A robust bank asset allocation model integrating credit-rating migration risk and capital adequacy ratio regulations

Abstract

In this paper, we consider a bank asset allocation problem with uncertain migration risk of credit ratings and capital adequacy ratio (CAR) regulations. In the practical scenarios, the future market values of each risky asset are largely affected by outer complex environments. We only observe the information about their first-moment and marginal second-moment of year-ahead market value of each loan asset. Based on these scenarios, we propose a new distributionally robust optimization model with the chance constraint characterized by uncertain CAR. Following the duality theory in infinite-dimensional optimization problem and the theory of conic linear optimization model, we can reformulate the original problem into a tractable linear deterministic semi-definite programming (SDP) model. By using this tractable linear SDP model, we can provide a robust asset allocation policy to bank managers. Further, we conduct a simulation study to illustrate the application of our method under two different economic conditions, a downward condition and an upward condition. Then a series of sensitivity tests is applied to examine the impacts of various factors, including safety probability, target CAR and recovery rate, on the optimal asset allocations. We also compare the performance of our model and the CVaR model. We demonstrate our model provides an efficient way to deal with the trade-off between expected return and CAR.

Appendices

Appendix A: Proof of Proposition 1

In order to keep a self-contained and convenient analysis in the context, we copy the left-hand-side item of inequality (17) with the set of probability distribution D (16) as follows.

$$ \mathop {\hbox{min} }\limits_{{F \in D(R^{n} ,\mu ,\rho )}} P_{F} (y_{0} (x) + y(x)^{T} \xi \le 0) $$

(17’)

where

$$ D(R^{n} ,\mu ,\rho ) = \{ F \in \varXi :P(\xi \in R^{n} ) = 1,E_{F} [\xi_{k} ] = \mu_{k} ,\;E_{F} [\xi_{k}^{2} ] = \rho_{k} ,\;k = 1,2, \ldots ,n\} $$

(16’)

By using the probability theory, we can equivalently transform the optimization problem (17’) under the constraint (16’) into the following optimization problem:

$$ \mathop { \hbox{min} }\limits_{F} {\kern 1pt} {\kern 1pt} \int_{{R^{n} }} {{\text{I}}_{( - \infty ,0]} \left( {y_{0} (x) + y(x)^{T} \xi } \right)} dF(\xi ) $$

(24a)

$$ s.t.\int_{{R^{n} }} {dF(\xi } ) = 1, $$

(24b)

$$ {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} \int_{{R^{n} }} {\xi dF(\xi } ) = \mu , $$

(24c)

$$ {\kern 1pt} {\kern 1pt} \langle e_{k} e_{k}^{T} ,\;\;\int_{{R^{n} }} {\xi \xi^{T} dF(\xi } )\rangle = \rho_{k} ,k = 1, \ldots ,n, $$

(24d)

$$ {\kern 1pt} dF(\xi ) \ge 0, $$

(24e)

where $\int_{{R^{n} }} {}$ represents n dimensional integral over real field. For the sake of simplicity, domain of the integral, Rⁿ is dropped in the remainder of this section. I_(-∞,0] in (24a) stands for the indicator function whose value equals 1 when the value of y₀(x) + y(x)^Tξ falls into(-∞,0] otherwise it equals 0. In (24d), $\left\langle \cdot \right\rangle$ refers to inner product between matrices and is equivalent to the trace of the product of two matrices; and e_k is a n × 1 unit vector where only the k-th entry is 1 and the remaining entries are 0.

Define the Lagrangian function of the optimization problem (24a)–(24e) as

$$ L(\xi ,\theta ,\alpha ,\beta ) = \theta \;\left( {1 - \int {dF(\xi )} } \right) + \alpha^{T} \left( {\mu - \int {\xi dF(\xi )} } \right) + \sum\limits_{k = 1}^{n} {\beta_{k} } \left( {\rho_{k} - \left\langle {e_{k} e_{k}^{T} ,\;\int {\xi \xi^{T} dF(\xi )} } \right\rangle } \right) + \int {1_{{\left( { - \infty ,0} \right]}} } \left( {y_{0} (x) + y(x)^{T} \xi } \right)dF(\xi ). $$

Base on the above Lagrangian function, we construct the following optimization problem

$$ \mathop {\hbox{min} }\limits_{dF(\xi ) \ge 0} \mathop {\hbox{max} }\limits_{\theta ,\alpha ,\beta } \;L(\xi ,\theta ,\alpha ,\beta ), $$

(25)

where θ,α,β are the lagrangian multipliers (θ is a real number; and α, β are n × 1 vectors).

According to Convex Analysis (Rockafellar 1970), we can easily know that the above problem (25) is equivalent to the problem (24a)–(24e). Indeed, problem (25) implies that constraints (24b)–(24d) must be satisfied. According to the dual theory from Convex Analysis (Rockafellar 1970), the dual problem for the optimization problem (25) can be written as the following form:

$$ \begin{aligned}&\mathop {\hbox{max} }\limits_{\theta ,\alpha ,\beta } \;\mathop {\hbox{min} }\limits_{dF(\xi ) \ge 0} \left\{ {\theta \;\left( {1 - \int {dF(\xi )} } \right) + \alpha^{T} \left( {\mu - \int {\xi dF(\xi )} } \right) + \sum\limits_{k = 1}^{n} {\beta_{k} } \left( {\rho_{k} - \left\langle {e_{k} e_{k}^{T} ,\;\int {\xi \xi^{T} dF(\xi )} } \right\rangle } \right) + } \right.\\ &\qquad \left. {\int {1_{{\left( { - \infty ,0} \right]}} } \left( {y_{0} (x) + y(x)^{T} \xi } \right)dF(\xi )} \right\}.\end{aligned} $$

(26)

Furthermore, the optimization problem (26) can be equivalently converted as

$$ \mathop {\hbox{max} }\limits_{\theta ,\alpha ,\beta } \,\,\,\,\theta + \alpha^{T} \mu + \beta^{T} \rho + \mathop {\hbox{min} }\limits_{dF(\xi ) \ge 0} \int { - (\theta + \alpha^{T} \xi + \xi^{T} Diag(\beta )\xi - 1_{{\left( { - \infty ,0} \right]}} (y_{0} (x) + y(x)^{T} \xi ))dF(\xi )} , $$

(27)

where Diag(β) denotes diagonal matrix with the vector of β being the diagonal elements. The optimization problem (27) can further be transformed to the following optimization problem:

$$ \mathop {\hbox{max} }\limits_{\theta ,\alpha ,\beta } \quad \theta + \alpha^{T} \mu + \beta^{T} \rho $$

(28)

$$ {\text{s}} . {\text{t}} .\quad {\text{I}}_{{\left( { - \infty ,0} \right]}} \left( {y_{0} (x) + y(x)^{T} \xi } \right) \ge \theta + \alpha^{T} \xi + \xi^{T} Diag(\beta )\xi , \, \;\forall \xi \in R^{n} . $$

(29)

Indeed, the constraint (29) can be equivalently transformed into the following forms:

$$ \begin{array}{*{20}c} { - \left( {\theta + \alpha^{T} \xi + \xi^{T} Diag(\beta )\xi - 1_{{\left( { - \infty ,0} \right]}} \left( {y_{0} (x) + y(x)^{T} \xi } \right)} \right) \ge 0,} & {\forall \xi \in R^{n} } \\ \end{array} $$

and

$$ - \left( {\theta + \alpha^{T} \xi + \xi^{T} Diag(\beta )\xi - 1_{{\left( { - \infty ,0} \right]}} \left( {y_{0} (x) + y(x)^{T} \xi } \right)} \right)dF(\xi ) \ge 0, $$

for any dF(ξ) ≥ 0,and hence the minimum of the inner problem of the above problem (27) is zero. Therefore, the objective function of the problem (27) becomes θ + α^Tμ + β^Tρ in this case.

Note that the semi-infinite constraint (29) can be easily transformed into the two equivalent semi-infinite constraints as follows:

$$ \theta + \alpha^{T} \xi + \xi^{T} Diag(\beta )\xi \le 1,{\kern 1pt} {\kern 1pt} \forall \xi \in R^{n} , $$

(30)

$$ \theta + \alpha^{T} \xi + \xi^{T} Diag(\beta )\xi \le 0\quad {\text{for}}\quad \xi :y_{0} (x) + y(x)^{T} \xi > 0. $$

(31)

Further, we can equivalently transform the inequality (30) into:

$$ \left[ {\begin{array}{*{20}c} {\theta - 1} & {\frac{1}{2}\alpha^{T} } \\ {\frac{1}{2}\alpha } & {Diag(\beta )} \\ \end{array} } \right] \prec 0, $$

(32)

where“≺0” stands for negative semi-definite, and equivalently transform the inequality (31) into:

$$ \mathop {\hbox{max} }\limits_{\xi } \left\{ {\theta + \alpha^{T} \xi + \xi^{T} Diag(\beta )\xi :y_{0} (x) + y(x)^{T} \xi > 0} \right\}\; \le 0. $$

(33)

The above constraint (32) implies that Diag(β) is a negative semi-definite matrix, hence the left-hand-side of the inequality (33) is a concave quadratic maximization problem. Furthermore, by introducing Lagrange multiplier λ and the dual problem of the maximization problem of the left hand side of the inequality (33), the inequality (33) has the following equivalent form

$$ \mathop {\hbox{min} }\limits_{\lambda \ge 0} \;\mathop {\hbox{max} }\limits_{\xi } \left\{ {\theta + \alpha^{T} \xi + \xi^{T} Diag(\beta )\xi + \lambda \left( {y_{0} (x) + y(x)^{T} \xi } \right)} \right\} \le 0, $$

(34a)

or

$$ \mathop {\hbox{min} }\limits_{\lambda \ge 0} \;\mathop {\hbox{max} }\limits_{\xi } \left\{ {\theta + \lambda y_{0} (x) + \left( {\alpha + \lambda y(x)} \right)^{T} \xi + \xi^{T} Diag(\beta )\xi } \right\} \le 0. $$

(34b)

The inequality (34b) implies that there exists some λ ≥ 0 and for arbitrary ξ the following inequality holds

$$ \theta + \lambda y_{0} (x) + \left( {\alpha + \lambda y(x)} \right)^{T} \xi + \xi^{T} Diag(\beta )\xi \le 0. $$

(35)

Furthermore, inequality (35) is equivalent to:

$$ \left[ {\begin{array}{*{20}c} {\theta + \lambda y_{0} (x)} & {\frac{1}{2}(\alpha + \lambda y(x))^{T} } \\ {\frac{1}{2}(\alpha + \lambda y(x))} & {Diag(\beta )} \\ \end{array} } \right] \prec 0, $$

(36)

for some λ ≥ 0. In (36), “≺0” stands for negative semi-definite. Through the above interpretation, the dual problem (26) or the problem (28)–(29) is equivalent to the following problem:

$$ \mathop {\hbox{max} }\limits_{{\theta \in R,\alpha \in R^{n} ,\beta \in R^{n} ,\lambda \ge 0}} \theta + \alpha^{T} \mu + \beta^{T} \rho $$

(37a)

$$ {\text{s}} . {\text{t}} .\quad \left[ {\begin{array}{*{20}c} {\theta - 1} & {\frac{1}{2}\alpha^{T} } \\ {\frac{1}{2}\alpha } & {Diag(\beta )} \\ \end{array} } \right] \prec 0 $$

(37b)

$$ \left[ {\begin{array}{*{20}c} {\theta + \lambda y_{0} (x)} & {\frac{1}{2}(\alpha + \lambda y(x))^{T} } \\ {\frac{1}{2}(\alpha + \lambda y(x))} & {Diag(\beta )} \\ \end{array} } \right] \prec 0. $$

(37c)

The remaining job in order to complete the proof of Proposition 1, is to prove that the primal problem (24a)–(24e) or (25) and its dual problem (26) is equivalent. We set φ = I_(-∞,0](y₀(x) + y(x)^Tξ), ψ₁(·) = 1, b₁ = 1, ψ₂(·) = ξ, b₂ = μ, ψ_i(·) = ξ ²_i-2 , b_i= ρ_i-2 (i = 3,…,n + 2). We can easily prove that functions ψ₂,.., ψ_n+2 are continuous and φ is upper semicontinuous. Further, we can know that the primal problem is consistent. According to the Corollary 3.1 of Shapiro (2001), we can prove that there is no gaps between the primal optimization problem (24a)–(24e) and it’s dual optimization problem (26). Based on the above analysis, we can prove that the chance constraint (17) with the uncertain set (16) is equivalent to

$$ \mathop {\hbox{max} }\limits_{\theta ,\alpha ,\beta ,\lambda \ge 0} \left\{ {\theta + \alpha^{T} \mu + \beta^{T} \rho :\left[ {\begin{array}{*{20}c} {\theta - 1} & {\frac{1}{2}\alpha^{T} } \\ {\frac{1}{2}\alpha } & {Diag(\beta )} \\ \end{array} } \right] \prec 0,\left[ {\begin{array}{*{20}c} {\theta + \lambda y_{0} (x)} & {\frac{1}{2}(\alpha + \lambda y(x))^{T} } \\ {\frac{1}{2}(\alpha + \lambda y(x))} & {Diag(\beta )} \\ \end{array} } \right] \prec 0} \right\} \ge 1 - \varepsilon . $$

(38)

Further, by using a simple analysis, the inequality (38) can be equivalently rewritten as follows: there exist θ∈R, α∈Rⁿ,β∈Rⁿ and λ ≥ 0 such that θ + α^Tμ+β^Tρ ≥ 1 − ε and constraints (37b)–(37c) are satisfied. Thus, the proof of Proposition 1 is completed.

Appendix B: The Proof of Theorem 1

Let θ^* = θ/λ, α^* = α/λ, β^* = β/λ and λ^* = 1/λ, (19a)–(19c) in proposition 1 are equivalently transformed to (21b)–(21d) in theorem 1 for the case of λ > 0 and x, θ^*, α^*,β^* and λ^*(> 0) become decision variables. In fact, the constraint (34a) or (34b) implies that there exists some λ > 0 such that

$$ \mathop {\hbox{max} }\limits_{\xi } \left\{ {\theta + \lambda y_{0} (x) + \left( {\alpha + \lambda y(x)} \right)^{T} \xi + \xi^{T} Diag(\beta )\xi } \right\} \le 0, $$

otherwise for any λ > 0,

$$ \mathop {\hbox{max} }\limits_{\xi } \left\{ {\theta + \lambda y_{0} (x) + \left( {\alpha + \lambda y(x)} \right)^{T} \xi + \xi^{T} Diag(\beta )\xi } \right\} > 0. $$

(39)

If (39) holds for any λ > 0, then the minimization of the problem described as the left-hand-side of inequality (34a) can be obtained only when λ = 0; thus inequality (34) is equivalent to

$$ \mathop {\hbox{max} }\limits_{\xi } \left\{ {\theta + \alpha^{T} \xi + \xi^{T} Diag(\beta )\xi } \right\} \le 0, $$

i.e.,

$$ \left[ {\begin{array}{*{20}c} \theta &\quad {\frac{1}{2}\alpha^{T} } \\ {\frac{1}{2}\alpha } &\quad {Diag(\beta )} \\ \end{array} } \right] \prec 0, $$

and the equivalent form (38) of CAR constraint (17) or (15) becomes

$$ \mathop {\hbox{max} }\limits_{\theta ,\alpha ,\beta } \left\{ {\theta + \alpha^{T} \mu + \beta^{T} \rho :\left[ {\begin{array}{*{20}c} \theta &\quad {\frac{1}{2}\alpha^{T} } \\ {\frac{1}{2}\alpha } &\quad {Diag(\beta )} \\ \end{array} } \right] \prec 0} \right\} \ge 1 - \varepsilon , $$

which implies CAR constraint has relation with lagrangian multipliers but no any relation with investment proportions x i.e., there is virtually no restriction on the optimal investment proportions of CAR constraint. Therefore, the constraint (19c) in Proposition 1 for some λ > 0 is required from practical perspective and it is equivalent to the constraint (21d). Besides, the equivalence of (19a) and (21b), (19b) and (21c) is obvious, if let θ^* = θ/λ, α^* = α/λ, β^* = β/λ and λ^* = 1/λ. Finally, combining the objective function (14), the transformed CAR constraints (21b)–(21d) and other constraints (20), the asset allocation optimization model is reformulated as tractably linear SDP problem as (21a)–(21g). The proof of Theorem 1 is completed.