Separation results for multi-product inventory hedging problems

Abstract

We analyze financial hedging tools for inventory management in a risk-averse corporation. We consider the problem of optimizing simultaneously over both the operational policy and the hedging policy of the corporation in a multi-product model. Our main contribution is a separation result such that for a corporation with multiple products and inventory departments, the inventory decisions of each department can be made independently of the other departments’ decisions. That is, no interaction needs to be considered among different products.

Appendix

All proofs are presented in this appendix.

Proof of Theorem 1

The proof follows closely the proof of Proposition 6.6.5 in Pham (2009). Recall that

$$\begin{aligned} A^{\gamma}(\lambda) =&\inf_{\theta\in\varTheta}E \bigl[ \bigl(\omega _T^{\gamma,\theta}-\lambda\bigr)^2 \bigr],\quad \lambda\in\mathbb{R}, \\ B^{\gamma}(m) =&\inf_{\theta\in\varTheta} \bigl\{ Var \bigl[\omega_T^{\gamma,\theta} \bigr] | E \bigl[\omega_T^{\gamma,\theta} \bigr]=m \bigr\} ,\quad m\in\mathbb{R}, \end{aligned}$$

with $\omega_{T}^{\gamma,\theta} = \omega_{T}$ given in (2). We shall first prove that

$$\begin{aligned} A^{\gamma}(\lambda) =&\inf_{m} \bigl[B^{\gamma }(m)+(m-\lambda)^2 \bigr],\quad \lambda\in \mathbb{R}, \end{aligned}$$

(25)

$$\begin{aligned} B^{\gamma}(m) =&\sup_{\lambda} \bigl[A^{\gamma }(\lambda )-(m-\lambda)^2 \bigr],\quad m\in \mathbb{R}. \end{aligned}$$

(26)

Take any $\lambda\in\mathbb{R}$. We start by noting that

$$\begin{aligned} E\bigl[\bigl(\omega_T^{\lambda,\theta}-\lambda\bigr)^2 \bigr]=Var\bigl[\omega_T^{\gamma ,\theta }\bigr]+\bigl(E\bigl[ \omega_T^{\gamma,\theta}\bigr]-\lambda\bigr)^2. \end{aligned}$$

Let $m \in\mathbb{R}$. Suppose first that $\{ \theta\,|\, E[\omega _{T}^{\gamma,\theta}] = m \} \neq\emptyset$. By definition of B ^γ(m), for each ϵ>0 we can find θ ^ϵ∈Θ with controlled diffusion $\omega^{\gamma,\theta^{\epsilon}}$, such that $E[\omega^{\gamma,\theta^{\epsilon}}_{T}]=m$ and $Var[\omega^{\gamma,\theta^{\epsilon}}_{T}]\leq B^{\gamma }(m)+\epsilon$, that is

$$\begin{aligned} E\bigl[\bigl(\omega^{\gamma,\theta^{\epsilon}}_T-\lambda\bigr)^2 \bigr]\leq B^{\gamma}(m)+(m-\lambda)^2+\epsilon \end{aligned}$$

and hence

$$\begin{aligned} A^{\gamma}(\lambda)=\inf_{\theta\in\varTheta}E \bigl[\bigl(\omega ^{\gamma ,\theta}_T-\lambda\bigr)^2\bigr] \leq B^{\gamma}(m)+(m-\lambda)^2. \end{aligned}$$

(27)

Next, if $\{ \theta\,|\, E[\omega_{T}^{\gamma,\theta}] = m \} = \emptyset$, then B ^γ(m)=∞. So we have (27) for all $m \in\mathbb{R}$. On the other hand, for each ϵ>0 we can find θ ^ϵ∈Θ with controlled diffusion $\omega^{\gamma,\theta ^{\epsilon }}$, such that

$$\begin{aligned} A^{\gamma}(\lambda) + \epsilon\geq E\bigl[\bigl(\omega^{\gamma,\theta ^{\epsilon }}_T- \lambda\bigr)^2\bigr] = Var\bigl[\omega^{\gamma,\theta^{\epsilon}}_T\bigr] + \bigl(E\bigl[ \omega^{\gamma ,\theta ^{\epsilon}}_T\bigr]-\lambda\bigr)^2. \end{aligned}$$

Setting $m_{\lambda,\epsilon} = E[\omega^{\gamma,\theta^{\epsilon }}_{T}]$, it follows that

$$A^{\gamma}(\lambda) \geq B^{\gamma}(m_{\lambda,\epsilon}) + (m_{\lambda ,\epsilon} - \lambda)^2 - \epsilon. $$

Together with (27) we obtain (25).

Also, since X↦Var[X] is convex in X, the function B ^γ(m) is convex in m, and since

$$\begin{aligned} A^{\gamma}(\lambda) =&\inf_{m} \bigl[B^{\gamma }(m)+(m-\lambda )^2\bigr], \\ \frac{(\lambda^2-A^{\gamma}(\lambda))}{2} =&\sup_m \biggl[m\lambda- \frac{B^{\gamma}(m)+m^2}{2} \biggr], \end{aligned}$$

the function $\lambda\mapsto\frac{\lambda^{2}-A^{\gamma}(\lambda)}{2}$ is the Fenchel-Legendre transform of the convex function $m\mapsto\frac{(B^{\gamma}(m)+m^{2})}{2}$. We then have the duality relation

$$\begin{aligned} \frac{(B^{\gamma}(m)+m^2)}{2}=\sup_{\lambda} \biggl[m\lambda - \frac{(\lambda^2-A^{\gamma}(\lambda))}{2} \biggr], \end{aligned}$$

which yields (26).

Finally, take any $m\in\mathbb{R}$, and let $\lambda_{m}\in\mathbb {R}$ be an argument maximum of B ^γ(m) in (26). Then m is an argument minimum of A ^γ(λ _m ) in (25). Since

$$\begin{aligned} m\mapsto B^{\gamma}(m)+(m-\lambda_m)^2\quad \mbox{is strictly convex}, \end{aligned}$$

this argument minimum is unique. Moreover, suppose that $\hat{\theta}_{\lambda_{m}}\in\varTheta$ with controlled diffusion $\hat{\omega}_{T}^{\gamma,\theta,\lambda_{m}}$ is an optimal control for A ^γ(λ _m ). Set $m_{\lambda _{m}}=E[\hat {\omega}_{T}^{\gamma,\theta,{\lambda_{m}}}]$. We obtain

$$\begin{aligned} A^{\gamma}({\lambda_m}) =&Var\bigl[\hat{\omega}_T^{\gamma,\theta ,{\lambda _m}} \bigr]+(m_{{\lambda_m}}-{\lambda_m})^2 \\ \geq&B^{\gamma}(m_{{\lambda_m}})+(m_{{\lambda_m}}-{ \lambda_m})^2. \end{aligned}$$

Therefore $m=m_{\lambda_{m}}=E[\hat{\omega}_{T}^{\gamma,\theta,\lambda_{m}}]$. Hence

$$\begin{aligned} B^{\gamma}(m)=A^{\gamma}(\lambda_m)-(m- \lambda_m)^2 =&E\bigl[\bigl(\hat{\omega}_T^{\gamma,\theta,\lambda_m} - \lambda _m\bigr)^2\bigr]- \bigl(E\bigl[\hat{ \omega}_T^{\gamma,\theta,\lambda_m}\bigr]-\lambda_m \bigr)^2\\ =&Var\bigl[\hat{\omega}_T^{\gamma,\theta,\lambda_m}\bigr] \end{aligned}$$

and so $\hat{\theta}_{\lambda_{m}}$ is a solution to B ^γ(m). □

For the proof of Theorem 3, we need the following result which provides the explicit expression for the optimal value in (13) by solving for the optimal hedging strategy. This is a straightforward extension of Theorem 2 in Caldentey and Haugh (2006) to the multi-product case.

Proposition 1

Let $K_{t}=\int_{0}^{t}\eta_{s}^{2}ds$. The value

$$\begin{aligned} A^{\gamma}(\lambda)=E\bigl[\bigl(V_T^{\gamma}+G_T^*+ \omega_0-\lambda\bigr)^2\bigr] \end{aligned}$$

is given by

$$\begin{aligned} A^{\gamma}(\lambda) =&e^{-K_T} \Biggl( \bigl(\omega_0+V_0^{\gamma}- \lambda \bigr)^2+\int_0^Te^{K_t} \sum_{j=1}^N E \bigl[\bigl( \delta_t^{\gamma_j}\bigr)^2 \bigr]dt \Biggr). \end{aligned}$$

The proof is the same as for Theorem 2 in Caldentey and Haugh (2006), and it is formulated for the multi-product case in Sun (2011). It can be seen from Proposition 1 that, given the Föllmer-Schweizer decomposition, the value function in (13) can be determined without explicit reference to the optimal hedging strategy.

Proof of Theorem 3

Let us first suppose K _T >0. By Theorem 1 we have

$$\begin{aligned} B^{\gamma}(m) &=\inf_{\theta\in\varTheta} \bigl\{ Var[ \omega _T]\bigm| E[\omega_T]=m \bigr\} =\sup_{\lambda}\bigl(A^{\gamma}(\lambda)-(m- \lambda)^2\bigr). \end{aligned}$$

The maximum in the last line is achieved for

$$\begin{aligned} 0=\frac{\partial}{\partial\lambda}A^{\gamma}(\lambda)+2(m-\lambda) \end{aligned}$$

where

$$\begin{aligned} A^{\gamma}(\lambda)=e^{-K_T} \Biggl(\bigl(\omega_0+V_0^{\gamma}- \lambda \bigr)^2+\int_0^Te^{K_t} \sum_{j=1}^nE\bigl[\bigl( \delta_t^{\gamma_j}\bigr)^2\bigr]dt \Biggr) \end{aligned}$$

by Proposition 1. So the optimizer λ=λ _m fulfills

$$\begin{aligned} &-2e^{-K_T}\bigl(\omega_0+V_0^{\gamma}- \lambda\bigr)+2(m-\lambda)=0 \\ &\quad{}\Leftrightarrow\quad\bigl(1-e^{-K_T}\bigr)\lambda=m-e^{-K_T}\bigl( \omega_0+V_0^{\gamma }\bigr) \\ &\quad{}\Leftrightarrow\quad\lambda=\frac{m-e^{-K_T}(\omega_0+V_0^{\gamma })}{1-e^{-K_T}}. \end{aligned}$$

Plugging this equation into (12), we obtain

$$\begin{aligned} B^{{\gamma}}(m)=A^{\gamma}(\lambda)-(m-\lambda)^2 =&e^{-K_T} \biggl(\frac{\omega_0+V_0^{\gamma}-m}{1-e^{-K_T}} \biggr)^2- \biggl( \frac{e^{-K_T}(\omega_0+V_0^{\gamma }-m)}{1-e^{-K_T}} \biggr)^2 \\ &+e^{-K_T}\int_0^Te^{K_t} \sum_{j=1}^n E\bigl[\bigl( \delta_t^{\gamma_j}\bigr)^2\bigr]dt. \end{aligned}$$

This proves (20) and (21). To solve the problem

$$\begin{aligned} U^{\gamma}=\sup_m\bigl(m-\kappa B^{\gamma}(m)\bigr), \end{aligned}$$

note that the first order condition for m=m _opt is

$$\begin{aligned} & 1-\kappa\frac{\partial}{\partial m}B^{\gamma}(m)=0 \\ &\quad{}\Leftrightarrow\quad 2\bigl(m-\omega_0-V_0^{\gamma} \bigr)\frac {e^{-K_T}}{1-e^{-K_T}}=\frac{1}{\kappa} \\ &\quad{}\Leftrightarrow\quad m=\frac{1}{2\kappa}\frac {e^{-K_T}}{1-e^{-K_T}}+\omega _0+V_0^{\gamma} \\ &\quad{}\Leftrightarrow\quad \bigl(m-\omega_0-V_0^{\gamma} \bigr)^2=\frac{1}{4\kappa ^2} \biggl(\frac{e^{-K_T}}{1-e^{-K_T}} \biggr)^2. \end{aligned}$$

Plugging the last two equations and (21) into (9), we obtain

$$\begin{aligned} U^{\gamma} =&m_{opt}-\kappa B^{\gamma}(m_{opt}) \\ =&\frac{1}{2\kappa}\frac{e^{-K_T}}{1-e^{-K_T}}+\omega _0+V_0^{\gamma }- \frac{1}{4\kappa}\frac{e^{-K_T}}{1-e^{-K_T}}-\kappa e^{-K_T}\int _0^Te^{K_t}\sum _{j=1}^NE\bigl[\bigl(\delta_t^{\gamma_j} \bigr)^2\bigr]dt \\ =&\omega_0+\sum_{j=1}^N V_0^{\gamma_j}+\frac{1}{4\kappa }\bigl(e^{K_T}-1\bigr)- \kappa e^{-K_T}\int_0^Te^{K_t} \sum_{j=1}^NE\bigl[\bigl( \delta_t^{\gamma_j}\bigr)^2\bigr]dt. \end{aligned}$$

This finishes the proof of (22) and (23).

Next suppose that K _T =0. Then $E[\omega_{T}] = \omega_{0} + V_{0}^{\gamma}$. If $m > \omega_{0} + V_{0}^{\gamma}$, the supremum in (12) is achieved for λ→∞, and if $m < \omega_{0} + V_{0}^{\gamma}$, the supremum in (12) is achieved for λ→−∞, and in these cases we have B ^γ(m)=∞. If $m = \omega _{0} + V_{0}^{\gamma}$, we have by Proposition 1

$$B^{\gamma}(m) = A^{\gamma}(\lambda) - (m-\lambda) = \int _0^Te^{K_t}\sum _{j=1}^nE\bigl[\bigl(\delta_t^{\gamma_j} \bigr)^2\bigr]dt $$

for all $\lambda\in\mathbb{R}$. Hence we have (20) and (21). From (9) we also obtain $m_{opt} = \omega_{0} + V_{0}^{\gamma}$, which yields (22) and (23). □

Proof of Theorem 4

For simplicity, we suppress the dependence on the index j in the proof. The problem we consider is

$$\begin{aligned} \max_{\gamma} \biggl(V_0^{\gamma}-\kappa e^{-K_T}\int _0^Te^{K_t}E\bigl[\bigl( \delta_t^{\gamma}\bigr)^2\bigr]dt \biggr) \end{aligned}$$

(28)

where $V_{0}^{\gamma}$ and $\delta_{s}^{\gamma}$ are given in (18) and (19). First, note that it follows directly from (19) and the boundedness of Δ^BS that $e^{-K_{T}}\int_{0}^{T}e^{K_{t}}E[(\delta_{t}^{\gamma})^{2}]dt$ is a bounded function in γ∈[0,∞). Next, we compute

$$\begin{aligned} \frac{\partial V_0^{\gamma}}{\partial \gamma} =& s-p + (R+q-s)\varPhi \biggl(\frac{\log\frac{\hat{D}_0}{\gamma } + \frac{1}{2}Y}{\sqrt{Y}} \biggr) \\ \frac{\partial^2 V_0^{\gamma}}{\partial \gamma^2} =& -(R+q-s) \frac{1}{\gamma\sqrt{Y}} \phi \biggl(\frac {\log\frac {\hat{D}_0}{\gamma} + \frac{1}{2}Y}{\sqrt{Y}} \biggr) <0 \end{aligned}$$

where ϕ(⋅) and Φ(⋅) are the pdf and cdf of the standard normal distribution, and $Y = c^{2} T + b^{2} \int_{0}^{T} \sigma_{t}^{2} dt$. Hence $V_{0}^{\gamma}$ is a strictly concave function in γ which satisfies

$$\begin{aligned} \lim_{\gamma\rightarrow+\infty}\frac{\partial V_0^{\gamma }}{\partial \gamma}=s-p<0,\qquad \lim_{\gamma\rightarrow0}\frac{\partial V_0^{\gamma}}{\partial \gamma }=R-p+q>0. \end{aligned}$$

The last inequality together with the boundedness of $e^{-K_{T}}\int_{0}^{T}e^{K_{t}}E[(\delta_{t}^{\gamma})^{2}]dt$ implies that there exists a solution to (28). Moreover, if κ is sufficiently small, then $| \frac{\partial^{2}}{\partial\gamma^{2}} ( \kappa e^{-K_{T}} \int_{0}^{T}e^{K_{t}}E[(\delta_{t}^{\gamma})^{2}]dt ) |$ is sufficiently small for γ inside a suitable compact interval. Then the objective function in (28) is strictly concave inside this interval, strictly increasing at the left end point, and strictly decreasing at the right end point. It then easily follows that the objective function has a unique maximizer. □

Remark

The objective function in (28) is in general not concave in γ on [0,∞).