Discretely monitored first passage problems and barrier options: an eigenfunction expansion approach

Abstract

This paper develops an eigenfunction expansion approach to solve discretely monitored first passage time problems for a rich class of Markov processes, including diffusions and subordinate diffusions with jumps, whose transition or Feynman–Kac semigroups possess eigenfunction expansions in $L^{2}$-spaces. Many processes important in finance are in this class, including OU, CIR, (JD)CEV diffusions and their subordinate versions with jumps. The method represents the solution to a discretely monitored first passage problem in the form of an eigenfunction expansion with expansion coefficients satisfying an explicitly given recursion. A range of financial applications is given, drawn from across equity, credit, commodity, and interest rate markets. Numerical examples demonstrate that even in the case of frequent barrier monitoring, such as daily, approximating discrete first passage time problems with continuous solutions may result in unacceptably large errors in financial applications. This highlights the relevance of the method to financial applications.

Appendix: Supplementary proofs

Proof of Proposition 2.4

It is easy to see that we have $\mathcal{P}^{\phi}_{t}\varphi_{n}(x)=e^{-\phi (\lambda_{n}^{d})t}\varphi_{n}(x)$ after subordination. Hence

$$ \operatorname {Tr}\mathcal{P}^{\phi}_{t}=\sum_{n=1}^{\infty}(\varphi_{n},\mathcal{P}^{\phi}_{t}\varphi_{n})=\sum_{n=1}^{\infty}e^{-\phi(\lambda_{n}^{d})t}. $$

We also note that

$$\begin{aligned} \mathcal{P}^{\phi}_{t}1_{A}(x)&=\int_{(0,\infty)}\mathcal{P}^{d}_{s}1_{A}(x)\pi _{t}(ds)=\int_{(0,\infty)}\int_{A}p^{d}_{s}(x,y)m(dy)\pi_{t}(ds)\\ &=\int_{A}\int_{(0,\infty)}p^{d}_{s}(x,y)\pi_{t}(ds)m(dy). \end{aligned}$$

Define $p^{\phi}_{t}(x,y):=\int_{(0,\infty)}p^{d}_{s}(x,y)\pi_{t}(ds)$. It is clear that $p^{\phi}_{t}(x,y)$ is symmetric. We want to show it is also jointly continuous in $x$ and $y$.

(i) $\gamma>0$. Since $\phi(\lambda)=\gamma\lambda+\int_{(0,\infty )}(1-e^{-\lambda s})\nu(ds)$, we have $\phi(\lambda_{n}^{d})\geq\lambda _{n}^{d}\gamma t$. Hence

$$ \operatorname {Tr}\mathcal{P}^{\phi}_{t}=\sum_{n=1}^{\infty}e^{-\phi(\lambda_{n}^{d})t}\leq \sum_{n=1}^{\infty}e^{-\lambda_{n}^{d}\gamma t}=\operatorname {Tr}\mathcal{P}^{d}_{\gamma t}< \infty. $$

Applying Theorem 2.3 to $(\mathcal{P}^{d}_{t})_{t\geq0}$, the eigenfunctions satisfy

$$ |\varphi_{n}(x)|\leq e^{\lambda^{d}_{n}\gamma t/3}\sqrt{p^{d}_{2\gamma t/3}(x,x)}. $$

(A.1)

(A.1) implies

$$\begin{aligned} \sum_{n=1}^{\infty}e^{-\phi(\lambda^{d}_{n})t}\left|\varphi_{n}(x)\varphi _{n}(y)\right|\leq\sqrt{p^{d}_{2\gamma t/3}(x,x)p^{d}_{2\gamma t/3}(y,y)}\sum_{n=1}^{\infty}e^{-\phi(\lambda^{d}_{n})t}e^{2\gamma t\lambda^{d}_{n}/3}. \end{aligned}$$

(A.2)

From the expression of $\phi$, $\sum_{n=1}^{\infty}e^{-\phi(\lambda ^{d}_{n})t}e^{2\gamma t\lambda^{d}_{n}/3}<\infty$. This shows that

$$ \int_{(0,\infty)}\sum_{n=1}^{\infty}e^{-\lambda^{d}_{n} s}|\varphi _{n}(x)\varphi_{n}(y)|\pi_{t}(ds)< \infty. $$

So we can apply the dominated convergence theorem and get

$$\begin{aligned} p^{\phi}_{t}(x,y)&=\int_{(0,\infty)}p^{d}_{s}(x,y)\pi_{t}(ds)\\ &=\int_{(0,\infty)}\sum_{n=1}^{\infty}e^{-\lambda^{d}_{n} s}\varphi _{n}(x)\varphi_{n}(y)\pi_{t}(ds)=\sum_{n=1}^{\infty}e^{-\phi(\lambda ^{d}_{n})t}\varphi_{n}(x)\varphi_{n}(y). \end{aligned}$$

(A.2) also implies that $\sum_{n=1}^{\infty}e^{-\phi (\lambda^{d}_{n})t}\varphi_{n}(x)\varphi_{n}(y)$ converges uniformly on compact sets to $p^{\phi}_{t}(x,y)$. Since $\varphi_{n}(x)$ is continuous by applying Theorem 2.3 to $(\mathcal{P}^{d}_{t})_{t\geq0}$, $p^{\phi}_{t}(x,y)$ is jointly continuous in $x$ and $y$.

(ii) The condition in (ii) implies that $\mathcal{P}^{\phi}_{t}$ is trace-class, and for $x$ and $y$ on compacts, $\sum_{n=1}^{\infty}e^{-\phi(\lambda^{d}_{n})t}|\varphi_{n}(x)\varphi_{n}(y)|<\infty$. This allows us to apply the dominated convergence theorem, and the rest of the proof is similar to the case $\gamma>0$. □

Proof of Theorem 2.7

Denote by $\|f\|_{\infty}$ the $L^{\infty}$-norm of $f$. For a given $t$, from (2.6) in Theorem 2.3, $|\varphi _{n}(x)|\leq e^{\lambda_{n}t'/2}\sqrt{p_{t'}(x,x)}$ for all $t'$ such that $0< t'< t$. Hence

$$\begin{aligned} |f_{n}|\leq\int_{I}|f(y)\varphi_{n}(y)|m(dy)\leq\|f\|_{\infty}e^{\lambda _{n}t'/2}\int_{\mathcal{S}_{f}}\sqrt{p_{t'}(y,y)}m(dy)< \infty. \end{aligned}$$

(A.3)

Together with the trace-class condition and our assumption, this implies that the quantity $\int_{I}|f(y)|\sum_{n=1}^{\infty}e^{-\lambda_{n}t}|\varphi_{n}(x)\varphi_{n}(y)|m(dy)$ is finite. So by the dominated convergence theorem,

$$\begin{aligned} \mathcal{P}_{t}f(x)&=\int_{I} f(y)p_{t}(x,y)m(dy)=\int_{I} f(y)\sum _{n=1}^{\infty}e^{-\lambda_{n}t}\varphi_{n}(x)\varphi_{n}(y)m(dy)\\ &=\sum_{n=1}^{\infty}e^{-\lambda_{n}t}\varphi_{n}(x)\int_{I} f(y)\varphi _{n}(y)m(dy)=\sum_{n=1}^{\infty}f_{n}e^{-\lambda_{n}t}\varphi_{n}(x). \end{aligned}$$

We note that

$$\begin{aligned} &\int_{I}\bigg(\sum_{n=1}^{\infty}|f_{n}|e^{-\lambda_{n}t}|\varphi_{n}(x)|\bigg)^{2}m(dx)\\ &=\sum_{n=1}^{\infty}f_{n}^{2}e^{-2\lambda_{n}t} \leq\|f\|_{\infty}^{2}\bigg(\int_{\mathcal{S}_{f}}\sqrt {p_{t'}(y,y)}m(dy)\bigg)^{2}\sum_{n=1}^{\infty}e^{-\lambda_{n}(2t-t')} \end{aligned}$$

which is finite by (A.3) and the trace-class condition. Hence

$$ \int_{I}\big(\mathcal{P}_{t}f(x)\big)^{2}m(dx)\leq\int_{I}\bigg(\sum _{n=1}^{\infty}|f_{n}|e^{-\lambda_{n}t}|\varphi_{n}(x)|\bigg)^{2}m(dx)< \infty, $$

i.e. $\mathcal{P}_{t}f\in L^{2}(I,m)$. Since

$$ \sum_{n=N}^{\infty}|f_{n}|e^{-\lambda_{n}t}|\varphi_{n}(x)|\leq\|f\|_{\infty }\sqrt{p_{t'}(x,x)}\int_{\mathcal{S}_{f}}\sqrt{p_{t'}(y,y)}m(dy)\sum _{n=N}^{\infty}e^{-\lambda_{n}(t-t')} $$

is finite and $p_{t'}(x,x)$ is continuous in $x$, the convergence in (2.7) is uniform on compact sets and $\mathcal {P}_{t}f(x)$ is continuous due to the continuity of $\varphi_{n}(x)$. We also note that $\|\sum_{n=N+1}^{\infty}f_{n}e^{-\lambda_{n}t}\varphi_{n}(x)\|=\sum _{n=N+1}^{\infty}f^{2}_{n}e^{-2\lambda_{n}t}\rightarrow0 \text{ as } N\rightarrow\infty$. Therefore, the convergence is also in $L^{2}$. □

Proof of Proposition 4.1

Note that $\varphi^{2}_{n}(x)=A^{\nu}|\mu+b|x^{2}\frac{n!}{\varGamma(\nu +n+1)}(L_{n}^{\nu}(x))^{2}$. From [1, 22.14.13], we have $|L^{\nu}_{n}(x)|\leq\frac{\varGamma(\nu+n+1)}{n!\varGamma(\nu+1)}e^{\frac {x}{2}}$ for $x\geq0$. Hence

$$ \varphi^{2}_{n}(x)\leq A^{\nu}|\mu+b|x^{2}\frac{\varGamma(\nu +n+1)}{n!(\varGamma(\nu+1))^{2}}e^{Ax^{-2\beta}}. $$

Therefore, we have

$$ \sqrt{p_{t}(x,x)}\leq\frac{A^{\frac{\nu}{2}}\sqrt{|\mu+b|}}{\varGamma(\nu +1)}\sqrt{\sum_{n=0}^{\infty}e^{-\lambda_{n}t}\frac{\varGamma(\nu +n+1)}{n!}}xe^{Ax^{-2\beta}/2}. $$

Since $\frac{\varGamma(\nu+n+1)}{n!}\sim n^{\nu}$ as $n\rightarrow\infty$, condition (4.4) implies $\sum_{n=0}^{\infty}e^{-\lambda_{n}t}\frac{\varGamma(\nu+n+1)}{n!}<\infty$ in the above. To prove the conclusion, note that when $\mu +b<0$, we obtain that $\int_{0}^{\infty}xe^{Ax^{-2\beta }/2}m(dx)<\infty$. When $\mu+b>0$, for any $S$ bounded above, we have $\int _{S}xe^{-Ax^{-2\beta}/2}m(dx)<\infty$. □

Proof of Proposition 4.3

The calculation is based on the following identities and then simplification:

$$\begin{gathered} \int_{0}^{x} y^{\gamma}e^{-ay^{\delta}}L^{(\alpha)}_{n}(ay^{\delta})dy=\frac {(\alpha+1)_{n}}{n!(\gamma+1)}x^{\gamma+1}\,_{2}F_{2}\bigg( \textstyle\begin{array}{ll}\alpha+n+1,&\frac{\gamma+1}{\delta}\\ \alpha+1,&\frac{\gamma+1}{\delta}+1 \end{array}\displaystyle ;-ax^{\delta}\bigg),\\ \int_{0}^{x} y^{\gamma}L^{(\alpha)}_{n}(ay^{\delta})dy=\frac{(\alpha +1)_{n}}{n!(\gamma+1)}x^{\gamma+1}\,_{2}F_{2}\bigg( \textstyle\begin{array}{ll}-n,&\frac{\gamma+1}{\delta}\\ \alpha+1,&\frac{\gamma+1}{\delta}+1 \end{array}\displaystyle ;ax^{\delta}\bigg). \end{gathered}$$

They can be obtained from [66, Eq. (1.14.3.3) and Eq. (1.14.3.7)] by a change of variable. In the simplification procedure, we use $_{2}F_{2}\left(a_{1},a;b_{1},a;x\right )= {}_{1}F_{1}\left(a_{1};b_{1};x\right)$ and $_{1}F_{1}(-n;\alpha+1;x)=\frac {\varGamma(n+1)}{(\alpha+1)_{n}}L_{n}^{(\alpha)}(x)$. □

Proof of Proposition 4.4

Define $S(x,h)=\mathbb{P}_{x}[\zeta>h]$. Then

$$D^{h}_{n}(0,u)=\int_{0}^{u}\varphi_{n}(x)m(dx)-\int_{0}^{u}S(x,h)\varphi_{n}(x)m(dx). $$

The first term is already calculated in Proposition 4.3. For the second term, we note that when $\mu+b<0$,

$$ S(x,h)=\frac{A^{\frac{\nu}{2}-\frac{c}{|\beta|}}\varGamma(\frac {c}{|\beta|}+1)}{\sqrt{|\mu+b|}}\sum_{n=0}^{\infty}e^{-\lambda_{n}h}\frac {(\frac{1}{2|\beta|})_{n}}{\sqrt{n!\varGamma(\nu+n+1)}}\varphi_{n}(x). $$

From this we can easily derive the expression for the second term by interchanging integration and summation, using the continuity of inner products. For $\mu+b>0$,

$$\begin{aligned} S(x,h) =&A^{\frac{1}{2|\beta|}}\frac{\varGamma(1+\frac{c}{|\beta |})}{\varGamma(\nu+1)}\\ &{}\times\sum_{m=0}^{\infty}e^{-(b+\omega m)h}\frac{(\frac{1}{2|\beta |})_{m}}{m!}xe^{-Ax^{-2\beta}}\,_{1}F_{1}\Big(1-m+\frac{c}{|\beta|};\nu +1;Ax^{-2\beta}\Big). \end{aligned}$$

Again the second term can be calculated by interchanging integration and summation since the above series converges uniformly on $(0,u)$. The only thing left to compute is $d_{m,n}(0,u)$. By change of variable,

$$\begin{aligned} d_{m,n}(0,u) =&A^{-\nu}\sqrt{\frac{n!}{|\mu+b|\varGamma(\nu+n+1)}}\\ &{}\times\int_{0}^{Au^{-2\beta}}y^{\nu}e^{-y}L^{(\nu)}_{n}(y)\,_{1}F_{1}\Big(1-m+\frac{c}{|\beta|};\nu+1;y\Big)dy. \end{aligned}$$

We denote the above integral by $I$. To calculate $I$, we use the following identities:

$$\begin{aligned} &\frac{d(\,_{1}F_{1}(a,b,y))}{dy}=\frac{a}{b}\,_{1}F_{1}(a+1;b+1;y), \end{aligned}$$

(A.4)

$$\begin{aligned} &\frac{d(y^{b-1}e^{-y}\,_{1}F_{1}(a,b,y))}{dy}=(b-1)e^{-y}y^{b-2}\, _{1}F_{1}(a-1;b-1;y), \end{aligned}$$

(A.5)

$$\begin{aligned} &\frac{dL^{(\nu)}_{n}(y)}{dy}=-L^{(\nu+1)}_{n-1}(y),\quad n\geq1, \end{aligned}$$

(A.6)

$$\begin{aligned} &\frac{d(e^{-y}y^{\nu}L^{(\nu)}_{n}(y))}{dy}=(n+1)e^{-y}y^{\nu-1}L^{(\nu -1)}_{n+1}(y). \end{aligned}$$

(A.7)

For $n=0$, by (A.5),

$$ I=\frac{1}{\nu+1}(Au^{-2\beta})^{\nu+1}e^{-Au^{-2\beta}}\,_{1}F_{1}\Big(2-m+\frac{c}{|\beta|};\nu+2;Au^{-2\beta}\Big). $$

For $n\geq1$, if $\frac{c}{|\beta|}=m-n-1$, $_{1}F_{1}(1-m+\frac{c}{|\beta |};\nu+1;y)=\frac{n!\varGamma(\nu+1)}{\varGamma(\nu+1+n)}L_{n}^{(\nu )}(y)$. Hence

$$ I=\frac{n!\varGamma(\nu+1)}{\varGamma(\nu+1+n)}\int_{0}^{Au^{-2\beta }}y^{\nu}e^{-y}\big(L^{(\nu)}_{n}(y)\big)^{2}dy=\varGamma(\nu+1)\pi_{n,n}(0,u). $$

If $\frac{c}{|\beta|}\neq m-n-1$, by (A.5), integration by parts and (A.6), we get

$$\begin{aligned} I =&\frac{1}{\nu+1}(Au^{-2\beta})^{\nu+1}e^{-Au^{-2\beta}}L_{n}^{(\nu )}(Au^{-2\beta})\,_{1}F_{1}\Big(2-m+\frac{c}{|\beta|};\nu+2;Au^{-2\beta}\Big) \\ &{}+\frac{1}{\nu+1}\int_{0}^{Au^{-2\beta}}y^{\nu+1}e^{-y}L_{n-1}^{(\nu +1)}(y)\,_{1}F_{1}\Big(2-m+\frac{c}{|\beta|};\nu+2;y\Big)dy. \end{aligned}$$

(A.8)

By (A.7), integration by parts and (A.4), we can also obtain

$$\begin{aligned} I =&\frac{1}{n}(Au^{-2\beta})^{\nu+1}e^{-Au^{-2\beta}}L_{n-1}^{(\nu +1)}(Au^{-2\beta})\,_{1}F_{1}\Big(1-m+\frac{c}{|\beta|};\nu+1;Au^{-2\beta }\Big) \\ &{}-\frac{1-m+\frac{c}{|\beta|}}{n(\nu+1)}\int_{0}^{Au^{-2\beta}}y^{\nu +1}e^{-y}L_{n-1}^{(\nu+1)}(y)\,_{1}F_{1}\Big(2-m+\frac{c}{|\beta|};\nu +2;y\Big)dy.\quad \end{aligned}$$

(A.9)

Multiplying (A.8) by $\frac{1-m+c/|\beta|}{n(\nu+1)}$, adding it to (A.9) and simplifying gives the expression for $I$. □

Proof of Proposition 4.5

$$\begin{aligned} f_{n}(\ell,u) =&\int_{\tilde{\ell}}^{\tilde{u}}\big(P(x,t,T)-K\big)\varphi _{n}(x)m(dx)\\ =&\int_{\tilde{\ell}}^{\tilde{u}}\sum_{m=0}^{\infty}p_{m}e^{-\lambda _{m}(T-t)}\varphi_{m}(x)\varphi_{n}(x)m(dx)-K\int_{\tilde{\ell}}^{\tilde {u}}\varphi_{n}(x)m(dx)\\ =&\sum_{m=0}^{\infty}p_{m}e^{-\lambda_{m}(T-t)}\int_{\tilde{\ell}}^{\tilde {u}}\varphi_{m}(x)\varphi_{n}(x)m(dx)\\ &{}-K\frac{\sigma}{\sqrt{2}}\left(\frac{2\gamma}{\sigma^{2}}\right)^{\frac {b}{2}}\int_{\tilde{\ell}}^{\tilde{u}}e^{(\kappa-\gamma)x/\sigma^{2}}\ell _{n}^{(b-1)}\bigg(\frac{2\gamma x}{\sigma^{2}}\bigg)\frac{2}{\sigma ^{2}}x^{b-1}e^{-\frac{2\kappa x}{\sigma^{2}}}dx\\ =&\sum_{m=0}^{\infty}p_{m}e^{-\lambda_{m}(T-t)}\pi^{(b-1)}_{m,n}(\tilde{\ell },\tilde{u})\\ &{}-K\frac{\sqrt{2}}{\sigma}\left(\frac{\sigma^{2}}{2\gamma}\right )^{b/2}\int_{\frac{2\gamma\tilde{\ell}}{\sigma^{2}}}^{\frac{2\gamma \tilde{u}}{\sigma^{2}}}e^{-\frac{\kappa+\gamma}{2\gamma}y}y^{b-1}\ell ^{(b-1)}_{n}(y)dy\\ =&\sum_{m=0}^{\infty}p_{m}e^{-\lambda_{m}(T-t)}\pi^{(b-1)}_{m,n}(\tilde{\ell },\tilde{u})\\ &{}-K\frac{\sqrt{2}}{\sigma}\left(\frac{\sigma^{2}}{2\gamma}\right)^{\frac {b}{2}}\left(\psi^{(b-1)}_{n}\left(\frac{\kappa+\gamma}{2\gamma},\frac {2\gamma\tilde{u}}{\sigma^{2}}\right)-\psi^{(b-1)}_{n}\bigg(\frac{\kappa +\gamma}{2\gamma},\frac{2\gamma\tilde{\ell}}{\sigma^{2}}\bigg)\right). \end{aligned}$$

The interchange of summation and integration in the second equality is justified by the continuity of inner products. The formula for $g^{i}_{n}((\ell,u)^{c})$ is proved similarly using the identity $\int _{0}^{\infty}e^{-sy}y^{\alpha}L_{n}^{(\alpha)}(y)dy=\frac{\varGamma(\alpha +n+1)(s-1)^{n}}{n!s^{\alpha+n+1}}$. The calculation of $\psi_{n}^{(\nu )}(s,x)$ can be done similarly as in [53]. □