Transcript where (x,t)

Week 7: Black Scholes
Models and Option Pricing
Binomial 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
r=0时，执行期为t个周期的期权价格
• 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
• [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?)
GBM
• 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
到期日的价值：
现价：
期权的价格
Brownian 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=jk-1 e(ti) t]2
= (k-j) t = tk – tj.
• For t1 < t2  t3 < t4, W(t4) - W(t3) is
uncorrelated with W(t2) - W(t1).
Simulation 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
normal 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
equation dW(t) = e(t)dt .
• In advanced courses in Probability, it will be
shown that this limiting operation is well-defined
and the limiting process is the Brownian motion.
Formally, we define a Brownian motion as
follows.
Wiener 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 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 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 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
integration.
Stock 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{0.01t + 0.01W(t)}. Notice the linear
growth of the mean, =.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(eW(t))
= z exp{(-2/2)t}E(e t ), ~N(0,1),
= z exp{(-2/2)t}exp{2 t/2}
=zet=S(0)et.
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 2t (exp{2 t}-1). (check)
Ito’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
extra correction term.
• Rule: Whenever dW(t) is involved, we need to
account 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)
to dS(t) is known as the Ito’s lemma.
• Remarks:
1. The above equation describes the dynamics of
the 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)et .
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
with lognormal coefficient. In practice, as long as t is
small, both expressions can be used to simulate S(t).
Ito’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
function 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)(at+bW)+(F/t)+(1/2)(2F/x2)(at+bW)2.
Consider this quadratic term, expanding
(at+bW)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 (at+bW)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)bW.
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,
2F/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.
•
1.
2.
3.
4.
Example.  W(s)dW(s) = ? To find out, first guess
an answer W2(t)/2. Is this correct? Use Ito’s
formula to check.
Let X(t)=W(t), dX(t)=dW(t) and identify a=0, b=1.
Let Y(t)=F(X(t))= W2(t)/2. Then F/W=W,
2F/W2=1, and F/t =0.
Recite Ito’s lemma:
dY(t) = [(F/x)a + F/t+(1/2)(2F/x2)b2]dt +
(F/x)bdW(t) so that
dY(t) = (1/2)dt + W(t)dW(t). Now integrate both
sides of this equation, we have
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 wasn’t correct. We need the extra
correction term from the Ito’s lemma.

Black-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
process 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
= (xtS + ytrB)dt + xtS 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
xtS + 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)
SW.
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)SW + (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  Ker(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=
 rKer(T  t). Substitute these quantities into the BS
equation, we have
f/t+(1/2)(2f/S2)2S2 +(f/S)rS =  rKer(T  t) + rS
=r(S  Ker(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 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{SK,0}) = E(S)(d1) K(d2),
where () denotes the cdf of a standard
normal 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(logSm). 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 sm), s=eq+m so that
dq=ds/(s). Consider
E(max{SK,0}) = 0 max{sK,0}g(s)ds
= K(sK)g(s)ds
=  (log Km)/(eq+m K)g(eq+m)sdq
=  (log Km)/(eq+m K) (q)dq = I  II, say.
Analyze I and II individually. Writing (q-)2=q2
–2q+ 2, consider I.
I =  (log Km)/eq+m (q)dq
=  (log Km)/ eq+m (2) 1/2 exp(q2/2)dq
=  (log Km)/ eq+m (2) 1/2 exp[–(q-)2/2]
exp(–q+2/2) dq
= exp(m+2/2)  (log Km)/ –  (q –)d(q–)
= exp(m+2/2) {1–[(log Km)/ –]}
= exp(m+2/2) [(– log K+m)/ +].
For II, observe that
II = K (log Km)/(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{SK,0}) =I–II =E(S) (d1) – K(d2). 
Black-Scholes Formula
Theorem. 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
geometric Brownian motion, dS = Sdt + S dW,
then EST = S0e 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
rT
The call price in the risk-neutral world must
satisfy C(S) = e rT E*(max{ST  K,0}). By the
lemma,
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+(r2/2)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+(r2/2)T + 2T]
= (T) –1 [ log (S0/K) + (r+2/2)T ].
Similarly, we can derive that
d2 = [log (S0/K) + (r2/2)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 riskneutral 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+(1p*)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
• In the risk neutral calculation of f, you don’t need
to know p, but you need to know p* =(erTd)/(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
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.