Hedging under multiple risk constraints

Abstract

Motivated by the asset–liability management problems under shortfall risk constraints, we consider in a general discrete-time framework the problem of finding the least expensive portfolio whose shortfalls with respect to a given set of stochastic benchmarks are bounded by a specific shortfall risk measure. We first show how the price of this portfolio may be computed recursively by dynamic programming for different shortfall risk measures, in complete and incomplete markets. We then focus on the specific situation where the shortfall risk constraints are imposed at each period on the next-period shortfalls, and obtain explicit results. Finally, we apply our results to a realistic asset–liability management problem of an energy company, and show how the shortfall risk constraints affect the optimal hedging of liabilities.

Appendices

Appendix A: Technical lemmas

Lemma A.1

Let $(\varOmega,\mathcal {F}, \mathbb{P})$ be a probability space, $X,Y \in \mathcal {F}$ and $\alpha,\beta\geq 0$. Then

$$\begin{aligned} V_{0} &= \inf_{m\in\mathcal {F}}\big\{ \mathbb{E}[m]: \mathbb {E}[(X-m)^{+}]\leq \alpha,\mathbb{E}[(Y-m)^{+}]\leq\beta\big\} \\ &= \max\big\{ \mathbb {E}[X-\alpha], \mathbb{E}[Y-\beta], \mathbb{E}[X\vee Y - \alpha- \beta]\big\} \end{aligned}$$

and

$$\begin{aligned} V'_{0} &= \inf_{m\in\mathcal {F}}\big\{ \mathbb{E}[m]: \mathbb{E}[\max\{(X-m)^{+}- \alpha,(Y-m)^{+}- \beta\}]\leq0 \big\} \\ &= \mathbb{E}[(X-\alpha)\vee(Y - \beta)]. \end{aligned}$$

Proof

We begin with $V_{0}$. By considering only one of the two constraints, we have clearly

$$V_{0} \geq\max\{\mathbb{E}[X-\alpha], \mathbb{E}[Y-\beta]\}. $$

In addition,

$$(X-m)^{+} + (Y-m)^{+} \geq X\vee Y - m $$

and so

$$V_{0} \geq\mathbb{E}[X\vee Y - \alpha- \beta]. $$

To finish the proof, we need to find $m^{*}$ which satisfies the constraint and realizes the equality.

Assume first that

$$\mathbb{E}[X\vee Y - \alpha- \beta] \geq\mathbb{E}[X-\alpha]\quad\text{and}\quad\mathbb{E}[X\vee Y - \alpha- \beta] \geq\mathbb{E}[Y-\beta]. $$

This is equivalent to

$$\mathbb{E}[(X-Y)^{+}]\geq\alpha\quad\text{and}\quad\mathbb {E}[(Y-X)^{+}]\geq \beta. $$

In this case, we can take

$$m^{*} = X \vee Y - \alpha\frac{(X-Y)^{+}}{\mathbb{E}[(X-Y)^{+}]} - \beta \frac{(Y-X)^{+}}{\mathbb{E}[(Y-X)^{+}]}. $$

If

$$\mathbb{E}[X-\alpha]\geq\mathbb{E}[X\vee Y - \alpha- \beta] \quad\text{and}\quad\mathbb{E}[X-\alpha] \geq\mathbb {E}[Y-\beta], $$

we take

$$m^{*} = X\wedge Y-\alpha+ \mathbb{E}[(X-Y)^{+}]. $$

If finally

$$\mathbb{E}[Y-\beta]\geq\mathbb{E}[X\vee Y - \alpha- \beta] \quad\text{and}\quad\mathbb{E}[Y-\beta] \geq\mathbb {E}[X-\alpha], $$

then we can take

$$m^{*} = X\wedge Y-\beta+ \mathbb{E}[(Y-X)^{+}]. $$

Now we turn to $V_{0}^{\prime}$. Since $m = (X-\alpha)\vee (Y-\beta)$ satisfies the constraint, it holds that $V'_{0} \leq\mathbb {E}[(X-\alpha)\vee (Y-\beta)]$. On the other hand,

$$\begin{aligned} V'_{0} &= \inf_{m\in\mathcal {F}}\big\{ \mathbb{E}[m]: \mathbb{E}[\max\{(X-m)^{+}- \alpha,(Y-m)^{+}- \beta\}]\leq0 \big\} \\ &\geq\inf_{m\in\mathcal {F}}\big\{ \mathbb{E}[m]: \mathbb{E}[\max\{X-m- \alpha,Y-m- \beta\}]\leq0 \big\} \\ &= \inf_{m\in\mathcal {F}}\big\{ \mathbb{E}[m]: \mathbb{E}[\max\{X- \alpha,Y- \beta\}]\leq m \big\} \\ & = \mathbb{E}[(X-\alpha)\vee (Y-\beta)]. \end{aligned}$$

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We now consider the case when $n$ is arbitrary, but the objectives are ordered.

Lemma A.2

Let $(\varOmega,\mathcal {F}, \mathbb{P})$ be a probability space, and let $Z_{1},\dots,Z_{n} \in\mathcal {F}$ with $Z_{1}\leq Z_{2}\leq\cdots\leq Z_{n}$ a.s. and $\alpha_{1},\dots,\alpha_{n} \geq 0$. Then

$$\inf_{M\in\mathcal {F}}\big\{ \mathbb{E}[M]: \mathbb{E}[(Z_{k}-M)^{+}]\leq \alpha_{k},k=1,\dots,n\big\} = \max_{k=1,\dots,n} \{\mathbb{E}[Z_{k}] - \alpha_{k}\}. $$

Proof

1) Assume that there exists $k\in\{1,\dots, n\}$ such that

$$\begin{aligned} \mathbb{E}[Z_{k}] - \alpha_{k} \geq \mathbb{E}[Z_{k+1}] - \alpha _{k+1}. \end{aligned}$$

(A.1)

Then the constraint $\mathbb{E}[(Z_{k} - m)^{+}] \leq\alpha_{k}$ implies $\mathbb{E}[(Z_{k+1} - m)^{+}] \leq\alpha_{k+1}$ since

$$\mathbb{E}[(Z_{k+1} - m)^{+}] \leq\mathbb{E}[(Z_{k} - m)^{+}] + \mathbb {E}[Z_{k+1}-Z_{k}] \leq\alpha_{k} + \alpha_{k+1} - \alpha_{k} = \alpha_{k+1}. $$

We can then remove the constraint $\mathbb{E}[(Z_{k+1} - m)^{+}] \leq\alpha_{k+1}$ without modifying the value function. Repeating the same argument for all other indices $k$ satisfying (A.1), we can assume without loss of generality that

$$\begin{aligned} \mathbb{E}[Z_{1}] - \alpha_{1} < \cdots< \mathbb{E}[Z_{n}] - \alpha_{n}. \end{aligned}$$

(A.2)

We then need to prove that

$$\inf_{m\in\mathcal {F}}\big\{ \mathbb{E}[m]: \mathbb{E}[(Z_{k}-m)^{+}]\leq \alpha_{k},\,k=1,\dots,n\big\} = \mathbb{E}[Z_{n}] - \alpha_{n}. $$

2) Removing all the constraints except the last one, it is easy to see that

$$\inf_{m\in\mathcal {F}}\big\{ \mathbb{E}[m]: \mathbb{E}[(Z_{k}-m)^{+}]\leq \alpha_{k},\,k=1,\dots,n\big\} \geq\mathbb{E}[Z_{n}] - \alpha_{n}. $$

To finish the proof, it is then enough to find $m$ which satisfies the constraints and is such that $\mathbb{E}[m] = \mathbb{E}[Z_{n}] - \alpha_{n}$.

3) If $\mathbb{E}[Z_{n} - Z_{1}]\geq\alpha_{n}$, we let

$$k^{*} = \max\{i: \mathbb{E}[Z_{n} - Z_{i}] \geq\alpha_{n}\} $$

and

$$m:= w Z_{k^{*}} + (1-w) Z_{k^{*}+1},\quad w = \frac{\mathbb{E}[Z_{k^{*}+1}] - \mathbb{E}[Z_{n}] + \alpha_{n}}{\mathbb{E}[Z_{k^{*}+1}] - \mathbb{E}[Z_{k^{*}}]} \in[0,1]. $$

Otherwise, let $k^{*} = 0$ and

$$m := Z_{1} - C,\quad C = \alpha_{n} - \mathbb{E}[Z_{n}-Z_{1}]. $$

By construction, we have $\mathbb{E}[m] = \mathbb{E}[Z_{n}] - \alpha_{n}$ and since $m \leq Z_{n}$, it also holds that $\mathbb{E}[(Z_{n} - m)^{+}] = \alpha_{n}$. To check the other constraints, observe that if $k\leq k^{*}$, then $m\geq Z_{k}$ a.s. and so $\mathbb{E}[(Z_{k} - m)^{+}] = 0$. On the other hand, if $k>k^{*}$, then $m\leq Z_{k}$ a.s. and so by (A.2),

$$\begin{aligned} \mathbb{E}[(Z_{k}-m)^{+}] = \mathbb{E}[Z_{k}] - \mathbb{E}[m] = \mathbb{E}[Z_{k}] - \mathbb{E}[Z_{n}] + \alpha_{n} < \alpha_{k}. \end{aligned}$$

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Appendix B: Explicit computations in the Black–Scholes model

In this appendix, we specialize the formulas of Propositions 4.2 and 4.3 to the Black–Scholes market with a risk-free asset evolving with zero interest rate and a risky asset which follows the Black–Scholes model

$$\frac{dX_{t}}{X_{t}} = \mu\,dt + \sigma\,dW_{t}, $$

where the rate of return $\mu$ of the risky asset is assumed to be positive.

Complete market

We assume that the agent can trade dynamically between the benchmark dates, which means that the market is complete, and the unique risk-neutral measure has density

$$Z_{T} = e^{-\frac{\mu}{\sigma}W_{T} - \frac{\mu^{2}}{2\sigma^{2}}T}. $$

In this case,

$$D_{k} = \frac{\mu^{2}}{2p\sigma^{2}}(t_{k} - t_{k-1}). $$

To compute the value function in the case of the entropic penalty, introduce the functions

$$\text{BSPut}(S,K,v) = \mathbb{E}\big[\big(K-S e^{vZ-\frac{v^{2}}{2}}\big)^{+}\big],\quad \text{and}\quad\text{LBPut}(S,K) = \mathbb{E}[(K-S Z)^{+}], $$

where $Z\sim N(0,1)$ and LB stands for the Louis Bachelier model. Then we obtain for $1\leq k \leq n-1$ that $V^{\mathit{NP}}_{k-1} = V^{\mathit{NP}}_{k}$ if $V^{\mathit{NP}}_{k} \geq S_{k} - \rho_{k}$ and

$$\begin{aligned} V^{\mathit{NP}}_{k-1}&= V^{\mathit{NP}}_{k} \\ &\phantom{=:}+ {{\text{LBPut}\left( \frac{|\mu|}{\sigma p} \sqrt {t_{k}-t_{k-1}},\, S_{k} - V^{\mathit{NP}}_{k} - \frac{1}{p}\log\tilde{\lambda}_{k-1}-\frac{\mu^{2} (t_{k}-t_{k-1})}{2\sigma^{2} p} \right)}} \end{aligned}$$

otherwise, where $\tilde{\lambda}_{k-1}$ is the solution to

$$\text{BSPut}\left({\tilde{\lambda}_{k-1}}, e^{-p(V^{\mathit{NP}}_{k}-S_{k})},\frac{|\mu|}{\sigma}\sqrt{t_{k}-t_{k-1}}\right) = e^{-p(V^{\mathit{NP}}_{k}-S_{k})} - e^{p\rho_{k}}. $$

The optimal strategy at date $n-1$ consists of replicating the claim

$$S_{n} - \rho_{n} - \frac{\log U_{n}}{p}. $$

This is achieved by investing a constant amount of ${\mu}/({\sigma^{2} p})$ into the risky asset between $t_{n-1}$ and $t_{n}$. The optimal strategy at date $k-1< n-1$ consists of replicating the claim with payoff at date $k$ given by

$$V^{\mathit{NP}}_{k} + \bigg(S_{k} - V^{\mathit{NP}}_{k} - \frac{{\log\tilde{\lambda}_{k-1}}}{p} - \frac{\mu(\mu-\sigma^{2})}{2p\sigma^{2}}(t_{k} - t_{k-1}) + \frac{\mu}{\sigma^{2} p} \log\frac{X_{t_{k}}}{X_{t_{k-1}}}\bigg)^{+}. $$

For the power loss function with $p=2$, at the last date,

$$\begin{aligned} V^{\mathit{NP}}_{n-1} = S_{n} - \sigma_{n} \sqrt{\mathbb{E}^{\mathbb{P}}[U_{n}^{2}]}, \end{aligned}$$

and at previous dates,

$$\begin{aligned} V^{\mathit{NP}}_{k-1} = V^{\mathit{NP}}_{k} + \mathbb{E}^{\mathbb{P}}[U_{k}^{2}]\, \text{BSPut}\bigg(-{{{\tilde{\lambda}_{k-1}}}},\frac{ S_{k}-V^{\mathit{NP}}_{k}}{\mathbb{E}^{\mathbb{P}}[U_{k}^{2}]}, \frac{|\mu|}{\sigma}\sqrt {t_{k}-t_{k-1}}\bigg) \end{aligned}$$

whenever $S_{k} - V^{\mathit{NP}}_{k} > \sigma_{k}$, and $V^{\mathit{NP}}_{k-1} = V^{\mathit{NP}}_{k}$ otherwise, where $\tilde{\lambda}_{k-1}<0$ is the solution of

$$\frac{(V^{\mathit{NP}}_{k}-S_{k})^{2} - \sigma_{k}^{2}}{\mathbb{E}^{\mathbb{P}}[U_{k}^{2}]} = \text{BSPut}\bigg({{{\tilde{\lambda}_{k-1}^{2}}}}, \frac{(V^{\mathit{NP}}_{k}- S_{k})^{2}}{\mathbb{E}^{\mathbb{P}}[U_{k}^{2}]}, 2\frac{|\mu|}{\sigma}\sqrt{t_{k}-t_{k-1}}\bigg). $$

The optimal strategy at date $n-1$ is to replicate the claim

$$S_{n} - \frac{\sigma_{n}}{\mathbb{E}^{\mathbb{P}}[U_{n}^{2}]^{\frac{1}{2}}} U_{n} = S_{n} - \sigma_{n} \bigg(\frac{X_{t_{k}}}{X_{t_{k-1}}}\bigg)^{-\frac{\mu }{\sigma^{2}}} \exp\big(-\mu(t_{k}-t_{k-1})/2\big), $$

which can be achieved with a constant proportion portfolio strategy. On the other hand, at date $k-1< n-1$, one should replicate the claim with the option-like payoff

$$V^{\mathit{NP}}_{k} + \bigg(S_{k} - V^{\mathit{NP}}_{k} + \tilde{\lambda}_{k-1} \Big(\frac{X_{t_{k}}}{X_{t_{k-1}}}\Big)^{-\frac{\mu}{\sigma^{2}}} \exp\Big(\frac{\mu}{2\sigma^{2}} (\mu-\sigma^{2}) (t_{k}-t_{k-1})\Big)\bigg)^{+}. $$

Incomplete market

Now we assume that the agent cannot modify her position between the benchmark dates. This means that the market is incomplete, and the return of the risky asset between date $t_{k-1}$ and $t_{k}$ is given by

$$R_{k} = \frac{X_{t_{k}} - X_{t_{k-1}}}{X_{t_{k-1}}} = e^{(\mu-\frac{\sigma ^{2}}{2})(t_{k} - t_{k-1}) + \sigma(W_{t_{k}} - W_{t_{k-1}})} -1. $$

Let $h = t_{k} - t_{k-1}$. Then

$$\begin{aligned} \mathbb{E}^{\mathbb{P}}\big[\big((1-\eta R_{k})^{+}\big)^{2}\big] &= 2(1+\eta) \text{BSPut} \big(\eta e^{\mu h}, 1+\eta, \sigma\sqrt {h}\big) \\ &\phantom{=:}- \text{BSPut}\big(\eta^{2} e^{2\mu h + \sigma^{2} h}, (1+\eta )^{2}, 2\sigma\sqrt{h}\big) \end{aligned}$$

for $\eta>0$,

$$\mathbb{E}^{\mathbb{P}}\big[\big((1-\eta R_{k})^{+}\big)^{2}\big] = (1+\eta )^{2} -2\eta(1+\eta) e^{\mu h} + \eta^{2} e^{2\mu h + \sigma^{2} h} $$

for $\eta\in[-1,0]$, and

$$\begin{aligned} \mathbb{E}^{\mathbb{P}}\big[\big((1-\eta R_{k})^{+}\big)^{2}\big] &=2(1+\eta) \text{BSCall} \big(-\eta e^{\mu h}, -1-\eta, \sigma\sqrt {h}\big) \\ &\phantom{=:}+ \text{BSCall}\big(\eta^{2} e^{2\mu h + \sigma^{2} h}, (1+\eta )^{2}, 2\sigma\sqrt{h}\big) \end{aligned}$$

for $\eta< -1$.