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On this page
  • Basic Wiener Process
  • Generalized Wiener Process
  • Ito Process
  • Ito's Lemma
  • Proof
  • Property of Stock Prices
  • The Process for A Stock Return
  • The Log Return
  • The Lognormal Property of Stock Prices
  • Black-Scholes Equation
  • Proof

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  1. Trading

The Deriving of Black-Scholes Equation

Basic Wiener Process

dz=εdt\mathrm{d}z = \varepsilon \sqrt{\mathrm{d}t}dz=εdt​

where ε\varepsilonε has a standard normal distribution ϕ(0,1)\phi(0, 1)ϕ(0,1).

Generalized Wiener Process

A generalized Wiener process for a variable xxx can be defined in terms of dx\mathrm{d} xdx

dx=adt+bdz\mathrm{d} x = a \mathrm{d} t + b\mathrm{d}zdx=adt+bdz

where a and b are constants.

Ito Process

An Ito process for a variable dx\mathrm{d}xdx can be written as

dx=a(x,t)dt+b(x,t)dz\mathrm{d}x = a( x, t) \mathrm{d}t + b( x, t )\mathrm{d}zdx=a(x,t)dt+b(x,t)dz

where a(x,t)a(x,t)a(x,t) and b(x,t)b(x, t)b(x,t) are functions of the value of the underlying variable x and time t, and dz is a Wiener process.

Ito's Lemma

Suppose that the value of a variable x\mathrm{x}x follows the Ito process, Ito's lemma shows that a function G of x and t follows the process

dG=(∂G∂xa(x,t)+∂G∂t+12∂2G∂x2b(x,t)2)dt+∂G∂xb(x,t)dzdG = \left ( \frac{\partial{G}}{\partial{x}}a(x,t)+\frac{\partial{G}}{\partial{t}}+\frac{1}{2} \frac{\partial^2{G}}{\partial{x^2}} b(x,t)^2 \right )\mathrm{d}t+\frac{\partial{G}}{\partial{x}}b(x,t)\mathrm{d}zdG=(∂x∂G​a(x,t)+∂t∂G​+21​∂x2∂2G​b(x,t)2)dt+∂x∂G​b(x,t)dz

Note, G also follows an Ito process.

Proof

First of all, let's solve the mean and variance of YYYgiven Y=X2Y=X^2Y=X2where X∼ϕ(0,σ2)X \sim \phi(0, \sigma^2)X∼ϕ(0,σ2).

Since σ2=E(X2)−E(X)2\sigma^2 = E(X^2) - E(X)^2σ2=E(X2)−E(X)2, and E(X)=0E(X) = 0E(X)=0, we get

E(X2)=σ2E(X^2) = \sigma^2E(X2)=σ2

And becuaseVarX2=EX4−(EX2)2VarX^2 = EX^4 - (EX^2)^2VarX2=EX4−(EX2)2, and

E(X4)=∫x42πσe−x22σ2dx=∫−σ2x32πσe−x22σ2d−x22σ2=∫−σ2x32πσde−x22σ2=12πσ(−σ2x3e−x22σ2∣−∞+∞+3σ2∫x2e−x22σ2dx)=12πσ(0−3σ4xe−x22σ2∣−∞+∞+3σ4∫e−x22σ2dx)=3σ4\begin{aligned} E(X^4) &= \int \frac{x^4}{\sqrt{2\pi}\sigma}e^{-\frac{x^2}{2\sigma^2}}dx \\ &=\int -\frac{\sigma^2x^3}{\sqrt{2\pi}\sigma}e^{-\frac{x^2}{2\sigma^2}}d-\frac{x^2}{2\sigma^2} \\&=\int -\frac{\sigma^2x^3}{\sqrt{2\pi}\sigma}de^{-\frac{x^2}{2\sigma^2}} \\&= \frac{1}{\sqrt{2\pi}\sigma}\left ( -\sigma^2x^3e^{-{\frac{x^2}{2\sigma^2}}}|_{-\infty }^{+\infty } + 3\sigma^2\int x^2e^{-{\frac{x^2}{2\sigma^2}}}dx \right ) \\ &= \frac{1}{\sqrt{2\pi}\sigma} \left ( 0 - 3\sigma^4xe^{-{\frac{x^2}{2\sigma^2}}}|_{-\infty }^{+\infty } + 3\sigma^4\int e^{-{\frac{x^2}{2\sigma^2}}}dx \right ) \\ &= 3\sigma^4 \end{aligned}E(X4)​=∫2π​σx4​e−2σ2x2​dx=∫−2π​σσ2x3​e−2σ2x2​d−2σ2x2​=∫−2π​σσ2x3​de−2σ2x2​=2π​σ1​(−σ2x3e−2σ2x2​∣−∞+∞​+3σ2∫x2e−2σ2x2​dx)=2π​σ1​(0−3σ4xe−2σ2x2​∣−∞+∞​+3σ4∫e−2σ2x2​dx)=3σ4​

Actually, the generalized form of E[X2n]E[X^{2n}]E[X2n]is,

E[X2n]=(2n−1)!!σ2nE\left [ X^{2n}\right ] = (2n - 1)!!\sigma^{2n}E[X2n]=(2n−1)!!σ2n,

Thus,

VarX2=EX4−(EX2)2=3σ4−σ4=2σ4\begin{aligned} VarX^2 &= EX^4 - (EX^2)^2 \\ &= 3\sigma^4 - \sigma^4 \\ &= 2\sigma^4 \end{aligned}VarX2​=EX4−(EX2)2=3σ4−σ4=2σ4​

Consider a continuous and differentiable function G of two variables x and t, a Taylor series expansion of ΔG\Delta GΔG is

ΔG=∂G∂xΔx+∂G∂tΔt+12∂2G∂x2Δx2+∂2G∂x∂tΔxΔt+12∂2G∂t2Δt2+…\Delta{G}=\frac{\partial{G}}{\partial{x}}\Delta{x} + \frac{\partial{G}}{\partial{t}}\Delta{t} + \frac{1}{2}\frac{\partial^2G}{\partial{x}^2}\Delta{x}^2 + \frac{\partial^2{G}}{\partial{x}\partial{t}}\Delta{x}\Delta{t}+\frac{1}{2}\frac{\partial^2{G}}{\partial{t}^2}\Delta{t}^2+\dotsΔG=∂x∂G​Δx+∂t∂G​Δt+21​∂x2∂2G​Δx2+∂x∂t∂2G​ΔxΔt+21​∂t2∂2G​Δt2+…

discretize the Ito process to

Δx=a(x,t)Δt+b(x,t)Δz=a(x,t)Δt+b(x,t)εΔt\begin{aligned} \mathrm{\Delta}x &= a( x, t) \mathrm{\Delta}t + b( x, t )\mathrm{\Delta}z \\ &= a(x, t)\Delta{t}+b(x,t)\varepsilon\sqrt{\Delta{t}} \end{aligned}Δx​=a(x,t)Δt+b(x,t)Δz=a(x,t)Δt+b(x,t)εΔt​​

then we have

Δx2=b2(x,t)ε2Δt+terms of higher order in Δt\Delta{x}^2 = b^2(x, t)\varepsilon^2\Delta{t} + \text{terms of higher order in } \Delta{t}Δx2=b2(x,t)ε2Δt+terms of higher order in Δt

It shows that the term involving Δx2\Delta{x}^2Δx2 in the Taylor series expansion cannot be ignored, and

E(X2)=σ2 and Var(X2)=2σ4E(X^2) = \sigma^2 \text{ and }Var{(X^2)} = 2\sigma^4E(X2)=σ2 and Var(X2)=2σ4

Since the variance of a standard normal distribution is 1, E(ε2Δt)=ΔtE(\varepsilon^2\Delta{t}) = \Delta{t}E(ε2Δt)=Δt and Var(ε2Δt)=2Δt2.Var(\varepsilon^2\Delta{t}) = 2\Delta{t}^2.Var(ε2Δt)=2Δt2.

We know that the variance of the change in a stochastic variable in time Δt\Delta{t}Δt is proportional to Δt\Delta{t}Δt, not Δt2\Delta{t}^2Δt2. The variance of ε2Δt\varepsilon^2\Delta{t}ε2Δt is therefore too small for it to have a stochastic component. As a result, we can treat it as nonstochastic and equal to its expected value, Δt\Delta{t}Δt, as Δt\Delta{t}Δt tends to zero.

Taking limits as Δx\Delta{x}Δx and Δt\Delta{t}Δt tend to zero, and using dx2=b2(x,t)dtdx^2 = b^2(x,t)dtdx2=b2(x,t)dt, we obtain

dG=∂G∂xdx+∂G∂tdt+12∂2G∂x2b2(x,t)dtdG = \frac{\partial{G}}{\partial{x}}dx + \frac{\partial{G}}{\partial{t}}dt + \frac{1}{2}\frac{\partial^2{G}}{\partial{x}^2}b^2(x,t)dtdG=∂x∂G​dx+∂t∂G​dt+21​∂x2∂2G​b2(x,t)dt

if we substitute the dx with dx=a(x,t)dt+b(x,t)dz\mathrm{d}x = a( x, t) \mathrm{d}t + b( x, t )\mathrm{d}zdx=a(x,t)dt+b(x,t)dz, equation becomes

dG=(∂G∂xa(x,t)+∂G∂t+12∂2G∂x2b2(x,t))dt+∂G∂xb(x,t)dzdG = \left ( \frac{\partial{G}}{\partial{x}}a(x,t) + \frac{\partial{G}}{\partial{t}} + \frac{1}{2}\frac{\partial^2G}{\partial{x}^2} b^2(x,t)\right )dt + \frac{\partial{G}}{\partial{x}}b(x,t)dzdG=(∂x∂G​a(x,t)+∂t∂G​+21​∂x2∂2G​b2(x,t))dt+∂x∂G​b(x,t)dz

Property of Stock Prices

The Process for A Stock Return

The most widely used model of stock return is

dSS=μdt+σdz and ΔSS∼ϕ(μΔt,σ2Δt)\frac{dS}{S} = \mu \mathrm{d}t + \sigma \mathrm{d}z \text{ and } \frac{\Delta{S}}{S} \sim \phi(\mu\Delta{t}, \sigma^2\Delta{t})SdS​=μdt+σdz and SΔS​∼ϕ(μΔt,σ2Δt)

where μ\muμ is the stock's expected rate of return, and σ\sigmaσ is the volatility of the stock price. It represents the stock return process in the real world. In a risk-neutral world, μ\muμ equals the risk-free rate rrr.

The Log Return

The Taylor series expansion of ln⁡(1+x)\ln{(1+x)}ln(1+x) is

ln⁡(1+x)=0+11+0×Δx−121(1+0)2×Δx2+⋯≈x\ln{(1+x)} = 0 + \frac{1}{1+0}\times \Delta{x} - \frac{1}{2}\frac{1}{(1+0)^2} \times \Delta{x}^2 + \dots \approx xln(1+x)=0+1+01​×Δx−21​(1+0)21​×Δx2+⋯≈x

for small Δx\Delta{x}Δx and ignore the terms of higher order in Δx\Delta{x}Δx.

Thus, ln⁡StS0=ln⁡(1+St−S0S0)≈ΔSS0 ONLY for small ΔS\ln{\frac{S_t}{S_0}} = \ln{(1 + \frac{S_t-S_0}{S_0})} \approx \frac{\Delta{S}}{S_0} \text{ ONLY for small } \Delta{S}lnS0​St​​=ln(1+S0​St​−S0​​)≈S0​ΔS​ ONLY for small ΔS, also note that StS_tSt​ here is instantaneous price.

The log rate of return is also called the continuously compounded rate of return.

The Lognormal Property of Stock Prices

Define

G=ln⁡SG = \ln{S}G=lnS

Where S follows dSS=μdt+σdz\frac{dS}{S} = \mu \mathrm{d}t + \sigma \mathrm{d}zSdS​=μdt+σdz.

Since

∂G∂S=1S,∂2G∂S2=−1S2,∂G∂t=0\frac{\partial{G}}{\partial{S}} = \frac{1}{S}, \frac{\partial^2G}{\partial{S^2}} = - \frac{1}{S^2}, \frac{\partial{G}}{\partial{t}} = 0∂S∂G​=S1​,∂S2∂2G​=−S21​,∂t∂G​=0

Using Ito's Lemma, we can get

dG=(μ−σ22)dt+σdzdG = \left ( \mu - \frac{\sigma^2}{2}\right)dt + \sigma dzdG=(μ−2σ2​)dt+σdz

Since μ\muμand σ\sigmaσare constant, this equation indicates that G=ln⁡SG = \ln{S}G=lnSfollows a generalized Wiener process. It has constant drift rate μ−σ2/2\mu - \sigma^2/2μ−σ2/2and constant variance rate σ2\sigma^2σ2. The change in ln⁡S\ln{S}lnSbetween time 0 and some future time TTTis therefore normally distributed, with mean (μ−σ2/2)\left( \mu-\sigma^2/2\right)(μ−σ2/2)and variance σ2T\sigma^2Tσ2T. This means that

ln⁡STS0∼ϕ[(u−σ2/2)T,σ2T]\ln\frac{S_T}{S_0} \sim \phi{\left [\left ( u - \sigma^2/2 \right )T, \sigma^2T \right ]}lnS0​ST​​∼ϕ[(u−σ2/2)T,σ2T]

Where STS_TST​is the stock price at time T, S0S_0S0​is the stock price at time 0.

Black-Scholes Equation

c=S0N(d1)−Ke−rTN(d2)c = S_0N(d_1) - Ke^{-rT}N(d_2)c=S0​N(d1​)−Ke−rTN(d2​)
p=Ke−rTN(−d2)−S0N(−d1)p = Ke^{-rT}N(-d_2)-S_0N(-d_1)p=Ke−rTN(−d2​)−S0​N(−d1​)

Where

d1=ln⁡(S0/K)+(r+σ2/2)TσTd_1 = \frac{\ln{\left ( S_0 / K \right )} + \left ( r + \sigma^2/2\right )T}{\sigma \sqrt{T}}d1​=σT​ln(S0​/K)+(r+σ2/2)T​
d2=ln⁡(S0/K)+(r−σ2/2)TσT=d1−σTd_2 = \frac{\ln{\left ( S_0/K\right )} +\left ( r - \sigma^2/2\right )T}{\sigma \sqrt T} = d_1 - \sigma \sqrt{T}d2​=σT​ln(S0​/K)+(r−σ2/2)T​=d1​−σT​

Proof

Key Result

If V is lognormally distributed, and the mean of ln⁡V\ln{V}lnVis mmm and the standard deviation of ln⁡V\ln{V}lnVis www, then

E(max⁡(V−K,0))=E(V)N(d1)−KN(d2)E( \max(V - K, 0)) = E(V)N(d_1) - KN(d_2)E(max(V−K,0))=E(V)N(d1​)−KN(d2​)

Where

d1=ln⁡[E(V)/K]+w2/2wd_1 = \frac{\ln{\left [ E(V)/K\right ]} + w^2/2}{w}d1​=wln[E(V)/K]+w2/2​
d2=ln⁡[E(V)/K]−w2/2wd_2 = \frac{\ln \left [ E(V) / K\right ] - w^2/2}{w}d2​=wln[E(V)/K]−w2/2​

Note that as to the call option,

  • N(d2)N(d_2)N(d2​)is the probability of exercise,

  • S0erTN(d1)/N(d2)S_0e^{rT}N(d_1)/N(d_2)S0​erTN(d1​)/N(d2​)is the expected stock price in a risk-neutral world if option is exercised

Similarly, as to the put option

  • N(−d2)N(-d_2)N(−d2​)is the probability of exercise,

  • S0N(−d1)/N(−d2)S_0N(-d_1)/N(-d_2)S0​N(−d1​)/N(−d2​)is the expected stock price in a risk-neutral world if option is exercised

Proof of Key Result

Define g(V)g(V)g(V)as the probability density function of V. It follows that

E[max⁡(V−K,0)]=∫K∞(V−K)g(V)dVE\left [ \max(V-K, 0) \right ] = \int_{K}^{\infty} (V-K)g(V)dVE[max(V−K,0)]=∫K∞​(V−K)g(V)dV
E(V)=em+12σ2E(V)=e^{m+\frac{1}{2}\sigma^2}E(V)=em+21​σ2
m=ln⁡E(V)−w2/2m = \ln{E(V)} - w^2/2m=lnE(V)−w2/2

Define a new variable

Q=f(V)=ln⁡V−mwQ = f(V) = \frac{\ln{V} - m}{w}Q=f(V)=wlnV−m​

It follows standard normal distribution,

Denote the density function of QQQby h(Q)h(Q)h(Q)so thath(Q)=12πe−Q2/2h(Q) = \frac{1}{\sqrt{2\pi}}e^{-Q^2/2}h(Q)=2π​1​e−Q2/2,

Support RQ={q=f(v):v∈RV}R_{Q} = \left \{ q = f(v) : v \in R_{V}\right \}RQ​={q=f(v):v∈RV​},

h(q)={g(f−1(q))df−1(q)dq,if q∈RQ0,ifq∉RQ\begin{aligned} h(q) = \left \{ \begin{aligned} &g(f^{-1}(q))\frac{df^{-1}(q)}{dq}, && \text{if}\ q \in R_{Q}\\ &0, && \text{if} q \notin R_{Q} \end{aligned} \right. \end{aligned}h(q)=⎩⎨⎧​​g(f−1(q))dqdf−1(q)​,0,​​if q∈RQ​ifq∈/RQ​​​
E[max⁡(V−K,0)]=∫K∞(V−K)g(V)dv=∫f(K)∞[f−1(Q)−K]g(f−1(Q))d(f−1(Q))dQdQ=∫(ln⁡K−m)/w∞(eQw+m−K)h(Q)dQ=∫(ln⁡K−m)/w∞eQw+mh(Q)dQ−K∫(ln⁡K−m)/2∞h(Q)d(Q)\begin{aligned} E\left [ \max(V-K, 0) \right ] &= \int_{K}^{\infty} (V-K)g(V)dv \\ &= \int_{f(K)}^{\infty} \left [f^{-1}(Q) - K\right ]g(f^{-1}(Q))\frac{d(f^{-1}(Q))}{dQ}dQ \\ &= \int_{(\ln{K}-m)/w}^{\infty}\left( e^{Qw+m} - K \right )h(Q)dQ \\ &= \int_{(\ln{K} -m)/w}^{\infty}e^{Qw+m}h(Q)dQ-K\int_{(\ln{K} - m)/2}^{\infty}h(Q)d(Q) \end{aligned}E[max(V−K,0)]​=∫K∞​(V−K)g(V)dv=∫f(K)∞​[f−1(Q)−K]g(f−1(Q))dQd(f−1(Q))​dQ=∫(lnK−m)/w∞​(eQw+m−K)h(Q)dQ=∫(lnK−m)/w∞​eQw+mh(Q)dQ−K∫(lnK−m)/2∞​h(Q)d(Q)​

Now

eQw+mh(Q)=12πe(−Q2+2Qw+2m)/2=12πe[−(Q−w)2+2m+w2]/2=em+w2/22πe[−(Q−w)2]/2=em+w2/2h(Q−w)\begin{aligned} e^{Qw+m}h(Q) &= \frac{1}{\sqrt{2\pi}}e^{(-Q^2+2Qw+2m)/2} \\ &=\frac{1}{\sqrt{2\pi}}e^{\left [ -(Q-w)^2+2m+w^2 \right ]/2} \\ &=\frac{e^{m+w^2/2}}{\sqrt{2\pi}}e^{\left [ -(Q-w)^2\right ]/2} \\ &= e^{m+w^2/2}h(Q-w) \end{aligned}eQw+mh(Q)​=2π​1​e(−Q2+2Qw+2m)/2=2π​1​e[−(Q−w)2+2m+w2]/2=2π​em+w2/2​e[−(Q−w)2]/2=em+w2/2h(Q−w)​

This means that

E(max(V−K,0))=em+w2/2∫(ln⁡K−m)/w∞h(Q−w)dQ−K∫(ln⁡K−m)/w∞h(Q)dQ=em+w2/2∫(ln⁡K−m)/w−w∞h(Q−w)d(Q−w)−K∫(ln⁡K−m)/w∞h(Q)dQ=em+w2/2[1−N(ln⁡K−mw−w)]−K[1−N(ln⁡K−mw)]=em+w2/2N(−ln⁡K+mw+w)−KN(−ln⁡K+mw)=E(V)N(−ln⁡K+ln⁡E(V)−w2/2w+w)−KN(−ln⁡K+ln⁡E(V)−w2/2w)=E(V)N(ln⁡E(V)K+w2/2w)−KN(ln⁡E(V)K−w2/2w)=E(V)N(d1)−KN(d2)\begin{aligned} E(max(V-K, 0)) &= e^{m+w^2/2}\int_{(\ln{K}-m)/w}^{\infty}h(Q-w)dQ - K\int_{(\ln{K}-m)/w}^{\infty}h(Q)dQ \\ &= e^{m+w^2/2}\int_{(\ln{K}-m)/w-w}^{\infty}h(Q-w)d(Q-w) - K\int_{(\ln{K}-m)/w}^{\infty}h(Q)dQ \\ &= e^{m+w^2/2}\left[1 - N\left (\frac{\ln{K}-m}{w}-w\right )\right] - K\left[1 - N\left(\frac{\ln{K-m}}{w}\right) \right] \\ &=e^{m+w^2/2} N\left (\frac{-\ln{K}+m}{w}+w \right) - K N\left ( \frac{-\ln{K}+m}{w}\right ) \\ &= E(V) N\left (\frac{-\ln{K}+\ln{E(V)}-w^2/2}{w}+w \right) - K N\left ( \frac{-\ln{K}+\ln{E(V)}-w^2/2}{w}\right ) \\ &= E(V) N\left (\frac{\ln{\frac{E(V)}{K}}+w^2/2}{w} \right) - K N\left ( \frac{\ln{\frac{E(V)}{K}}-w^2/2}{w}\right ) \\ &= E(V) N\left (d_1\right) - K N\left ( d_2\right ) \end{aligned}E(max(V−K,0))​=em+w2/2∫(lnK−m)/w∞​h(Q−w)dQ−K∫(lnK−m)/w∞​h(Q)dQ=em+w2/2∫(lnK−m)/w−w∞​h(Q−w)d(Q−w)−K∫(lnK−m)/w∞​h(Q)dQ=em+w2/2[1−N(wlnK−m​−w)]−K[1−N(wlnK−m​)]=em+w2/2N(w−lnK+m​+w)−KN(w−lnK+m​)=E(V)N(w−lnK+lnE(V)−w2/2​+w)−KN(w−lnK+lnE(V)−w2/2​)=E(V)N(wlnKE(V)​+w2/2​)−KN(wlnKE(V)​−w2/2​)=E(V)N(d1​)−KN(d2​)​

Where

d1=ln⁡E(V)K+w2/2wd_1 = \frac{\ln{\frac{E(V)}{K}}+w^2/2}{w}d1​=wlnKE(V)​+w2/2​
d2=ln⁡E(V)K−w2/2wd_2 = \frac{\ln{\frac{E(V)}{K}}-w^2/2}{w}d2​=wlnKE(V)​−w2/2​

The Black-Scholes Result

We now consider a call option on a non-dividend-paying stock maturing at time T. The strike price is K, the risk-free rate is r, the current stock price is S0S_0S0​, the stock price at time T is STS_{T}ST​ and the volatility is σ\sigmaσ. The call price is given by

c=e−rTEˆ(max⁡(ST−K,0))c = e^{-rT}\^{E}(\max(S_{T}-K, 0))c=e−rTEˆ(max(ST​−K,0))

Where Eˆ\^{E}Eˆdenotes the expectation in a risk-neutral world. As mentioned above, under the stochastic process assumed by Black-Scholes, STS_{T}ST​is lognormal. Based on the deduction above, this means that

ln⁡STS0∼ϕ[(r−σ2/2)T,σ2T]\ln\frac{S_T}{S_0} \sim \phi{\left [\left ( r - \sigma^2/2 \right )T, \sigma^2T \right ]}lnS0​ST​​∼ϕ[(r−σ2/2)T,σ2T]

Based on the property of lognormal, Eˆ(STS0)=erT\^{E}(\frac{S_T}{S_0})=e^{rT}Eˆ(S0​ST​​)=erT, and Var(STS0)=e2rT(eσ2T−1)Var\left( \frac{S_T}{S_0}\right)=e^{2rT}\left(e^{\sigma^2T}-1 \right)Var(S0​ST​​)=e2rT(eσ2T−1), from the key result just proved, it implies

c=e−rT[S0erTN(d1)−KN(d2)]=S0N(d1)−Ke−rTN(d2)c = e^{-rT}\left [ S_{0}e^{rT}N(d_1)-KN(d_2) \right ] = S_0N(d_1)-Ke^{-rT}N(d_2)c=e−rT[S0​erTN(d1​)−KN(d2​)]=S0​N(d1​)−Ke−rTN(d2​)

Where

d1=ln⁡[Eˆ(ST)/K]+σ2T/2σT=ln⁡(S0/K)+(r+σ2/2)TσTd_1 = \frac{\ln{\left [ \^{E}(S_T)/K\right ]+\sigma^2T/2}}{\sigma \sqrt{T}} = \frac{\ln(S_0/K)+(r+\sigma^2/2)T}{\sigma \sqrt{T}}d1​=σT​ln[Eˆ(ST​)/K]+σ2T/2​=σT​ln(S0​/K)+(r+σ2/2)T​
d2=ln⁡[Eˆ(ST)/K]−σ2T/2σT=ln⁡(S0/K)+(r−σ2/2)TσTd_2 = \frac{\ln{\left [ \^{E}(S_T)/K\right ]-\sigma^2T/2}}{\sigma \sqrt{T}} = \frac{\ln(S_0/K)+(r-\sigma^2/2)T}{\sigma \sqrt{T}}d2​=σT​ln[Eˆ(ST​)/K]−σ2T/2​=σT​ln(S0​/K)+(r−σ2/2)T​

Proof of the Put Option

We can do it in the same way as above. However, here we will do it in another way.

We know that as put option, N(−d2)N(-d_2)N(−d2​) is the probability of exercise,

Eˆ(max⁡(K−ST,0))=∫−∞−d2(K−S0e[(r−12σ2)T+σTε])f(ε)dε=KN(−d2)−∫−∞−d2S0erT12πe−(ε−σT)22dε=KN(−d2)−∫−∞−d2−σTS0erT12πe−(ε−σT)22d(ε−σT)=KN(−d2)−∫−∞−d1S0erT12πe−(ε−σT)22d(ε−σT)=KN(−d2)−S0erTN(−d1)\begin{aligned} \^{E}(\max(K-S_T, 0))&=\int_{-\infty}^{-d_2}\left ( K - S_0e^{\left [(r-\frac{1}{2}\sigma^2)T+\sigma \sqrt{T}\varepsilon\right ]}\right )f(\varepsilon)d\varepsilon \\ &=KN(-d_2) - \int_{-\infty}^{-d_2}S_0e^{rT}\frac{1}{\sqrt{2\pi}}e^{-\frac{(\varepsilon-\sigma \sqrt{T})^2}{2}}d\varepsilon \\ &=KN(-d_2) - \int_{-\infty}^{-d_2-\sigma \sqrt{T}}S_0e^{rT}\frac{1}{\sqrt{2\pi}}e^{-\frac{(\varepsilon-\sigma \sqrt{T})^2}{2}}d(\varepsilon-\sigma \sqrt{T}) \\ &= KN(-d_2) - \int_{-\infty}^{-d_1}S_0e^{rT}\frac{1}{\sqrt{2\pi}}e^{-\frac{(\varepsilon-\sigma \sqrt{T})^2}{2}}d(\varepsilon-\sigma \sqrt{T}) \\ &= KN(-d_2)-S_0e^{rT}N(-d_{1}) \end{aligned}Eˆ(max(K−ST​,0))​=∫−∞−d2​​(K−S0​e[(r−21​σ2)T+σT​ε])f(ε)dε=KN(−d2​)−∫−∞−d2​​S0​erT2π​1​e−2(ε−σT​)2​dε=KN(−d2​)−∫−∞−d2​−σT​​S0​erT2π​1​e−2(ε−σT​)2​d(ε−σT​)=KN(−d2​)−∫−∞−d1​​S0​erT2π​1​e−2(ε−σT​)2​d(ε−σT​)=KN(−d2​)−S0​erTN(−d1​)​

where

  • ε\varepsilonε has a standard normal distribution ϕ(0,1)\phi(0, 1)ϕ(0,1)

  • f(ε)=12πe−ε22f(\varepsilon) = \frac{1}{\sqrt{2 \pi}}e^{-\frac{\varepsilon^2}{2}}f(ε)=2π​1​e−2ε2​


Strictly speaking, this is not a deduction of the Black-Scholes equation, because we were using the result from the equation that N(−d2)N(-d_2)N(−d2​)is the probability of exercise. And if the function underneath the expectation has been changed, the probability of exercise could change along with that.

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where!!!!!! is the .

From ,

And since f(V)f(V)f(V)is strictly increasing, we can use .

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double factorial
the properties of the lognormal distribution
the formula for the density of a strictly increasing function