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反弹走廊

Bouncing Corridor

专题
Finance / 金融
难度
L4

题目详情

Let BtB_{t} be a Brownian Motion and uu and dd two positive real numbers. We consider an option which pays 1 if BtB_{t} reaches uu and touched the down barrier before. The option is knocked out and pays zero if it touches the up barrier first.

pays 1 if t0:Bt0=u;t1[0,t0]:Bt1=d\mathrm{pays~1~if~}\exists t_{0}:B_{t_{0}} = u;\exists t_{1}\in [0,t_{0}]:B_{t_{1}} = -d pays 0 if t0:Bt0=u;t[0,t0],Bt>d\mathrm{pays~0~if~}\exists t_{0}:B_{t_{0}} = u;\forall t\in [0,t_{0}],B_{t} > - d

payment is made when the barrier is touched. Calculate the price of this option with rates r>0r > 0 . Generalize to an option paying 1 after nn bounces (pays 1 if the

option touches uu then uu then d- d etc... n times, pays 0 if uu is touched first).

解析

payoff:若先碰到 d-d,之后再到 uu 则支付 1(到达 uu 时支付),若先到 uu 则 0。

由强马尔可夫性质:先到 d-d 的贴现概率乘以上移后的单侧命中贴现概率(从 d-duu 等价于从 0 到 u+du+d),得到

价格=e2r(u+d)sinh(2rd)sinh(2r(u+d)).\boxed{\text{价格}=e^{-\sqrt{2r}(u+d)}\cdot\frac{\sinh(\sqrt{2r}\,d)}{\sinh(\sqrt{2r}(u+d))}}.

更多 bounces 可用同样的分解反复迭代,得到几何型衰减因子(每多一次往返通常再乘一个 e2r(u+d)e^{-\sqrt{2r}(u+d)} 量级的因子)。