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Outside barrier lookback options with floating strike


This paper introduces a new class of exotic options, lookback outside barrier options with two underlying assets, which combine lookback and barrier options by incorporating active barrier conditions into lookback payoff. Bermin (Essays on lookback and barrier options: A Malliavin calculus approachs, Lund University, Department of Economics, PhD thesis, 1998) proposed lookback-barrier options to compensate for the shortcomings of the high-priced lookback option, and later several applications were made by either mitigating the conditions or adding other conditions. In this paper, we extend the existing idea to lookback-barrier options with two underlying assets. To reduce the burden of the computational process and simplify the conditions, the life of the options was divided into two non-overlapping intervals. The structure of the lookback outside barrier option with two underlying assets is as follows: whether the barrier conditions are satisfied or not in the first subinterval, and the amount of payoff is determined in the second subinterval if the condition is met. In traditional lookback-barrier options, barrier conditions and payoff are determined by one asset, while in a lookback-barrier option with two underlying assets, one of the two assets determines the barrier conditions and the other is responsible for payoff. Our main study develops a complete valuation framework that allows for closed-form pricing formulas under the Black-Scholes model. This closed pricing formula for lookback outside barrier option with two underlying assets offers significant advantages over Monte Carlo simulation methods, as a large number of simulations may be required for accurate computation. Complexities involved in the derivation process would be resolved by the Esscher transform and the reflection principle of Brownian motion. We illustrate our results with numerical examples, showing the effects of different values of correlation coefficients between two underlying asset prices.

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Appendix A: Proof of Lemma 1

(i) We assume that c, d, k, and \(\xi \) are constants that satisfy \(d\ge 0\), \(k\ge 0\), and \(c+\xi \ne 0\). We will divide our proof into two conditions, whether \(X(t)+k<d\) or \(X(t)+k>d\) as follows:

$$\begin{aligned} P&r(M(t)> d, M(t)> X(t)+k) \nonumber \\&=Pr(M(t)>d, M(t)>X(t)+k,X(t)+k<d)\nonumber \\&\quad +Pr(M(t)>d, M(t)>X(t)+k,X(t)+k>d)\nonumber \\&=Pr(M(t)>d, X(t)+k<d)\nonumber \\&\quad +Pr(M(t)>X(t)+k,X(t)+k>d). \end{aligned}$$

Then using (21), (42) will be

$$\begin{aligned} e&^{\xi d}Pr(X(t) \le -(d+k)) + Pr(M(t)>X(t)+k, k-d>X(0)-X(t)) \nonumber \\&=e^{\xi d}Pr(X(t) \le -(d+k)) + Pr(M^{*}(t)>k, k-d>X^{*}(t)), \end{aligned}$$

where \(X^*(s)=X(t-s)-X(t)\) and \(M^*(t)=\max \{X^*(\tau ):0\le \tau \le t\}\). By applying (21) to the second term on the right side of (43) again, the above probability becomes

$$\begin{aligned} e&^{\xi d}Pr(X(t) \le -(d+k)) + e^{-\xi k}Pr(X^{*}(t)+2k< k-d) \nonumber \\&=e^{\xi d}Pr(X(t) \le -(d+k)) + e^{-\xi k}Pr(-X(t)< -(d+k)) \nonumber \\&=e^{\xi d}Pr(X(t) \le -(d+k)) + e^{-\xi k}Pr(X(t)> -(d+k)). \end{aligned}$$

(ii) Let us express (27) as the derivative of the survival distribution function of X(t) and M(t), we obtain

$$\begin{aligned} E&[e^{cM(t)}I(M(t)>d, M(t)>X(t)+k)] \nonumber \\&= -\int _{m>d}e^{cm}d(Pr(M(t)>m,M(t)>X(t)+k)). \end{aligned}$$

By the result from (26), the survival distribution will be

$$\begin{aligned}&-\int _{m>d}e^{cm}d(e^{\xi m}Pr(X(t)\le -(m+k))+e^{-\xi k}Pr(X(t)>m+k)) \nonumber \\&= -\int _{m>d}e^{cm}d(e^{\xi m}Pr(X(t)\le -(m+k))-\int _{m>d}e^{cm}d(e^{-\xi k}Pr(X(t)>m+k))) \nonumber \\&= -\int _{m>d}e^{(c+\xi )m}d(Pr(X(t)\le -(m+k)))-\xi \int _{m>d}e^{(c+\xi )m}Pr(X(t)\le -(m+k))dm \nonumber \\&\quad -e^{-\xi k}\int _{m>d}e^{cm}d(Pr(X(t)>m+k)). \end{aligned}$$

Now the second term on the last equation of (46) can be handled via integration by parts as follows.

$$\begin{aligned} E&[e^{(c+\xi )(k+X(t))}I(X(t)\le -(d+k))]+\tfrac{\xi }{c+\xi }e^{(c+\xi )d}Pr(X(t)\le -(d+k)) \nonumber \\&\qquad +\tfrac{\xi }{c+\xi }\int _{m>d}e^{(c+\xi )m}d(Pr(X(t)\le -(m+k))) \nonumber \\&\qquad +e^{-\xi k}E[e^{c(X(t)-k)}I(X(t)>(d+k)))] \nonumber \\&\quad =\tfrac{c}{c+\xi }e^{-(c+\xi )k}E[e^{-(c+\xi )X(t)}I(X(t)\le -(d+k))] \nonumber \\&\qquad +\tfrac{\xi }{c+\xi }e^{(c+\xi )d}Pr(X(t)\le -(d+k)) \nonumber \\&\qquad +e^{-(c+\xi )k}E[e^{cX(t)}I(X(t)>(d+k))] \nonumber \\&\quad =\tfrac{2c+\xi }{c+\xi }e^{-(c+\xi )k}E[e^{cX(t)}I(X(t)>(d+k))] \nonumber \\&\qquad +\tfrac{\xi }{c+\xi }e^{(c+\xi )d}Pr[X(t)\le -(d+k)). \end{aligned}$$

This result might be proved by using the joint density function of X(t) and M(t) as usual.

Appendix B: Proof of Proposition 1

(i) Assuming that c, d, k, m, and \(\xi _i\) are constants that satisfy \(d\ge 0\), \(k\ge 0\), and \(c+\xi _1\ne 0\), we apply the properties of the Brownian motion for \(X_i(t)\), i.e., the independent and stationary increments for \(0\le t\le T\) and \(i=1,2\).

$$\begin{aligned} E&[e^{cM_1(t,T)}I(M_2(t)>m, M_1(t,T)>X_1(T)+k)] \nonumber \\&=E[e^{cX_1(t)}I(M_2(t)>m) \nonumber \\&\quad \times E[e^{c(M_1(t,T)-X_1(t))}I(M_1(t,T)-X_1(t)>X_1(T)-X_1(t)+k)|X_1(t)]] \nonumber \\&=E[e^{cX_1(t)}I(M_2(t)>m)] \nonumber \\&\quad \times E[e^{c(M_1(t,T)-X_1(t))}I(M_1(t,T) -X_1(t)>X_1(T)-X_1(t)+k)] \nonumber \\&=E[e^{cX_1(t)}]Pr(M_2(t)>m;(c,0)) \nonumber \\&\quad \times E[e^{cM_1(T-t)}I(M_1(T-t)>X_1(T-t)+k)] \nonumber \\&=E[e^{cX_1(t)}]Pr(M_2(t)>m;(c,0)) \nonumber \\&\quad \times [\tfrac{\xi _1}{c+\xi _1}Pr(X_1(T-t)\le -k)+\tfrac{2c+\xi _1}{c+\xi _1}e^{-(c+\xi _1)k}E[e^{cX(T-t}I(X_1(T-t)>k]]. \end{aligned}$$

(ii) The expected value can be expressed as a conditional mean given \(X_1(t)\) as follows:

$$\begin{aligned} E&[e^{cM_1(t, T)}I(M_2(t)> m, M_1(t, T)> d, M_1(t, T)> X_1(T) + k)] \nonumber \\&= E[e^{cX_1(t)}I(M_2(t)> m) E[e^{c[M_1(t, T)-X_1(t)]}I(M_1(t, T) {-} X_1(t)> d {-} X_1(t), \nonumber \\&\qquad M_1(t, T)-X_1(t)> X_1(T)-X_1(t)+k|X_1(t)]] \nonumber \\&= E[e^{cX_1(t)}I(M_2(t)> m, X_1(t) < d) E[e^{c[M_1(t, T)-X_1(t)]}I(M_1(t, T) {-} X_1(t)> \nonumber \\&\qquad d {-} X_1(t), M_1(t, T)-X_1(t)> X_1(T)-X_1(t) +k|X_1(t)]] \nonumber \\&\quad + E[e^{cX_1(t)}I(M_2(t)> m, X_1(t)> d) E[e^{c[M_1(t, T)-X_1(t)]} \nonumber \\&\qquad I(M_1(t, T) - X_1(t)> 0, M_1(t, T)-X_1(t) > X_1(T)-X_1(t) +k|X_1(t)]]. \end{aligned}$$

By applying the techniques used in (45), the first term on the last equation of (49) can be written as

$$\begin{aligned}&E[e^{cX_1(t)}I(M_2(t)> m, X_1(t)< d) \\&\quad \{\tfrac{\xi _1}{c+\xi _1}e^{(c+\xi _1)(d-X_1(t))}E[I(X_1(T)-X_1(t)\le -(d {-} X_1(t) +k))|X_1(t)] \\&\quad +\tfrac{2c+\xi _1}{c+\xi _1}e^{-(c+\xi _1)k} E[e^{c(X_1(T)-X_1(t))} I(X_1(T)-X_1(t)> (d {-} X_1(t)+k)|X_1(t)]\}] \\&=\tfrac{\xi _1}{c+\xi _1}e^{(c+\xi _1)d} E[e^{-\xi _1 X_1(t)} I(M_2(t)>m, X_1(t)<d) \\&\qquad \times E[I(X_1(T)-X_1(t) \le -(d {-} X_1(t) +k))|X_1(t)]] \\&\quad +\tfrac{2c+\xi _1}{c+\xi _1}e^{-(c+\xi _1)k} E[e^{cX_1(T)}I(M_2(t)> m, X_1(t)< d) \\&\qquad \times E[e^{c(X_1(T)-X_1(t))} I(X_1(T)-X_1(t) >(d-X_1(t)+k)|X_1(t)]]. \\ \end{aligned}$$

By using (16), the above expectation becomes

$$\begin{aligned}&\tfrac{\xi _1}{c+\xi _1}e^{(c+\xi _1)d}E[e^{-\xi _1 X_1(t)} \nonumber \\&\qquad \times I(M_2(t)> m, X_1(t)< d, X_1(T)-X_1(t)\le -(d {-} X_1(t) +k))] \nonumber \\&\quad +\tfrac{2c+\xi _1}{c+\xi _1}e^{-(c+\xi _1)k} E[e^{cX_1(T)}I(M_2(t)> m, X_1(t)< d, X_1(T)> (d + k)] \nonumber \\&= \tfrac{\xi _1}{c+\xi _1}e^{(c+\xi _1)d} E[e^{-\xi _1 X_1(t)} ] \nonumber \\&\qquad \times E[I(M_2(t)> m, X_1(t)< d, -(X_1(T)-X_1(t)) \le -(d {-} X_1(t) +k)); (-\xi _1 ,0)] \nonumber \\&\quad + \tfrac{2c+\xi _1}{c+\xi _1}e^{-(c+\xi _1)k} E[e^{c X_1(T)}]Pr(M_2(t)> m, X_1(t)< d, X_1(T)> (d + k); (c,0)] \nonumber \\&= \tfrac{\xi _1}{c+\xi _1}e^{(c+\xi _1)d} Pr(M_2(t)> m, X_1(t)< d, X_1(T)> (d + k)); (-\xi _1 ,0)] \nonumber \\&\quad + \tfrac{2c+\xi _1}{c+\xi _1}e^{-(c+\xi _1)k} E[e^{cX_1(T)}]Pr(M_2(t)> m, X_1(t) < d, X_1(T) > (d + k); (c,0)]. \end{aligned}$$

In the last equation of (50), by dividing it across two conditions, whether \(X_2(t)<m\) or \(X_2(t)>m\), we then obtain

$$\begin{aligned}&\tfrac{\xi _1}{c+\xi _1}e^{(c+\xi _1)d} e^{2m(\mu _2-2\rho \tfrac{\sigma _2}{\sigma _1}\mu _1)/\sigma _2^2}Pr(X_1(t) +2m\rho \tfrac{\sigma _1}{\sigma _2}< d, \nonumber \\&\qquad X_1(T)+2m\rho \tfrac{\sigma _1}{\sigma _2}> d + k, X_2(t)<-m ; ( -\xi _1,0)) \nonumber \\&\quad +\tfrac{\xi _1}{c+\xi _1}e^{(c+\xi _1)d} Pr(X_1(t)<d, X_1(T)>d+k,X_2(t)>m;(-\xi _1,0) \nonumber \\&\quad +\tfrac{2c+\xi _1}{c+\xi _1}e^{-(c+\xi _1)k}e^{2m(\mu _2 +c\rho \sigma _1\sigma _2)/\sigma _2^2} E[e^{cX_1(t)}] \nonumber \\&\qquad \times Pr(X_1(t) +2m\rho \tfrac{\sigma _1}{\sigma _2}< d, X_1(T)+2m\rho \tfrac{\sigma _1}{\sigma _2}> (d + k), X_2(t)<-m; (c,0)) \nonumber \\&\quad +\tfrac{2c+\xi _1}{c+\xi _1}e^{-(c+\xi _1)k} E[e^{cX_1(T)}]Pr(X_1(t)<d,X_1(T)>d+k,X_2(t)>m;(c,0)). \end{aligned}$$

By Eq. (45), the second term on the right side of (49) proceeds as

$$\begin{aligned}&E[e^{cX_1(t)}I(M_2(t)> m, X_1(t)> d) \nonumber \\&\qquad \times E[e^{c[M_1(t, T)-X_1(t)]}I(M_1(t, T)-X_1(t)> X_1(T)-X_1(t) +k|X_1(t)]] \nonumber \\&= E[e^{cX_1(t)}I(M_2(t)> m, X_1(t)> d)] \nonumber \\&\qquad \times E[e^{c[M_1(t, T)-X_1(t)]}I(M_1(t, T)-X_1(t)> X_1(T)-X_1(t) +k)] \nonumber \\&= E[e^{cX_1(t)}]Pr(M_2(t)> m, X_1(t)> d ; (c,0)) \nonumber \\&\qquad \times E[e^{cM_1(T-t)}I(M_1(T-t)> X_1(T-t) +k)] \nonumber \\&= E[e^{cX_1(t)}]Pr(M_2(t)> m, X_1(t)> d; (c,0))[\tfrac{\xi _1}{c+\xi _1}Pr(X_1(T-t) \le -k) \nonumber \\&\qquad + \tfrac{2c+\xi _1}{c+\xi _1}e^{-(c+\xi _1)k}E[e^{cX_1(T-t)} I(X_1(T-t) > k)]]. \end{aligned}$$

Note that the first probability term on the last equation in (52) can be calculated as follows:

$$\begin{aligned}&Pr(M_2(t)> m, X_1(t)> d) \nonumber \\&= Pr(M_2(t)> m, X_2(t)>m, X_1(t)> d) \nonumber \\&\qquad + Pr(M_2(t)> m, X_2(t)<m, X_1(t)> d) \nonumber \\&= Pr(X_2(t)>m, X_1(t)>d) + e^{\xi _2 m}Pr(X_2(t)+2m<m, X_1(t)+2m\rho \tfrac{\sigma _1}{\sigma _2} > d). \end{aligned}$$

(iii) Since \(M_1(t,T)<d\), \(X_1(t)\) is always less than d, it can be proved by using the complement.

$$\begin{aligned}&Pr(M_2(t)>m, M_1(t,T)<d, X_1(T)<d-k) \nonumber \\&= Pr(M_2(t)>m, M_1(t,T)<d, X_1(t)<d, X_1(T)<d-k) \nonumber \\&= Pr(M_2(t)>m, X_1(t)<d, X_1(T)<d-k) \nonumber \\&\qquad - Pr(M_2(t)>m, M_1(t,T)>d, X_1(t)<d, X_1(T)<d-k). \end{aligned}$$

The first term of the last equation in (54) is

$$\begin{aligned}&Pr(M_2(t)>m, X_1(t)<d, X_1(T)<d-k) \nonumber \\&= Pr(M_2(t)>m, X_2(t)>m,X_1(t)<d, X_1(T)<d-k) \nonumber \\&+ Pr(M_2(t)>m, X_2(t)<m,X_1(t)<d, X_1(T)<d-k). \end{aligned}$$

The second term on the right side of (55) can be written as

$$\begin{aligned}&Pr(M_2(t)>m, X_2(t)<m,X_1(t)<d, X_1(T)<d-k) \nonumber \\&=E[E[I(M_2(t)>m, X_2(t)<m, \rho \tfrac{\sigma _1}{\sigma _2}X_2(t)+Z(t)<d, \nonumber \\&\quad \rho \tfrac{\sigma _1}{\sigma _2}X_2(T)+Z(T)<d-k)|Z(t),Z(T)]], \end{aligned}$$

where \(Z(t)=X_1(t)- \rho \tfrac{\sigma _1}{\sigma _2}X_2(t)\) for any time t. By the reflection principle, (56) becomes

$$\begin{aligned}&E[e^{\xi _2 m} Pr(X_2(t)+2m<m, \rho \tfrac{\sigma _1}{\sigma _2}(X_2(t)+2m)+Z(t)<d, \nonumber \\&\rho \tfrac{\sigma _1}{\sigma _2}(X_2(T)+2m)+Z(T)<d-k) | Z(t), Z(T)] \nonumber \\&= e^{\xi _2 m}E[Pr(X_2(t) +2m<m, \rho \tfrac{\sigma _1}{\sigma _2}(X_2(t)+2m)+Z(t)<d, \nonumber \\&\rho \tfrac{\sigma _1}{\sigma _2}(X_2(T)+2m)+Z(T)<d-k) | Z(t), Z(T)] \nonumber \\&=e^{\xi _2m} Pr(X_2(t)< -m, X_1(t)+ 2m\rho \tfrac{\sigma _1}{\sigma _2}<d, X_1(T)+ 2m\rho \tfrac{\sigma _1}{\sigma _2} <d-k). \end{aligned}$$

Therefore, the first term of the last equation in (54) can be expressed as follows:

$$\begin{aligned} Pr&(M_2(t)>m, X_1(t)<d, X_1(T)<d-k) \nonumber \\&=Pr(M_2(t)>m, X_2(t)>m,X_1(t)<d,X_1(T)<d-k) \nonumber \\&\quad + Pr(M_2(t)>m, X_2(t)<m,X_1(t)<d, X_1(T)<d-k) \nonumber \\&=Pr(X_2(t)>m, X_1(t)<d, X_1(T)<d-k) \nonumber \\&\quad +e^{\xi _2m}Pr(X_2(t)<-m,X_1(t)+ 2m\rho \tfrac{\sigma _1}{\sigma _2}<d,X_1(T)+ 2m\rho \tfrac{\sigma _1}{\sigma _2} <d-k). \end{aligned}$$

The second term of the last equation in (54) is separated into two parts either \(X_2(t)>m\) or \(X_2(t)<m\).

$$\begin{aligned}&Pr(M_2(t)>m, M_1(t,T)>d, X_1(t)<d, X_1(T)<d-k) \nonumber \\&= Pr(M_2(t)>m, X_2(t)>m,M_1(t,T)>d, X_1(t)<d, X_1(T)<d-k) \nonumber \\&\quad + Pr(M_2(t)>m, X_2(t)<m, M_1(t,T)>d, X_1(t)<d, X_1(T)<d-k). \end{aligned}$$

By applying a conditional mean given \(X_1(t)\), the first term on the right side of (59) is

$$\begin{aligned}&Pr(M_2(t)>m, X_2(t)>m, M_1(t,T)>d, X_1(t)<d, X_1(T)<d-k) \nonumber \\&=E[I(X_2(t)>m, X_1(t)<d) \nonumber \\&\quad \times Pr(M_1(t,T)-X_1(t)>d-X_1(t), X_1(T)-X_1(t)< d-k {-}X_1(t) |X_1(t))] \nonumber \\&= E[I(X_2(t)>m, X_1(t)<d)e^{\xi _1(d-X_1(t))} \nonumber \\&\quad \times Pr(X_1(T)-X_1(t) +2(d-X_1(t))<d-k {-}X_1(t) |X_1(t))] \nonumber \\&= e^{\xi _1 d}E[e^{-\xi _1 X_1(t)} I(X_2(t)>m, X_1(t)<d, \nonumber \\&\quad X_1(T)-X_1(t) +2(d-X_1(t))<d-k {-}X_1(t))] \nonumber \\&= e^{\xi _1 d}E[I(X_2(t)>m, X_1(t)<d, \nonumber \\&\quad -(X_1(T)-X_1(t)) +2(d-X_1(t))<d-k {-}X_1(t)); (-\xi _1 ,0)] \nonumber \\&= e^{\xi _1 d}Pr(X_2(t)>m, X_1(t)<d, -X_1(T) +2d < d-k; (-\xi _1 ,0)). \end{aligned}$$

Again, by applying a conditional mean given \(X_1(t)\) and using reflection principle, the second term on the right side of (59) becomes

$$\begin{aligned}&Pr(M_2(t)>m, X_2(t)<m, M_1(t,T)>d, X_1(t)<d, X_1(T)<d-k) \nonumber \\&= E[I(X_2(t)<m, M_2(t)>m, X_1(t)<d) \nonumber \\&\quad \times Pr(M_1(t,T)-X_1(t)>d-X_1(t), X_1(T)-X_1(t)< d-k {-}X_1(t) |X_1(t))] \nonumber \\&= E[I(X_2(t)<m, M_2(t)>m, X_1(t)<d)e^{\xi _1(d-X_1(t))} \nonumber \\&\quad \times Pr(X_1(T)-X_1(t) +2(d-X_1(t))< d-k {-}X_1(t) |X_1(t))] \nonumber \\&=e^{\xi _1 d} E[e^{-\xi _1 X_1(t)} I(X_2(t)<m, M_2(t)>m, \nonumber \\&\quad X_1(t)<d, X_1(T)-X_1(t) +2(d-X_1(t)) < d-k {-}X_1(t))]. \nonumber \\ \end{aligned}$$

By using (16), we obtain the probability as follows:

$$\begin{aligned} e&^{\xi _1 d} E[e^{-\xi _1 X_1(t)}]E[I(X_2(t)<m, M_2(t)>m, X_1(t)<d, \nonumber \\&\quad -(X_1(T)-X_1(t)) +2(d-X_1(t))< d-k {-}X_1(t)); (-\xi _1,0)] \nonumber \\&=e^{\xi _1 d} Pr(X_2(t)<m, M_2(t)>m, X_1(t)<d, -X_1(T) +2d< d-k; (-\xi _1,0)) \nonumber \\&=e^{\xi _1 d}e^{2m(\mu _2-\tfrac{2\rho \sigma _2}{\sigma _1}\mu _1) /\sigma _2^2}Pr(X_2(t) +2m<m, X_1(t)+ 2m\tfrac{\rho \sigma _1}{\sigma _2}<d, \nonumber \\&\quad -(X_1(T) + 2m\tfrac{\rho \sigma _1}{\sigma _2} ) +2d < d-k; (-\xi _1 ,0)). \end{aligned}$$

Check that \(E[e^{-\xi _1 X_1(t)}]=1\) in (61).

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Lee, G., Lee, H. & Choi, Y.H. Outside barrier lookback options with floating strike. J. Korean Stat. Soc. 50, 1259–1286 (2021).

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