Duration gap with multiple liabilities for nonparallel shifts

Abstract

An important problem in finance is how to invest a portfolio of bonds in order to fund an exogenous stream of multiple liabilities. We prove that minimizing a risk measure, which generalizes Fong and Vasicek’s M-squared, subject to a cash flow dispersion condition, ensures that an asset portfolio dedicated to an exogenous stream of multiple liabilities is fully protected against a parallel shift and has the best worst-case sensitivity to a nonparallel interest rate shift. Further, we show how the new risk measure can be used as part of a risk minimizing strategy for multiple-liability cash flows. The resulting portfolio is fully protected against a parallel interest rate shock of any magnitude and has minimum sensitivity to the worst-case interest rate shock.

Appendices

Appendix A: Proof of Lemma 1

Assume $V=0$ and $G=0$. Integrating by parts the left-hand side of Eq. (13) and making use of $\beta \left( 0\right) =0$ and $\beta (T)=V=0$, we obtain:

$$\begin{aligned} \int _{0}^{T}u(t)d\beta (t)=-\int _{0}^{T}u^{\prime }(t)\beta (t)dt. \end{aligned}$$

A second integration by parts yields:

$$\begin{aligned} \int _{0}^{T}u(t)d\beta (t)=-u^{\prime }(T)\int _{0}^{T}\beta (t)dt+ \int _{0}^{T}u^{\prime \prime }(t)\left[ \int _{0}^{t}\beta (s)ds\right] dt. \end{aligned}$$

(18)

Integrating by parts the first integral on the right-hand side of Eq. (18) and making use of the condition that $V=\beta (T)=0$, we obtain:

$$\begin{aligned} \int _{0}^{T}\beta (t)dt=-\int _{0}^{T}td\beta (t)=-G \end{aligned}$$

(19)

Since G equals zero, the first term in Eq. (18) is zero:

$$\begin{aligned} \int _{0}^{T}u(t)d\beta (t)=\int _{0}^{T}u^{\prime \prime }(t)\int _{0}^{t} \beta (s)dsdt. \end{aligned}$$

Appendix B: DDC and immunization for a single-liability portfolio

Assume a single liability is due at date q. We want to show the the DDC is satisfied Eq. (12) when $A=L$ and $G=0$:

$$\begin{aligned} \int _{0}^{t}F^{+}(s)ds\ge \int _{0}^{t}F^{-}(s)ds \text { for all }t\in (0,T]. \end{aligned}$$

Note that by the properties of a distribution function $F^{+}(T)=F^{-}(T)=1$ and $F^{+}(s),F^{-}(s)\le 1$. By the assumption that $G=0$, a single liability is due at date $q=\int _{0}^{T}sdF^{+}$. Then $F^{-}(s)=U(t-q),$ where U is a unit step function. Clearly, for $0<t<q$, $\int _{0}^{t}F^{+}(s)ds>\int _{0}^{t}F^{-}(s)=0$. So examine the case where $q\le t\le T$:

$$\begin{aligned} \int _{0}^{t}F^{+}(s)ds-\int _{0}^{t}F^{-}(s)ds&=tF^{+}\left( t\right) -\int _{0}^{t}sdF^{+}(s)-\left( t-q\right) \\&=t(F^{+}\left( t\right) -1)-\int _{0}^{t}sdF^{+}(s)+q \\&=t\left( F^{+}\left( t\right) -1\right) -\int _{0}^{t}sdF^{+}(s)+\int _{0}^{T}sdF^{+}(s) \\&=t\left( F^{+}\left( t\right) -1\right) +\int _{t}^{T}sdF^{+}(s) \\&=t\left( F^{+}\left( t\right) -1\right) +TF^{+}(T)-tF^{+}(t)-\int _{t}^{T}F^{+}(s)ds \\&=T-t-\int _{t}^{T}F^{+}(s)ds\ge T-t-\int _{t}^{T}ds=0 \end{aligned}$$

The inequality holds because $F^{+}(s)\le 1$.

Appendix C: Proof of Theorem 2

Assume $V=0$, $G=0$, $DDC$, and forward rate restriction Eq. (14) all hold. Let $u(t)=H(t)$. Then by Lemma 1 with $u^{\prime }(t)=h(t)$ and $u^{\prime \prime }(t)=h^{\prime }(t)$, the interest rate sensitivity $D(h)=-\int _{0}^{T}H(t)d\beta (t)$ from Eq. (5) can be expressed as

$$\begin{aligned} D(h)&=-\int _{0}^{T}h^{\prime }(t)\left( \int _{0}^{t}\beta (s)ds\right) dt \end{aligned}$$

(20)

$$\begin{aligned}&=-\int _{0}^{T}h^{\prime }(t)\left( \int _{0}^{t}\beta (s)ds\right) dt. \end{aligned}$$

(21)

If both the DDC and forward rate restriction Eq. (14) hold, the worst-case value of Eq. (20) occurs when $h^{\prime }(t)=K$:

$$\begin{aligned} W=-K\int _{0}^{T}\int _{0}^{t}\beta (s)dsdt\le 0. \end{aligned}$$

(22)

Integrating by parts, results in:

$$\begin{aligned} \int _{0}^{T}\int _{0}^{t}\beta (s)dsdt&=T\int _{0}^{T}\beta (s)ds-\int _{0}^{T}t\beta (t)dt \\&=T\int _{0}^{T}\beta (t)dt-\int _{0}^{T}t\beta (t)dt \\&=\int _{0}^{T}(T-t)\beta (t)\,dt. \end{aligned}$$

Again, integrating parts and given that $V=\beta \left( T\right) =0$, results in:

$$\begin{aligned} \int _{0}^{T}(T-t)\beta (t)dt&=\frac{1}{2}T^{2}\beta (T)+\int _{0}^{T}\left( \frac{1}{2}t^{2}-Tt\right) d\beta (t) \\&=\int _{0}^{T}\left( \frac{1}{2}t^{2}-Tt\right) d\beta (t). \end{aligned}$$

Setting the money duration gap G from Eq. (9) to zero, results in

$$\begin{aligned} \int _{0}^{T}\int _{0}^{t}\beta (s)dsdt=\frac{1}{2}\int _{0}^{T}t^{2}d\beta (t)\ge 0\text {.} \end{aligned}$$