Week 7: Black Scholes ModelWeek 7: Black Scholes Models and Option Pricings and Option Pricing

Binomial treeBinomial tree

• 优点：直观简单，能够很好地解析无套利定价和利用关于风险中性测度的期望值来计算期权价格的思想。

• 缺点 : 不符合股票市场时时交易的特性 .

• 如何更好地刻画现实中的期权价格：

（ a ）多周期二叉树模型 （ b ）连续时间模型

多步二叉树多步二叉树

r=0r=0 时，执行期为时，执行期为 tt 个周期的期权价个周期的期权价格格

• Pt 为时间 t 时刻的股票价格，风险中性概率为 1/2,

Pt=P0+10{2(W1+W2+…..+Wt)-t}, Wi 以 1/2 概率取 0 或 1 。

则期权价格为 C=E{(Pt-E)+}

无穷多步期权定价无穷多步期权定价• [0, 1] 分成 n 个区间， 股票价格以 1/2 概率

上涨或下跌 σ/n1/2, t=m/n 时刻的价格为

pm:= Pt=P0+ σ/n1/2 {2(W1+W2+…..+Wm)-m},

则 (1) {pm} 为一 鞅（？） ; (2) E(Pt|P0)=P0;

(3) Var(Pt|P0)=t σ2.

当 n 充分大时， Pt 趋于一连续的随机过程 P0+ σBt (Bt 为 BM, BM 的性质 ?)

GBMGBM

• 利用 random-walk (BM) 来刻画期权价格的缺点：可能使得价格取到负值，这是不合理的。

• 我们利用 GBM 来克服该缺点。

期权价值期权价值

到期日的价值：

现价：

期权的价格期权的价格

Brownian MotionsBrownian Motions

• W(tk+1)= W(tk) + e(tk) t, where t = tk+1 – tk, k=0,…,N, t0 = 0, and e(tk) iid N(0,1).

• For j<k, W(tk) - W(tj) = i=jk-1 e(ti) t.

• The right-hand side is normally distributed, so is the left-hand side.

• Clearly, E(W(tk) - W(tj) ) = 0.• Var (W(tk) - W(tj)) = E [i=j

k-1 e(ti) t]2

= (k-j) t = tk – tj.• For t1 < t2 t3 < t4, W(t4) - W(t3) is uncorrelate

d with W(t2) - W(t1).

Simulation of Brownian MotionSimulation of Brownian Motion• Partition [0,1] into n subintervals each with length 1/n. For

each t in [0,1], let [nt] denote the greatest integer part of the number nt. For example, if n=10 and t=1/3, then [nt] =[10/3]=3.

• For each t in [0,1], define a stochastic process S[nt] = i=1[nt]

e(i)/n, e(i) iid N(0,1).

• Clearly,S[nt] = S[nt]-1 + e([nt])/ n, a special form of the additive model defined at the beginning with t =1/n and W(t)= S[nt].

• At time t=1, S[nt] = Sn = i=1n e(i)/n, which has a standard n

ormal distribution. Even if e(i)s are not normally distributed, CLT shows that Sn will still be normally distributed as long as n is big. This is the key idea in constructing a Brownian motion.

• By letting n goes to infinity (t goes to 0), Donsker (1953) proved that the stochastic process S[nt] constructed in this way tends to a Brownian motion. This result is known as the Functional Central Limit Theorem or the Invariance Principle. Therefore, the above equation provides a means to simulate the path of a Brownian motion. All we have to do is to iterate the equation

S[nt] = S[nt]-1 + e([nt])/ n by taking n bigger and bigger and in the limit, we have a Brownian motion.

• When t goes to 0, the discrete-time random walk equation

S[nt] = S[nt]-1+e([nt])/n, can be approximated by the continuous-time eq

uation dW(t) = e(t)dt . • In advanced courses in Probability, it will be sh

own that this limiting operation is well-defined and the limiting process is the Brownian motion. Formally, we define a Brownian motion as follows.

Wiener ProcessWiener Process

Definition. A Wiener process (standard Brownian motion) is a stochastic process which satisfies the following properties:

• For s<t, W(t) – W(s) is a N(0, t-s) random variable.

• For t1 < t2 t3 < t4, W(t4) - W(t3) is uncorrelated with W(t2) - W(t1). This is known as the independent increment property.

• W(t0) = 0 with probability one.

Properties of Wiener ProcessProperties of Wiener Process

• For each t, W(t) is normally distributed.

• For t<s, E(W(s)|W(t)) = E(W(s)-W(t)+W(t)|W(t))=W(t), hence a martingale.

• With probability 1, W(t) is nowhere differentiable. E[W(s)-W(t)/(s-t)]2 =1/(s-t) which tends to infinity when s tends to t. Consequently, the process (t) = dW(t)/dt is undefined mathematically.

Diffusion ProcessesDiffusion Processes

• Consider dX(t) = dt dW(t), integrating this equation, we get

• X(t) = X(0) + t W(t). In other words, the process X(t) is the solution to the integral equation dX(t)= dt + dW(t).

• X(t) defined in this way is known a diffusion process. It can be expressed explicitly in terms of W(t) and the constants and . To generalize, we have:

Generalized Wiener ProcessGeneralized Wiener Process

• Definition. An Ito’s process or a generalized Wiener process is a stochastic process which satisfies the following stochastic differential equation

dX(t) = (x,t)dt + (x,t) dW(t), where (x,t) is known as the drift function and

(x,t) is known as the diffusion function.• This equation has to be interpreted through integ

ration.

Stock ReturnsStock Returns

• Recall the multiplicative model log S(k+1) = log S(k)+w(k), w(k) iid N(2).

• The continuous-time approximation is dlog S(t) = dt + dW(t). Integrating this equation, we have

log S(t) = log S(0) + t + W(t). The process S(t) is the geometric Brownian motion.

• Definition. Let X(t) be a Brownian motion with drift and diffusion coefficient (variance) 2 , that is, dX(t) = dt + dW(t).

• The stochastic process S(t)= exp(X(t)) is said to be a geometric Brownian motion with drift parameter and variance 2, where = +2/2. Or equivalently, the process X(t) =logS(t) is a Brownian motion such that

dlogS(t) = (-2/2)dt + dW(t).

The following figure are simulations of S(t)=exp{0.0The following figure are simulations of S(t)=exp{0.01t + 0.01W(t)}. Notice the linear growth of the mea1t + 0.01W(t)}. Notice the linear growth of the mea

n, n, =.01.=.01.

• Let S(0)=z. An equivalent way of defining a geometric Brownian motion is that the process S(t) satisfies S(t) = z e X(t) = z exp{t + W(t)}

=z exp{(-2/2)t + W(t)}.

• Consider the successive ratios S(t1)/S(t0), S(t2)/S(t1), …,S(tn)/S(tn-1). Since the Brownian motion has independent increments, these ratios are independent random variables.

• log S(t) = log S(0)+X(t) ~ N(logS(0) + t, 2 t).

• As S(t)=ze X(t), to find the moments of S(t), consider

E(S(t)) = E(E(S(t)|S(0)=z))

=E(E(z exp{(-2/2)t +W(t)}|S(0)=z))

= z exp{(-2/2)t}E(eW(t))

= z exp{(-2/2)t}E(e t ), ~N(0,1),

= z exp{(-2/2)t}exp{2 t/2}

=zet=S(0)et.

For >0, E(S(t)) tends to infinity as t gets large. But for 0< < 2/2, the process X(t) = X(0) + (-2/2)t + W(t) has a negative drift, which means X(t) tends to negative infinity as t gets bigger. Consequently, the price S(t) = S(0)eX(t) tends to zero as t tends to infinity. In other words, the geometric Brownian motion S(t) drifts to zero but at the same time, its mean is drifting to infinity. Using the mean value can be misleading in describing the process.

Similarly, we can show that

Var S(t) = S(0)2 e 2t (exp{2 t}-1). (check)

Ito’s FormulaIto’s Formula

• In the preceding section, we define the geometric Brownain motion in terms of dlogS(t) = dt + dW(t).

• From ordinary calculus, dS(t)/S(t)=dlogS(t) Guess: dS(t)/S(t) = dt + dW(t). • Unfortunately, this is not correct. We need an e

xtra correction term.• Rule: Whenever dW(t) is involved, we need to a

ccount for a correction term. This is the essence of the Ito’s calculus.

• The correct formula is dS(t)/S(t) = (2/2)dt + dW(t) = dt + dW(t), as =-2/2.• The extra term 2/2 when transforming dlogS(t) t

o dS(t) is known as the Ito’s lemma.• Remarks:1. The above equation describes the dynamics of t

he instantaneous return process dS(t)/S(t).2. When =0, the process is deterministic and the

above equation leads back to S(t)=S(0)et .

3. One can simulate S(t) from the above equation,

S(tk+1) – S(tk) = S(tk)t + S(tk)(tk)t, or

S(tk+1) =[1+ t + (tk)t]S(tk), (tk) iid N(0,1). This is a multiplicative model in S(t), but the coefficient

is normal, not lognormal

4. Alternatively, we can discretize dlogS(t) = dt+dW(t) to simulate logS(tk+1)- logS(tk)=t+(tk)t, or

S(tk+1) = exp{t+(tk)t}S(tk). This time we have a multiplicative model again, but wit

h lognormal coefficient. In practice, as long as t is small, both expressions can be used to simulate S(t).

Ito’s LemmaIto’s Lemma

• Theorem. Let x(t) be a diffusion process that satisfies

dx(t) = a(x,t)dt + b(x,t)dW(t). Let the process y(t) = F(x,t) for some functi

on F. Then y(t) satisfies the Ito’s equation dy(t) = [(F/x)a + F/t + (1/2)(2F/x2)b2]

dt + (F/x)bdW(t).

Proof. It will only be a sketch. Recall from calculus that for a function of two variables y(t)=F(x,t),

dy(t) = (F/x) dx + (F/t) dt

= (F/x)(adt + bdW) + (F/t) dt.

Comparing this with the Ito’s equation, we see that there is the extra correction term (1/2)(2F/x2)b2 in front of the coefficient dt in the Ito’s equation. The reason for this correction is due to the fact that dW is of order dt, and as dx is expressed in terms of dW, it is also of order dt. To see how this is done, consider the Taylor’s expansion of F up to the order of dt. Specifically, calculate the expression

F(x+x, t+t) = F(x,t) + (F/x)x + (F/t)t + (1/2)(2F/x2)(x)2

=F(x,t)+(F/x)(at+bW)+(F/t)+(1/2)(2F/x2)(at+bW)2.

Consider this quadratic term, expanding (at+bW)2 = a2 (t)2 + 2ab (t)(W) + b2(W)2.Note that both the first term and the second term of the right hand side are of higher order than t. Recall the fact that since dW ~ e(t)dt, W is of order t and thus, (W)2 ~ t. Using this fact and remembering that we are expanding y only up to order t (i.e. terms of order higher than t can be ignored), we get (at+bW)2 ~ b2 t. Substituting this into the pink term, F(x+x, t+t) = F(x,t) + [(F/x)a + (F/t) + (1/2)(2F/x2)b2] t + (F/x)bW.Taking limit t goes to zero, the Ito’s formula is obtained.

Example. Consider the geometric Brownian motion S(t) that satisfies : dS(t) = S(t)dt + S(t)dW(t).

Consider the process y(t)=F(S(t))=logS(t).

What kind of equation does y satisfy? To find out, use Ito’s lemma.

1. Identify a= S and b= S.

2. We know F/S=1/S and 2F/S2=-1/S2.

3. Plug these into Ito’s formula,

dlogS =[a/S – (1/2)(b2/S2]dt + (b/S)dW

=(- 2/2)dt + dW,

which agrees with the earlier derivation.

• Example. sdW(s) =? First a guess,

tW(t) - W(s)ds. To check,

1. Let X(t)=W(t), dX(t)=dW(t) and identify a=0, b=1.

2. Let Y(t)=F(W(t))=tW(t). Then F/W=t, 2

F/W2=0, and F/t =W(t).

3. Substitute these into the Ito’s lemma, dY(t)=tdW(t) + W(t)dt. Integrating,

4. Y(t) =tW(t)= sdW(s) + W(s)ds, that is,

sdW(s) = tW(t) - W(s)ds as expected.

• Example. W(s)dW(s) = ? To find out, first guess an answer W2(t)/2. Is this correct? Use Ito’s formula to check.

1. Let X(t)=W(t), dX(t)=dW(t) and identify a=0, b=1.2. Let Y(t)=F(X(t))= W2(t)/2. Then F/W=W, 2F/W2=1,

and F/t =0.3. Recite Ito’s lemma: dY(t) = [(F/x)a + F/t+(1/2)(2F/x2)b2]dt + (F/x)bd

W(t) so that dY(t) = (1/2)dt + W(t)dW(t). Now integrate both sides of

this equation, we have4. W2(t)/2 = Y(t) = t/2 + W(s)dW(s), that is, W(s)dW(s) = W2(t)/2 - t/2!!! This time, our initial guess w

asn’t correct. We need the extra correction term from the Ito’s lemma.

Black-Scholes EquationBlack-Scholes Equation

• There are two securities: a stock that follows the geometric Brownian motion, dS = Sdt + S dW, and a bond that follows dB=rBdt.

• Consider a contingent claim (call option, say) to S. This derivative has a price, which is a function of S and t, f(S,t).

• Goal: To find an equation that describes the behavior of f(S,t). This goal is achieved by the Black-Scholes equation.

Theorem. With the notation just defined, then the price of the derivative satisfies:

f/t +(f/S)rS + (1/2)(2f/S2)2S2 = rf.

Proof. (Analytical). The idea of this proof is the same as in the binomial case. We try to construct a portfolio which replicates the characteristics of the contingent claim at each instant.

1. Recall from Ito’s lemma that since f is a function of S,

df = [(f/S)S + f/t + (1/2)(2f/S2)2S2]dt + (f/S) S dW. This equation shows that f is also a diffusion pr

ocess with drift […] and diffusion coefficient (…).

2. Construct a portfolio that replicates f. At each time t, select xt of stock and yt of bond, giving a total portfolio value G(t) = xtS(t) + ytB(t).

3. The instantaneous gain in value of this portfolio comes from changes in S and B so that

dG = xtdS + ytdB

= xt (Sdt + S dW) + ytrBdt

= (xtS + ytrB)dt + xtS dW.

4. We want this to be like df, match the coefficient of dt and dW between the preceding equation and that of df. From the coefficient of dW, we identify xt = f/S.

5. From G(t) = xtS(t) + ytB(t) and using G=f (our goal),

6. yt = [G(t) -xtS(t)]/B(t) = [f(S,t) – (f/S) S(t)]/B(t).

7. Substituting this expression into the equation of dG and matching it with the coefficient of dt in the equation of df, we have

xtS + ytrB

= (f/S)S + f/t+(1/2)(2f/S2)2S2.

That is,

(f/S)S +[1/B(t) ] [f(S,t) – (f/S) S(t)] r B(t) =

(f/S)S + f/t +(1/2)(2f/S2)2S2.

Consequently,

f/t + (f/S) S(t)r + (1/2)(2f/S2)2S2 =rf.

• If f(S,t) = S, then f/t =0, f/S=1, and 2f/S2=0 so that the Black-Scholes equation becomes rS=rS. Clearly, f=S is a trivial solution to the Black-Scholes (BS) equation.

• Similarly, if f(S,t) = ert, it is easily shown that it is also a trivial solution to the BS (check!!).

• For a European call option with strike price K and maturity T, f(S,t)=C(S,t) with boundary conditions C(0,t)=0 and C(S,T)= max{S-K,0}.

• For a European put option with strike price K and maturity T, f(S,t)=P(S,t) with boundary conditions P(,t)=0 and P(S,T) = max{K-S,0}.

• For an American call option that allows early exercise, we have the extra boundary condition max{0,S-K}C(S,t).

• Notice that the BS is a partial differential equation. There is no guarantee that it has a solution. As a matter of fact, except in simple cases such as a European call or put option, one cannot solve the BS analytically. As a result, either simulations or numerical solutions are possible alternatives.

• One can also derive the BS through a delta hedging argument. To see how it works, construct a portfolio that consists of shorting one derivative (call option, say) and longing f/S shares of the underlying stock. Let the value of this portfolio be . Then = f + (f/S)S. The change of the portfolio value in the time interval t is given by = f + (f/S) S.

• Since S follows a Geometric BM, S = S t + S W.

Recall from the Ito’s lemma, the discrete version of df is

f =[(f/S)S+f/t+(1/2)(2f/S2)2S2]t+(f/S) SW.

Substitute these expressions into the equation

= f + (f/S) S, we obtain

=[(f/S)S+f/t+(1/2)(2f/S2)2S2]t (f/S)SW + (f/S) (S t + S W)

=[ f/t (1/2)(2f/S2)2S2]t.

This portfolio has no random element since W has been eliminated by construction. In other words, regardless of the outcome of the stock prices, the change in the portfolio value is nonrandom. It must equal to the risk-free rate since if otherwise, there will be arbitrage opportunities (check!!).

Consequently, = r t. That is, [ f/t (1/2)(2f/S2)2S2]t = r[f + (f/S)S] t, f/t + (1/2)(2f/S2)2S2 + (f/S)rS = rf, which is just the Black-Scholes equation again.

Example. Let f be the price of a forward contract of a non-dividend-paying stock with delivery price K and maturity T. Then the price at time t is given by f(S,t) = S Ker(T t). To check the validity of this pricing formula, we need to check if it satisfies the BS equation. Observe that f/S=1, 2f/S2=0, and f/t= rKer(T t). Substitute these quantities into the BS equation, we have

f/t+(1/2)(2f/S2)2S2 +(f/S)rS = rKer(T t) + rS =r(S Ker(T t)) =rf. Therefore, the given formula satisfies the BS and thus

is the correct formula of the price for the forward contract.

Black-Scholes formulaBlack-Scholes formula

Lemma. Let S be a lognormally distributed random variable such that log S ~ N(m, 2) and let K>0 be a given constant. Then

E(max{SK,0}) = E(S)(d1) K(d2), where () denotes the cdf of a standard n

ormal random variable and

d1 = 1(logK+m+2) = 1(log E(S/K) + 2/2),

d2 = 1(logK+m) = 1(log E(S/K) 2/2).

Proof. Recall that since log S ~N(m,2), 1. E(S) = exp(m+2/2) so that log E(S) = m+2/

2. 2. Let g(s) denote the pdf of S and let Q = 1(lo

gSm). Then Q~N(0,1) with pdf (q)=(2) 1/2 exp(q2/2). 3. Also, g(s) = ((log s m)/)/(s). (check !!) 4. Observe that q= 1(log sm), s=eq+m so that

dq=ds/(s). Consider

E(max{SK,0}) = 0 max{sK,0}g(s)ds

= K(sK)g(s)ds

= (log Km)/(eq+m K)g(eq+m)sdq

= (log Km)/(eq+m K) (q)dq = I II, say. Analyze I

and II individually. Writing (q-)2=q2 –2q+ 2, consider I.

I = (log Km)/eq+m (q)dq

= (log Km)/ eq+m (2) 1/2 exp(q2/2)dq

= (log Km)/ eq+m (2) 1/2 exp[–(q-)2/2]

exp(–q+2/2) dq

= exp(m+2/2) (log Km)/ – (q –)d(q–)

= exp(m+2/2) {1–[(log Km)/ –]}

= exp(m+2/2) [(– log K+m)/ +].

For II, observe that

II = K (log Km)/(q)dq = K [(– log K+m)/].

I–II = exp(m+2/2) [(– log K+m)/ +] – K[(– log K+m)/].Recall that log E(S/K) = – log K + log E(S) = – log K + m+2/2,

so that (– log K+m)/ + = –1[log E(S/K)+2/2]=d1.

Similarly, d2 = –1(–log K+m). (check !!) Combining all these with the fact that E(S) = exp(m+2/2), we have

E(max{SK,0}) =I–II =E(S) (d1) – K(d2).

Black-Scholes FormulaBlack-Scholes FormulaTheorem. Consider a European call option with

strike price K and maturity T. Assume that the underlying stock pays no dividends during the time [0,T] and assume that there is a continuously compounded risk-free rate r. Then the price of this contingent claim at time 0, f(S,0)=C(S), is given by the formula

C(S) = S(d1) – Ke –rT (d2), where

d1= (T) –1[log (S/K) + (r+2/2)T],

d2= (T) –1[log (S/K) + (r–2/2)T] = d1 – T.

• Proof. The proof is based on the risk-neutral valuation. • Recall that if the price of the stock follows the geometri

c Brownian motion, dS = Sdt + S dW, then EST = S0

e T . Now in the risk neutral world, the stock should grow according to the risk free rate so that E*ST = S0e rT, where E* denotes the expectation is taken under the risk neutral probability.

• Now mimic the case of the geometic Brownian motion. Define a new Brownian motion W* in the risk-neutral world so that the price follows a new geometric Brownian motion dS = rSdt + S dW*, driven by W*. Note that E* denotes the expectation is taken with respect to W*. In this risk-neutral world, we have E*ST = S0e rT.

The call price in the risk-neutral world must

satisfy C(S) = e rT E*(max{ST K,0}). By thelemma,

E*(max{ST K,0}) = E*(ST)(d1)–K(d2).

It remains to identify E*(ST), d1, and d2.

1. E*(ST) = S0 erT, by virtue of the risk-neutral principle.

2. Since S follows a geometric Brownian motion under W*, a simple application of Ito’s lemma gives d log St = dt +dW*t with =r 2/2.

3. m=E*(log ST) = log S0+T = log S0+(r2/2)T.

4. 2 = Var*(log ST) = 2T.

5. In other words, St ~N(m, 2) in the risk-neutral world.

6. Recite from the lemma that

d1 = 1(log K+m+2)

= (T) –1[log K+ log S0+(r2/2)T + 2T]

= (T) –1 [ log (S0/K) + (r+2/2)T ]. Similarly, we can derive that

d2 = [log (S0/K) + (r2/2)T ] = d1 – T. This completes the proof of the Black-Scholes

formula.

Example. Consider a 5-month call option on a stock with a current price $62, volatility 20% per year, strike price $60 and the risk-free rate is 10% per year. Therefore, S=62, K=60, r=0.1, =0.2, and T=5/12. Using the BS formula,

d1= [.2(5/12)]1[log(62/60) + (0.1+0.22/2)(5/12)] =0.6413,

d2= d1 0.2 (5/12) = 0.5122,

so that (d1) = 0.7393 and (d2) = 0.6957. The value of this call option is C = (62)(0.7393) (60) e (0.1)(5/12) (0.6957) = 5.798.

• One can establish the BS formula by solving the PDE of the BS equation. But the risk-neutral valuation is more intuitive.

• For a European put option, the corresponding formula is P(S) = Ke –rT (– d2) – S(– d1).

• Suppose S>K, then both d1and d2 tend to as T tends to 0 so that C=S – K and P = 0. Is this reasonable? If S>K and at time t=0, the call should be worth S – K and the put is worthless. Thus, the BS formula is consistent with the boundary condition. A similar argument can be deduced for the case S<K.

• When T tends to , again both d1and d2 also tend to . In this case, C=S from the BS formula. This is known as a perpetual call. If we held the call for a long time, the stock increases to a very large value in probability, so that the strike price K is irrelevant. Hence, if we own the call and hold on to it, we could obtain the stock for free. As a result, the price of the call must equal to that of the stock at time 0.

• The BS formula is derived under the assumption that no dividends is paid during the time period. For dividend paying stocks, a similar formula can also be deduced with slight modifications, see Hull (2000).

• For an American option, exact analytic formula such as the BS cannot be derived. One has to resort to numerical methods, see Hull (2000).

• What about the parameter in the BS formula? This is the volatility factor. We can estimate it from historical data and substitute this estimate into the BS equation and apply the BS equation to price a contingent claim. This is known as the historical volatility approach.

• Alternatively, we can substitute the observed prices of a derivative into the BS formula and then solve for the parameter from the BS formula. This is known as the implied volatility approach. This quantity can be applied to monitor the market’s opinion about the volatility of a particular stock. Analysts often calculate implied volatilities from an actively traded derivative of a certain stock and plug in the calculated implied volatilities into the BS to calculate the price of a less traded derivative of the same stock.

• Note that the risk neutral valuation states that to evaluate the value of the option f, there is no need to have knowledge of the probability of the stock moving up in the real world. All one has to know is the number p*, which is the probability of the stock moving up in the risk neutral world. The probability of the stock moving up in the real world plays absolutely no role in this calculation.

• Suppose you play a game by tossing a coin. When the toss is a head, you win $22, when it’s a tail, you win $18. How much, S, should you pay for this game? A simple calculation shows that the expected payoff for this game is 22p + 18(1-p), where p is the probability of getting a head in the toss. When p=1/2, the expected payoff is $20. When p=0.9, the expected payoff becomes $21.6.

• Note that you never know your payoff until the game is played. All you know is that with probability p, your payoff is $(22-S) and with probability (1-p), your payoff is $(18-S). But the outcome is undetermined until the game is played.

• When S=20, then with probability ½, your expected payoff is $2 and with probability ½, your expected payoff is -$2.

• When S=21.6, with probability 0.9, your expected payoff is $0.4 and with probability 0.1, your expected payoff is -$3.6.

• The expected payoff only stipulates that if you play a large number of these games independently, on the average, that’s the amount you will get.

• But for each individual game, you may win or you may lose depending on the outcome of the toss.

• The expected payoff depends on the value p, which is the real probability of getting a head.

• When p=1/2, which is the toss of a fair coin, and if you set S=$20, which is the expected price, then on the average, you expect to win $2 half of the time and loss $2 half of the time.

• Contrast this with a simple example. Suppose you are a merchant and you want to determine the price of an umbrella now. You decided that if it rains tomorrow, you will sell the umbrella for $22. If it doesn’t rain, you will sell it for $18. After consulting with your friend who has a Ph.D. in meteorology, you come to the conclusion that the probability of raining tomorrow is p=½. How can you make a profit? An obvious choice is to set

• S0 =18(1-p) + 22p = E(S1), which is your expected payoff for tomorrow. But again, you won’t know the result until tomorrow. And the knowledge of p in the real world is critical for calculating S0 .

• Suppose now that you want to ensure a fixed payoff by tomorrow, regardless of the weather. How can you achieve this?

• Form a trading strategy, buy umbrella and sell one call option of an umbrella for a value f with a strike price $21 maturing tomorrow.

• The payoff of your trade by tomorrow is 22 -1 if it rains, and 18 if it doesn’t. Since your goal is to ensure a fixed payoff regardless of the weather, these two numbers must be equal, that is, 22 - 1= 18 , which implies =1/4. In this case, the fixed payoff by tomorrow is $4.5, regardless of the weather. Note that p is irrelevant in this calculation.

• But what about the value of your portfolio today? It equals to S0 - f. Assuming no arbitrage and the interest rate for one day is zero, this value must be equal to the payoff for tomorrow. That is,

S0/4 - f = 4.5. Now we have one equation but two unknowns, f and S0.

• In the real world, the value S0 is available. Once this is done, then f can be determined. For example, if S0=20, then the above equation would mean that f = 0.5. Again, the value of f is determined without any relevance to p, hence risk neutral.

• To determine the risk neutral probability p*, we invoke one more fact that p*=E*(f1) = f, that is p*=p*x1+(1-p*)x0 = 0.5. This risk neutral probability stipulates that under this probability, the value of f is 0.5 and you will be ensure a profit of $4.5 by tomorrow, regardless of the weather.

• It happens that p* equals to the probability of your meteorologist friends suggestion that p=1/2 in this example, which also happens to match the fair game idea.

• Instead of ½, suppose your friend said p=0.9. Everything being the same, with S0=20, you perform the same calculation mistakenly by letting p*=0.9 (the physical probability, not the risk neutral one). Then f= E*(f1)=0.9. Your portfolio has a value of 5-0.9=4.1 today, which would grow to 4.5 by tomorrow regardless of the weather according to your construction. There is clear arbitrage opportunity when p* is set to anything but ½.

• In the risk neutral calculation of f, you don’t need to know p, but you need to know p* =(erTd)/(u–d). As long as you know the interest rate r and the possible outcomes u and d, then everything is determined.

• If you evaluate f by using p, then it is no longer a risk neutral valuation. You are in fact using the so-called objective probability. This objective probability p may be a statistical estimate based on the history of the stock. This could lead to mispricing of the contingent claim.

• In the previous example, we assume r=12% and get p* = 0.6523, f = 0.633. In this calculation, when the probability of an upward movement in the risk-neutral world is 0.6523, the expected return of the stock is therefore the risk free rate 12%. If it happens that the real world probability is also p=0.6523, then the calculation matches with the risk neutral valuation.

• Suppose now that the expected return of the stock is 16% instead and let p denote the probability of an upward movement in the real world under this assumption. Then 22p + 18(1-p) = 20e0.16x3/12 so that p= 0.7041. The expected payoff of the option in the real world is px1+(1-p)x0 =p=0.7041, which is different from the risk neutral valuation where f=0.633. To calculate the value of the option today, you need to discount this number as f = 0.7041 e r1 x

3/12, where r1 denotes the discount rate of the

option. But it will be difficult to determine r1. An option is riskier than stock, so r1>16%, but by how much? Without the knowledge of r1, it is impossible to calculate f in the real world.

• The risk neutral valuation solves this problem. In the risk neutral world, the expected return of all assets (including option) equals to the risk free rate.

• Notice that in the whole set up, the risk neutral probability is derived based on a delta hedging argument. In other words, the delta hedging procedure takes into account of both risks of the stock and the option. It is a clever device that hedges all the risk away and as a result, quantities like f can be evaluated under the risk neutral world.