On the existence and uniqueness of the optimal central bank intervention policy in a forex market with jumps

Abstract

We study a central bank intervention (CBI) problem in the foreign exchange market when the exchange rate follows a jump-diffusion process and show that the optimal CBI policy is a control-band policy. Our main contribution is a rigorous proof of the existence and uniqueness of the optimal CBI policy.

Appendix: Further generalization of the problem

In our original problem in Section 2, we are only allowed to give impulses $\zeta$ with values in ${\mathcal {Z}}:=(0,\infty )$. However, in some applications, we may also have to bring the exchange rate upward by applying impulses with negative values. Therefore, we now assume that we are allowed to give impulses $\zeta$ with values in ${\mathcal {Z}}:=(-\infty ,\infty )$. Again, if we apply an impulse control $\nu = (\tau _{1},\tau _{2},\ldots ,\tau _{j},\ldots ;\zeta _{1},\zeta _{2},\ldots ,\zeta _{j},\ldots )$ to $X_{x}(t)$, the resulting intervened process $X_{x}^{(\nu )}(t)$ can be defined by $X_{x}^{(\nu )}(t)=X_{x}(t)+\sum _{\tau _{k} \le t} \zeta _{k}$. The cost of intervention in this case is defined by $K(\zeta )=C+\lambda |\zeta |$. Then, the expected total discounted cost associated with the given impulse control is

$$\begin{aligned} J^{(\nu )}(x)=E^{x}\left[ \int _{0}^{\infty } \hbox {e}^{-rt}(X_{x}^{(v)})^{2}\hbox {d}t+\sum _{k=1}^{N} \hbox {e}^{-r\tau _{k}}(C+\lambda |\zeta _{k}|)\right] . \end{aligned}$$

We again seek $\varPhi ^{*}(x)$ and $\nu ^{*}= (\tau _{1}^{*},\ldots ,\tau _{j}^{*},\ldots ;\zeta _{1}^{*},\ldots ,\zeta _{j}^{*},\ldots )$ such that

$$ \varPhi ^{*}(x)=\inf _{\nu }J^{(\nu )}(x) = J^{(\nu ^{*})}(x). $$

Here, we assume that the corresponding Lévy measure $\eta$ is symmetric, i.e., $\eta (H)=\eta (-H)$ for all $H\subset {\mathbb {R}}\setminus \{0\}$. By symmetry, we expect the continuation region to be of the form ${\mathbb {C}}=(-\bar{x},\bar{x})$, for some $\bar{x}>0$ to be determined. As soon as $X_{x}(t)$ reaches level $\bar{x}$ (or $-\bar{x}$), there is an intervention and $X_{x}(t)$ is brought down (or up) to a certain value $\hat{x}$ (or $-\hat{x}$) where $-\bar{x}<-\hat{x}<0<\hat{x}<\bar{x}$. We determine $\bar{x}$ and $\hat{x}$ in the following way.

In the continuation region ${\mathbb {C}}$, we have $A\varPhi +f=0$. Therefore we have

$$\begin{aligned} \mu \varPhi '(x)&+\frac{\sigma ^{2}}{2}\varPhi ''(x)-r\varPhi (x)\\&+\int _{{\mathbb {R}}}\{ \varPhi (x+\theta z)-\varPhi (x)-\theta z \varPhi '(x)\}\nu (\hbox {d}z)+x^{2}=0. \end{aligned}$$

Here, we try a solution of the form

$$ \varPhi (x)=C \cosh (\alpha x)+\dfrac{1}{r} x^{2}+\dfrac{1}{r^{2}}(2\mu x+b)+\dfrac{2\mu ^{2}}{r^{3}}, $$

where C is a constant (to be determined), $b=\sigma ^{2}+\int _{{\mathbb {R}}}\theta ^{2} z^{2} \nu (\hbox {d}z)$, and $\alpha >0$ is the positive solution of the equation $F(\alpha )=-r+\mu r+\frac{\sigma ^{2}}{2}\alpha ^{2}+\int _{{\mathbb {R}}}\{\hbox {e}^{\alpha \theta z}-1-\alpha \theta z \}\nu (\hbox {d}z)$. Note that if we make no intervention at all, then the cost is

$$\begin{aligned} \phi (x)&= E^{x}\left[ \int _{0}^{\infty }\hbox {e}^{-rt}(X_{x}^{(\nu )}(t))^{2}\hbox {d}t\right] \\&=E^{x}\left[ \int _{0}^{\infty }\hbox {e}^{-rt}(x+\sigma B(t)+\int _{0}^{t}\int _{{\mathbb {R}}}\theta z \widetilde{N}(\hbox {d}s,\hbox {d}z))^{2}\hbox {d}t\right] \\&=\int _{0}^{\infty }\hbox {e}^{-rt}\left( x^{2}+\mu ^{2}t^{2}+\left( 2\mu x+b\right) t \right) \hbox {d}t\\&=\hbox {e}^{-rs}\left( \dfrac{x^{2}}{r}+\dfrac{b+2\mu x}{r^{2}}+\dfrac{2\mu ^{2}}{r^{3}}\right) . \end{aligned}$$

By the definition of $\varPhi (x)$, we have $0 \le \varPhi (x) \le \phi (x)$ for all $-\bar{x}\le x \le \bar{x}$. Then, we get

$$\begin{aligned} 0\le C \cosh (\alpha x)+\dfrac{1}{r} x^{2}&+\dfrac{1}{r^{2}}(2\mu x+b)+\dfrac{2\mu ^{2}}{r^{3}}\\&\le \left( \dfrac{x^{2}}{r}+\dfrac{b+2\mu x}{r^{2}}+\dfrac{2\mu ^{2}}{r^{3}}\right) \end{aligned}$$

for all $-\bar{x}\le x\le \bar{x}$. From the right-hand side of the above inequality, we get $C \cosh (\alpha x)\le 0$ for all $-\bar{x}\le x\le \bar{x}$. Hence, $C<0$. We now let $C=-a$, where $a>0$, and define

$$ \varPhi _{0} (x)=\dfrac{x^{2}}{r} +\dfrac{b+2\mu x}{r^{2}}+\dfrac{2\mu ^{2}}{r^{3}}-a\cosh (\alpha x), $$

and let $\varPhi (x):=\varPhi _{0} (x)$ for $x\in {\mathbb {C}}$. The intervention operator in this case is

$$ {\mathcal {M}}\varPhi (x)=\inf \left\{ \varPhi (x+\zeta )+C+\lambda |\zeta | \; ; \; \zeta \in {\mathbb {R}}\right\} . $$

The first-order condition for a minimum $\hat{\zeta }=\hat{\zeta }(x)$ of the function

$$\begin{aligned} G(x) = \left\{ \begin{array}{ll} \varPhi (x+\zeta )+C+\lambda \zeta , &{}\quad {\text {if }}\; \zeta >0,\\ \ \varPhi (x+\zeta )+C-\lambda \zeta , &{}\quad {\text {if }}\; \zeta <0, \end{array} \right. \end{aligned}$$

is the following

$$\begin{aligned}&(I) \;\zeta >0 ; \; \varPhi '(x_{\hat{\zeta }}+\hat{\zeta })+\lambda =0 \quad {\text {or}} \quad \varPhi '(x_{\hat{\zeta }}+\hat{\zeta })=-\lambda ,\\&(II) \;\zeta <0; \; \varPhi '(x_{\hat{\zeta }}+\hat{\zeta })-\lambda =0 \quad {\text {or}} \quad \varPhi '(x_{\hat{\zeta }}+\hat{\zeta })=\lambda . \end{aligned}$$

Therefore, we look for points $\hat{x}$ and $\bar{x}$ such that $0<-\bar{x}<-\hat{x}<0<\hat{x}<\bar{x}$, $\varPhi '(-\hat{x})=-\lambda$ and $\varPhi '(\hat{x})=\lambda$. Note that, since $\hat{x}<\bar{x}$ and $-\hat{x}>-\bar{x}$, $\varPhi '(\hat{x})=\varPhi '_{0}(\hat{x})$ and $\varPhi '(-\hat{x})=\varPhi '_{0}(-\hat{x})$. We also have that, if $\zeta >0$, then $x+\hat{\zeta }=-\hat{x}$ or $\hat{\zeta }=-\hat{x}-x=-(\hat{x}+x)$ and if $\zeta <0$, then $x+\hat{\zeta }=\hat{x}$ or $\hat{\zeta }=\hat{x}-x$.

For $x > \bar{x}$ or $x < -\bar{x}$, we have $\varPhi (x)={\mathcal {M}}\varPhi (x)$ since $x\notin {\mathbb {C}}$. Therefore, if $\zeta >0$ (i.e., $x<-\bar{x}$), then

$$\varPhi (x)=\varPhi (x+\hat{\zeta })+C+\lambda \hat{\zeta }= \varPhi (-\hat{x})+C-\lambda ( x+\hat{x}).$$

Equivalently, we have

$$ \varPhi (x)=\varPhi _{0}(-\hat{x})+C-\lambda (\hat{x}+x) \quad {\text {if}}\; x<-\bar{x}. $$

Similarly, if $\zeta <0$ (i.e., $x>\bar{x}$), then

$$\varPhi (x)=\varPhi (x+\hat{\zeta })+C-\lambda \hat{\zeta }= \varPhi (\hat{x})+C+\lambda ( x-\hat{x}).$$

Equivalently, we have

$$ \varPhi (x)=\varPhi _{0}(\hat{x})+C+\lambda (x-\hat{x}) \quad {\text {if}}\; x>\bar{x}. $$

Consequently, we have

$$\begin{aligned} \varPhi (x) = \left\{ \begin{array}{ll} \varPhi _{0}(-\hat{x})+C-\lambda (x+\hat{x}); & \quad {\text{if}}\; x <-\bar{x}\\ \varPhi _{0}(x); &{}\quad {\text {if}}\; -\bar{x} \le x \le \bar{x}\\ \varPhi _{0}(\hat{x})+C+\lambda (x-\hat{x}); & \quad {\text{if}}\; x >\bar{x}. \end{array} \right. \end{aligned}$$

Differentiability at $x=\bar{x}$ gives the equation $\varPhi '_{0}(\bar{x})=\lambda$. Thus, we have

$$ \dfrac{2\bar{x}}{r}-a\alpha \sinh (\alpha \bar{x})=\lambda -\dfrac{2\mu }{r^{2}}. $$

(15)

We also have that $\varPhi '(\hat{x})=\lambda$. That is $\varPhi '_{0}(\hat{x})=\lambda$. Equivalently, we have

$$ \dfrac{2\hat{x}}{r}-a\alpha \sinh (\alpha \hat{x})=\lambda -\dfrac{2\mu }{r^{2}}. $$

(16)

Continuity at $x=\bar{x}$ gives the equation

$$\varPhi _{0}(\hat{x})+C+\lambda (\bar{x}-\hat{x})=\varPhi _{0}(\bar{x}).$$

Substituting for $\varPhi _{0}$ in the above equation, we get

$$\begin{aligned}&\dfrac{\hat{x}^{2}}{r} +\dfrac{b+2\mu \hat{x}}{r^{2}}+\dfrac{2\mu ^{2}}{r^{3}}-a\cosh (\alpha \hat{x})+C+\lambda (\bar{x}-\hat{x})\\&\quad =\dfrac{\bar{x}^{2}}{r} +\dfrac{b+2\mu \bar{x}}{r^{2}}+\dfrac{2\mu ^{2}}{r^{3}}-a\cosh (\alpha \bar{x}). \end{aligned}$$

Then, we have

$$ a \cosh (\alpha \bar{x})-a \cosh (\alpha \hat{x})+\frac{1}{r}(\hat{x}^{2}-\bar{x}^{2})+\lambda -\frac{2\mu }{r^{2}} (\bar{x}-\hat{x})+C=0. $$

(17)

Again, the solution to the CBI problem is uniquely determined by the system (15)–(17). We can prove the existence and uniqueness of the solution in this case by using a similar approach to Section 3.