Game theoretic valuation of deposit insurance under jump risk: from too small to survive to too big to fail

Abstract

This study examines the valuation problem in deposit insurance as a game option between the deposit insurer and the insured bank with asymmetric bankruptcy costs. The asset-to-deposit ratio of the insured bank is modeled as an exponential Lévy process with a spectrally negative jump. The study examines a wide range of scenarios in which the optimal closure policies of both parties are fully characterized. Explicit solutions are derived under the exponential jump diffusion case. This model captures several important issues in banking supervision, including the too big to fail and too small to survive phenomena, bank reorganization, and regulatory forbearance.

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Notes

  1. 1.

    Unless the asset-to-deposit begins at the self-closure region, which implies that the bank will go bankrupt immediately.

References

  1. 1.

    Allen, L., Saunders, A.: Forbearance and valuation of deposit insurance as a callable put. J. Bank. Financ. 17(4), 629–643 (1993)

    Article  Google Scholar 

  2. 2.

    Avram, F., Chan, T., Usabel, M.: On the valuation of constant barrier options under spectrally one-sided exponential Lévy models and Carr’s approximation for American puts. Stoch. Process. Appl. 100(1–2), 75–107 (2002)

    Article  Google Scholar 

  3. 3.

    Avram, F., Kyprianou, A.E., Pistorius, M.R.: Exit problems for spectrally negative Lévy processes and applications to (Canadized) Russian options. Ann. Appl. Probab. 14(1), 215–238 (2004)

    MathSciNet  Article  Google Scholar 

  4. 4.

    Avram, F., Palmowski, Z., Pistorius, M.R.: On the optimal dividend problem for a spectrally negative Lévy process. Ann. Appl. Probab. 17(1), 156–180 (2007)

    MathSciNet  Article  Google Scholar 

  5. 5.

    Baurdoux, E.J., Kyprianou, A.E.: The McKean stochastic game driven by a spectrally negative Lévy process. Electron. J. Probab. 13, 173–197 (2008)

    MathSciNet  Article  Google Scholar 

  6. 6.

    Baurdoux, E.J., Kyprianou, A.E.: The Shepp–Shiryaev stochastic game driven by a spectrally negative Lévy process. Theory Probab. Appl. 53(3), 481–499 (2009)

    MathSciNet  Article  Google Scholar 

  7. 7.

    Baurdoux, E.J., Kyprianou, A.E., Pardo, J.C.: The Gapeev–Kühn stochastic game driven by a spectrally positive Lévy process. Stoch. Process. Appl. 121(6), 1266–1289 (2011)

    Article  Google Scholar 

  8. 8.

    Baurdoux, E.J., Van Schaik, K.: Further calculations for the McKean stochastic game for a spectrally negative Lévy process: from a point to an interval. J. Appl. Probab. 48(1), 200–216 (2011)

    MathSciNet  Article  Google Scholar 

  9. 9.

    Bichteler, K., Klaus, B.: Stochastic Integration with Jumps, vol. 89. Cambridge University Press, Cambridge (2002)

    Google Scholar 

  10. 10.

    Dynkin, E.: Game variant of a problem on optimal stopping. Sov. Math. Dokl. 10, 270–274 (1969)

    MATH  Google Scholar 

  11. 11.

    Egami, M., Leung, T., Yamazaki, K.: Default swap games driven by spectrally negative Lévy processes. Stoch. Process. Appl. 123(2), 347–384 (2013)

    Article  Google Scholar 

  12. 12.

    Egami, M., Oryu, T.: An excursion-theoretic approach to regulators bank reorganization problem. Op. Res. 63(3), 527–539 (2015)

    MathSciNet  Article  Google Scholar 

  13. 13.

    Garrido, J., Morales, M.: On the expected discounted penalty function for Lévy risk processes. North Am. Actuar. J. 10(4), 196–216 (2006)

    MathSciNet  Article  Google Scholar 

  14. 14.

    Hwang, D.Y., Shie, F.S., Wang, K., Lin, J.C.: The pricing of deposit insurance considering bankruptcy costs and closure policies. J. Bank. Financ. 33(10), 1909–1919 (2009)

    Article  Google Scholar 

  15. 15.

    Kifer, Y.: Game options. Financ. Stoch. 4(4), 443–463 (2000)

    MathSciNet  Article  Google Scholar 

  16. 16.

    Kuznetsov, A., Kyprianou, A.E., Rivero, V.: The theory of scale functions for spectrally negative Lévy processes. In: Morel, J.-M., Teissier, B. (eds.) Lévy Matters II, pp. 97–186. Springer, Berlin (2012)

    Google Scholar 

  17. 17.

    Kyprianou, A.E.: Some calculations for Israeli options. Financ. Stoch. 8(1), 73–86 (2004)

    MathSciNet  Article  Google Scholar 

  18. 18.

    Kyprianou, A.E.: Introductory Lectures on Fluctuations of Lévy Processes with Applications. Springer, Berlin (2006)

    Google Scholar 

  19. 19.

    Merton, R.C.: An analytic derivation of the cost of deposit insurance and loan guarantees an application of modern option pricing theory. J. Bank. Financ. 1(1), 3–11 (1977)

    Article  Google Scholar 

  20. 20.

    Mordecki, E.: Optimal stopping and perpetual options for Lévy processes. Financ. Stoch. 6(4), 473–493 (2002)

    Article  Google Scholar 

  21. 21.

    Nivorozhkin, E.: Market discipline of subordinated debt in banking: the case of costly bankruptcy. Eur. J. Op. Res. 161(2), 364–376 (2005)

    Article  Google Scholar 

  22. 22.

    Nivorozhkin, E.: The informational content of subordinated debt and equity prices in the presence of bankruptcy costs. Eur. J. Op. Res. 163(1), 94–101 (2005)

    Article  Google Scholar 

  23. 23.

    Surya, B.A., Yamazaki, K.: Optimal capital structure with scale effects under spectrally negative Lévy models. Int. J. Theor. Appl. Financ. 17(02), 1450013 (2014)

    Article  Google Scholar 

  24. 24.

    Wang, L., Jin, Z.: Valuation of game options in jump-diffusion model and with applications to convertible bonds. J. Appl. Math. Decis. Sci. (2009). https://doi.org/10.1155/2009/945923

    MathSciNet  Article  MATH  Google Scholar 

  25. 25.

    Wong, T.W., Fung, K.W.T., Leung, K.S.: Strategic bank closure and deposit insurance valuation. Eur. J. Op. Res. (2018). https://doi.org/10.1016/j.ejor.2018.09.032

    Article  Google Scholar 

  26. 26.

    Yagi, K., Sawaki, K.: The pricing and optimal strategies of callable warrants. Eur. J. Op. Res. 206(1), 123–130 (2010)

    MathSciNet  Article  Google Scholar 

  27. 27.

    Yamazaki, K.: Cash management and control band policies for spectrally one-sided Lévy processes. In: Recent Advances in Financial Engineering 2014: Proceedings of the TMU Finance Workshop 2014, pp. 99–215 (2016)

  28. 28.

    Yamazaki, K.: Inventory control for spectrally positive Lévy demand processes. Math. Op. Res. 42(1), 212–237 (2016)

    Article  Google Scholar 

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Appendices

Appendix A: Proof of Theorem 2

Before presenting the proof of Theorem 2, we need to state some preliminary results from [5] with slight modifications.

Lemma 1

Consider the stochastic game (1). For \(\tau ^*, \varsigma ^* \in {\mathcal {T}}^+\) let

$$\begin{aligned} V^*(x) = {\mathbb {E}}_x [ e^{-r \tau ^*} (1-k_1e^{X_{\tau ^*}})^+ {\varvec{1}}_{\{\tau ^* \le \varsigma ^*\}} + e^{-r \varsigma ^*} ((1-k_2e^{X_{\varsigma ^*}})^+ + \delta ) {\varvec{1}}_{\{\varsigma ^* < \tau ^* \}}]. \end{aligned}$$

Then the triple \((V^*, \tau ^*, \varsigma ^*)\) is a solution to (1) if

  1. (i)

    \(V^*(x) \ge (1-k_1 e^x)^+\) on \({\mathbb {R}}\),

  2. (ii)

    \(V^*(x) \le (1-k_2 e^x)^+ + \delta \) on \(\{x \in {\mathbb {R}} | (1-k_1 e^x)^+ \le (1-k_2 e^x)^+ + \delta \}\),

  3. (iii)

    \(V^*(X_{\tau ^*}) = (1-k_1 e^{X_{\tau ^*}})^+\) almost surely on \(\{\tau ^* < \infty \}\),

  4. (iv)

    \(V^*(X_{\varsigma ^*}) = (1-k_2 e^{X_{\varsigma ^*}})^+ + \delta \) almost surely on \(\{\varsigma ^* < \infty \}\),

  5. (v)

    \(\{ e^{-r (t \wedge \tau ^*)} V^*(X_{t \wedge \tau ^*}) \}_{t \ge 0}\) is a right continuous submartingale, and

  6. (vi)

    \(\{ e^{-r (t \wedge \varsigma ^*)} V^*(X_{t \wedge \varsigma ^*}) \}_{t \ge 0}\) is a right continuous supermartingale.

The proof of this Lemma is the same as the proof of Lemma 5 in [5], except for slight changes caused by the payoff functions, and is therefore omitted. Theorem 2 consists of many sub-cases; we consider them individually in the following subsections.

Proof for Case 1A

We want to show that it is never optimal for the deposit insurer to exercise the call provision when \(\delta \ge \delta _0\). Consider the stopping time pair

$$\begin{aligned} \tau ^{*}= & {} \inf \{t \ge 0 | X_t \le x^{*}_N \} ~\text {and} \\ \varsigma ^{*}= & {} \infty , \end{aligned}$$

and thus \(V(x) = U(x)\), as defined in (4). Note that \(\delta _0 = 1 - k_1 e^{x^*_N} = U(x^*_N)\), and hence for any \(x > x^*_N\), \(U(x) \le U(x^*_N) = \delta _0 \le (1-k_2 e^x)^+ + \delta \) for all \(\delta \ge \delta _0\). We can see that conditions (i)–(iv) in Lemma 1 are satisfied.

For (v) and (vi) in Lemma 1, U(x) satisfies the variational inequalities

$$\begin{aligned} \left\{ \begin{array}{rcll} ({\mathcal {L}} - r) U(x) &{} = &{} 0, &{} \quad x \in (x^*_N, \infty ), \\ U(x) &{} = &{} 1 - k_1 e^x, &{} \quad x \in (-\infty , x^*_N], \end{array} \right. \end{aligned}$$

where \({\mathcal {L}}\) is the infinitesimal generator of the Lévy process X defined as

$$\begin{aligned} {\mathcal {L}} f = \mu \frac{\partial f}{\partial x} + \frac{\sigma ^2}{2} \frac{\partial ^2 f}{\partial x^2} + \int _{-\infty }^{\infty } \left( f(x+y) - f(x) - y\frac{\partial f}{\partial x} {\varvec{1}}_{\{y > -1\}}\right) {\varPi }(dy). \end{aligned}$$

For \(x \le x^*_N\), we have

$$\begin{aligned} ({\mathcal {L}} - r) U(x)= & {} ({\mathcal {L}} - r) (1 - k_1 e^x) \\= & {} (k_1 e^x - 1) r - \psi (1) e^x \\\le & {} (k_1 e^{x^*_N} - 1) r - \psi (1) e^x \\\le & {} - \psi (1) e^x \\\le & {} 0, \end{aligned}$$

because of the assumption that \(0 \le \psi (1) \le r\). This implies the supermartinagle property (vi). By Itô’s formula for the Lévy process (see Theorem 4.3 of [18]), we can write

$$\begin{aligned} e^{-r t} U(X_t) = U(X_0) + \int _0^t e^{-r s} ({\mathcal {L}} - r) U(X_s) ds + M_t, \end{aligned}$$

where \(M_t\) is a \({\mathbb {P}}\)-martingale. The fact that \(({\mathcal {L}} - r) U(X_s) = 0\) for \(s \le \tau ^{*}\) implies the (sub)martingale property (v).

Proof for Case 1B

Consider the stopping time pair

$$\begin{aligned} \tau= & {} \inf \{t \ge 0 | X_t < x^* \} ~\text {and} \\ \varsigma= & {} \inf \{t \ge 0 | x^* \le X_t \le y \}, \end{aligned}$$

where \(y \ge x^*\). If both parties adopt the above strategies, the continuity of \(V(\cdot )\) at \(x^*\) implies that \(1 - k_1 e^{x^*} = \delta \), and hence we have \(x^{*(1)}_C = \tfrac{1 - \delta }{k_1}\). For \(x > y \ge x^{*(1)}_C\), define

$$\begin{aligned} h^{(1)}(x,y) = {\mathbb {E}}_x [e^{-r \tau ^-_{y}} w^{(1)}_{\delta }(X_{\tau ^-_{y}}) ], \end{aligned}$$
(30)

where \(w^{(1)}_{\delta }(\cdot )\) is given in (8). \(h^{(1)}(x,y)\) can be interpreted as the optimal value the insured bank can obtain from the deposit insurance when the deposit insurer enforces closure at the interval \([x^{*(1)}_C, y]\). Denote by \(u^{(r)}(s,t)\) the resolvent density of X starting at \(s > 0\) and killed at the first passage below 0 (see Sect. 8.4 of [18]). We can write

$$\begin{aligned} h^{(1)}(x, y)= & {} \delta {\mathbb {E}}_x [e^{-r \tau ^-_{y}}] + \int _{-\infty }^0 \int _{-\infty }^t (w^{(1)}_{\delta }(t+y) - \delta ) u^{(r)}(x-y,t-u) {\varPi }(du) dt \\= & {} \delta \left[ Z^{(r)}(x-y) - \frac{r}{{\varPhi }(r)} W^{(r)}(x-y) \right] \\&+ \int _{-\infty }^0 \int _{-\infty }^t (w^{(1)}_{\delta }(t+y) - \delta ) \\&\left[ e^{-{\varPhi }(r) (t-u)} W^{(r)}(x-y) - W^{(r)}(x-y-t+u) \right] {\varPi }(du) dt . \end{aligned}$$

As X is of unbounded variation, we can compute

$$\begin{aligned} \frac{\partial h^{(1)}(x, y)}{\partial x}= & {} \delta \left[ W^{(r)}(x-y) - \frac{r}{{\varPhi }(r)} W^{(r)}{}'(x-y) \right] \\&+ \int _{-\infty }^0 \int _{-\infty }^t (w^{(1)}_{\delta }(t+y) - \delta ) \\&\left[ e^{-{\varPhi }(r) (t-u)} W^{(r)}{}'(x-y) - W^{(r)}{}'(x-y-t+u) \right] {\varPi }(du) dt . \end{aligned}$$

Let \(x \downarrow y+\) and note that \(W^{(r)}{}'(0+) = \tfrac{2}{\sigma ^2}\), we have

$$\begin{aligned} \frac{\partial h^{(1)}(y+, y)}{\partial x} = \left( \int _{-\infty }^0 \int _{-\infty }^t (w^{(1)}_{\delta }(t+y) - \delta ) e^{-{\varPhi }(r) (t-u)} {\varPi }(du) dt - \frac{r\delta }{{\varPhi }(r)} \right) \frac{2}{\sigma ^2}. \end{aligned}$$

Following similar arguments as in Theorem 5 of [8], we take \(\delta ^{*(1)}\) to be a solution to \(\frac{\partial h^{(1)}}{\partial x} (x^{*(1)}_C(\delta )+, x^{*(1)}_C(\delta )) = 0\), which implies (7). The smooth pasting condition \(V'(y^{*(1)}+) = 0\) (cf. Sect. 9 of [5]) implies \(\frac{\partial h^{(1)}}{\partial x} (y^{*(1)}+, y^{*(1)})= 0\), which gives (11).

When \(\delta \ge \delta ^{*(1)}\), we have \(\frac{\partial h^{(1)}}{\partial x} (x^{*(1)}_C(\delta )+, x^{*(1)}_C(\delta )) < 0\) and it is not optimal for the deposit insurer to exercise at any \(x> x^{*(1)}_C(\delta )\). We propose that the recall region of the deposit insurer should be a singleton that matches the upper bound of the self-closure region. The value function is then given by

$$\begin{aligned} V(x) = h^{(1)}(x,x^{*(1)}_C) = {\mathbb {E}}_x \left[ e^{-r \tau ^-_{x^{*(1)}_C}} \left( 1- k_1 e^{X_{\tau ^-_{x^{*(1)}_C}}}\right) \,{\varvec{1}}_{\tau ^-_{x^{*(1)}_C} < \infty }\right] , \end{aligned}$$

which gives (9).

When \(\delta < \delta ^{*(1)}\), we have \(\frac{\partial h^{(1)}}{\partial x} (x^{*(1)}_C(\delta )+, x^{*(1)}_C(\delta )) > 0\), and we can find \(y^{*(1)} > x^{*(1)}_C\) by solving (11), such that \(\frac{\partial h^{(1)}}{\partial x} (y^{*(1)}+, y^{*(1)})= 0\). In this case, the deposit insurer should exercise on \([x^{*(1)}_C, y^{*(1)}]\), such that \(V(x) = \delta \) on \([x^{*(1)}_C, y^{*(1)}]\). Hence, we have \(V(x) = h^{(1)}(x, y^{*(1)})\), where \(y^{*(1)}\) solves (11). On \((y^{*(1)}, \infty )\), we have

$$\begin{aligned} V(x) = \delta Z^{(r)} (x - y^{*(1)}) - \int _{-\infty }^0 \int _{-\infty }^t (w^{(1)}_{\delta }(t+y) - \delta ) W^{(r)}(x- y^{*(1)} - t + u) {\varPi }(du) dt, \end{aligned}$$

which gives (10).

For the optimality of \(\tau ^*\) and \(\varsigma ^*\) defined in Case 1B, note that conditions (i)–(iv) in Lemma 1 are trivially satisfied in view of the fact that \(V(x) = h^{(1)}(x,y^{*(1)})\) and the definition of \(h^{(1)}(x,y)\) in (30). V(x) satisfies the variational inequalities

$$\begin{aligned} \left\{ \begin{array}{llll} ({\mathcal {L}} - r) V(x) &{} = &{} 0, &{} \quad x \in (y^{*(1)}, \infty ), \\ V(x) &{} = &{} \delta , &{} \quad x \in (x^{*(1)}_C, y^{*(1)}], \\ V(x) &{} = &{} 1 - k_1 e^x, &{} \quad x \in (-\infty , x^{*(1)}_C], \end{array} \right. \end{aligned}$$

where \(y^{*(1)} = x^{*(1)}_C\) when \(\delta \ge \delta ^{*(1)}\). Using similar arguments as in Case 1A, we have \(({\mathcal {L}} - r) V(x) \le 0\) for \(x \le x^{*(1)}_C\). With \(X_0 > y^{*(1)}\), \(e^{-rt} V(X_t)\) is a martingale up to the stopping time \(\tau ^-_{y^{*(1)}}\), as \(({\mathcal {L}} - r) V(x) = 0\) on \(\{t < \tau ^-_{y^{*(1)}}\}\). We see that condition (vi) in Lemma 1 is satisfied.

Under Case 1B(i), for which \(y^{*(1)} = x^{*(1)}_C\), an application of Itô’s formula for the Lévy process gives

$$\begin{aligned} e^{-r t} V(X_t)= & {} V(X_0) + \int _0^t e^{-r s} ({\mathcal {L}} - r) V(X_s) ds + M_t. \end{aligned}$$

\(({\mathcal {L}} - r) V(X_s) = 0\) for \(s \le \varsigma ^{*}\) implies the submartingale property (v). Note that V(x) is smooth at \(x^{*(1)}_C\).

Under Case 1B(ii), for which \(y^{*(1)} > x^{*(1)}_C\), pick two arbitrary points, ab, such that \(x^{*(1)}_C< a< b < y^{*(1)}\). By recognizing that

$$\begin{aligned} V(x) = h^{(1)}(x, y^{*(1)}) = \inf \limits _{\varsigma \in {\mathcal {T}}^+} {\mathbb {E}}_x \left[ e^{-r \varsigma \wedge \tau ^-_{x^{*(1)}_C}} w^{(1)}_{\delta }(X_{\varsigma \wedge \tau ^-_{x^{*(1)}_C}}) \right] \end{aligned}$$

(see Lemma 7(v) in [8]), we see \(\{e^{-rt} V(X_t) | t \le \tau ^+_a \wedge \tau ^-_{x^{*(1)}_C} \}\) is a submartingale, and hence \(({\mathcal {L}} - r) V(x) > 0\) for \(x \in (x^{*(1)}_C, a)\). For \(X_0 \le a\) and \(t < \tau ^+_b \wedge \tau ^-_{x^{*(1)}_C}\), \(\{ e^{-rt}V(X_t) | t \le 0 \}\) is a submartingale. Moreover, for \(X_0 \ge b\) and \(t < \tau ^-_a \wedge \tau ^-_{x^{*(1)}_C}\), \(\{ e^{-rt}V(X_t) | t \le 0 \}\) is also a submartingale. Combined these with the fact that \(({\mathcal {L}} - r) V(X_s) = 0\) for \(s < \varsigma ^{*}\), this implies the submartingale property (v).

Proof for Case 2A

Similar to Case 1A, we consider the stopping time pair

$$\begin{aligned} \tau ^{*}= & {} \inf \{t \ge 0 | X_t \le x^{*}_N \} ~\text {and} \\ \varsigma ^{*}= & {} \infty . \end{aligned}$$

Note that \(\delta _1 = U(\log \frac{1}{k_2})\), and hence for any \(x > \log \frac{1}{k_2}\), \(U(x) \le U(\log \frac{1}{k_2}) = \delta _1 \le (1-k_2 e^x)^+ + \delta \) for all \(\delta \ge {\overline{\delta }} = \max (\delta _1, \delta _2)\). The rest of the proof is the same as in Case 1A and is omitted.

Proof for Case 2B

Consider the stopping time pair

$$\begin{aligned} \tau ^*= & {} \inf \{ t> 0 | X_t < x^*\} ~\text {and} \\ \varsigma ^*= & {} \inf \left\{ t > 0 | \log \tfrac{1}{k_2} \le X_t \le y \right\} , \end{aligned}$$

where \(y \ge x^*\). For \(x \le \log \frac{1}{k_2}\), we can evaluate the following expectation according to the strategy specified above.

$$\begin{aligned}&{\mathbb {E}}_x \left[ e^{-r \tau ^+_{\log \frac{1}{k_2}}} \delta {\varvec{1}}_{\left\{ \tau ^+_{\log \frac{1}{k_2}} < \tau ^-_{x^*} \right\} } + e^{-r \tau ^-_{x^*}} \left( 1 - k_1 e^{X_{\tau ^-_{x^*}}}\right) {\varvec{1}}_{\left\{ \tau ^+_{\log \frac{1}{k_2}} \ge \tau ^-_{x^*} \right\} } \right] \\&\quad = \delta \frac{W^{(r)}(x - x^*)}{W^{(r)} \left( \log \frac{1}{k_2} - x^*\right) } + Z^{(r)}(x - x^*) - \frac{W^{(r)}(x - x^*) Z^{(r)} \left( \log \frac{1}{k_2} - x^*\right) }{W^{(r)} \left( \log \frac{1}{k_2} - x^*\right) } \\&\qquad -\, k_1 e^x {\mathbb {E}}_x^1 \left[ e^{-(r - \psi (1)) \tau ^-_{x^*} } \right] \\&\quad = \delta \frac{W^{(r)}(x - x^*)}{W^{(r)} \left( \log \frac{1}{k_2} - x^*\right) } + Z^{(r)}(x - x^*) - \frac{W^{(r)}(x - x^*) Z^{(r)} \left( \log \frac{1}{k_2} - x^*\right) }{W^{(r)} \left( \log \frac{1}{k_2} - x^*\right) } \\&\qquad -\, k_1 e^x \left( Z^{(r - \psi (1))}_1 (x - x^*) - \frac{W^{(r - \psi (1))}_1(x - x^*) Z^{(r - \psi (1))}_1 \left( \log \frac{1}{k_2} - x^*\right) }{W^{(r - \psi (1))}_1 \left( \log \frac{1}{k_2} - x^*\right) } \right) \\&\quad = \frac{W^{(r)}(x - x^*)}{W^{(r)} \left( \log \frac{1}{k_2} - x^*\right) } \left[ \delta - Z^{(r)} \left( \log \tfrac{1}{k_2} - x^*\right) + \tfrac{k_1}{k_2} Z^{(r - \psi (1))}_1 \left( \log \tfrac{1}{k_2} - x^*\right) \right] \\&\qquad +\, Z^{(r)}(x - x^*) - k_1 e^x Z^{(r - \psi (1))}_1(x - x^*), \end{aligned}$$

where \({\mathbb {E}}_x^1\) is the expectation under \({\mathbb {P}}_x^1\). If we take \(x^* = x^{*(2)}_C\) as a solution to (12), the above expression simplifies into

$$\begin{aligned} Z^{(r)}(x - x^{*(2)}_C) - k_1 e^x Z^{(r - \psi (1))}_1(x - x^{*(2)}_C) = w^{(2)}_{\delta }(x), \end{aligned}$$

where \(w^{(2)}_{\delta }(\cdot )\) is defined in (14). Note that \(x^{*(2)}_C\) defines the self-closure boundary. For this boundary to be valid, we must have \(x^{*(2)}_C \in (x^*_N, \log \tfrac{1}{k_2})\), as an insured bank would always declare self-closure if its log-asset-to-deposit ratio is less than \(x^*_N\) and an insured bank would always undergo enforced closure if its log-asset-to-deposit ratio is \(\log \tfrac{1}{k_2}\). This implies

$$\begin{aligned}&\left[ \delta - Z^{(r)} \left( \log \tfrac{1}{k_2} - x^*_N\right) + \tfrac{k_1}{k_2} Z^{(r - \psi (1))}_1 \left( \log \tfrac{1}{k_2} - x^*_N\right) \right] \left[ \delta - 1 + \tfrac{k_1}{k_2} \right] \\&\quad = (\delta - \delta _1) (\delta - \delta _2) < 0, \end{aligned}$$

and hence \({\underline{\delta }}< \delta < {\overline{\delta }}\).

For \(x > y \ge x^{*(2)}_C\), consider

$$\begin{aligned} h^{(2)}(x,y) = {\mathbb {E}}_x [e^{-r \tau ^-_{y}} w^{(2)}_{\delta }(X_{\tau ^-_{y}}) ]. \end{aligned}$$
(31)

Using similar arguments as in Case 1B, we have

$$\begin{aligned} \frac{\partial h^{(2)}(y+, y)}{\partial x} = \left( \int _{-\infty }^0 \int _{-\infty }^t (w^{(2)}_{\delta }(t+y) - \delta ) e^{-{\varPhi }(r) (t-u)} {\varPi }(du) dt - \frac{r\delta }{{\varPhi }(r)} \right) \frac{2}{\sigma ^2}. \end{aligned}$$

We take \(\delta ^{*(2)}\) to be a solution to \(\frac{\partial h^{(2)}}{\partial x} (\log \tfrac{1}{k_2}+, \log \tfrac{1}{k_2}) = 0\), which gives (13) and \(y^{*(2)}\) as a solution to \(\frac{\partial h^{(2)}}{\partial x} (y^{*(2)}+, y^{*(2)})= 0\), which gives (18).

When \(\delta \ge \delta ^{*(2)}\), the recall region is a singleton at \(\log \tfrac{1}{k_2}\). We can evaluate the following expectation by using the fact that \({\varPhi }_1(r - \psi (1)) = {\varPhi }(r) - 1\).

$$\begin{aligned}&{\mathbb {E}}_x \left[ e^{-r \tau ^-_{x^{*(2)}_C}} \left( 1 - k_1 e^{X_{\tau ^-_{x^{*(2)}_C}}}\right) {\varvec{1}}_{\left\{ \tau ^-_{x^{*(2)}_C}< T_{\log \frac{1}{k_2}} \right\} } + e^{-r T_{\log \frac{1}{k_2}}} \delta {\varvec{1}}_{\left\{ T_{\log \frac{1}{k_2}} < \tau ^-_{x^{*(2)}_C} \right\} } \right] \\&\quad = \left( \frac{Z^{(r)} \left( \log \frac{1}{k_2} - x^{*(2)}_C\right) }{W^{(r)} \left( \log \frac{1}{k_2} - x^{*(2)}_C\right) } - \frac{r}{{\varPhi }(r)} \right) e^{{\varPhi }(r) \left( \log \frac{1}{k_2} - x^{*(2)}_C\right) } W^{(r)}\left( x - \log \tfrac{1}{k_2}\right) \\&\qquad +\, Z^{(r)}(x - x^{*(2)}_C) - \frac{Z^{(r)} \left( \log \frac{1}{k_2} - x^{*(2)}_C\right) }{W^{(r)} \left( \log \frac{1}{k_2} -x^{*(2)}_C\right) } W^{(r)}(x - x^{*(2)}_C) \\&\qquad -\, k_1 e^x \left[ \left( \frac{Z_1^{(r-\psi (1))} \left( \log \frac{1}{k_2} - x^{*(2)}_C\right) }{W_1^{(r-\psi (1))} \left( \log \frac{1}{k_2} - x^{*(2)}_C\right) } - \frac{r - \psi (1)}{{\varPhi }(r) - 1} \right) \right. \\&e^{({\varPhi }(r) - 1) \left( \log \frac{1}{k_2} - x^{*(2)}_C\right) } W_1^{(r-\psi (1))} \left( x - \log \tfrac{1}{k_2}\right) + Z_1^{(r-\psi (1))}(x - x^{*(2)}_C) \\&\left. -\, \frac{Z_1^{(r-\psi (1))} \left( \log \frac{1}{k_2} - x^{*(2)}_C\right) }{W_1^{(r-\psi (1))} \left( \log \frac{1}{k_2} - x^{*(2)}_C\right) } W_1^{(r-\psi (1))}(x - x^{*(2)}_C) \right] \\&\qquad \delta \left[ \frac{W^{(r)}(x - x^{*(2)}_C)}{W^{(r)}\left( \log \frac{1}{k_2} - x^{*(2)}_C\right) } - e^{{\varPhi }(r) \left( \log \frac{1}{k_2} - x^{*(2)}_C\right) } \frac{W^{(r)}\left( x - \log \tfrac{1}{k_2}\right) }{W^{(r)}\left( \log \tfrac{1}{k_2} - x^{*(2)}_C\right) } \right] . \end{aligned}$$

Applying (3) and noticing that \(\delta \) is a solution to (12), the above expression can be simplified into

$$\begin{aligned} Z^{(r)}(x - x^{*(2)}_C) - k_1 e^x Z^{(r - \psi (1))}_1(x - x^{*(2)}_C) + \alpha e^{{\varPhi }(r) \left( \log \frac{1}{k_2} - x^{*(2)}_C\right) } W^{(r)}\left( x - \log \tfrac{1}{k_2}\right) , \end{aligned}$$

where \(\alpha \) is defined in (16). When \(\delta < \delta ^{*(2)}\), we have \(y^{*(2)} > \log \tfrac{1}{k_2}\) and \(V(x) = \delta \) on \([\log \tfrac{1}{k_2}, y^{*(2)}]\). Noticing that \(V(x) = h^{(2)}(x, y^{*(2)})\) and \(y^{*(2)}\) solves (18), we have

$$\begin{aligned} V(x) = \delta Z^{(r)} (x - y^{*(2)}) - \int _{-\infty }^0 \int _{-\infty }^t (w^{(2)}_{\delta }(t+y) - \delta ) W^{(r)}(x- y^{*(2)} - t + u) {\varPi }(du) dt \end{aligned}$$

on \((y^{*(2)}, \infty )\), which gives (17).

For the optimality of \(\tau ^*\) and \(\varsigma ^*\) defined in Case 2B, conditions (i)–(iv) in Lemma 1 are easily established in view of the equality \(V(x) = h^{(2)}(x, y^{*(2)})\). V(x) satisfies the variational inequalities

$$\begin{aligned} \left\{ \begin{array}{llll} ({\mathcal {L}} - r) V(x) &{} = &{} 0, &{} \quad x \in \left( x^{*(2)}_C, \log \frac{1}{k_2}\right) \cup (y^{*(2)}, \infty ), \\ V(x) &{} = &{} \delta , &{} \quad x \in \left[ \log \frac{1}{k_2}, y^{*(2)}\right] , \\ V(x) &{} = &{} 1 - k_1 e^x, &{} \quad x \in (-\infty , x^{*(2)}_C], \end{array} \right. \end{aligned}$$

where \(y^{*(2)} = \log \frac{1}{k_2}\) when \(\delta \ge \delta ^{*(2)}\). Similar to Case 1B, \(({\mathcal {L}} - r) V(x) \le 0\) for \(x \le x^{*(2)}_C\) implies the supermartinagle property (vi).

Under Case 2B(i), for which \(y^{*(2)} = \log \frac{1}{k_2}\), an application of Itô’s formula for the Lévy process gives

$$\begin{aligned} e^{-r t} V(X_t)= & {} V(X_0) + \int _0^t e^{-r s} ({\mathcal {L}} - r) V(X_s) ds + M_t \\&+ \int _0^t e^{-r s} \left( V'\left( \log \tfrac{1}{k_2}+\right) - V'\left( \log \tfrac{1}{k_2}-\right) \right) dL^{\log \frac{1}{k_2}}_s, \end{aligned}$$

where \(L^{\log \frac{1}{k_2}}_s\) is a semi-martingale local time. The integral with respect to \(L^{\log \frac{1}{k_2}}_s\) is always non-negative because

$$\begin{aligned} V'\left( \log \tfrac{1}{k_2}+\right) - V'\left( \log \tfrac{1}{k_2}-\right) = \frac{2}{\sigma ^2} \alpha e^{{\varPhi }(r)\left( \log \frac{1}{k_2} - x^{*(2)}_C\right) }, \end{aligned}$$

in which \(\alpha \) as defined in (16) satisfies

$$\begin{aligned} \alpha = -\frac{r}{{\varPhi }(r)} + k_1 e^{x^{*(2)}_C}\frac{r - \psi (1)}{{\varPhi }(r)-1} > -\frac{r}{{\varPhi }(r)} + k_1 e^{x^{*}_N} \frac{r - \psi (1)}{{\varPhi }(r)-1} = 0. \end{aligned}$$

This inequality and the fact that \(({\mathcal {L}} - r) V(X_s) = 0\) for \(s < \varsigma ^{*}\) implies the submartingale property (v).

Under Case 2B(ii), for which \(y^{*(2)} > \log \frac{1}{k_2}\), recognizing

$$\begin{aligned} V(x) = h^{(2)}(x, y^{*(2)}) = \inf \limits _{\varsigma \in {\mathcal {T}}^+} {\mathbb {E}}_x \left[ e^{-r \varsigma \wedge \tau ^-_{x^{*(2)}_C}} w^{(2)}_{\delta }(X_{\varsigma \wedge \tau ^-_{x^{*(2)}_C}}) \right] \end{aligned}$$

and using similar arguments in Case 1B(ii) gives the submartingale property (v).

Proof for Case 2C

The proof is the same as in Case 1B and is therefore omitted.

Appendix B: Proof of Proposition 2

Refering to (20) for \(c = 0\), (22), and Proposition 1 (i), the polynomial \((\theta + z)(\psi (z) - r)\) has three roots \(\beta _1< \beta _2< 0 < \beta _3 = 1\), and we can write

$$\begin{aligned} (\theta + z)(\psi (z) - r) = \frac{\sigma ^2}{2} (z - \beta _1)(z - \beta _2)(z - 1). \end{aligned}$$

Dividing both sides by \(z - 1\) and taking limit for \(z \rightarrow 1\), we arrive at

$$\begin{aligned} (\theta + 1)\psi '(1) = \frac{\sigma ^2}{2} (1 - \beta _1)(1 - \beta _2). \end{aligned}$$
(32)

Using (32) and Proposition 1 (ii), we have

$$\begin{aligned} \delta _0= & {} 1 - \tfrac{\psi (1)}{\psi '(1)} = 1 - \tfrac{2(\theta +1)r}{\sigma ^2 (1-\beta _1)(1-\beta _2)}, \\ \delta _1= & {} r \sum \limits _{i=1}^3 \tfrac{C_i}{\beta _i} \left( \tfrac{k_1 }{k_2} \tfrac{\sigma ^2 (1-\beta _1)(1-\beta _2)}{2(\theta +1)r} \right) ^{\beta _i} - \tfrac{k_1}{k_2}, ~\text {and} \\ \delta _2= & {} \tfrac{k_2 - k_1}{k_2}. \end{aligned}$$

Note that under the risk neutral setting \(r - \psi (1) = 0\), we have \(Z^{(r - \psi (1))}_1(x) \equiv 1\) and thus

$$\begin{aligned} \delta _1 = Z^{(r)} \left( x^*_N - \log \tfrac{1}{k_2} \right) - \tfrac{k_1}{k_2} > Z^{(r)} (0) - \tfrac{k_1}{k_2} = \delta _2. \end{aligned}$$

This enable us to identify \({\overline{\delta }} = \delta _1\) and \({\underline{\delta }} = \delta _2\). The proofs for (i)–(ix) are given below.

  1. (i)

    The expression for \(x^*_N\) can be obtained by (6) and (32); (23) is a direct application of Proposition 1 (ii).

  2. (ii)

    According to (7), \(\delta ^{*(1)}\) is a solution to the following equation in \(\delta \).

    $$\begin{aligned} r \delta= & {} \int _{-\infty }^0 \int _{-\infty }^t \left[ w^{(1)}_{\delta }\left( t + x^{*(1)}_C(\delta ) \right) - \delta \right] e^{-(t-u)} \lambda \theta e^{\theta u} du ~ dt \\= & {} \frac{\lambda \theta }{\theta + 1} \int _{-\infty }^0 (1 - k_1 e^{t + x^{*(1)}_C(\delta )} - \delta ) e^{\theta t} dt \\= & {} \frac{\lambda (1-\delta )}{(\theta + 1)^2}. \end{aligned}$$

    Solving for \(\delta \) gives \(\delta ^{*(1)} = \tfrac{\lambda }{\lambda + r(\theta + 1)^2}\). For \(\delta < \delta ^{*(1)}\), we know from Theorem 2 that \(y^{*(1)}\) solves (11) and the expression can be evaluated explicitly as follows.

    $$\begin{aligned} r \delta= & {} \int _{-\infty }^0 \int _{-\infty }^t \left[ w^{(1)}_{\delta }\left( t + y^{*(1)} \right) - \delta \right] e^{-(t-u)} \lambda \theta e^{\theta u} du ~ dt \\= & {} \frac{\lambda \theta }{\theta + 1} \int _{-\infty }^{x^{*(1)}_C - y^{*(1)}} (1 - k_1 e^{t + y^{*(1)}} - \delta ) e^{\theta t} dt \\= & {} \frac{\lambda \theta e^{-\theta y^{*(1)}}}{\theta + 1} \left( \frac{1 - \delta }{\theta } e^{\theta x^{*(1)}_C} - \frac{k_1}{\theta + 1} e^{(\theta + 1)x^{*(1)}_C} \right) . \end{aligned}$$

    Recall that \(x^{*(1)}_C = \tfrac{1 - \delta }{k_1}\) and rearranging terms gives \(e^{\theta y^{*(1)}} = \tfrac{\lambda (1-\delta )^{\theta +1)}}{r \delta (\theta +1)^2 k_1^{\theta }}\).

  3. (iii)

    It is a direct application of Proposition 1 (ii).

  4. (iv)

    According to Case 1B(ii) of Theorem 2, we have for \(x \ge y^{*(1)}\)

    $$\begin{aligned} V(x)= & {} r \delta \sum \limits _{i=1}^3 \frac{C_i}{\beta _i} e^{\beta _i (x - y^{*(1)})} \nonumber \\&- \int _{-\infty }^0 \int _{-\infty }^t \left[ w^{(1)}_{\delta }\left( t + y^{*(1)} \right) - \delta \right] \sum \limits _{j=1}^3 C_j e^{\beta _j (x - y^{*(1)} - t + u)} \lambda \theta e^{\theta u} du ~ dt \nonumber \\= & {} r \delta \sum \limits _{i=1}^3 \frac{C_i}{\beta _i} e^{\beta _i (x - y^{*(1)})} \nonumber \\&- \lambda \theta \left( \sum \limits _{j=1}^3 \frac{C_j e^{\beta _j (x - y^{*(1)})}}{\beta _j + \theta } \right) \int _{-\infty }^0 \left[ w^{(1)}_{\delta }\left( t + y^{*(1)} \right) - \delta \right] e^{\theta t} dt. \end{aligned}$$
    (33)

    Recall that \(y^{*(1)}\) solves

    $$\begin{aligned} \int _{-\infty }^0 \left[ w^{(1)}_{\delta }\left( t + y^{*(1)} \right) - \delta \right] e^{\theta t} dt = \frac{r \delta (\theta + 1)}{\lambda \theta }, \end{aligned}$$

    and \(\beta _3 = 1\), thus (33) becomes

    $$\begin{aligned} V(x)= & {} r \delta \sum \limits _{i=1}^3 \frac{C_i e^{\beta _i (x - y^{*(1)})}}{\beta _i} - r \delta (\theta + 1) \left( \sum \limits _{i=1}^3 \frac{C_i e^{\beta _j (x - y^{*(1)})}}{\beta _i + \theta } \right) \\= & {} r \delta \sum \limits _{i=1}^2 \frac{C_i e^{\beta _i (x - y^{*(1)})}}{\beta _i} - r \delta (\theta + 1) \left( \sum \limits _{i=1}^2 \frac{C_i e^{\beta _j (x - y^{*(1)})}}{\beta _i + \theta } \right) \\= & {} \frac{\delta }{\beta _2 - \beta _1}\left( \beta _2 e^{\beta _1 (x - y^{*(1)})} - \beta _1 e^{\beta _2 (x - y^{*(1)})}\right) , \end{aligned}$$

    where the last equality follows from (21).

  5. (v)

    It is a direct application of Proposition 1 (ii).

  6. (vi)

    According to (13), \(\delta ^{*(2)}\) is a solution to the following equation in \(\delta \).

    $$\begin{aligned} r \delta= & {} \int _{-\infty }^0 \int _{-\infty }^t \left[ w^{(2)}_{\delta }\left( t + \log \tfrac{1}{k_2} \right) - \delta \right] e^{-(t-u)} \lambda \theta e^{\theta u} du ~ dt \\= & {} \frac{\lambda \theta }{\theta + 1} \int _{-\infty }^0 \left[ w^{(2)}_{\delta }\left( t + \log \tfrac{1}{k_2} \right) - \delta \right] e^{\theta t} dt \\= & {} \frac{\lambda \theta }{\theta + 1} \Bigg ( \int _{-\infty }^{x^{*(2)}_C - \log \tfrac{1}{k_2}} e^{\theta t} dt + \int _{x^{*(2)}_C - \log \tfrac{1}{k_2}}^0 r \sum \limits _{i=1}^3 \frac{C_i}{\beta _i} e^{\beta _i \left( t + \log \tfrac{1}{k_2} - x^{*(2)}_C\right) + \theta t} dt \\&-\int _{-\infty }^0 k_1 e^{t + log\tfrac{1}{k_2} + \theta t} dt - \delta \int _{-\infty }^0 e^{\theta t} dt \Bigg ) \\= & {} \frac{\lambda \theta }{\theta + 1} \Bigg [ \frac{e^{\theta x^{*(2)}_C(\delta )} k_2^{\theta }}{\theta } + r \sum _{i=1}^3 \frac{C_i e^{-\beta _i x^{*(2)}_C(\delta )}}{\beta _i (\beta _i + \theta ) k_2^{\beta _i}} - r \sum _{i=1}^3 \frac{C_i e^{\theta x^{*(2)}_C(\delta )} k_2^{\theta }}{\beta _i (\beta _i + \theta )} \\&-\, \frac{k_1}{k_2(\theta + 1)} -\frac{\delta }{\theta } \Bigg ]. \end{aligned}$$

    Rearranging the terms gives (24).

  7. (vii)

    For \(\delta < \delta ^{*(2)}\), we know from Theorem 2 that \(y^{*(2)}\) solves (18), in which the expression can be evaluated explicitly as follows.

    $$\begin{aligned} r \delta= & {} \int _{-\infty }^0 \int _{-\infty }^t \left[ w^{(2)}_{\delta }\left( t + y^{*(2)} \right) - \delta \right] e^{-(t-u)} \lambda \theta e^{\theta u} du ~ dt \\= & {} \frac{\lambda \theta }{\theta + 1} \int _{-\infty }^{0} (w^{(2)}_{\delta }\left( t + y^{*(2)} \right) - \delta ) e^{\theta t} dt \\= & {} \frac{\lambda \theta }{\theta + 1} \Bigg [ \int _{-\infty }^{x^{*(2)}_C - y^{*(2)}} e^{\theta t} dt + \int _{x^{*(2)}_C - y^{*(2)}}^{\log \frac{1}{k_2} - y^{*(2)}} r \sum \limits _{i=1}^3 \frac{C_i}{\beta _i} e^{\beta _i(t + y^{*(2)} - x^{*(2)}_C) + \theta t} dt \\&- \int _{-\infty }^{\log \frac{1}{k_2} - y^{*(2)}} k_1 e^{y^{*(2)} + (\theta +1)t} dt - \delta \int _{-\infty }^{\log \frac{1}{k_2} - y^{*(2)}} e^{\theta t} dt \Bigg ] \\= & {} \frac{\lambda \theta e^{-\theta y^{*(2)}}}{\theta + 1} \Bigg [ \frac{e^{\theta x^{*(2)}_C}}{\theta } + r \sum _{i=1}^3 \frac{C_i e^{-\beta _i x^{*(2)}_C}}{\beta _i (\beta _i + \theta )} \left( \frac{1}{k_2^{\beta _i + \theta }} - e^{(\beta _i + \theta ) x^{*(2)}_C} \right) \\&-\, \frac{k_1}{k_2^{\theta + 1} (\theta +1)} - \frac{\delta }{\theta k_2^\theta } \Bigg ]. \end{aligned}$$

    \(y^{*(2)}\) can then be solved from the above and results follows.

  8. (viii)

    We focus on \(x \in [\log \tfrac{1}{k_2}, \infty )\). Note that for \(\psi (1) = r\) we have from (16) that \(\alpha = k_1 e^{x^{*(2)}_C} \psi '(1) - \psi (1)\). According to Case 2B(i) of Theorem 2, on \(x \in [\log \tfrac{1}{k_2}, \infty )\) we have

    $$\begin{aligned} V(x)= & {} r \sum \limits _{i=1}^3 \frac{C_i}{\beta _i} e^{x - x^{*(2)}_C} - k_1 e^x + \frac{k_1 e^{x^{*(2)}_C} \psi '(1) - \psi (1) }{k_2e^{ x^{*(2)}_C}} \sum \limits _{i=1}^3 C_i e^{\beta _i \left( x - \log \tfrac{1}{k_2}\right) } \\= & {} \sum \limits _{i=1}^3 C_i \left[ \frac{r}{\beta _i e^{\beta _i x^{*(2)}_C}} + k_2^{\beta _i - 1} \left( k_1 \psi '(1) - r e^{-x^{*(2)}_C} \right) \right] e^{\beta _i x} - k_1 e^x \\= & {} \sum _{i=1}^2 C_i \left[ \frac{r}{\beta _i e^{\beta _i x^{*(2)}_C}} + k_2^{\beta _i - 1} \left( k_1 \psi '(1) - r e^{-x^{*(2)}_C} \right) \right] e^{\beta _i x}, \end{aligned}$$

    where the last equality follows by noting that \(e^{x^*_N} = \frac{\psi (1)}{k_1 \psi '(1)}\) and hence

    $$\begin{aligned}&C_3 \left[ \frac{r}{e^{x^{*(2)}_C}} + \left( k_1 \psi '(1) - r e^{-x^{*(2)}_C} \right) \right] e^{x} - k_1 e^x \\&\quad = C_3 r e^{x - x^*_N} - k_1 e^x \\&\quad = \left( \frac{2r (\theta + 1) e^{-x^*_N}}{\sigma ^2 (1-\beta _1)(1-\beta _2)} - k_1 \right) e^x \\&\quad = 0. \end{aligned}$$
  9. (ix)

    The part for \(x < y^{*(2)}\) is straightforward. For \(x \ge y^{*(2)}\), recall that \(y^{*(2)}\) solves

    $$\begin{aligned} \int _{-\infty }^0 \left[ w^{(2)}_{\delta }\left( t + y^{*(2)} \right) - \delta \right] e^{\theta t} dt = \frac{r \delta (\theta + 1)}{\lambda \theta }. \end{aligned}$$

    The results are obtained by using similar derivations as in (iv).

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Wong, T.W. Game theoretic valuation of deposit insurance under jump risk: from too small to survive to too big to fail. Math Finan Econ 14, 67–95 (2020). https://doi.org/10.1007/s11579-019-00245-x

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Keywords

  • Game option
  • Deposit insurance
  • Bankruptcy cost
  • Lévy process

JEL Classification

  • C73
  • G280