptstyle\ell \neq n}^{m}\frac{1}{z_n-z_\ell}. \end{aligned}$$

Now we prove that p(y)≡0. Indeed, for any z _i , 0≤i≤m−1, we have

$$\begin{aligned} \prod_{\scriptstyle\ell=0 \atop\scriptstyle\ell\neq n}^{m-1}\frac {z_i-z_\ell}{z_n-z_\ell}= \begin{cases} 1,& n=i,\\ 0,& n\neq i. \end{cases} \end{aligned}$$

Therefore, p(z _i )=0, and z ₀,…,z _m−1 are m distinct roots of p, which is a polynomial of degree at most m−1. Thus, p≡0, and this proves (A.3). □

Proof of Lemma 3.14

From the definitions (3.31) and (3.32) we have

$$\begin{aligned} b_{m,n}v_n(j)= \begin{cases} (\prod_{\ell=j}^{n-1} \frac{i({\xi}+\ell i)}{f_n-f_\ell}) ( \prod_{\ell =n}^{m-1} \frac{i({\xi}+\ell i)}{f_n-f_{\ell+1}} ), & j \leq n, \\ 0, & \mbox{otherwise}, \end{cases} \end{aligned}$$

for any m≥n≥0. Therefore, by the relations in Lemma 3.7, for a suitable constant C>0,

$$\begin{aligned} \Vert b_{m,n}v_n\Vert^2&=\sum_{j=0}^n e^{-f((j+1) i)\log{(j+1)}} |b_{m,n}v_n(j)|^2\\ & = \bigg( \prod_{\ell=n}^{m-1} \frac{|{\xi}+\ell i|^2}{|f_n-f_{\ell +1}|^2}\bigg) \sum_{j=0}^n e^{-f((j+1) i)\log{(j+1)}} \prod_{\ell =j}^{n-1} \frac{|{\xi}+\ell i|^2}{|f_n-f_\ell|^2} \\ &\leq\bigg( \prod_{\ell=n}^{m-1} \frac{|{\xi}+\ell i|^2}{|f_n-f_{\ell +1}|^2} \bigg) n C^n \prod_{\ell=0}^{n-1} \frac{|{\xi}+\ell i|^2}{|f_n-f_\ell|^2}\\ &=\bigg( \prod_{\ell=0}^{m-1} |{\xi}+\ell i|^2 \bigg) \bigg( \prod_{\ell =n+1}^{m} \frac{1}{|f_n-f_{\ell}|^2} \bigg) n C^n \prod_{\ell=0}^{n-1} \frac{1}{|f_n-f_\ell|^2}. \end{aligned}$$

Now, setting

$$g(n):=n^{1/2} C^{n/2} \prod_{\ell=0}^{n-1} \frac{1}{|f_n-f_\ell|}, $$

we get

$$\begin{aligned} \Vert b_{m,n}v_n \Vert\leq g(n) \bigg( \prod_{\ell=0}^{m-1} |{\xi }+\ell i| \bigg) \bigg( \prod_{\ell=n+1}^{m} \frac{1}{|f_n-f_{\ell}|} \bigg) \end{aligned}$$

for any m≥n≥0. Finally, by (3.28) we obtain

$$\begin{aligned} \prod_{\ell=0}^{n-1} \frac{1}{|f_n-f_\ell|} \leq\prod_{l=0}^{n-1} \frac{1}{| |f_n|-|f_\ell||} \leq\prod_{\ell=0}^{n-1} \frac{1}{| |f_n|-|f_{n-1}||} & \leq\prod_{\ell=0}^{n-1} e^{- p \bar{z} (n-1)} \\ &= e^{- p \bar{z} (n^2+n)} \end{aligned}$$

for any n greater than a suitable $\bar{n}\in{\mathbb{N}}_{0}$, which yields the desired result. □

Proof of Lemma 3.15

From the well-known identity $\sum_{k=0}^{m} {m \choose k} p^{k} (1-p)^{m-k} = 1$, we have

$${m \choose k} p^k (1-p)^{m-k} \leq1\quad\forall k \leq m \in {\mathbb{N}}_0,\ \forall p \in(0,1). $$

By letting p=1/2 we obtain ${m \choose k} \leq2^{m}$, which means m!≤2^m k!(m−k)!. Using this fact, we have

$$\begin{aligned} \sum_{j=0}^{\infty} C^j \left( j+n-1\right)! e^{- p \bar{z} j(j+1)/2} \leq& \sum_{j=0 }^{\infty} C^j 2^{j+n-1} n! (j - 1)! e^{- p \bar{z} j(j+1)/2} \\ = & 2^{n-1} n! \sum_{j=0}^{\infty} (2 C )^j (j - 1)! e^{- p \bar {z} j(j+1)/2}. \end{aligned}$$

The latter sum converges and does not depend on n; so the desired result follows. □