이번 포스팅은 Binomial option pricing and Black-Scholes (John Thickstun)의 내용도 참고하여 작성하였습니다.
블랙-숄즈 모형
현재 주가가 S S S K K K t t t
C = S ⋅ N ( d 1 ) − K e − r t ⋅ N ( d 2 ) C=S\cdot N(d_1)-Ke^{-rt}\cdot N(d_2)\\ C = S ⋅ N ( d 1 ) − K e − r t ⋅ N ( d 2 ) C = S ⋅ B ( a ; n t , ρ ) − K e − r t ⋅ B ( a ; n t , π ) C=S\cdot B(a;nt,\rho)-Ke^{-rt}\cdot B(a;nt,\pi) C = S ⋅ B ( a ; n t , ρ ) − K e − r t ⋅ B ( a ; n t , π ) 여기서 N ( ⋅ ) N(\cdot) N ( ⋅ ) 누적 분포 함수(Cumulative Distribution Function)입니다. 누적 분포 함수는 확률 변수가 어떤 값보다 작거나 같은 확률을 알려 주는 함수입니다.
N ( d 1 ) = p ( z ≤ d 1 ) N ( d 2 ) = p ( z ≤ d 2 ) \begin{aligned} N(d_1)&=p(z\leq d_1)\\ N(d_2)&=p(z\leq d_2) \end{aligned} N ( d 1 ) N ( d 2 ) = p ( z ≤ d 1 ) = p ( z ≤ d 2 ) 그리고 d 1 d_1 d 1 d 2 d_2 d 2
d 1 = ln ( S / K ) + ( r + σ 2 / 2 ) t σ t d 2 = ln ( S / K ) + ( r − σ 2 / 2 ) t σ t \begin{aligned} d_1&=\frac{\ln(S/K)+(r+\sigma^2/2)t}{\sigma\sqrt{t}}\\ d_2&=\frac{\ln(S/K)+(r-\sigma^2/2)t}{\sigma\sqrt{t}}\\ \end{aligned} d 1 d 2 = σ t ln ( S / K ) + ( r + σ 2 / 2 ) t = σ t ln ( S / K ) + ( r − σ 2 / 2 ) t n n n N ( ⋅ ) N(\cdot) N ( ⋅ ) B ( ⋅ ) B(\cdot) B ( ⋅ )
lim n → ∞ B ( a ; n t , ρ ) = N ( d 1 ) lim n → ∞ B ( a ; n t , π ) = N ( d 2 ) \begin{aligned} \lim_{n\rightarrow\infty}B(a;nt,\rho)=N(d_1)\\ \lim_{n\rightarrow\infty}B(a;nt,\pi)=N(d_2) \end{aligned} n → ∞ lim B ( a ; n t , ρ ) = N ( d 1 ) n → ∞ lim B ( a ; n t , π ) = N ( d 2 ) 일단 드므와브르-라플라스 정리에 따라 n n n
j ∼ B ( n t , p ) → d j ∼ N ( n t p , n t p ( 1 − p ) ) j \sim B(nt,p) \xrightarrow{d} j \sim N(ntp,ntp(1-p)) j ∼ B ( n t , p ) d j ∼ N ( n t p , n t p ( 1 − p ) ) 그리고 j j j z z z
z = j − n t p n t p ( 1 − p ) z=\frac{j-ntp}{\sqrt{ntp(1-p)}} z = n t p ( 1 − p ) j − n t p z ∼ N ( 0 , 1 ) z \sim N(0,1) z ∼ N ( 0 , 1 ) 따라서 동전을 n t nt n t a a a
B ( a ; n t , p ) = p ( j ≥ a ) = p ( z ≥ a − n t p n t p ( 1 − p ) ) = p ( z ≤ − a + n t p n t p ( 1 − p ) ) = N ( − a + n t p n t p ( 1 − p ) ) = N ( d ? ) \begin{aligned} B(a;nt,p)&=p(j\geq a)\\ &=p\left(z\geq \frac{a-ntp}{\sqrt{ntp(1-p)}}\right)\\ &=p\left(z\leq \frac{-a+ntp}{\sqrt{ntp(1-p)}}\right)\\ &=N\left(\frac{-a+ntp}{\sqrt{ntp(1-p)}}\right)\\ &=N(d_?) \end{aligned} B ( a ; n t , p ) = p ( j ≥ a ) = p ( z ≥ n t p ( 1 − p ) a − n t p ) = p ( z ≤ n t p ( 1 − p ) − a + n t p ) = N ( n t p ( 1 − p ) − a + n t p ) = N ( d ? ) 괄호 안의 값을 d ? d_? d ? p = ρ p=\rho p = ρ d ? = d 1 d_?=d_1 d ? = d 1 p = π p=\pi p = π d ? = d 2 d_?=d_2 d ? = d 2
d ? = − a + n t p n t p ( 1 − p ) d_?=\frac{-a+ntp}{\sqrt{ntp(1-p)}} d ? = n t p ( 1 − p ) − a + n t p 기하 브라운 운동
증명에 들어 가기 전에 잠시 기하 브라운 운동을 복습해 보겠습니다. 주가가 기하 브라운 운동을 한다면 주식의 (로그) 수익율은 산술 브라운 운동을 합니다. 동전을 던져서 앞면이 나오면 x i = 1 x_i=1 x i = 1 x i = − 1 x_i=-1 x i = − 1
ln ( S i / S i − 1 ) = μ ⋅ Δ t + σ ⋅ x i ⋅ Δ t \ln(S_{i}/S_{i-1})=\mu\cdot\Delta t+\sigma\cdot x_i\cdot\sqrt{\Delta t} ln ( S i / S i − 1 ) = μ ⋅ Δ t + σ ⋅ x i ⋅ Δ t E ( ln ( S i / S i − 1 ) ) = E ( μ ⋅ Δ t + σ ⋅ x i ⋅ Δ t ) = μ ⋅ Δ t + σ ⋅ Δ t ⋅ E ( x i ) = μ ⋅ Δ t ∵ E ( x i ) = 0 \begin{aligned} E(\ln(S_{i}/S_{i-1}))&=E(\mu\cdot\Delta t+\sigma\cdot x_i\cdot\sqrt{\Delta t})\\ &=\mu\cdot \Delta t+\sigma\cdot\sqrt{\Delta t}\cdot E(x_i)\\ &=\mu\cdot \Delta t\quad\because E(x_i)=0 \end{aligned} E ( ln ( S i / S i − 1 ) ) = E ( μ ⋅ Δ t + σ ⋅ x i ⋅ Δ t ) = μ ⋅ Δ t + σ ⋅ Δ t ⋅ E ( x i ) = μ ⋅ Δ t ∵ E ( x i ) = 0 V a r ( ln ( S i / S i − 1 ) ) = V a r ( μ ⋅ Δ t + σ ⋅ x i ⋅ Δ t ) = σ 2 ⋅ Δ t ⋅ V a r ( x i ) = σ 2 ⋅ Δ t ∵ V a r ( x i ) = 1 \begin{aligned} Var(\ln(S_{i}/S_{i-1}))&=Var(\mu\cdot\Delta t+\sigma\cdot x_i\cdot\sqrt{\Delta t})\\ &=\sigma^2\cdot\Delta t\cdot Var(x_i)\\ &=\sigma^2\cdot \Delta t\quad\because Var(x_i)=1 \end{aligned} V a r ( ln ( S i / S i − 1 ) ) = V a r ( μ ⋅ Δ t + σ ⋅ x i ⋅ Δ t ) = σ 2 ⋅ Δ t ⋅ V a r ( x i ) = σ 2 ⋅ Δ t ∵ V a r ( x i ) = 1 1년에 동전을 n n n Δ t = 1 / n \Delta t=1/n Δ t = 1 / n t t t n t nt n t t t t
ln ( S n t S ) = ln ( S 1 S ⋅ S 2 S 1 ⋅ . . . ⋅ S n t S n t − 1 ) = ln ( S 1 S ) + ln ( S 2 S 1 ) + . . . + ln ( S n t S n t − 1 ) = ∑ i = 1 n t ( ln ( S i / S i − 1 ) ) ∴ E ( ln ( S n t S ) ) = E ( ∑ i = 1 n t ( ln ( S i / S i − 1 ) ) ) = ∑ i = 1 n t E ( ln ( S i / S i − 1 ) ) ) = n t ⋅ μ ⋅ Δ t = μ t V a r ( ln ( S n t S ) ) = V a r ( ∑ i = 1 n t ( ln ( S i / S i − 1 ) ) ) = ∑ i = 1 n t V a r ( ln ( S i / S i − 1 ) ) ) = n t ⋅ σ 2 ⋅ Δ t = σ 2 t \begin{aligned} \ln\left(\frac{S_{nt}}{S}\right)&=\ln\left(\frac{S_1}{S}\cdot\frac{S_2}{S_1}\cdot...\cdot\frac{S_{nt}}{S_{nt-1}}\right)\\ &=\ln\left(\frac{S_{1}}{S}\right)+\ln\left(\frac{S_{2}}{S_1}\right)+...+\ln\left(\frac{S_{nt}}{S_{nt-1}}\right)\\ &=\displaystyle\sum_{i=1}^{nt}(\ln(S_{i}/S_{i-1}))\\ \therefore E\left(\ln\left(\frac{S_{nt}}{S}\right)\right)&=E\left(\displaystyle\sum_{i=1}^{nt}(\ln(S_{i}/S_{i-1}))\right)\\ &=\displaystyle\sum_{i=1}^{nt}E\left(\ln(S_{i}/S_{i-1}))\right)\\ &=nt\cdot\mu\cdot \Delta t\\ &=\mu t\\ Var\left(\ln\left(\frac{S_{nt}}{S}\right)\right)&=Var\left(\displaystyle\sum_{i=1}^{nt}(\ln(S_{i}/S_{i-1}))\right)\\ &=\displaystyle\sum_{i=1}^{nt}Var\left(\ln(S_{i}/S_{i-1}))\right)\\ &=nt\cdot\sigma^2\cdot \Delta t\\ &=\sigma^2 t \end{aligned} ln ( S S n t ) ∴ E ( ln ( S S n t ) ) V a r ( ln ( S S n t ) ) = ln ( S S 1 ⋅ S 1 S 2 ⋅ . . . ⋅ S n t − 1 S n t ) = ln ( S S 1 ) + ln ( S 1 S 2 ) + . . . + ln ( S n t − 1 S n t ) = i = 1 ∑ n t ( ln ( S i / S i − 1 ) ) = E ( i = 1 ∑ n t ( ln ( S i / S i − 1 ) ) ) = i = 1 ∑ n t E ( ln ( S i / S i − 1 ) ) ) = n t ⋅ μ ⋅ Δ t = μ t = V a r ( i = 1 ∑ n t ( ln ( S i / S i − 1 ) ) ) = i = 1 ∑ n t V a r ( ln ( S i / S i − 1 ) ) ) = n t ⋅ σ 2 ⋅ Δ t = σ 2 t 한편 로그 수익율이 산술 브라운 운동을 한다면 기대 수익율이 수익율의 기대값 보다 크다는 것을 기억하고 계실 겁니다.
ln ( E ( S n t S ) ) = ( μ + σ 2 2 ) t > μ t \ln\left(E\left(\frac{S_{nt}}{S}\right)\right)=\left(\mu+\frac{\sigma^2}{2}\right)t > \mu t ln ( E ( S S n t ) ) = ( μ + 2 σ 2 ) t > μ t 이항 모형의 극한
주가의 움직임을 기하 브라운 운동으로 가정한다면 이항 모형의 상승 계수 u u u d d d
ln ( u S / S ) = ln ( u ) = μ Δ t + σ Δ t ln ( d S / S ) = ln ( d ) = μ Δ t − σ Δ t \begin{aligned} \ln(uS/S)=\ln(u)&=\mu\Delta t+\sigma\sqrt{\Delta t}\\ \ln(dS/S)=\ln(d)&=\mu\Delta t-\sigma\sqrt{\Delta t}\\ \end{aligned} ln ( u S / S ) = ln ( u ) ln ( d S / S ) = ln ( d ) = μ Δ t + σ Δ t = μ Δ t − σ Δ t 그런데 기하 브라운 운동과는 달리 이항 모형에서는 실제 동전을 던져서 앞면이 나올 확률은 전혀 의미가 없습니다. 이미 살펴 보았듯이 이항 모형 그 어디에서도 실제 확률은 찾을 수 없습니다. 대신에 위험 중립 확률이라는 가상의 확률이 사용됩니다. 동전을 던져서 앞면이 나올 이 가상의 확률을 p p p
E ( ln ( S i / S i − 1 ) ) = p ⋅ ( μ Δ t + σ Δ t ) + ( 1 − p ) ( μ Δ t − σ Δ t ) = μ Δ t + ( 2 p − 1 ) σ Δ t = : μ ′ Δ t ∴ p = μ ′ Δ t − μ Δ t + σ Δ t 2 σ Δ t \begin{aligned} E(\ln(S_{i}/S_{i-1}))&=p\cdot(\mu \Delta t+\sigma\sqrt{\Delta t})+(1-p)(\mu\Delta t-\sigma\sqrt{\Delta t})\\ &=\mu \Delta t+(2p-1)\sigma\sqrt{\Delta t}\\ &=:\mu'\Delta t\\ \\ \therefore p&=\frac{\mu'\Delta t-\mu\Delta t+\sigma\sqrt{\Delta t}}{2\sigma\sqrt{\Delta t}} \end{aligned} E ( ln ( S i / S i − 1 ) ) ∴ p = p ⋅ ( μ Δ t + σ Δ t ) + ( 1 − p ) ( μ Δ t − σ Δ t ) = μ Δ t + ( 2 p − 1 ) σ Δ t = : μ ′ Δ t = 2 σ Δ t μ ′ Δ t − μ Δ t + σ Δ t 그리고 만기가 되었을 때 수익율의 기대값은 이렇게 됩니다(주석 (2)).
E ( ln ( S n t / S ) ) = n t ( p ⋅ ln ( u / d ) + ln ( d ) ) = n t ( p ⋅ 2 σ Δ t + μ Δ t − σ Δ t ) = n t μ ′ Δ t = μ ′ t ∵ ln ( u / d ) = ln ( u ) − ln ( d ) = μ Δ t + σ Δ t − ( μ Δ t − σ Δ t ) = 2 σ Δ t \begin{aligned} E\left(\ln\left({S_{nt}}/{S}\right)\right)&=nt\left(p\cdot\ln\left(u/d\right)+\ln(d)\right)\\ &=nt(p\cdot 2\sigma\sqrt{\Delta t}+\mu\Delta t-\sigma\sqrt{\Delta t})\\ &=nt\mu'\Delta t\\ &=\mu't\\ \\ \because \ln(u/d)&=\ln(u)-\ln(d)\\ &=\mu\Delta t+\sigma\sqrt{\Delta t}-(\mu\Delta t-\sigma\sqrt{\Delta t})\\ &=2\sigma\sqrt{\Delta t} \end{aligned} E ( ln ( S n t / S ) ) ∵ ln ( u / d ) = n t ( p ⋅ ln ( u / d ) + ln ( d ) ) = n t ( p ⋅ 2 σ Δ t + μ Δ t − σ Δ t ) = n t μ ′ Δ t = μ ′ t = ln ( u ) − ln ( d ) = μ Δ t + σ Δ t − ( μ Δ t − σ Δ t ) = 2 σ Δ t 복잡해 보이기는 하지만 1-기간 동안 수익율의 분산은 이렇게 계산됩니다(주석 (2)).
V a r ( ln ( S n t / S ) ) = n t p ( 1 − p ) ⋅ ( ln ( u / d ) ) 2 = n t ⋅ μ ′ Δ t − μ Δ t + σ Δ t 2 σ Δ t ⋅ ( 1 − μ ′ Δ t − μ Δ t + σ Δ t 2 σ Δ t ) ⋅ ( 2 σ Δ t ) 2 = − n t ⋅ ( ( μ ′ Δ t − μ Δ t ) + σ Δ t ) ( ( μ ′ Δ t − μ Δ t ) − σ Δ t ) = n t ⋅ σ 2 Δ t − n t ⋅ ( μ ′ − μ ) 2 Δ t 2 = σ 2 t − t ⋅ ( μ ′ − μ n ) 2 \begin{aligned} Var\left(\ln\left({S_{nt}}/{S}\right)\right)&=ntp(1-p)\cdot\left(\ln\left(u/d\right)\right)^2\\ &=nt\cdot\frac{\mu'\Delta t-\mu\Delta t+\sigma\sqrt{\Delta t}}{2\sigma\sqrt{\Delta t}}\cdot\left(1-\frac{\mu'\Delta t-\mu\Delta t+\sigma\sqrt{\Delta t}}{2\sigma\sqrt{\Delta t}}\right)\cdot(2\sigma\sqrt{\Delta t})^2\\ &=-nt\cdot((\mu'\Delta t-\mu\Delta t)+\sigma\sqrt{\Delta t})((\mu'\Delta t-\mu\Delta t)-\sigma\sqrt{\Delta t})\\ &=nt\cdot\sigma^2\Delta t-nt\cdot(\mu'-\mu)^2{\Delta t}^2\\ &=\sigma^2t-t\cdot\left(\frac{\mu'-\mu}{\sqrt{n}}\right)^2\\ \end{aligned} V a r ( ln ( S n t / S ) ) = n t p ( 1 − p ) ⋅ ( ln ( u / d ) ) 2 = n t ⋅ 2 σ Δ t μ ′ Δ t − μ Δ t + σ Δ t ⋅ ( 1 − 2 σ Δ t μ ′ Δ t − μ Δ t + σ Δ t ) ⋅ ( 2 σ Δ t ) 2 = − n t ⋅ ( ( μ ′ Δ t − μ Δ t ) + σ Δ t ) ( ( μ ′ Δ t − μ Δ t ) − σ Δ t ) = n t ⋅ σ 2 Δ t − n t ⋅ ( μ ′ − μ ) 2 Δ t 2 = σ 2 t − t ⋅ ( n μ ′ − μ ) 2 n n n σ 2 t \sigma^2t σ 2 t
∴ lim n → ∞ V a r ( ln ( S n t / S ) ) = σ 2 t \therefore \lim_{n\rightarrow\infty}Var\left(\ln\left({S_{nt}}/{S}\right)\right)=\sigma^2t ∴ n → ∞ lim V a r ( ln ( S n t / S ) ) = σ 2 t 그렇다면 수익율의 기대값도 기하 브라운 운동일 때와 같아야 될 것 같습니다.
∴ lim n → ∞ E ( ln ( S n t / S ) ) = μ t \therefore \lim_{n\rightarrow\infty}E\left(\ln\left({S_{nt}}/{S}\right)\right)=\mu t ∴ n → ∞ lim E ( ln ( S n t / S ) ) = μ t 그런데 이렇게 되려면 p p p
∴ lim n → ∞ p = 1 2 \therefore \lim_{n\rightarrow\infty}p=\frac{1}{2} ∴ n → ∞ lim p = 2 1 증명 1 단계
a a a
a = ln ( K / S ) − n t ln ( d ) ln ( u / d ) + ζ a=\frac{\ln\left(K/S\right)-nt\ln(d)}{\ln\left(u/d\right)}+\zeta a = ln ( u / d ) ln ( K / S ) − n t ln ( d ) + ζ a a a d ? d_? d ?
d ? = − a + n t p n t p ( 1 − p ) = − ln ( K / S ) − n t ln ( d ) ln ( u / d ) + ζ + n t p n t p ( 1 − p ) = ln ( S / K ) + n t ln ( d ) + n t p ln ( u / d ) ln ( u / d ) n t p ( 1 − p ) + ζ n t p ( 1 − p ) = ln ( S / K ) + n t ln ( d ) + n t p ln ( u / d ) ln ( u / d ) n t p ( 1 − p ) ∵ lim n → ∞ ζ = 0 \begin{aligned} d_?=\frac{-a+ntp}{\sqrt{ntp(1-p)}}&=\frac{-\frac{\ln\left(K/S\right)-nt\ln(d)}{\ln\left(u/d\right)}+\zeta+ntp}{\sqrt{ntp(1-p)}}\\ &=\frac{\ln\left(S/K\right)+nt\ln(d)+ntp\ln\left(u/d\right)}{\ln\left(u/d\right)\sqrt{ntp(1-p)}}+\frac{\zeta}{\sqrt{ntp(1-p)}}\\ &=\frac{\ln\left(S/K\right)+nt\ln(d)+ntp\ln\left(u/d\right)}{\ln\left(u/d\right)\sqrt{ntp(1-p)}}\quad\because \lim_{n\rightarrow\infty}{\zeta}=0 \end{aligned} d ? = n t p ( 1 − p ) − a + n t p = n t p ( 1 − p ) − l n ( u / d ) l n ( K / S ) − n t l n ( d ) + ζ + n t p = ln ( u / d ) n t p ( 1 − p ) ln ( S / K ) + n t ln ( d ) + n t p ln ( u / d ) + n t p ( 1 − p ) ζ = ln ( u / d ) n t p ( 1 − p ) ln ( S / K ) + n t ln ( d ) + n t p ln ( u / d ) ∵ n → ∞ lim ζ = 0 이항 모형에서 주식 수익율의 기대값과 분산은 다음과 같습니다(주석 2). 그리고 이 값들이 분모, 분자에 숨어 있었습니다.
E ( ln ( S n t / S ) ) = n t ( p ⋅ ln ( u / d ) + ln ( d ) ) V a r ( ln ( S n t / S ) ) = n t p ( 1 − p ) ⋅ ( ln ( u / d ) ) 2 \begin{aligned} E\left(\ln\left({S_{nt}}/{S}\right)\right)&=nt\left(p\cdot\ln\left(u/d\right)+\ln(d)\right)\\ Var\left(\ln\left({S_{nt}}/{S}\right)\right)&=ntp(1-p)\cdot\left(\ln\left(u/d\right)\right)^2 \end{aligned} E ( ln ( S n t / S ) ) V a r ( ln ( S n t / S ) ) = n t ( p ⋅ ln ( u / d ) + ln ( d ) ) = n t p ( 1 − p ) ⋅ ( ln ( u / d ) ) 2 이 값들을 이용하여 마저 정리하면 d ? d_? d ?
d ? = ln ( S / K ) + E ( ln ( S n t / S ) ) V a r ( ln ( S n t / S ) ) = ln ( S / K ) + E ( ln ( S n t / S ) ) σ t ∵ lim n → ∞ V a r ( ln ( S n t / S ) ) = σ 2 t \begin{aligned} d_?&=\frac{\ln\left(S/K\right)+E\left(\ln\left({S_{nt}}/{S}\right)\right)}{\sqrt{Var\left(\ln\left({S_{nt}}/{S}\right)\right)}}\\ &=\frac{\ln\left(S/K\right)+E\left(\ln\left({S_{nt}}/{S}\right)\right)}{\sigma\sqrt{t}}\quad\because \lim_{n\rightarrow\infty}Var\left(\ln\left({S_{nt}}/{S}\right)\right)=\sigma^2t \end{aligned} d ? = V a r ( ln ( S n t / S ) ) ln ( S / K ) + E ( ln ( S n t / S ) ) = σ t ln ( S / K ) + E ( ln ( S n t / S ) ) ∵ n → ∞ lim V a r ( ln ( S n t / S ) ) = σ 2 t 증명 2 단계
먼저 p = π p=\pi p = π d ? = d 2 d_?=d_2 d ? = d 2 π \pi π
E ( S i ) = π u S i − 1 + d S i − 1 ( 1 − π ) = S i − 1 e r Δ t E(S_i)=\pi uS_{i-1}+dS_{i-1}(1-\pi)=S_{i-1}e^{r\Delta t} E ( S i ) = π u S i − 1 + d S i − 1 ( 1 − π ) = S i − 1 e r Δ t E ( S i ) / S i − 1 = E ( S i / S i − 1 ) = e r Δ t E(S_i)/S_{i-1}=E(S_i/S_{i-1})=e^{r\Delta t} E ( S i ) / S i − 1 = E ( S i / S i − 1 ) = e r Δ t ∴ ln ( E ( S i / S i − 1 ) ) = r Δ t \therefore \ln(E(S_i/S_{i-1}))=r\Delta t ∴ ln ( E ( S i / S i − 1 ) ) = r Δ t 그래서 t t t
ln ( E ( S n t / S ) ) = ln ( E ( S 1 S ⋅ S 2 S 1 ⋅ . . . ⋅ S n t S n t − 1 ) ) = ln ( E ( S 1 S ) ⋅ E ( S 2 S 1 ) ⋅ . . . ⋅ E ( S n t S n t − 1 ) ) = ∑ i = 0 n t ln ( E ( S i S i − 1 ) ) = n t ⋅ r Δ t = r t \begin{aligned} \ln(E(S_{nt}/S))&=\ln\left(E\left(\frac{S_1}{S}\cdot\frac{S_2}{S_1}\cdot...\cdot\frac{S_{nt}}{S_{nt-1}}\right)\right)\\ &=\ln\left( E\left(\frac{S_1}{S}\right)\cdot E\left(\frac{S_2}{S_1}\right) \cdot...\cdot E\left(\frac{S_{nt}}{S_{nt-1}}\right) \right)\\ &=\displaystyle\sum_{i=0}^{nt}\ln\left( E\left(\frac{S_i}{S_{i-1}}\right)\right)\\ &=nt\cdot r\Delta t\\ &=rt \end{aligned} ln ( E ( S n t / S ) ) = ln ( E ( S S 1 ⋅ S 1 S 2 ⋅ . . . ⋅ S n t − 1 S n t ) ) = ln ( E ( S S 1 ) ⋅ E ( S 1 S 2 ) ⋅ . . . ⋅ E ( S n t − 1 S n t ) ) = i = 0 ∑ n t ln ( E ( S i − 1 S i ) ) = n t ⋅ r Δ t = r t 그런데 무한의 세계에서는 이항 모형이 기하 브라운 운동이 된다고 했으니 기대 수익율도 같아야 될 겁니다.
lim n → ∞ ln ( E ( S n t / S ) ) = ( μ + σ 2 2 ) t = μ t + σ 2 2 t = E ( ln ( S n t / S ) ) + σ 2 2 t ∵ lim n → ∞ E ( ln ( S n t / S ) ) = μ t = r t ∴ E ( ln ( S n t / S ) ) = ( r − σ 2 2 ) t \begin{aligned} \lim_{n\rightarrow\infty}\ln(E(S_{nt}/S))&=\left(\mu+\frac{\sigma^2}{2}\right)t\\ &=\mu t+\frac{\sigma^2}{2}t\\ &=E\left(\ln\left({S_{nt}}/{S}\right)\right)+\frac{\sigma^2}{2}t\quad\because \lim_{n\rightarrow\infty}E\left(\ln\left({S_{nt}}/{S}\right)\right)=\mu t \\ &=rt\\ \therefore E\left(\ln\left({S_{nt}}/{S}\right)\right)&=\left(r-\frac{\sigma^2}{2}\right)t \end{aligned} n → ∞ lim ln ( E ( S n t / S ) ) ∴ E ( ln ( S n t / S ) ) = ( μ + 2 σ 2 ) t = μ t + 2 σ 2 t = E ( ln ( S n t / S ) ) + 2 σ 2 t ∵ n → ∞ lim E ( ln ( S n t / S ) ) = μ t = r t = ( r − 2 σ 2 ) t 따라서 p = π p=\pi p = π d ? = d 2 d_?=d_2 d ? = d 2
d ? = ln ( S / K ) + E ( ln ( S n t / S ) ) σ t = ln ( S / K ) + ( r − σ 2 / 2 ) t σ t = d 2 \begin{aligned} d_?&=\frac{\ln\left(S/K\right)+E\left(\ln\left({S_{nt}}/{S}\right)\right)}{\sigma\sqrt{t}}\\ &=\frac{\ln\left(S/K\right)+(r-\sigma^2/2)t}{\sigma\sqrt{t}}=d_2 \end{aligned} d ? = σ t ln ( S / K ) + E ( ln ( S n t / S ) ) = σ t ln ( S / K ) + ( r − σ 2 / 2 ) t = d 2 증명 3 단계
이제 p = ρ p=\rho p = ρ d ? = d 1 d_?=d_1 d ? = d 1 ρ \rho ρ 1 / u 1/u 1 / u 1 / d 1/d 1 / d
S i − 1 = ρ S i − 1 + ( 1 − ρ ) S i − 1 = ρ ⋅ 1 u ⋅ u S i − 1 + ( 1 − ρ ) ⋅ 1 d ⋅ d S i − 1 = ρ ⋅ 1 u ⋅ E ( S i ) + ( 1 − ρ ) ⋅ 1 d ⋅ E ( S i ) = π ⋅ u ⋅ e − r Δ t u ⋅ E ( S i ) + ( 1 − π ) ⋅ d ⋅ e − r Δ t d ⋅ E ( S i ) = E ( S i ) ⋅ e − r Δ t \begin{aligned} S_{i-1}&=\rho S_{i-1}+(1-\rho)S_{i-1}\\ &=\rho \cdot \frac{1}{u} \cdot uS_{i-1}+(1-\rho) \cdot \frac{1}{d} \cdot dS_{i-1}\\ &=\rho \cdot \frac{1}{u} \cdot E(S_{i})+(1-\rho) \cdot \frac{1}{d} \cdot E(S_{i})\\ &=\frac{\pi\cdot u\cdot e^{-r\Delta t}}{u} \cdot E(S_{i})+\frac{(1-\pi)\cdot d\cdot e^{-r\Delta t}}{d} \cdot E(S_{i})\\ &=E(S_{i})\cdot e^{-r\Delta t} \end{aligned} S i − 1 = ρ S i − 1 + ( 1 − ρ ) S i − 1 = ρ ⋅ u 1 ⋅ u S i − 1 + ( 1 − ρ ) ⋅ d 1 ⋅ d S i − 1 = ρ ⋅ u 1 ⋅ E ( S i ) + ( 1 − ρ ) ⋅ d 1 ⋅ E ( S i ) = u π ⋅ u ⋅ e − r Δ t ⋅ E ( S i ) + d ( 1 − π ) ⋅ d ⋅ e − r Δ t ⋅ E ( S i ) = E ( S i ) ⋅ e − r Δ t S i − 1 / E ( S i ) = E ( S i − 1 / S i ) = e − r Δ t S_{i-1}/E(S_i)=E(S_{i-1}/S_i)=e^{-r\Delta t} S i − 1 / E ( S i ) = E ( S i − 1 / S i ) = e − r Δ t ∴ ln ( E ( S i − 1 / S i ) ) = − r Δ t \therefore \ln(E(S_{i-1}/S_i))=-r\Delta t ∴ ln ( E ( S i − 1 / S i ) ) = − r Δ t 그리고 t t t
ln ( E ( S / S n t ) ) = − r t \ln(E(S/S_{nt}))=-rt ln ( E ( S / S n t ) ) = − r t 한편 시간을 거꾸로 돌리면 로그 수익율은 이렇게 산술 브라운 운동을 하게 됩니다.
ln ( S / u S ) = ln ( 1 / u ) = − μ Δ t − σ Δ t ln ( S / d S ) = ln ( 1 / d ) = − μ Δ t + σ Δ t \begin{aligned} \ln(S/uS)=\ln(1/u)&=-\mu\Delta t-\sigma\sqrt{\Delta t}\\ \ln(S/dS)=\ln(1/d)&=-\mu\Delta t+\sigma\sqrt{\Delta t}\\ \end{aligned} ln ( S / u S ) = ln ( 1 / u ) ln ( S / d S ) = ln ( 1 / d ) = − μ Δ t − σ Δ t = − μ Δ t + σ Δ t n n n
lim n → ∞ V a r ( ln ( S / S n t ) ) = σ 2 t lim n → ∞ E ( ln ( S / S n t ) ) = − μ t lim n → ∞ ln ( E ( S / S n t ) ) = ( − μ + σ 2 2 ) t \begin{aligned} &\lim_{n\rightarrow\infty}Var\left(\ln\left({S}/{S_{nt}}\right)\right)=\sigma^2t\\ &\lim_{n\rightarrow\infty}E\left(\ln\left({S}/{S_{nt}}\right)\right)=-\mu t\\ &\lim_{n\rightarrow\infty}\ln(E(S/S_{nt}))=\left(-\mu+\frac{\sigma^2}{2}\right)t\\ \end{aligned} n → ∞ lim V a r ( ln ( S / S n t ) ) = σ 2 t n → ∞ lim E ( ln ( S / S n t ) ) = − μ t n → ∞ lim ln ( E ( S / S n t ) ) = ( − μ + 2 σ 2 ) t 마찬가지로 시간을 거꾸로 돌려도 기대 수익율은 기하 브라운 운동일 때와 같아야 될 겁니다.
lim n → ∞ ln ( E ( S / S n t ) ) = ( − μ + σ 2 2 ) t = − μ t + σ 2 2 t = E ( ln ( S / S n t ) ) + σ 2 2 t ∵ lim n → ∞ E ( ln ( S / S n t ) ) = − μ t = − r t ∴ E ( ln ( S / S n t ) ) = ( − r − σ 2 2 ) t = − E ( ln ( S n t / S ) ) \begin{aligned} \lim_{n\rightarrow\infty}\ln(E(S/S_{nt}))&=\left(-\mu+\frac{\sigma^2}{2}\right)t\\ &=-\mu t+\frac{\sigma^2}{2}t\\ &=E\left(\ln\left({S}/{S_{nt}}\right)\right)+\frac{\sigma^2}{2}t\quad\because \lim_{n\rightarrow\infty}E\left(\ln\left({S}/{S_{nt}}\right)\right)=-\mu t \\ &=-rt\\ \therefore E\left(\ln\left({S}/{S_{nt}}\right)\right)&=\left(-r-\frac{\sigma^2}{2}\right)t\\ &=-E\left(\ln\left({S_{nt}}/{S}\right)\right) \end{aligned} n → ∞ lim ln ( E ( S / S n t ) ) ∴ E ( ln ( S / S n t ) ) = ( − μ + 2 σ 2 ) t = − μ t + 2 σ 2 t = E ( ln ( S / S n t ) ) + 2 σ 2 t ∵ n → ∞ lim E ( ln ( S / S n t ) ) = − μ t = − r t = ( − r − 2 σ 2 ) t = − E ( ln ( S n t / S ) ) 따라서 p = ρ p=\rho p = ρ d ? = d 1 d_?=d_1 d ? = d 1
d ? = ln ( S / K ) + E ( ln ( S n t / S ) ) σ t = ln ( S / K ) + ( r + σ 2 / 2 ) t σ t = d 1 \begin{aligned} d_?&=\frac{\ln\left(S/K\right)+E\left(\ln\left({S_{nt}}/{S}\right)\right)}{\sigma\sqrt{t}}\\ &=\frac{\ln\left(S/K\right)+(r+\sigma^2/2)t}{\sigma\sqrt{t}}=d_1 \end{aligned} d ? = σ t ln ( S / K ) + E ( ln ( S n t / S ) ) = σ t ln ( S / K ) + ( r + σ 2 / 2 ) t = d 1 주석
(1) 앞면이 적어도 나와야 할 회수
만기에 옵션이 가치를 가지려면 만기 주가가 행사 가격 보다는 높아야 합니다.
u a d n t − a S > K ln ( u a d n t − a S ) > ln ( K ) a ln ( u ) + ( n t − a ) ln ( d ) + ln ( S ) > ln ( K ) a ( ln ( u ) − ln ( d ) ) > ln ( K ) − ln ( S ) − n t ln ( d ) ∴ a > ln ( K / S ) − n t ln ( d ) ln ( u / d ) \begin{aligned} u^ad^{nt-a}S&>K\\ \ln(u^ad^{nt-a}S)&>\ln(K)\\ a\ln(u)+(nt-a)\ln(d)+\ln(S) &> \ln(K)\\ a(\ln(u)-\ln(d)) &> \ln(K)-\ln(S)-nt\ln(d)\\ \therefore a &> \frac{\ln\left(K/S\right)-nt\ln(d)}{\ln\left(u/d\right)} \end{aligned} u a d n t − a S ln ( u a d n t − a S ) a ln ( u ) + ( n t − a ) ln ( d ) + ln ( S ) a ( ln ( u ) − ln ( d ) ) ∴ a > K > ln ( K ) > ln ( K ) > ln ( K ) − ln ( S ) − n t ln ( d ) > ln ( u / d ) ln ( K / S ) − n t ln ( d ) a a a ζ \zeta ζ
a = ln ( K / S ) − n t ln ( d ) ln ( u / d ) + ζ , 0 ≤ ζ < 1 a=\frac{\ln\left(K/S\right)-nt\ln(d)}{\ln\left(u/d\right)}+\zeta,\quad 0\leq\zeta < 1 a = ln ( u / d ) ln ( K / S ) − n t ln ( d ) + ζ , 0 ≤ ζ < 1 n n n S S S K K K n n n ζ \zeta ζ
(2) 주식 수익율의 기대값과 분산
이항 모형에서 콜 옵션의 만기가 되었을 때 주식의 로그 수익율은 이렇게 됩니다.
ln ( S n t / S ) = ln ( u j d n t − j S S ) = j ln ( u ) + ( n t − j ) ln ( d ) = j ln ( u / d ) + n t ln ( d ) \begin{aligned} \ln\left({S_{nt}}/{S}\right)&=\ln\left(\frac{u^jd^{nt-j}S}{S}\right)\\ &=j\ln(u)+(nt-j)\ln(d)\\ &=j\ln\left(u/d\right)+nt\ln(d) \end{aligned} ln ( S n t / S ) = ln ( S u j d n t − j S ) = j ln ( u ) + ( n t − j ) ln ( d ) = j ln ( u / d ) + n t ln ( d ) 그리고 기대값과 분산은 다음과 같습니다.
E ( ln ( S n t / S ) ) = E ( j ln ( u / d ) + n t ln ( d ) ) = E ( j ) ⋅ ln ( u / d ) + n t ln ( d ) = n t p ⋅ ln ( u / d ) + n t ln ( d ) = n t ( p ⋅ ln ( u / d ) + ln ( d ) ) \begin{aligned} E\left(\ln\left({S_{nt}}/{S}\right)\right)&=E\left(j\ln\left(u/d\right)+nt\ln(d)\right)\\ &=E(j)\cdot\ln\left(u/d\right)+nt\ln(d)\\ &=ntp\cdot\ln\left(u/d\right)+nt\ln(d)\\ &=nt\left(p\cdot\ln\left(u/d\right)+\ln(d)\right) \end{aligned} E ( ln ( S n t / S ) ) = E ( j ln ( u / d ) + n t ln ( d ) ) = E ( j ) ⋅ ln ( u / d ) + n t ln ( d ) = n t p ⋅ ln ( u / d ) + n t ln ( d ) = n t ( p ⋅ ln ( u / d ) + ln ( d ) ) V a r ( ln ( S n t / S ) ) = V a r ( j ln ( u / d ) + n t ln ( d ) ) = V a r ( j ) ⋅ ( ln ( u / d ) ) 2 = n t p ( 1 − p ) ⋅ ( ln ( u / d ) ) 2 \begin{aligned} Var\left(\ln\left({S_{nt}}/{S}\right)\right)&=Var\left(j\ln\left(u/d\right)+nt\ln(d)\right)\\ &=Var(j)\cdot\left(\ln\left(u/d\right)\right)^2\\ &=ntp(1-p)\cdot\left(\ln\left(u/d\right)\right)^2 \end{aligned} V a r ( ln ( S n t / S ) ) = V a r ( j ln ( u / d ) + n t ln ( d ) ) = V a r ( j ) ⋅ ( ln ( u / d ) ) 2 = n t p ( 1 − p ) ⋅ ( ln ( u / d ) ) 2 