Quanto option期权公式推导（更新中）

\qquad Quanto期权不同于以标的股票或资产的计价货币支付最终收益的普通欧式期权，它以另一种货币按预先确定的汇率支付最后的收益。

\qquad 以股票B的计价货币来计价的quanto看涨期权的收益为 F 0 × max ⁡ ( S a ( T ) − K , 0 ) × S b ( T ) F_0 \times \max(S_a(T)-K, 0) \times S_b(T) F0​×max(Sa​(T)−K,0)×Sb​(T) 其中 S a S_a Sa​以货币A计价， S b S_b Sb​以货币B计价， F 0 F_0 F0​是预先固定的汇率.

\qquad 假设上述股票均遵循几何布朗运动，可以得到以下恒等式：

d S a ( t ) / S a ( t ) = ( r d − r a ) d t + σ a d W a d S b ( t ) / S b ( t ) = ( r d − r b ) d t + σ b d W b d W a d W b = ρ d t \begin{aligned} \qquad & dS_a(t) / S_a(t) = (r_d - r_a) dt + \sigma_a dW_a \\ \qquad & dS_b(t) / S_b(t) = (r_d - r_b) dt + \sigma_b dW_b \\ \qquad & dW_a dW_b = \rho dt \\ \end{aligned} ​dSa​(t)/Sa​(t)=(rd​−ra​)dt+σa​dWa​dSb​(t)/Sb​(t)=(rd​−rb​)dt+σb​dWb​dWa​dWb​=ρdt​

\qquad 我们需要推导出Quanto-European看涨期权的闭式解是：

S b ( 0 ) F e − r b T ( S a ( 0 ) e ( r d − r a + ρ σ a σ b ) T N ( d + ) − K N ( d − ) ) d ± = ln ⁡ S a ( 0 ) K + ( r d − r a + ρ σ a σ b ± 1 2 σ a 2 ) T σ a T \begin{aligned} \qquad & S_b(0) F e^{-r_bT}\left( S_a(0) e^{(r_d - r_a + \rho \sigma_a \sigma_b)T} N(d_+) - K N(d_-) \right) \\ \qquad & d_{\pm} = \frac{\ln \frac{S_a(0)}{K} + (r_d - r_a + \rho \sigma_a \sigma_b \pm \frac{1}{2}\sigma_a^2)T}{\sigma_a \sqrt{T}} \end{aligned} ​Sb​(0)Fe−rb​T(Sa​(0)e(rd​−ra​+ρσa​σb​)TN(d+​)−KN(d−​))d±​=σa​T ​lnKSa​(0)​+(rd​−ra​+ρσa​σb​±21​σa2​)T​​

一、 *十分重要的引理：

\qquad 引理的证明可以参考这篇文章：/doi/abs/10.1002/9780470061602.eqf06006

\qquad 根据文章所述，用 S a , S b S_a, S_b Sa​,Sb​生成一个新的变量 :

S a n e w = S a S b S_a^{new} = \frac{S_a}{S_b} Sanew​=Sb​Sa​​

\qquad 从上面的三角形可以得知, 我们此处的 S a S_a Sa​ 相当于’XAU’, S b S_b Sb​ 相当于’USD’（文章中有所提及）, 此时，使用伊藤定理可得：

d 1 S b ( t ) = − 1 S b ( t ) 2 d S b ( t ) 2 + 1 2 ∗ 2 1 S b ( t ) 3 ( d S b ( t ) 2 ) 2 = ( r b − r a + σ b 2 ) 1 S b ( t ) 2 d t − σ 2 1 S b ( t ) d W b \begin{aligned} \qquad d\frac{1}{S_b(t)} & = -\frac{1}{S_b(t)^2}dS_b(t)^2+\frac{1}{2}*2\frac{1}{S_b(t)^3} (dS_b(t)^2)^2\\ & = (rb - ra + \sigma_b^2)\frac{1}{S_b(t)^2}dt-\sigma_2\frac{1}{S_b(t)}dW_b\\ \end{aligned} dSb​(t)1​​=−Sb​(t)21​dSb​(t)2+21​∗2Sb​(t)31​(dSb​(t)2)2=(rb−ra+σb2​)Sb​(t)21​dt−σ2​Sb​(t)1​dWb​​

\qquad 之后，我们假设相关关系如下：

− ρ i j = c o s ( π − ρ i j ) \qquad −\rho_{ij} =cos(π−\rho_{ij}) −ρij​=cos(π−ρij​). ( ρ 12 = S a \rho_{12} = S_a ρ12​=Sa​ 和 S b S_b Sb​之间的 ρ \rho ρ)

d S a n e w ( t ) = − 1 S b ( t ) d S a ( t ) + S a ( t ) d 1 S b ( t ) + d S a ( t ) d 1 S b ( t ) ( d S b ( t ) 2 ) 2 = S a ( t ) S b ( t ) ( r d − r a ) d t + S a ( t ) S b ( t ) σ a d W a + S a ( t ) S b ( t ) ( r b − r d + σ b ) d t − S a ( t ) S b ( t ) σ b d W b + S a ( t ) S b ( t ) ρ σ a σ b d t = ( r d − r a + σ b 2 + ρ a , n e w σ a s i g m a b ) S a n e w ( t ) d t + S a n e w ( t ) ( σ a d W a − σ b d W b ) = ( r d − r a + σ b 2 + ρ a , n e w σ a s i g m a b ) S a n e w ( t ) d t + S a n e w ( t ) d W n e w \begin{aligned} \qquad dS_a^{new}(t) & = -\frac{1}{S_b(t)}dS_a(t)+S_a(t)d\frac{1}{S_b(t)}+dS_a(t)d\frac{1}{S_b(t)} (dS_b(t)^2)^2\\ \qquad & = \frac{S_a(t)}{S_b(t)}(rd-ra)dt + \frac{S_a(t)}{S_b(t)}\sigma_adW_a + \frac{S_a(t)}{S_b(t)}(rb-rd+\sigma_b)dt - \frac{S_a(t)}{S_b(t)}\sigma_bdW_b + \frac{S_a(t)}{S_b(t)}\rho \sigma_a \sigma_b dt \\ & = (rd-ra+\sigma_b^2+\rho_{a,new} \sigma_a \ sigma_b)S_a^{new}(t)dt + S_a^{new}(t)(\sigma_a dW_a - \sigma_b dW_b) \\ & = (rd-ra+\sigma_b^2+\rho_{a,new} \sigma_a \ sigma_b)S_a^{new}(t)dt + S_a^{new}(t)dW_{new} \\ \end{aligned} dSanew​(t)​=−Sb​(t)1​dSa​(t)+Sa​(t)dSb​(t)1​+dSa​(t)dSb​(t)1​(dSb​(t)2)2=Sb​(t)Sa​(t)​(rd−ra)dt+Sb​(t)Sa​(t)​σa​dWa​+Sb​(t)Sa​(t)​(rb−rd+σb​)dt−Sb​(t)Sa​(t)​σb​dWb​+Sb​(t)Sa​(t)​ρσa​σb​dt=(rd−ra+σb2​+ρa,new​σa​sigmab​)Sanew​(t)dt+Sanew​(t)(σa​dWa​−σb​dWb​)=(rd−ra+σb2​+ρa,new​σa​sigmab​)Sanew​(t)dt+Sanew​(t)dWnew​​

\qquad 接着，使用余弦定理可得：

σ a 2 = σ n e w 2 + σ b 2 − ρ σ n e w σ b σ a 2 = σ a 2 + σ b 2 − ρ a , b σ a σ b \begin{aligned} \qquad \sigma_a^2 & = \sigma_{new}^2 + \sigma_b^2 - \rho\sigma_{new}\sigma_{b} \\ \sigma_a^2 & = \sigma_{a}^2 + \sigma_b^2 - \rho_{a,b}\sigma_{a}\sigma_{b} \\ \end{aligned} σa2​σa2​​=σnew2​+σb2​−ρσnew​σb​=σa2​+σb2​−ρa,b​σa​σb​​

\qquad 最后，我们有如下结果：

σ b 2 + ρ a , b σ a σ b = ρ a , n e w σ n e w σ b d S a n e w ( t ) = ( r d − r a + ρ σ n e w s i g m a b ) S a n e w ( t ) d t + S a n e w ( t ) d W n e w \begin{aligned} \qquad & \sigma_b^2 +\rho_{a,b}\sigma_{a}\sigma_{b} = \rho_{a,new}\sigma_{new}\sigma_{b} \\ & dS_a^{new}(t)= (rd-ra+\rho\sigma_{new}\ sigma_b)S_a^{new}(t)dt + S_a^{new}(t)dW_{new} \\ \end{aligned} ​σb2​+ρa,b​σa​σb​=ρa,new​σnew​σb​dSanew​(t)=(rd−ra+ρσnew​sigmab​)Sanew​(t)dt+Sanew​(t)dWnew​​

二、后续证明

\qquad 根据引理， S b ( t ) S_b(t) Sb​(t), S a ( t ) S_a(t) Sa​(t)可以被写作:

d S b ( t ) / S b ( t ) = ( r d − r b ) d t + σ b d W b d S a ( t ) / S a ( t ) = ( r d − r a + ρ σ a σ b ) d t + σ a d W a \begin{aligned} \qquad & dS_b(t) / S_b(t) = (r_d - r_b) dt + \sigma_b dW_b \\ \qquad & dS_a(t) / S_a(t) = (r_d - r_a + \rho \sigma_a \sigma_b) dt + \sigma_a dW_a \\ \end{aligned} ​dSb​(t)/Sb​(t)=(rd​−rb​)dt+σb​dWb​dSa​(t)/Sa​(t)=(rd​−ra​+ρσa​σb​)dt+σa​dWa​​

\qquad 使用伊藤公式后，我们得到：

l n S b ( t ) = l n S b ( 0 ) + ( r d − r b ) t + σ b W b l n S a ( t ) = l n S a ( 0 ) + ( r d − r a + ρ σ a σ b ) t + σ a W a l n S a ∽ ϕ ( μ 1 , σ 1 ) , l n S b ∽ ϕ ( μ 2 , σ 2 ) μ 1 = l n S a 0 + ( r d − r a − 0.5 σ a 2 + ρ σ a σ b ) T , σ 1 = σ a T μ 2 = l n S b 0 + ( r d − r b − 0.5 σ b 2 ) T , σ 2 = σ b T f ( b ) = 1 2 π σ b e x p [ ( l n b − μ 2 ) 2 σ b 2 ] f ( a ) = 1 2 π σ a e x p [ ( l n a − μ 1 ) 2 σ a 2 ] \begin{aligned} \qquad & lnS_b(t) = lnS_b(0) + (r_d - r_b) t + \sigma_b W_b \\ & lnS_a(t) = lnS_a(0) + (r_d - r_a + \rho \sigma_a \sigma_b) t + \sigma_a W_a \\ & lnS_a \backsim \phi(\mu_1, \sigma_1), lnS_b \backsim \phi(\mu_2, \sigma_2)\\ & \mu_1 = lnS_a0 + (r_d - r_a -0.5\sigma_a^2 + \rho\sigma_a\sigma_b)T, \sigma_1 = \sigma_a\sqrt{T} \\ & \mu_2 = lnS_b0 + (r_d - r_b -0.5\sigma_b^2)T, \sigma_2 = \sigma_b\sqrt{T}\\ & f(b) = \frac{1}{\sqrt{2\pi}\sigma_b}exp[\frac{(lnb-\mu_2)^2}{\sigma_b^2}] \\ \qquad & f(a) = \frac{1}{\sqrt{2\pi}\sigma_a}exp[\frac{(lna-\mu_1)^2}{\sigma_a^2}] \\ \end{aligned} ​lnSb​(t)=lnSb​(0)+(rd​−rb​)t+σb​Wb​lnSa​(t)=lnSa​(0)+(rd​−ra​+ρσa​σb​)t+σa​Wa​lnSa​∽ϕ(μ1​,σ1​),lnSb​∽ϕ(μ2​,σ2​)μ1​=lnSa​0+(rd​−ra​−0.5σa2​+ρσa​σb​)T,σ1​=σa​T ​μ2​=lnSb​0+(rd​−rb​−0.5σb2​)T,σ2​=σb​T ​f(b)=2π ​σb​1​exp[σb2​(lnb−μ2​)2​]f(a)=2π ​σa​1​exp[σa2​(lna−μ1​)2​]​

\qquad 现在这两支股票是“不相关”的了, 根据Quanto期权的计算式，有:

E [ F ∗ S b ( T ) ∗ m a x ( S a ( T ) − K ) , 0 ] = F ∫ − ∞ ∞ ∫ K ∞ b ( a − K ) f ( a ) f ( b ) d a d b = F ∫ − ∞ ∞ ∫ K ∞ a b f ( a ) f ( b ) d a d b − F ∫ − ∞ ∞ ∫ K ∞ − K b f ( a ) f ( b ) d a d b \begin{aligned} \qquad & E[F*S_b(T)*max(S_a(T)-K),0] \\ & = F\int_{-\infty}^\infty\int_K^\infty {b(a-K)f(a)f(b)} \,{\rm d}a{\rm d}b \\ & = F\int_{-\infty}^\infty\int_K^\infty {abf(a)f(b)} \,{\rm d}a{\rm d}b - F\int_{-\infty}^\infty\int_K^\infty {-Kbf(a)f(b)} \,{\rm d}a{\rm d}b\\ \end{aligned} ​E[F∗Sb​(T)∗max(Sa​(T)−K),0]=F∫−∞∞​∫K∞​b(a−K)f(a)f(b)dadb=F∫−∞∞​∫K∞​abf(a)f(b)dadb−F∫−∞∞​∫K∞​−Kbf(a)f(b)dadb​

\qquad 第一部分：

∫ − ∞ ∞ ∫ K ∞ a b f ( a ) f ( b ) d a d b = ∫ K ∞ a f ( a ) ∫ − ∞ ∞ b f ( b ) d a d b = ∫ K ∞ a 2 π σ 2 e x p [ ( l n a − μ 1 ) 2 σ 1 2 ] ∫ − ∞ ∞ b 2 π σ 2 e x p [ ( l n b − μ 2 ) 2 σ 2 2 ] d a d b u s e t 1 = l n a , t 2 = l n b t o r e p l a c e , = ∫ l n K ∞ e t 2 2 π σ 2 e x p [ ( t 1 − μ 1 ) 2 σ 1 2 ] ∫ − ∞ ∞ e t 2 2 π σ 2 e x p [ ( t 2 − μ 2 ) 2 σ 2 2 ] d t 2 d t 1 ( ∗ ) \begin{aligned} \int_{-\infty}^\infty\int_K^\infty {abf(a)f(b)} \,{\rm d}a{\rm d}b & = \int_K^\infty af(a)\int_{-\infty}^\infty {bf(b)} \,{\rm d}a{\rm d}b \\ & = \int_{K}^\infty {\frac{a}{\sqrt{2\pi}\sigma_2}exp[\frac{(lna-\mu_1)^2}{\sigma_1^2}]}\int_{-\infty}^\infty{\frac{b}{\sqrt{2\pi}\sigma_2}exp[\frac{(lnb-\mu_2)^2}{\sigma_2^2}]} \,{\rm d}a{\rm d}b \\ \qquad use\;t_1 = lna, t_2 = lnb\;to\;replace,\\ & = \int_{lnK}^\infty {\frac{e^{t_2}}{\sqrt{2\pi}\sigma_2}exp[\frac{(t_1-\mu_1)^2}{\sigma_1^2}]}\int_{-\infty}^\infty{\frac{e^{t_2}}{\sqrt{2\pi}\sigma_2}exp[\frac{(t_2-\mu_2)^2}{\sigma_2^2}]} \,{\rm d}t_2{\rm d}t_1 (*)\\ \end{aligned} ∫−∞∞​∫K∞​abf(a)f(b)dadbuset1​=lna,t2​=lnbtoreplace,​=∫K∞​af(a)∫−∞∞​bf(b)dadb=∫K∞​2π ​σ2​a​exp[σ12​(lna−μ1​)2​]∫−∞∞​2π ​σ2​b​exp[σ22​(lnb−μ2​)2​]dadb=∫lnK∞​2π ​σ2​et2​​exp[σ12​(t1​−μ1​)2​]∫−∞∞​2π ​σ2​et2​​exp[σ22​(t2​−μ2​)2​]dt2​dt1​(∗)​

\qquad 我们将使用以下方法进行后续简化：

e t 1 e x p ( t 1 − μ 1 ) 2 σ 1 2 = e μ 1 + σ 1 2 2 e x p [ t 1 − ( μ 1 + σ 1 2 ) ] 2 2 σ 1 2 t 1 = t 1 − ( μ 1 + σ 1 2 ) σ 1 t 2 = t 2 − ( μ 1 + σ 2 2 ) σ 2 ( ∗ ) = e μ 1 + σ 1 2 2 { 1 − N [ l n K − ( μ 1 + σ 1 2 ) σ 1 ] } e μ 2 + σ 2 2 2 = e μ 1 + σ 1 2 2 + μ 2 + σ 2 2 2 N [ − l n K − ( μ 1 + σ 1 2 ) σ 1 ] } \begin{aligned} \qquad & {e^{t_1}exp{\frac{(t_1-\mu_1)^2}{\sigma_1^2}}} = e^{\mu_1+\frac{\sigma_1^2}{2}}exp{\frac{[t_1-(\mu_1+\sigma_1^2)]^2}{2\sigma_1^2}} \\ & t_1 = \frac{t_1-(\mu_1+\sigma_1^2)}{\sigma_1} \\ & t_2 = \frac{t_2-(\mu_1+\sigma_2^2)}{\sigma_2} \\ & (*) = e^{\mu_1+\frac{\sigma_1^2}{2}}\{1-N[\frac{lnK-(\mu_1+\sigma_1^2)}{\sigma_1}]\}e^{\mu_2+\frac{\sigma_2^2}{2}} \\ & = e^{\mu_1+\frac{\sigma_1^2}{2}+\mu_2+\frac{\sigma_2^2}{2}}N[-\frac{lnK-(\mu_1+\sigma_1^2)}{\sigma_1}]\} \\ \end{aligned} ​et1​expσ12​(t1​−μ1​)2​=eμ1​+2σ12​​exp2σ12​[t1​−(μ1​+σ12​)]2​t1​=σ1​t1​−(μ1​+σ12​)​t2​=σ2​t2​−(μ1​+σ22​)​(∗)=eμ1​+2σ12​​{1−N[σ1​lnK−(μ1​+σ12​)​]}eμ2​+2σ22​​=eμ1​+2σ12​​+μ2​+2σ22​​N[−σ1​lnK−(μ1​+σ12​)​]}​

\qquad 第二部分：

= ∫ − ∞ ∞ ∫ K ∞ − K b f ( a ) f ( b ) d a d b = ∫ K ∞ − K f ( a ) d a ∫ − ∞ ∞ b f ( b ) d b = ∫ K ∞ − K f ( a ) d a = − K ∫ K ∞ a 2 π σ 2 e x p [ ( l n a − μ 1 ) 2 σ 1 2 ] d a = − K ∫ l n K ∞ e t 1 2 π σ 2 e x p [ ( t 1 − μ 1 ) 2 σ 1 2 ] d t 1 = K [ 1 − N ( l n K − μ 1 σ 1 ) ] = K N ( − l n K − μ 1 σ 1 ) \begin{aligned} \qquad & = \int_{-\infty}^\infty\int_K^\infty {-Kbf(a)f(b)} \,{\rm d}a{\rm d}b\\ & = \int_K^\infty-Kf(a)da\int_{-\infty}^\infty bf(b) \,{\rm d}b = \int_K^\infty-Kf(a)da\\ & = -K\int_K^\infty{\frac{a}{\sqrt{2\pi}\sigma_2}exp[\frac{(lna-\mu_1)^2}{\sigma_1^2}]}da \\ & = -K\int_{lnK}^\infty{\frac{e^{t1}}{\sqrt{2\pi}\sigma_2}exp[\frac{(t1-\mu_1)^2}{\sigma_1^2}]}dt_1 \\ & = K[1-N(\frac{lnK-\mu_1}{\sigma_1})] = KN(-\frac{lnK-\mu_1}{\sigma_1})\\ \end{aligned} ​=∫−∞∞​∫K∞​−Kbf(a)f(b)dadb=∫K∞​−Kf(a)da∫−∞∞​bf(b)db=∫K∞​−Kf(a)da=−K∫K∞​2π ​σ2​a​exp[σ12​(lna−μ1​)2​]da=−K∫lnK∞​2π ​σ2​et1​exp[σ12​(t1−μ1​)2​]dt1​=K[1−N(σ1​lnK−μ1​​)]=KN(−σ1​lnK−μ1​​)​

三、结果整理

\qquad 引入 μ 1 , σ 1 \mu_1, \sigma_1 μ1​,σ1​:

− l n K − ( μ 1 + σ 1 2 ) σ 1 = ln ⁡ S a ( 0 ) K + ( r d − r a + ρ σ a σ b + 1 2 σ a 2 ) T σ a T = d + − l n K − μ 1 σ 1 = ln ⁡ S a ( 0 ) K + ( r d − r a + ρ σ a σ b − 1 2 σ a 2 ) T σ a T = d − ( ∗ ) = S b ( 0 ) S a ( 0 ) e ( r d − r a + ρ σ a σ b ) T N ( d + ) E [ F ∗ S b ( T ) ∗ m a x ( S a ( T ) − K ) , 0 ] = S b ( 0 ) F e − r b T ( S a ( 0 ) e ( r d − r a + ρ σ a σ b ) T N ( d + ) − K N ( d − ) ) . \begin{aligned} -\frac{lnK-(\mu_1+\sigma_1^2)}{\sigma_1} & = \frac{\ln \frac{S_a(0)}{K} + (r_d - r_a + \rho \sigma_a \sigma_b + \frac{1}{2}\sigma_a^2)T}{\sigma_a \sqrt{T}} = d_{+} \\ -\frac{lnK-\mu_1}{\sigma_1}& = \frac{\ln \frac{S_a(0)}{K} + (r_d - r_a + \rho \sigma_a \sigma_b - \frac{1}{2}\sigma_a^2)T}{\sigma_a \sqrt{T}} = d_{-} \\ (*) & = S_b(0)S_a(0) e^{(r_d - r_a + \rho \sigma_a \sigma_b)T} N(d_+) \\ \qquad E[F*S_b(T)*max(S_a(T)-K),0] & = S_b(0) F e^{-r_bT}\left( S_a(0) e^{(r_d - r_a + \rho \sigma_a \sigma_b)T} N(d_+) - K N(d_-) \right). \end{aligned} −σ1​lnK−(μ1​+σ12​)​−σ1​lnK−μ1​​(∗)E[F∗Sb​(T)∗max(Sa​(T)−K),0]​=σa​T ​lnKSa​(0)​+(rd​−ra​+ρσa​σb​+21​σa2​)T​=d+​=σa​T ​lnKSa​(0)​+(rd​−ra​+ρσa​σb​−21​σa2​)T​=d−​=Sb​(0)Sa​(0)e(rd​−ra​+ρσa​σb​)TN(d+​)=Sb​(0)Fe−rb​T(Sa​(0)e(rd​−ra​+ρσa​σb​)TN(d+​)−KN(d−​)).​

\qquad 大概的思路暂时是这些，如果大家发现问题，可以留言一起讨论，我后续还会修改的。