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Market-reaction-adjusted optimal central bank intervention policy in a forex market with jumps


Impulse control with random reaction periods (ICRRP) is used to derive a country’s optimal foreign exchange (forex) rate intervention policy when the forex market reacts to the interventions. This paper extends the previous work on ICRRP by incorporating a multi-dimensional jump diffusion process to model the state dynamics, and hence, enhance the viability of the extant model for applications. Furthermore, we employ a novel minimum cost operator that simplifies the computations of the optimal solutions. Finally, we demonstrate the efficacy of our framework by finding a market-reaction-adjusted optimal central bank intervention (CBI) policy for a country. Our numerical results suggests that market reactions and the jumps in the forex market are complements when the reactions increase the forex rate volatility; otherwise, they are substitutes.

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  1. 1.

    Empirical evidence on market reactions to central bank interventions can be found in Beine et al. (2002), Beine et al. (2003), Bonser-Neal and Tanner (1996), Caporale and Doroodian (2001), Dominguez (1998), Hung (1997), Mundaca (2001), Mundaca (2011) and Wilfling (2009).

  2. 2.

    Note that each column \(\gamma ^{j}\) of the \(k\times l\) matrix \(\gamma =[\gamma _{ij}]\) depends on z only through the jth coordinate \(z_{j}\), i.e.

    $$\begin{aligned} \gamma ^{(j)}(t,z)=\gamma ^{(j)}(t,z_{j}), \ j=1,\ldots , l;\quad z=(z_{1},\ldots ,z_{l})^{T}\in \mathbb {R}^{l}. \end{aligned}$$

    Thus the integral on the right-hand side of (1) is just a shorthand matrix notation, i.e.,

    $$\begin{aligned}&\int _{\mathbb {R}^l} \gamma (X(t-),z)\widetilde{N}(dt,dz)\\&\quad =\left( \sum _{j=1}^{l}\int _{\mathbb {R}} \gamma _{1j}(X(t-),z_{j})\widetilde{N}_{j}(dt,dz_{j}),\ldots \,,\sum _{j=1}^{l}\int _{\mathbb {R}} \gamma _{kj}(X(t-),z_{j})\widetilde{N}_{j}(dt,dz_{j})\right) ^{T}. \end{aligned}$$

    In component form (1) becomes: \( dX_{d}(t)=\mu _{d}(X(t))dt+\sum _{j=1}^{m} \sigma _{dj}(X(t))dB_{j}(t)+\sum _{j=1}^{l}\int _{\mathbb {R}} \gamma _{dj}(X(t-),z_{j})\widetilde{N}_{j}(dt,dz_{j});\qquad 1\le d\le k. \)

  3. 3.

    The gradient operator \(\nabla \phi (x)\) denotes \(\left( \frac{\partial \phi }{\partial x_1},\frac{\partial \phi }{\partial x_2},\ldots ,\frac{\partial \phi }{\partial x_k}\right) \), and the dot product is defined by \(x\cdot y=\sum _{i=1}^{k}x_i y_i\) for \(x=(x_1,x_2,\ldots ,x_k)\) and \(y=(y_1,y_2,\ldots ,y_k)\).

  4. 4.

    The number of domestic currency units per unit of foreign currency or a given basket of foreign currencies.

  5. 5.

    Note that, from \(\tau _j\) to \(\tau _j+t\), the process \(X^{(u)}\) follows \(X^{i_j}\).


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The authors thank the guest editors Jun-ya Gotoh and Stan Uryasev for their contribution, and the anonymous reviewers for their valuable comments and suggestions. Moreover, the authors are grateful to Alain Bensoussan for his thoughtful suggestions that resulted in the improvement of this paper. We also would like to thank participants of the session of OM-Finance interface at INFORMS Annual Meeting 2011 in Charlotte, session of Financial Services Section (Best Student Research Paper Competition) at INFORMS Annual Meeting 2012 in Phoenix, OM seminar of the Jindal School of Management at the University of Texas Dallas, Brown Bag Lunch Seminar of the Swiss Finance Institute, and Advances of Computational Economics & Finance Seminar of the Institute of Operations Research at the University of Zurich for their helpful comments.

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Correspondence to Sandun Perera.

Appendix: Proof of the verification theorem

Appendix: Proof of the verification theorem

Let \(\phi \) be a solution of the QIVI; then, \(\phi \in C^{1}\). However, using (i)–(iii) and the Approximation Theorem of Øksendal and Sulem (2007), we can and will assume that \(\phi \in C^{2}\). Now, let \(u= (\tau _{1},\tau _{2},\ldots \,,\tau _{j},\ldots \,;(i_{1},\zeta _{1}),(i_{2},\zeta _{2}),\ldots \,,(i_{j},\zeta _{j}),\ldots )\in \mathcal {U}\) with \(\tau _{0}=0\), and define \(\phi _{d}\left( X_{x}^{(u)}(s)\right) =e^{-rs}\phi \left( X_{x}^{(u)}(s)\right) \) for \(s \ge 0\). Then, since \(T^{i_j}_{j}\) is independent of \(X_{x}^{(u)}\) for \(j \ge 1\) and \(i_j \in \{1,2,\ldots ,n\}\), we have

$$\begin{aligned} E^{x}\left[ \phi _{d}\left( X^{(u)}(\tau _{j}+T^{i_{j}}_{j})\right) \bigg \vert T^{i_{j}}_{j}=t\right] =E^{x}\left[ \phi _{d}\left( X^{(u)}(\tau _{j}+t)\right) \right] , \end{aligned}$$

where \(t \ge 0\).

Next, we derive an expression for the right-hand side of (24). For this, consider \(g(s,X_{x}^{(u)}(s)):=e^{-rs}\phi \left( X_{x}^{(u)}(s)\right) =\phi _{d}\left( X_{x}^{(u)}(s)\right) \) and observe that g(sy) is \(C^1\) w.r.t. \(s \ge 0\) (i.e., for a given \(y \in \mathbb {R}^{k}\), \(g(\cdot ,y)\) is a continuous real-valued function on \([0,\infty )\) with continuous first order derivatives) and \(C^2\) w.r.t. \(y \in \mathbb {R}^{k}\) (i.e., for a given \(s\ge 0\), \(g(s,\cdot )\) is a continuous real-valued function on \(\mathbb {R}^{k}\) with continuous second order derivatives). Therefore, applying the multi-dimensional Itô ’s formula (cf. Øksendal and Sulem (2007)) for g from \(\tau _{j}\) to \(\tau _{j}+t\), we obtainFootnote 5

$$\begin{aligned}&\phi _{d}\left( X_{x}^{(u)}(\tau _{j}+t)\right) =\phi _{d}\left( X_{x}^{(u)}(\tau _{j})\right) -r\int _{\tau _{j}}^{\tau _{j}+t}\phi _{d}\left( X_{x}^{(u)}(s)\right) ds\\&\quad + \int _{\tau _{j}}^{\tau _{j}+t} \sum _{p=1}^{k}\frac{\partial \phi _{d}}{\partial x_{p}}\left( X_{x}^{(u)}(s)\right) \left[ \mu ^{i_{j}}_{p}\left( X_{x}^{(u)}(s)\right) ds+\sigma ^{i_{j}}_{p}\left( X_{x}^{(u)}(s)\right) dB(s)\right] \\&\quad +\frac{1}{2} \int _{\tau _{j}}^{\tau _{j}+t} \sum _{p,q=1}^{k}(\sigma ^{i_{j}} (\sigma ^{i_{j}})^{T})_{pq}\left( X_{x}^{(u)}(s)\right) \frac{\partial ^{2} \phi _{d}}{\partial x_{p} \partial x_{q}}\left( X_{x}^{(u)}(s)\right) ds\\&\quad +\int _{\tau _{j}}^{\tau _{j}+t} \sum _{p=1}^{l} \int _{\mathbb {R}} \{ \phi _{d}\left( X_{x}^{(u)}(s-)+(\gamma ^{i_{j}})^{(p)}(X_{x}^{(u)}(s-),z_{p})\right) -\phi _{d}\left( X_{x}^{(u)}(s-)\right) \\&\qquad \qquad \quad - \sum _{q=1}^{k} (\gamma ^{i_{j}})_{q}^{(p)}\left( X_{x}^{(u)}(s-),z_{p}\right) \frac{\partial \phi _{d}}{\partial x_{q}}\left( X_{x}^{(u)}(s-)\right) \} \nu _{p}(dz_{p})ds\\&\quad + \int _{\tau _{j}}^{\tau _{j}+t}\sum _{p=1}^{l}\int _{\mathbb {R}} \left\{ \phi _{d}\left( X_{x}^{(u)}(s-)+(\gamma ^{i_{j}})^{(p)}(X_{x}^{(u)}(s-),z_{p})\right) \right. \\&\qquad \qquad \quad \left. -\,\phi _{d}\left( X_{x}^{(u)}(s-)\right) \right\} {\widetilde{N}_{p}}(ds,dz_{p}). \end{aligned}$$

Rearranging the terms in the above equation then yields

$$\begin{aligned} \phi _{d}\left( X_{x}^{(u)}(\tau _{j}+t)\right)= & {} \phi _{d}\left( X_{x}^{(u)}(\tau _{j})\right) + \int _{\tau _{j}}^{\tau _{j}+t} \sum _{p=1}^{k}\frac{\partial \phi _{d}}{\partial x_{p}}\left( X_{x}^{(u)}(t)\right) \sigma ^{i_{j}}_{p}\left( X_{x}^{(u)}(s)\right) dB(s)\nonumber \\&+\int _{\tau _{j}}^{\tau _{j}+t}A_{i_{j}}\phi _{d}\left( X_{x}^{(u)}(s)\right) ds\nonumber \\&+ \int _{\tau _{j}}^{\tau _{j}+t}\sum _{p=1}^{l}\int _{\mathbb {R}} \left\{ \phi _{d}\left( X_{x}^{(u)}(s-)+(\gamma ^{i_{j}})^{(p)}(X_{x}^{(u)}(s-),z_{p})\right) \right. \nonumber \\&\qquad \qquad \quad \left. -\,\phi _{d}\left( X_{x}^{(u)}(s-)\right) \right\} {\widetilde{N}_{p}}(ds,dz_{p}), \end{aligned}$$

where, for \(x \in \mathbb {R}^{k}\),

$$\begin{aligned} A_{i_{j}}\phi _{d}(x):= & {} \displaystyle \sum _{p=1}^{k} \mu ^{i_{j}}_{p}(x)\frac{\partial \phi _{d}}{\partial x_{p}}(x)+\frac{1}{2} \displaystyle \sum _{p,q=1}^{k}(\sigma ^{i_{j}} (\sigma ^{i_{j}})^{T})_{pq}(x) \frac{\partial ^{2} \phi _{d}}{\partial x_{p} \partial x_{q}}(x)-r\phi _{d}(x)\\&+\displaystyle \sum _{p=1}^{l} \int _{\mathbb {R}} \{ \phi _{d}(x\!+\!(\gamma ^{i_{j}})^{(p)} (x,z_{j}))\!-\!\phi _{d}(x)\!-\!\nabla \phi _{d}(x)\cdot (\gamma ^{i_{j}})^{(p)}(x,z_{p})\}\nu _{p}(dz_{p}). \end{aligned}$$

Thus, since

$$\begin{aligned} E^{x}\left[ \int _{\tau _{j}}^{\tau _{j}+t} \sum _{p=1}^{k}\frac{\partial \phi _{d}}{\partial x_{p}}\left( X^{(u)}(t)\right) \sigma ^{i_{j}}_{p}\left( X^{(u)}(s)\right) dB(s)\right] =0 \end{aligned}$$


$$\begin{aligned}&E^{x}\left[ \int _{\tau _{j}}^{\tau _{j}+t}\sum _{p=1}^{l}\int _{\mathbb {R}} \left\{ \phi _{d}( X^{(u)}(s-)+(\gamma ^{i_{j}})^{(p)}(X^{(u)}(s),z_{p}))\right. \right. \\&\left. \left. \qquad -\,\phi _{d}\left( X^{(u)}(s-)\right) \right\} {\widetilde{N}_{p}}(ds,dz_{p}) \right] =0, \end{aligned}$$

taking the expectation of both sides of (25), we have

$$\begin{aligned} E^{x}\left[ \phi _{d}\left( X^{(u)}(\tau _{j}+t)\right) \right] =E^{x}\left[ \phi _{d}\left( X^{(u)}(\tau _{j})\right) \right] +E^{x}\left[ \int _{\tau _{j}}^{\tau _{j}+t}A_{i_{j}}\phi _{d}\left( X^{(u)}(s)\right) ds\right] . \end{aligned}$$

Now, note that, from \(\tau _j+t\) to \(\tau _{j+1}\), the process \(X^{(u)}\) follows the original process X. Therefore, using an argument similar to that leading up to (26), it can be easily shown that

$$\begin{aligned} E^{x}\left[ \phi _{d}\left( \check{X}^{(u)}(\tau _{j+1}-)\right) \right] =E^{x}\left[ \phi _{d}\left( X^{(u)}(\tau _{j}+t)\right) \right] +E^{x}\left[ \int _{\tau _{j}+t}^{\tau _{j+1}}A\phi _{d}\left( X^{(u)}(s)\right) ds\right] , \end{aligned}$$

where \(\check{X}_{x}^{(u)}(\tau _{j+1}-)=X_{x}^{(u)}(\tau _{j+1}-)+\triangle _{N}X_{x}(\tau _{j+1})\).

It follows from (26) and (27) that

$$\begin{aligned} E^{x}\left[ \phi _{d}\left( X^{(u)}(\tau _{j})\right) -\phi _{d}\left( \check{X}^{(u)}(\tau _{j+1}-)\right) \right]&= -E^{x}\left[ \int _{\tau _{j}}^{\tau _{j}+t}A_{i_{j}}\phi _{d}\left( X^{(u)}(s)\right) ds\right] \nonumber \\&\quad \,\,-E^{x}\left[ \int _{\tau _{j}+t}^{\tau _{j+1}}A\phi _{d}\left( X^{(u)}(s)\right) ds\right] . \end{aligned}$$

Next, observe that

$$\begin{aligned}&\int _{0}^{\infty }E^{x}\left[ \phi _{d}\left( X^{(u)}(\tau _{j})\right) -\phi _{d}\left( \check{X}^{(u)}(\tau _{j+1}-)\right) \right] dF_{i_{j}}(t)\\&\quad =E^{x}\left[ \phi _{d}\left( X^{(u)}(\tau _{j})\right) -\phi _{d}\left( \check{X}^{(u)}(\tau _{j+1}-)\right) \right] \int _{0}^{\infty }dF_{i_{j}}(t)\\&\quad =E^{x}\left[ \phi _{d}\left( X^{(u)}(\tau _{j})\right) -\phi _{d}\left( \check{X}^{(u)}(\tau _{j+1}-)\right) \right] , \end{aligned}$$

where the first equality is due the independence. Therefore, integrating (28) with respect to the probability measure induced by \(T^{i_{j}}\) on \([0,\infty )\) first and then summing from \(j=1\) to \(j=m\), we obtain

$$\begin{aligned}&\sum _{j=1}^{m}E^{x}\left[ \phi _{d}( X^{(u)}(\tau _{j}))-\phi _{d}\left( \check{X}^{(u)}(\tau _{j+1}-)\right) \right] \nonumber \\&\quad =-\sum _{j=1}^{m}\int _{0}^{\infty }E^{x}\left[ \int _{\tau _{j}}^{\tau _{j}+t}A_{i_{j}}\phi _{d}\left( X^{(u)}(s)\right) ds\right] dF_{i_{j}}(t)\nonumber \\&\qquad -\sum _{j=1}^{m}\int _{0}^{\infty }E^{x}\left[ \int _{\tau _{j}+t}^{\tau _{j+1}}A\phi _{d}\left( X^{(u)}(s)\right) ds\right] dF_{i_{j}}(t). \end{aligned}$$

Moreover, using an argument similar to that leading up to (26) again, it can be easily observed that

$$\begin{aligned} E^{x}\left[ \phi _{d}\left( \check{X}^{(u)}(\tau _{1}-)\right) \right]&=E^{x}\left[ \phi _{d}\left( X_x^{(u)}(0)\right) \right] + E^{x}\left[ \int _{0}^{\tau _{1}}A\phi _{d}\left( X^{(u)}(s)\right) ds\right] \nonumber \\&=\phi \left( x\right) + E^{x}\left[ \int _{0}^{\tau _{1}}A\phi _{d}\left( X^{(u)}(s)\right) ds\right] . \end{aligned}$$

Combining (29) and (30) yield

$$\begin{aligned}&\phi \left( x\right) +\sum _{j=1}^{m}E^{x}\left[ \phi _{d}( X^{(u)}(\tau _{j}))-\phi _{d}\left( \check{X}^{(u)}(\tau _{j+1}-)\right) \right] -E^{x}\left[ \phi _{d}\left( \check{X}^{(u)}(\tau _{1}-)\right) \right] \\&\quad =-\sum _{j=1}^{m}\int _{0}^{\infty }E^{x}\left[ \int _{\tau _{j}}^{\tau _{j}+t}A_{i_{j}}\phi _{d}\left( X^{(u)}(s)\right) ds\right] dF_{i_{j}}(t)-E^{x}\left[ \int _{0}^{\tau _{1}}A\phi _{d}\left( X^{(u)}(s)\right) ds\right] \\&\qquad \quad -\sum _{j=1}^{m}\int _{0}^{\infty }E^{x}\left[ \int _{\tau _{j}+t}^{\tau _{j+1}}A\phi _{d}\left( X^{(u)}(s)\right) ds\right] dF_{i_{j}}(t). \end{aligned}$$

Equivalently, we have

$$\begin{aligned}&\phi (x)+\sum _{j=1}^{m} E^{x}\left[ \phi _{d}(X^{(u)}(\tau _{j}))-\phi _{d}(\check{X}^{(u)}(\tau _{j}-)) \right] -E^{x}\left[ \phi _{d}(\check{X}^{(u)}(\tau _{m+1}-))\right] \nonumber \\&\quad =\sum _{j=1}^{m}\int _{0}^{\infty }E^{x}\left[ \int _{\tau _{j}}^{\tau _{j}+t}\left\{ -A_{i_{j}}\phi _{d}\left( X^{(u)}(s)\right) \right\} ds\right] dF_{i_{j}}(t)\nonumber \\&\qquad +E^{x}\left[ \int _{0}^{\tau _{1}}\left\{ -A\phi _{d}\left( X^{(u)}(s)\right) \right\} ds\right] \nonumber \\&\qquad +\sum _{j=1}^{m}\int _{0}^{\infty }E^{x}\left[ \int _{\tau _{j}+t}^{\tau _{j+1}}\left\{ -A\phi _{d}\left( X^{(u)}(s)\right) \right\} ds\right] dF_{i_{j}}(t). \end{aligned}$$

Now, since \(\phi \) satisfies the QIVI, we have \(f \ge -A\phi \). Therefore, \(f_{d} \ge -A\phi _{d}\) where \(f_{d}( X^{(u)}(s))=e^{-rs}f(X^{(u)}(s))\). Then, it follows from (31) that

$$\begin{aligned}&\phi (x)+\sum _{j=1}^{m} E^{x}\left[ \phi _{d}(X^{(u)}(\tau _{j}))-\phi _{d}(\check{X}^{(u)}(\tau _{j}-)) \right] -E^{x}\left[ \phi _{d}(\check{X}^{(u)}(\tau _{m+1}-))\right] \nonumber \\&\quad \le \sum _{j=1}^{m}\int _{0}^{\infty }E^{x}\left[ \int _{\tau _{j}}^{\tau _{j}+t}\left\{ -A_{i_{j}}\phi _{d}\left( X^{(u)}(s)\right) \right\} ds\right] dF_{i_{j}}(t)+E^{x}\left[ \int _{0}^{\tau _{1}}f_{d}\left( X^{(u)}(s)\right) ds\right] \nonumber \\&\qquad +\sum _{j=1}^{m}\int _{0}^{\infty }E^{x}\left[ \int _{\tau _{j}+t}^{\tau _{j+1}}f_{d}\left( X^{(u)}(s)\right) ds\right] dF_{i_{j}}(t). \end{aligned}$$

Next, since \(T^{i_j}\) is independent of \(X_{x}^{i_j}\) for \(i_j \in \{1,2,\ldots ,n\}\), by the definition of \(\mathcal {M}_{r}\), we have

$$\begin{aligned} \mathcal {M}_{r}\phi (\check{X}_{x}^{(u)}(\tau _{j}-))\le & {} K_{i_{j}}(\check{X}_{x}^{(u)}(\tau _{j}-),\zeta _{j})\\&+\int _{0}^{\infty }E^{\Gamma _{i_{j}}(\check{X}_{x}^{(u)}(\tau _{j}-),\zeta _{j})}\left[ \int _{0}^{t}e^{-rs}f\left( X^{i_{j}}(s)\right) ds+e^{-rt}\phi \left( X^{i_{j}}(t)\right) \right] dF_{i_{j}}(t) \end{aligned}$$

But \(X_{x}^{(u)}(\tau _{j})=\Gamma _{i_{j}}(\check{X}_{x}^{(u)}(\tau _{j}-),\zeta _{j})\), therefore

$$\begin{aligned} \mathcal {M}_{r}\phi (\check{X}_{x}^{(u)}(\tau _{j}\!-\!))\!\le & {} \! K_{i_{j}}(\check{X}_{x}^{(u)}(\tau _{j}-),\zeta _{j})\!+\!\int _{0}^{\infty }E^{X_{x}^{(u)}(\tau _{j})}\left[ \int _{0}^{t}e^{-rs}f\left( X^{i_{j}}(s)\!\right) ds\right] dF_{i_{j}}(t)\nonumber \\&+\int _{0}^{\infty }E^{X_{x}^{(u)}(\tau _{j})}\left[ e^{-rt}\phi \left( X^{i_{j}}(t)\right) \right] dF_{i_{j}}(t). \end{aligned}$$

Now, using the strong Markov property and the fact that \(X_{x}^{(u)}(s)=X^{i_{j}}_{X_{x}^{(u)}(\tau _{j})}(s)\) for \(\tau _{j}\le s \le \tau _{j}+t\), we observe that

$$\begin{aligned}&\int _{0}^{\infty }E^{X_{x}^{(u)}(\tau _{j})}\left[ \int _{0}^{t}e^{-rs}f\left( X^{i_{j}}(s)\right) ds\right] dF_{i_{j}}(t)\\&\quad = \int _{0}^{\infty }\left\{ \int _{0}^{t}e^{-rs}E^{X_{x}^{(u)}(\tau _{j})}\left[ f\left( X^{i_{j}}(s)\right) \right] ds\right\} dF_{i_{j}}(t) \\&\quad = \int _{0}^{\infty }\left\{ \int _{0}^{t}e^{-rs}E^{x}\left[ f\left( X^{(u)}(\tau _{j}+s)\right) \,\bigg \vert \,\mathcal {F}_{\tau _{j}}\right] ds\right\} dF_{i_{j}}(t) \\&\quad = \int _{0}^{\infty }E^{x}\left[ \int _{0}^{t}e^{-rs}f\left( X^{(u)}(\tau _{j}+s)\right) ds\, \bigg \vert \, \mathcal {F}_{\tau _{j}}\right] dF_{i_{j}}(t) \\&\quad = \int _{0}^{\infty } e^{r \tau _{j}}E^{x}\left[ \int _{\tau _{j}}^{\tau _{j}+t}e^{-rs}f\left( X^{(u)}(s)\right) ds \,\bigg \vert \,\mathcal {F}_{\tau _{j}}\right] dF_{i_{j}}(t) \end{aligned}$$


$$\begin{aligned}&\int _{0}^{\infty }E^{X_{x}^{(u)}(\tau _{j})}\left[ e^{-rt}\phi \left( X^{i_{j}}(t)\right) \right] dF_{i_{j}}(t)\\&\quad = \int _{0}^{\infty }E\left[ e^{-rt}\phi \left( X^{i_{j}}_{X^{(u)}_{x}(\tau _{j})}(\tau _{j}+t)\right) \,\bigg \vert \,\mathcal {F}_{\tau _{j}}\right] dF_{i_{j}}(t)\\&\quad = \int _{0}^{\infty }E^{x}\left[ e^{-rt}\phi \left( X^{(u)}(\tau _{j}+t)\right) \,\bigg \vert \,\mathcal {F}_{\tau _{j}}\right] dF_{i_{j}}(t). \end{aligned}$$

Then, from (33), we have

$$\begin{aligned} \mathcal {M}_{r}\phi (\check{X}_{x}^{(u)}(\tau _{j}-))\le & {} K_{i_{j}}(\check{X}_{x}^{(u)}(\tau _{j}-),\zeta _{j})\nonumber \\&+\int _{0}^{\infty }e^{r\tau _{j}} E^{x}\left[ \int _{\tau _{j}}^{\tau _{j}+t}e^{-rs}f\left( X^{(u)}(s)\right) ds\,\bigg \vert \,\mathcal {F}_{\tau _{j}}\right] dF_{i_{j}}(t)\\&+\int _{0}^{\infty }E^{x}\left[ e^{-rt}\phi \left( X^{(u)}(\tau _{j}+t)\right) \,\bigg \vert \,\mathcal {F}_{\tau _{j}}\right] dF_{i_{j}}(t). \end{aligned}$$

Multiplying the above equation by \(e^{-r \tau _{j}}\) yields

$$\begin{aligned} \mathcal {M}_{r}\phi _{d}(\check{X}_{x}^{(u)}(\tau _{j}-))\le & {} e^{-r \tau _{j}}K_{i_{j}}(\check{X}_{x}^{(u)}(\tau _{j}-),\zeta _{j})\nonumber \\&+\int _{0}^{\infty }E^{x}\left[ \int _{\tau _{j}}^{\tau _{j}+t}f_{d}\left( X^{(u)}(s)\right) ds \,\bigg \vert \, \mathcal {F}_{\tau _{j}}\right] dF_{i_{j}}(t)\nonumber \\&+\int _{0}^{\infty }E^{x}\left[ \phi _{d}\left( X^{(u)}(\tau _{j}+t)\right) \,\bigg \vert \, \mathcal {F}_{\tau _{j}}\right] dF_{i_{j}}(t). \end{aligned}$$

Next, taking the conditional expectation of both sides of (25) first and then using an argument similar to that leading up to (26) again, we obtain

$$\begin{aligned}&E^{x}\left[ \phi _{d}\left( X^{(u)}(\tau _{j}+t)\right) \,\bigg \vert \,\mathcal {F}_{\tau _{j}}\right] \nonumber \\&\quad =E^{x}\left[ \phi _{d}\left( X_{x}^{(u)}(\tau _{j})\right) \,\bigg \vert \,\mathcal {F}_{\tau _{j}}\right] + E^{x}\left[ \int _{\tau _{j}}^{\tau _{j}+t}A_{i_{j}}\phi _{d}\left( X^{(u)}(s)\right) ds\,\bigg \vert \,\mathcal {F}_{\tau _{j}}\right] .\nonumber \\&\quad =\phi _{d}\left( X_{x}^{(u)}(\tau _{j})\right) + E^{x}\left[ \int _{\tau _{j}}^{\tau _{j}+t}A_{i_{j}}\phi _{d}\left( X^{(u)}(s)\right) ds\,\bigg \vert \,\mathcal {F}_{\tau _{j}}\right] , \end{aligned}$$

where the second equality is due to the \(\mathcal {F}_{\tau _{j}}\)-measurability of \(\phi _{d}\left( X_{x}^{(u)}(\tau _{j})\right) \).

Integrating (35) with respect to the probability measure induced by \(T^{i_{j}}\) on \([0,\infty )\), we have

$$\begin{aligned}&\int _{0}^{\infty }E^{x}\left[ \phi _{d}\left( X^{(u)}(\tau _{j}+t)\right) \,\bigg \vert \,\mathcal {F}_{\tau _{j}}\right] dF_{i_{j}}(t)\nonumber \\&\quad =\phi _{d}\left( X_{x}^{(u)}(\tau _{j})\right) +\int _{0}^{\infty }E^{x}\left[ \int _{\tau _{j}}^{\tau _{j}+t}A_{i_{j}}\phi _{d}\left( X^{(u)}(s)\right) ds\,\bigg \vert \,\mathcal {F}_{\tau _{j}}\right] dF_{i_{j}}(t). \end{aligned}$$

Substituting (36) into (34) and subtracting \(\phi _{d}(\check{X}_{x}^{(u)}(\tau _{j}-))\) from both sides then yield

$$\begin{aligned} \mathcal {M}_{r}\phi _{d}(\check{X}_{x}^{(u)}(\tau _{j}-))-\phi _{d}(\check{X}_{x}^{(u)}(\tau _{j}-))\le & {} e^{-r \tau _{j}} K_{i_{j}}(\check{X}_{x}^{(u)}(\tau _{j}-),\zeta _{j})\\&+\int _{0}^{\infty }E^{x}\left[ \int _{\tau _{j}}^{\tau _{j}+t}f_{d}\left( X^{(u)}(s)\right) ds\,\bigg \vert \,\mathcal {F}_{\tau _{j}}\right] dF_{i_{j}}(t)\\&+\int _{0}^{\infty }E^{x}\left[ \int _{\tau _{j}}^{\tau _{j}+t}A_{i_{j}}\phi _{d}\left( X^{(u)}(s)\right) ds\,\bigg \vert \,\mathcal {F}_{\tau _{j}}\right] dF_{i_{j}}(t)\\&+\,\phi _{d}\left( X_{x}^{(u)}(\tau _{j})\right) -\phi _{d}(\check{X}_{x}^{(u)}(\tau _{j}-)). \end{aligned}$$

Now, taking the expectation of both sides of the above inequality first and then summing the resulting inequality from \(j=1\) to \(j=m\), we obtain

$$\begin{aligned}&\sum _{j=1}^{m}E^{x}\left[ \mathcal {M}_{r}\phi _{d}(\check{X}_{x}^{(u)}(\tau _{j}-))-\phi _{d}(\check{X}_{x}^{(u)}(\tau _{j}-))\right] \nonumber \\&\quad \le E^{x}\left[ \sum _{j=1}^{m}e^{-r \tau _{j}}K_{i_{j}}(\check{X}_{x}^{(u)}(\tau _{j}-),\zeta _{j})\right] \nonumber \\&\qquad +\sum _{j=1}^{m}\int _{0}^{\infty }E^{x}\left[ \int _{\tau _{j}}^{\tau _{j}+t}f_{d}\left( X^{(u)}(t)\right) dt\right] dF_{i_{j}}(t)\nonumber \\&\qquad +\sum _{j=1}^{m}\int _{0}^{\infty }E^{x}\left[ \int _{\tau _{j}}^{\tau _{j}+t}A_{i_{j}}\phi _{d}\left( X^{(u)}(t)\right) dt\right] dF_{i_{j}}(t)\nonumber \\&\qquad +\sum _{j=1}^{m} E^{x}\left[ \phi _{d}(X^{(u)}(\tau _{j}))-\phi _{d}(\check{X}^{(u)}(\tau _{j}-)) \right] . \end{aligned}$$

Finally, it follows from (32) and (37) that

$$\begin{aligned}&\phi (x)+\sum _{j=1}^{m} E^{x}\left[ \mathcal {M}_{r}\phi _{d}(\check{X}^{(u)}(\tau _{j}-))-\phi _{d}(\check{X}^{(u)}(\tau _{j}-))\right] \nonumber \\&\quad \le E^{x}\left[ \phi _{d}(\check{X}^{(u)}(\tau _{m+1}-))\right] -\sum _{j=1}^{m} E^{x}\left[ \phi _{d}(X^{(u)}(\tau _{j}))-\phi _{d}(\check{X}^{(u)}(\tau _{j}-)) \right] \nonumber \\&\quad \quad -\sum _{j=1}^{m}\int _{0}^{\infty }E^{x}\left[ \int _{\tau _{j}}^{\tau _{j}+t}A_{i_{j}}\phi _{d}\left( X^{(u)}(s)\right) ds\right] dF_{i_{j}}(t)+E^{x}\left[ \int _{0}^{\tau _{1}}f_{d}\left( X^{(u)}(s)\right) ds\right] \nonumber \\&\quad \quad +\sum _{j=1}^{m}\int _{0}^{\infty }E^{x}\left[ \int _{\tau _{j}+t}^{\tau _{j+1}}f\left( X^{(u)}(t)\right) dt\right] dF_{i_{j}}(t)\nonumber \\&\quad \quad +E^{x}\left[ \sum _{j=1}^{m}e^{-r\tau _{j}}K_{i_{j}}(\check{X}^{(u)}(\tau _{j}-),\zeta _{j})\right] \!+\! \sum _{j=1}^{m}\int _{0}^{\infty }E^{x}\left[ \int _{\tau _{j}}^{\tau _{j}+t}f\left( \!X^{(u)}(s)\right) ds\right] dF_{i_{j}}(t)\nonumber \\&\quad \quad +\sum _{j=1}^{m}\int _{0}^{\infty }E^{x}\left[ \int _{\tau _{j}}^{\tau _{j}+t}A_{i_{j}}\phi _{d}\left( X^{(u)}(s)\right) ds\right] dF_{i_{j}}(t)\nonumber \\&\qquad +\sum _{j=1}^{m} E^{x}\left[ \phi _{d}(X^{(u)}(\tau _{j}))-\phi _{d}(\check{X}^{(u)}(\tau _{j}-)) \right] \nonumber \\&\quad = E^{x}\left[ \int _{0}^{\tau _{m+1}}f_{d}(X^{(u)}(s))ds\right] +E^{x}\left[ \phi _{d}(\check{X}^{(u)}(\tau _{m+1}-))\right] \nonumber \\&\qquad +E^{x}\left[ \sum _{j=1}^{m}e^{-r\tau _{j}}K_{i_{j}}(\check{X}^{(u)}(\tau _{j}-),\zeta _{j})\right] \nonumber \\&\quad = E^{x}\left[ \int _{0}^{\tau _{m+1}}e^{-rs}f(X^{(u)}(s))ds+e^{-r\tau _{m+1}}\phi (\check{X}^{(u)}(\tau _{m+1}-))\right. \nonumber \\&\qquad +\left. \sum _{j=1}^{m}e^{-r\tau _{j}}K_{i_{j}}(\check{X}^{(u)}(\tau _{j}-),\zeta _{j}) \right] .\qquad \end{aligned}$$

Since the second term on the left-hand side of (38) is non-negative, letting \(m\rightarrow \infty \) in (38) and using conditions (iv)-(v), we have

$$\begin{aligned} \phi (x)\le E^{x}\left[ \int _{0}^{\infty }e^{-rt}f(X^{(u)}(t))dt+\sum _{j=1}^{\infty }e^{-r\tau _{j}}K_{i_{j}}(\check{X}^{(u)}(\tau _{j}-),\zeta _{j}) \right] =J^{(u)}(x). \end{aligned}$$

Hence, we have that \(\phi (x)\le \varPhi (x)=\inf \{ J^{(u)}(x); \ u \in \mathcal{U}\}\).

Moreover, if the QIVI-control \(\hat{u}\) corresponding to \(\phi \) is admissible, then we can apply the above argument to \(\hat{u}= (\hat{\tau }_{1},\hat{\tau }_{2},\ldots \,,\hat{\tau }_{j},\ldots \,;(\hat{i}_{1},\hat{\zeta }_{1}),(\hat{i}_{2},\hat{\zeta }_{2}),\ldots \,,(\hat{i}_{j},\hat{\zeta }_{j}),\ldots )\). Now, since \(A\phi +f=0\) on the boundary of \(\mathbb {C}\), we obtain the equality in (32) and by the choices of \(\zeta _{i}=\hat{\zeta }_{i}\) and \(i_{j}=\hat{i}_{j}\), we also have the equality in (34) and (37). Then, since condition (i) implies that the second term on the left-hand side of (38) goes to zero as \(m \rightarrow \infty \), we achieve the desired equality in (39). Therefore, we have \(\phi (x)=J^{(\hat{u})}(x)\). Hence, \(\phi (x)=\varPhi (x)\) and \(u^{*}=\hat{u}\). \(\square \)

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Perera, S., Buckley, W. & Long, H. Market-reaction-adjusted optimal central bank intervention policy in a forex market with jumps. Ann Oper Res 262, 213–238 (2018).

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  • Optimal central-bank/government intervention policy
  • Financial market reactions
  • Jump diffusions
  • Stochastic control