Pricing and hedging basis risk under no good deal assumption

Abstract

We consider the problem of explicitly pricing and hedging an option written on a non-exchangeable asset when trading in a correlated asset is possible. This is a typical case of incomplete market where it is well known that the super-replication concept provides generally too high prices. We study several prices and in particular the instantaneous no-good-deal price (see Cochrane and Saa-Requejo in J Polit Econ 108(1):79–119, 2001) and the global one. We show numerically that the global no-good-deal price can be significantly higher that the instantaneous one. We then propose several hedging strategies and show numerically that the mean-variance hedging strategy can be efficient.

Appendices

Appendices

1.1 Technical results

1.1.1 On Black-Scholes formula

We recall the following formula which is analogous to the Black-Scholes formula. All proofs are omitted since they are completely similar to the one of Black-Scholes model which can be found for example in Musiela and Rutkowski (2007) (starting from page 94).

Let $Y$ be a geometric Brownian motion, with drift $\eta$ and volatility $\varphi$, i.e.

$$\begin{aligned} Y_t = Y_0 \exp \left( \left( \eta - \frac{\varphi ^2}{2} \right) t + \varphi W_t \right) . \end{aligned}$$

Then, the function $BS(Y_t,T-t,K,\eta ,\varphi )$ defined by

$$\begin{aligned} BS\left( Y_t,T-t,K,\eta ,\varphi \right) = {\mathbb {E}}\left[ (Y_T-K)_+ \mid \mathcal{F}_{t}\right] , \end{aligned}$$

(80)

can be explicitly expressed as

$$\begin{aligned} BS(Y_t,T-t,K,\eta ,\varphi )&= Y_t e^{\eta (T-t)} \mathcal{N}\left( d_1(Y_t,T-t,K,\eta ,\varphi )\right) \nonumber \\&- K \mathcal{N}\left( d_0(Y_t,T-t,K,\eta ,\varphi )\right) \end{aligned}$$

(81)

where⁸

$$\begin{aligned} d_1(Y_t,T-t,K,\eta ,\varphi )&= \frac{ \ln \left( \frac{Y_t}{K} \right) + \left( \eta + \frac{\varphi ^2}{2} \right) (T-t) }{\varphi \sqrt{T-t}} ;\\ d_0(Y_t,T-t,K,\eta ,\varphi )&= d_1 -\varphi \sqrt{T-t}. \end{aligned}$$

By Ito’s Formula, one can show for any $x>0$ that:

$$\begin{aligned} d BS(x,T-t,K,\eta ,\varphi )&= \varphi e^{\eta (T-t)} \mathcal{N}(d_1(x,T-t,K,\eta ,\varphi ))xdW_t. \end{aligned}$$

(82)

$$\begin{aligned} \frac{\partial BS}{\partial \eta } (x,T-t,K,\eta ,\varphi )&= (T-t)K \mathcal{N}(d_{0}(x,T-t,K,\eta ,\varphi ))>0 \end{aligned}$$

(83)

$$\begin{aligned} \frac{\partial BS}{\partial \varphi } (x,T-t,K,\eta ,\varphi )&= \sqrt{T-t} x \frac{e^{\eta (T-t)}}{\sqrt{2\pi }}e^{-\frac{d_{1}^{2}(x,T-t,K,\eta ,\varphi )}{2}} > 0, \end{aligned}$$

(84)

(83) and (84) implies that the Black-Scholes function $BS$ is increasing with respect to $\eta$ and $\varphi$. Moreover, the function BS satisfies the following linearity property:

$$\begin{aligned} \forall \alpha \in {\mathbb {R}}_{+}, \qquad BS(\alpha Y_t,T-t,\alpha K,\eta ,\varphi ) = \alpha BS(Y_t,T-t,K,\eta ,\varphi ). \end{aligned}$$

(85)

1.1.2 Proof of Lemma 1

Proof

Let $\lambda \in \varLambda$. Since $Z_{T}^\lambda$ is a martingale, one has ${\mathbb {E}}Z_T^\lambda =Z_{0}^{\lambda }=1$. We define the following process

$$\begin{aligned} \bar{Z}_{t}^\lambda = \exp \left( -2h_S W_t - 2h_S^2 t +2\int \limits _0^t \lambda _s dW^*_s - 2 \int \limits _0^t \lambda _s^2 ds\right) . \end{aligned}$$

It is a Doléans-Dade process and thus a continuous local martingale (see Karatzas and Shreve 1991, page 191). It is easy to see from the maximal inequality that $\bar{Z}^\lambda$ is in fact a martingale (since it is of class (DL)).

Then, we can define the following probability measure

$$\begin{aligned} d\tilde{{\mathbb {Q}}}/d{\mathbb {P}}= \bar{Z}_{T}^\lambda . \end{aligned}$$

Using Bayes Formula

$$\begin{aligned} {\mathbb {E}}\left( \left( \frac{Z_T^\lambda }{Z_t^\lambda }\right) ^2 \mid \mathcal{F}_t\right) = e^{h_S^2(T-t)}{\mathbb {E}}^{\tilde{{\mathbb {Q}}}}\left( e^{\int _t^T \lambda _s^2ds}\mid \mathcal{F}_t \right) . \end{aligned}$$

$\square$

1.1.3 Proof of Lemma 5

We recall the following lemma for ease of exposition. Let $X \in L^2,\,X \ge 0$ such that $X {\mathbf 1}_{X>0}$ has a density with respect to Lebesgue measure and $\gamma$ be a positive number. One has

$$\begin{aligned} p_0^{opt} = {\mathop {\mathop {\sup }\limits _{Y \ge 0, \; {\mathbb {E}}Y =1 }}\limits _{{\mathbb {E}}Y^2 \le 1 +\gamma ^2}} {\mathbb {E}}\left[ Y X \right] \end{aligned}$$

(86)

Lemma 6

The optimal solution of problem (86) is given as follows:

• 1. If $1-\gamma \frac{{\mathbb {E}}\left( X\right) }{\sqrt{\mathop {\mathrm{Var}}\nolimits {X}}} \ge 0$ then one has $Y^{opt} = 1+\gamma \frac{X-{\mathbb {E}}\left( X\right) }{\sqrt{\mathop {\mathrm{Var}}\nolimits {X}}}$ and $p_0^{opt} = {\mathbb {E}}\left( X\right) + \gamma \sqrt{\mathop {\mathrm{Var}}\nolimits {X}}$.

• 2. If $1-\gamma \frac{{\mathbb {E}}\left( X\right) }{\sqrt{\mathop {\mathrm{Var}}\nolimits {X}}}< 0$ then there exists a positive number $\alpha$, such that⁹

  $$\begin{aligned} \frac{{\mathbb {E}}\left( X-\alpha \right) ^2_+}{{\mathbb {E}}^2\left( X-\alpha \right) _+}&= 1+\gamma ^2, \end{aligned}$$

  (87)

  and one has $Y^{opt} = \frac{\left( X-\alpha \right) _+}{{\mathbb {E}}\left( X-\alpha \right) _+}$ and $p_0^{opt} = \alpha + (1+\gamma )^2{\mathbb {E}}\left( X-\alpha \right) _+$.

Proof

1. If $1-\gamma \frac{{\mathbb {E}}\left( X\right) }{\sqrt{\mathop {\mathrm{Var}}\nolimits {X}}} \ge 0$, then it is straightforward that ${\mathbb {E}}Y^{opt} X={\mathbb {E}}\left( X\right) + \gamma \sqrt{\mathop {\mathrm{Var}}\nolimits {X}}$. Let $Y$ such that $Y \ge 0,\,{\mathbb {E}}Y =1$ and ${\mathbb {E}}Y^2 \le 1 +\gamma ^2$ then $\mathop {\mathrm{Var}}\nolimits Y \le \gamma ^2$ and by the Cauchy-Schwarz inequality

$$\begin{aligned} {\mathbb {E}}(Y X)&\le {\mathbb {E}}\left( (Y-{\mathbb {E}}Y) (X-{\mathbb {E}}X) \right) + {\mathbb {E}}X \le \sqrt{\mathop {\mathrm{Var}}\nolimits X} \sqrt{\mathop {\mathrm{Var}}\nolimits Y} + {\mathbb {E}}X \\&\le \gamma \sqrt{\mathop {\mathrm{Var}}\nolimits X} + {\mathbb {E}}X ={\mathbb {E}}[ Y^{opt} X]. \end{aligned}$$

To prove that $Y^{opt}$ is the optimal solution of (86), it remains to verify that it satisfies the constraints of (86). One has $Y^{opt} = 1-\gamma \frac{ {\mathbb {E}}X}{\sqrt{\mathop {\mathrm{Var}}\nolimits {X}}} +\gamma \frac{X}{\sqrt{\mathop {\mathrm{Var}}\nolimits {X}}} \ge 0$ by assumption (recall that $X \ge 0$). The two others constraints are straightforward.

2. If $1-\gamma \frac{{\mathbb {E}}\left( X\right) }{\sqrt{\mathop {\mathrm{Var}}\nolimits {X}}} < 0$, assume that there exists $\alpha$ such that condition (87) is satisfied. Then it is straightforward that

$$\begin{aligned} {\mathbb {E}}\left[ Y^{opt} X \right]&= \frac{{\mathbb {E}}\left[ \left( X-\alpha \right) _+(X-\alpha +\alpha ) \right] }{{\mathbb {E}}\left( X-\alpha \right) _+} \\&= \alpha + \frac{{\mathbb {E}}\left( X-\alpha \right) ^2_+}{{\mathbb {E}}\left( X-\alpha \right) _+} = \alpha + (1+\gamma ^2){\mathbb {E}}\left( X-\alpha \right) _+ \end{aligned}$$

using condition (87). Let $Y$ such that $Y \ge 0,\,{\mathbb {E}}Y =1$ and ${\mathbb {E}}Y^2 \le 1 +\gamma ^2$. Then by the Cauchy-Schwarz inequality and condition (87)

$$\begin{aligned} {\mathbb {E}}[Y X]&= {\mathbb {E}}\left( Y\left( X-\alpha \right) _+ \right) + \alpha + {\mathbb {E}}\left( Y\left( X-\alpha \right) {\mathbf 1}_{X < \alpha } \right) \\&\le \sqrt{{\mathbb {E}}{\left( X-\alpha \right) ^2_+}} \sqrt{{\mathbb {E}}Y^2} + \alpha \\&\le \sqrt{1 +\gamma ^2}\sqrt{1 +\gamma ^2} {\mathbb {E}}{\left( X-\alpha \right) _+} + \alpha ={\mathbb {E}}\left[ Y^{opt} X\right] . \end{aligned}$$

The solution $Y^{opt}$ is thus optimal for the optimization problem (86) since it satisfies the constraints (see condition (87)).

It remains to prove that there exists some $\alpha$ such that condition (87) is satisfied. Let $f(x)= \frac{{\mathbb {E}}\left( X-x\right) ^2_+}{{\mathbb {E}}^2\left( X-x\right) _+}$ then $f(0)=\frac{\mathop {\mathrm{Var}}\nolimits X}{{\mathbb {E}}^2 X} + 1 < 1+ \gamma ^2$ by assumption. Below we show that there exists $\alpha _0>0$ such that $f(\alpha _0) \ge 1+ \gamma ^2$, thus by continuity of $f$ there will exist some $\alpha >0$ such that $f(\alpha ) = 1+ \gamma ^2$. We prove first that there exist $\alpha _0$ such that ${\mathbb {P}}\left( X>\alpha _0\right) =\frac{1}{1+ \gamma ^2}$. Such an $\alpha _0$ exists since

$$\begin{aligned} 1-\gamma \frac{{\mathbb {E}}\left( X\right) }{\sqrt{\mathop {\mathrm{Var}}\nolimits {X}}} < 0&\Leftrightarrow \mathop {\mathrm{Var}}\nolimits \left[ X {\mathbf 1}_{X>0}\right] <\gamma ^2 {\mathbb {E}}^2\left[ X {\mathbf 1}_{X>0}\right] \\&\Leftrightarrow {\mathbb {E}}\left[ X^2 {\mathbf 1}_{X>0}\right] <(\gamma ^2+1) {\mathbb {E}}^2\left[ X \left( \sqrt{{\mathbf 1}_{X>0}}\right) ^2\right] \\&\Rightarrow {\mathbb {E}}\left[ X^2 {\mathbf 1}_{X>0}\right] <(\gamma ^2+1) {\mathbb {E}}\left[ X^2 {\mathbf 1}_{X>0}\right] {\mathbb {P}}(X>0), \end{aligned}$$

by the Cauchy-Schwarz inequality. Thus one has ${\mathbb {P}}(X=0)\le \frac{\gamma ^2}{1+ \gamma ^2}$ and by continuity of $x\rightarrow {\mathbb {P}}(X\le x)$, for $x>0$ there exists $\alpha _0$ such that ${\mathbb {P}}(X\le \alpha _0)=\frac{\gamma ^2}{1+ \gamma ^2}$. Then, by the Cauchy-Schwarz inequality, one has

$$\begin{aligned} {\mathbb {E}}^2\left( X\!-\!\alpha _0\right) _+&= {\mathbb {E}}^2( X {\mathbf 1}_{X >\alpha _0}) -2 \alpha _0 {\mathbb {P}}(X>\alpha _0) {\mathbb {E}}(X {\mathbf 1}_{X >\alpha _0})+\alpha _0^2 {\mathbb {P}}^2(X>\alpha _0) \\&\le {\mathbb {P}}(X\!>\!\alpha _0) {\mathbb {E}}(X^2 {\mathbf 1}_{X >\alpha _0}) \!-\!2 \alpha _0 {\mathbb {P}}(X\!>\!\alpha _0){\mathbb {E}}(X {\mathbf 1}_{X >\alpha _0})\!+\!\alpha _0^2 {\mathbb {P}}^2(X\!>\!\alpha _0) \\&\le \frac{1}{1+ \gamma ^2} \left( {\mathbb {E}}X^2 {\mathbf 1}_{X >\alpha _0} - 2 \alpha _0 {\mathbb {E}}X {\mathbf 1}_{X >\alpha _0}+\alpha _0^2 {\mathbb {P}}(X>\alpha _0) \right) \\&\le \frac{1}{1+ \gamma ^2} {\mathbb {E}}\left( X-\alpha _0\right) ^2_+. \end{aligned}$$

Thus one has $f(\alpha _0) \ge 1+ \gamma ^2$, which concludes the proof. $\square$

1.1.4 Proof of Theorem 6

Proof

In order to apply the results of Schweizer (1992), we need to compute the decomposition of $e^{-rt}{p}^{0}_{t}$ w.r.t. $W^{0}_{t}$ and $W^{0,*}_{t}$. The process $V$ under the probability ${\mathbb {Q}}^{0}$ is a geometric Brownian motion and we can achieve these decomposition using Itô’s formula. Under ${\mathbb {Q}}^0,\,V$ satisfies the following SDE (see (8))

$$\begin{aligned} \frac{dV_t}{V_t} = \eta ^0 dt + \sigma _V \left( \rho dW^{0}_t + \sqrt{1- \rho ^2} dW^{*,0}_{t}\right) , \end{aligned}$$

see (9) for the definition of $\eta ^0$ and (6) for the definition of the ${\mathbb {Q}}^{0}$-Brownian motions $W^{0}$ and $W^{0,*}$. From (72) and (80), we obtain that

$$\begin{aligned} e^{-rt}{p}^{0}_{t} = e^{-rT} {\mathbb {E}}^{{\mathbb {Q}}^0}\left( (V_T-K)_{+}\right) =e^{-rT}BS \left( V_t,T-t,K,\eta ^0,\sigma _{V}\right) . \end{aligned}$$

Equations (82) and (64) imply

$$\begin{aligned} d BS\left( V_t,T-t,K,\eta ^0,\sigma _{V}\right) =e^{r(T-t)} L_{t}^{0} \left( \rho dW^{0}_t + \sqrt{1- \rho ^2} dW^{*,0}_{t}\right) , \end{aligned}$$

where we used the shorter notation $d_1$ for $d_1(V_t,T-t,K,\eta ^0 ,\sigma _V)$. Thus,

$$\begin{aligned} d (e^{-rt}{p}^{0}_{t})&= e^{-rt} L^0_{t}\left( \rho dW^{0}_t + \sqrt{1- \rho ^2} dW^{*,0}_t\right) \\&= e^{-rt} L^0_{t}\left( \frac{\rho e^{rt}}{\sigma _{S}S_{t}} d(e^{-rt}S_{t}) + \sqrt{1- \rho ^2} dW^{*,0}_t\right) . \end{aligned}$$

We now apply the theorem stated on page 175 of Schweizer (1992) and his Eq. (2.4). Let $G^{*}$ be the solution of the SDE

$$\begin{aligned} dG^{*}_{t} = \varPhi (G^{*}_{t}) \sigma _{S} e^{-rt}S_{t}dW^{0}_{t}, \quad G^{*}_{0}=0, \end{aligned}$$

(88)

where

$$\begin{aligned} \varPhi (x) = \frac{\rho L^0_{t}}{\sigma _{S}S_{t}} + \frac{h_{S} e^{rt}}{\sigma _{S} S_{t}} \left( e^{-rt}{p}^{0}_{t} -X_{0} - x \right) . \end{aligned}$$

(89)

Then the hedging strategy $\varPhi (G^{*}_{t})$ solves the optimal solution for problem (68). It remains to compute $G^{*}_{t}$, i.e. solve the SDE (88). If we set

$$\begin{aligned} \bar{G}^{*}_{t} = e^{-rt}{p}^{0}_{t} -X_{0} -G^{*}_{t}, \end{aligned}$$

(90)

the process $\bar{G}^{*}_{t}$ is solution of the following SDE:

$$\begin{aligned} d\bar{G}^{*}_{t}&= d\left( e^{-rt} {p}^{0}_{t}\right) - dG^{*}_{t} = e^{-rt} L^0_{t}\left( \rho dW^{0}_t + \sqrt{1- \rho ^2} dW^{*,0}_t\right) \\&- \varPhi (G^{*}_{t}) \sigma _{S}S_{t} e^{-rt} dW^{0}_{t} \\&= e^{-rt} L^0_{t}\left( \rho dW^{0}_t + \sqrt{1- \rho ^2} dW^{*,0}_t\right) \\&- \rho e^{-rt} L^0_{t} dW^{0}_{t} - h_{S} \left( e^{-rt}{p}^{0}_{t} -X_{0} - G^{*}_{t} \right) d W^{0}_{t}\\&= e^{-rt} L^0_{t} \sqrt{1- \rho ^2} dW^{*,0}_t - h_{S} \bar{G}^{*}_{t}d W^{0}_{t}, \end{aligned}$$

with the initial condition $\bar{G}^{*}_{0} = {p}_{0}^{0} -X_{0}$. The solution of this SDE is given by:

$$\begin{aligned} \bar{G}^{*}_{t}&= e^{-h_{S} W^{0}_{t} - \frac{h_{S}^{2}}{2}t} \left( {p}_{0}^{0} -X_{0} + \int \limits _{0}^{t} \sqrt{1-\rho ^{2}} L^0_{s} e^{h_{S} W^{0}_{s} +\frac{h_{S}^{2}}{2}s-rs} dW^{0,*}_{s} \right) \\&= U_te^{-rt} \left( {p}_{0}^{0} -X_{0} + \int \limits _{0}^{t} \sqrt{1-\rho ^{2}} \frac{L^0_{s}}{U_s} dW^{0,*}_{s} \right) , \end{aligned}$$

since by (69) $U_{t} = e^{-h_SW_t +(r -3/2h_S^2)t}= e^{rt}e^{-h_SW^0_t -1/2h_S^2t}$ (recall that $W^0_t= W_t +h_St$ and $W^{0,*}_t= W^*_t$). Thus using (90) we obtain that:

$$\begin{aligned} G^{*}_{t} = e^{-rt}{p}^{0}_{t} -X_{0} - U_te^{-rt} \left( {p}_{0}^{0} -X_{0} + \int \limits _{0}^{t} \sqrt{1-\rho ^{2}} \frac{L^0_{s}}{U_s} dW^{0,*}_{s}\right) . \end{aligned}$$

(91)

Using (89) we obtain that:

$$\begin{aligned} \varPhi (G^{*}_{t}) = \frac{\rho L^0_{t}}{\sigma _{S}S_{t}} + \frac{h_{S}U_{t}}{\sigma _{S} S_{t}} \left( {p}_{0}^{0} -X_{0} + \sqrt{1-\rho ^{2}} \int \limits _{0}^{t} \frac{L^0_{s} }{U_{s}}dW^{*}_{s} \right) \end{aligned}$$

and (73) is proved. To obtain (74), we rewrite the problem (68). Using (88), we obtain that $\int _0^T\varPhi (G^{*}_{t})d(e^{-rt}S_t)=\int _0^T\varPhi (G^{*}_{t})S_te^{-rt} \sigma _SdW^0_t =G^{*}_{T}$. Thus using (91), we obtain that

$$\begin{aligned} v^{X_{0}}(H)&= e^{2rT} {\mathbb {E}}\left[ U_Te^{-rT} \left( {p}_{0}^{0} -X_{0} + \int \limits _{0}^{T} \sqrt{1-\rho ^{2}} \frac{L^0_{t}}{U_t} dW^{0,*}_{t}\right) \right] ^{2} \\&= {\mathbb {E}}\left[ U_{T}^{2} \right] {\mathbb {E}}^{{\mathbb {Q}}^{U}} \left[ {p}_{0}^{0} -X_{0} + \sqrt{1-\rho ^{2}}\int \limits _{0}^{T} \frac{ L^0_{t} }{U_{t}} dW^{U,*}_{t} \right] ^{2}, \end{aligned}$$

since $W^{0,*}=W^{U,*}$ is a Brownian motion under ${\mathbb {Q}}^{U}$ (see (71)) and since ${\mathbb {E}}\left[ U_{T}^{2} \right] =e^{(2r-h_{S}^{2})T}$: (74) follows. $\square$