OPTIONS
1. THE TWO BASIC OPTIONS - PUT AND CALL
•
A call (put) is the right to buy (sell) an asset.
•
Most other options are just combinations of these.
•
Options are “derivatives” and other derivatives
may include options
•
The price of an option is called a “premium”
because options are equivalent to insurance and
the price of insurance is called a premium.
2. For most of this lecture we will assume that the option
a. Can only be exercised at maturity (called
European). An American option, which is the most
common type should behave similarly because, in
most cases, American options are not exercised
until maturity. They are almost always worth more
left unexercised so very few are exercised. If they
are never exercised before expiration, there
should be no difference in value between an
American and European option.
b. Pays no dividends – most options aren’t
dividend protected so dividends will affect price.
Lower Bound for the European Call
Option Value is also a Lower Bound
for the American Call Option Value
1.
Many of the early results in options were derived using
arbitrage arguments. We will not cover all of them here
but the lower bound is one example.
Consider 2 portfolios A and B:
A. Purchase a European call for C with exercise price E and (zero
coupon, default free) bonds that will have a value of E at the
maturity of the option, i.e., purchase E/(1 + r) amount of bonds
B. Purchase the stock at P0 at time 0. Price at expiration is P1.
Portfolio Action
A
Buy call
Buy bonds
TOTALS
B
P1
Buy stock
if P1 > E
Investment
Value at Expiration
C
E / (1+r)
C + E / (1+r)
P1 > E
P1 -E
E
P1
P0
P1  E
0
E
E
P1
then both have same value
if P1  E then Portfolio A is more valuable thus the
investment in A must exceed the investment in B or else
arbitrage is possible, thus:
C
E
 P0
(1  r )
C  P0 
so
E
(1  r )
CALL OPTION CONTRACT
Definition: The right to purchase 100 shares of a security at
a specified exercise price (Strike) during a specific period.
EXAMPLE: A January 60 call on Microsoft (at 7 1/2)
This means the call is good until the third Friday of January
and gives the holder the right to purchase the stock from the
writer at $60 / share for 100 shares.
Cost is $7.50 / share x 100 shares = $750 premium or option
contract price.
PUT OPTION CONTRACT
Definition: The right to sell 100 shares of a security at a
specified exercise price during a specific period.
EXAMPLE: A January 60 put on Microsoft (at 14 1/4)
This means the put is good until the third Friday of January
and gives the holder the right to sell the stock to the writer
for $60 / share for 100 shares.
Cost $14.25 / share x 100 shares = $1425 premium.
INTRINSIC AND TIME VALUE
•AN OPTION'S INTRINSIC VALUE IS ITS VALUE IF IT
WERE EXERCISED IMMEDIATELY.
•AN OPTION'S TIME VALUE IS ITS COST ABOVE ITS
INTRINSIC VALUE.
Microsoft Stock Price = 53 1/4 at the time - October 1987
QUESTION: Which Microsoft option has greater intrinsic
value? - put
QUESTION: Which Microsoft option has greater time value?
– call
QUESTION: Which option is a better deal?
Look at Options Quotes: www.bigcharts.com
COMBINED STRATEGIES
•
•
•
•
•
straddle - 1 call and 1 put
strip - 1 call, 2 puts
strap - 2 calls, 1 put
money spread
time spread
OPTIONS EXAMPLES
Call - option to buy another company or company's line.
Call - capital expenditures on R & D and marketing. Give
an option to make further investments if promising.
Call - buy car at the end of the lease
Call - rain check at a grocery store
Put - abandonment
Put - agreement to buy company but only if loan losses are
less than 50 million (WCIS).
Put - guarantees - government price supports - consider
farmer's incentives
Note: The option pricing model can be used to price any
asset for which we can obtain the required inputs. Although
stock options are more common, many assets can be priced
– often called “real” options.
Call - A stock is a call option on the value of the firm if there is
debt in the capital structure. The value of the debt is the strike
price. Shareholders exercise their option to own the firm if the
firm's value exceeds the value of debt, otherwise, they default
and give the firm to the debt-holders. This means that many
stock options are actually options on options – pricing is more
complex than Black-Scholes.
Firm value
Debt value
EQUITY value
10mm
5mm
5mm
3mm
3mm
0mm
This situation can also be consider from the bondholders
perspective which involves a put option.
Put – Bondholders have the equivalent of a risk-free bond and
a short position in a put on the company that is given to
shareholders. If the value of the firm drops below 5mm, then
shareholders put the company to the bondholders so that
Risk-free Debt value
Short put value
DEBT value
5mm
0mm
5mm
5mm
-2mm
3mm
USE OPTIONS TO CUT UP PRICE
DISTRIBUTIONS - NOW CALLED
"FINANCIAL ENGINEERING"
Stock Price Distribution
Present
Stock Price
Financial engineering involves combining purchases and sales of
stocks, options or futures to capture only the parts of a stock price
distribution that you w ant. The possibilities are endless. If you buy
only the stock you ow n the w hole price distribution.
Put
Exercise
Price
Present
Stock Price
Call
Exercise
Price
If you believe the stock w ill soon rise (fall) sharply, buy the right (lef t)
tail by purchasing a call (put). If you believe that the stock w ill either
rise or fall sharply, but don’t know w hich, buy both tails by purchasing
a call and a put. This combination is called a straddle. Selling both
tails is called a strangle.
Put
Exercise
Price
Present
Stock Price
If you are longterm bullish on a stock but w ish to avoid any near-term
sharp decline, buy the stock and a put. This is called a protective put.
Of course, the put costs money so your net gains w ill be smaller.
Put
Exercise
Price
Present
Stock Price
Call
Exercise
Price
One w ay to pay the additional costs of the put in a protective put is to
also sell a call of similar value. This is called a collar. Of course, by
selling the call you give up some gains if the stock price rises above
the call’s exercise price.
BLACK - SCHOLES MODEL
CRUCIAL INSIGHT - it is possible to replicate the payoff to
an option by some investment strategy involving the
underlying asset and lending or borrowing - like the stock of a
leveraged company.
Therefore, we should be able to derive the value of an option
from the asset price and the interest rate.
Bullish --> Long position --> buy 100 shares of stock for,
say $10,000 or buy Call option on 100 shares for
$1000 and receive interest on $9000.
Here the option position has an advantage over the
stock position of earning interest.
Bearish --> Short position --> sell 100 shares short, put up
$1000 margin and earn interest on short sale
proceeds or buy Put for $1000.
Here the short stock position has an advantage
over the option position of earning interest.
THERE IS A HEDGE RATIO
BETWEEN THE CALL AND STOCK
THAT ESSENTIALLY ALLOWS ONE
TO EXACTLY REPLICATE AN
OPTION
For a simplified approach to replication one can use the
Binomial Model.
1. Assume that
S = Stock price today
C = Call option price today
r = risk-free rate
q = the probability the stock price will increase
(1-q)= the probability the stock price will decrease
u = the multiplicative increase (u > 1 + r > 1)
d = the multiplicative decrease (0 < d < 1 < 1 + r )
Cu = call price if stock price increases
Cd = call price if stock decreases
Each period the stock can take on only two
values; the stock can move up to uS or down to
dS.
2. Construct a risk-free hedge portfolio composed of one
share of stock and m call options written against the stock.
This means the payoffs in the up or down moves will be the
same so that
uS – mCu = dS – mCd
Solve for m, the hedge ratio of calls to be written on stock
m = S(u – d)/(Cu - Cd )
3. Because we constructed the portfolio to be risk-free, then
(1 + r)(S – mC) = uS – mCu
Or
C
S [(1  r )  u ]  m Cu
m(1  r )
4. Substituting for the hedge ratio m,
  (1  r )  d 
 u  (1  r ) 
C

C

d
 u u  d 
u

d




C
1 r
Or simplify let
(1  r )  d
p
ud
and
So
C = [pCu + (1 - p)Cd] / (1 + r)
u  (1  r )
1 p 
ud
Compare
C = [pCu + (1 - p)Cd] / (1 + r)
to
C = [πCu + (1 - π)Cd] / (1 + RAR)
Here, p is called the hedging probability, also called the
risk-neutral probability. The potential option payoffs Cu
and Cd are multiplied by the risk neutral probabilities
and the sum is discounted at the risk-free rate. Note:
the risk-neutral probability for the up (down) move is
less (more) than the objective probability, π, that
would be used if we discounted the payoffs with a riskadjusted rate (RAR from a CAPM rate based on option
beta) because the value in the numerator must be
smaller if we are discounting at the smaller risk-free
rate r.
Relation between
That is, p is the value q would take if all investors are risk
neutral and thus value the stock as follows,
(1 + r)S = quS + (1 – q)dS
Solving for q gives
(1  r )  d
q p
ud
Now this probability can be used to discount the payoffs for
the call option Cu and Cd.
5. Newer techniques for solving for derivatives prices often
employ the risk neutral approach. The risk neutral approach
takes advantage of the fact that we can value an option in two
ways. One is based upon the assumption that a stock will
follow some process where its price drifts upward over time at
a rate equal to its expected return. The option is then priced
off the stock. The problem is that we don’t have good models
of expected returns for the stock or the option (APT and
CAPM don’t work well). Instead, one can assume that the
stock will increase at the risk-free rate and use the risk-neutral
probabilities and discount the future option payoffs at the riskfree rate. That is, we use the wrong expected return for the
stock (a risk-free return instead of the risky expected return)
and the wrong probabilities (risk neutral probabilities) for the
option payoffs, but the errors always just offset.
Relation between the risk
neutral and true (physical)
probabilities
Start as before but with a risk averse investor, and the stock
return is R = r + e, where e is the excess return.
(1 + R)S = quS + (1 – q)dS
Or
(1 + r + e)S = quS + (1 – q)dS
Solving for q gives
(1  r  e)  d
e
q
 p
ud
ud
This shows that the true probability q is larger than the risk
neutral probability p.
To get a more formal expression for the relation between p
and q, recall that the standard deviation of a binomial is
σ = [q(1-q)].5(u-d)
Also E(R) = qu + (1-q)d – 1
And B = [E(R) - r]/σ
Plug these in above and get
p = q - B[q(1-q)].5
So again p < q for risk averse investors, and for risk neutral
investors B = 0 so p = q.
Note: The exact call option payoffs can be duplicated with a
portfolio of [(Cu - Cd)/(u – d)S] (inverse of m) shares of stock
and borrowing [(uCd - dCu)/(u – d)(1 + r)] at the risk-free rate.
6. Interesting features of call formula.
a. It does not depend on the probability of an
upward movement q (or (1-q)), so heterogeneous
expectations by investors about q are no problem because the
stock price aggregates heterogeneous expectations about q
and the option price is simply derived from the stock price. If
the probability of an upward move is large, the current stock
price will be higher, all else equal.
b. Investors risk preferences are irrelevant to call
price derivation since, again the stock price reflects it.
c. The only security relevant to the option is the
underlying stock; the market portfolio or factors are irrelevant.
Again, the stock is priced according to these.
d. We are able to perfectly hedge the option with
the stock because the returns for the two are perfectly
correlated.
7. To get the call price for a two period model, just apply the
one period model twice to get.
C = [p2Cuu + p(1 - p)Cud + p(1 - p)Cdu + (1 - p)2Cdd ] / (1 + r)2
Where Cuu, Cud = Cdu and Cdd are the three possible values for
the option after two periods. For more periods, just use
binomial expansion.
Example: Suppose that a stock’s price is S=100 and it can
increase by 100% or decrease by 50%. If the risk-free rate is
8% and the exercise price for a call is $125, find the price of
the call, the hedge ratio, the risk-neutral probabilities and the
amount of stock and risk-free borrowing one needs to
replicate call the option.
  (1  .08)  .5   2  (1  .08) 
  0

75
2

.
5
2

.
5

 

C
 26.85
1  .08
m = 100(2 – .5)/(75 - 0) = 2
Here, the option price moves half as much as the stock’s.
Therefore, if you own one share of the stock in this example,
you can hedge, that is, eliminate your risk, by selling two calls.
p
(1  .08)  .5
 .386667
2  .5
And 1 – p = .61333
Duplicate call with
[(Cu - Cd)/(u – d)S] = (75 – 0)/(2 - .5)100 = .5 shares of stock
And borrow [(uCd - dCu)/(u – d)(1 + r)] =
(2(0) - .5(.75))/(2 - .5)(1 + .08) = 23.15
1. To see how hedging works, form a hedged portfolio by
buying one share and selling 2 options and find its risk-free
end-of-period value.
Own the stock
Sold 2 options
Stock goes:
Down
Up
50
200
0
-150
50
50
Whatever happens, you get 50 so it is risk-free portfolio, i.e.,
perfectly hedged.
Find the present value of the portfolio’s end value by
discounting at the risk-free rate.
In this case, 50/(1+.08)=46.30.
You borrow this amount of money and add (S – 46.30) =
(100 – 46.30) = $53.70 of your own money to buy one
share. This leveraged position in the stock should give the
same return as owning two calls.
To see this note that in one year you pay off the loan and
you will have
150
or 0
[= 200 - 46.30(1+.08)] if stock goes to 200
[= 50 - 46.30(1+.08)] if the stock goes to 50.
Set the present value of the hedged portfolio equal to its
discounted risk-free value and solve for C.
Here, S - 2 C = 100 - 2C = 46.30 => C=26.85.
2. Another way to see this for the data above is to consider
that
a.
the stock payoff in one period is the combination of a
risk free 50 dollars (if you own the stock you know you
will have at least 50 in one period) plus the risky
payoff of 0 or 150 dollars.
b.
You can finance the position by borrowing 50/(1+r) =
50/(1.08) =46.30 at the risk free rate. A bank holding
your security would do this because they know the
security will be worth at least 50 in one year to cover
the loan.
c.
Of course, to buy the stock at 100 you had to put up
100 – 46.30 = 53.70 of your own money.
d.
We have m=2 as before, which means that the value
of your stock plus your borrowing portfolio will move
twice as much as the call. This means that you can
perfectly hedge a call with half of your stock plus
borrowing portfolio.
C = (1/m)[S - dS/(1 + r)]
= (1/2)[100 – 50/(1.08)] = 26.85
3. The binomial can also be illustrated using state-specific
securities. Note that there are only two states, up or down.
The payoffs can be used to determine the two state security
prices and then the option can be priced by multiplying the
state prices times the units of state securities that make up
the option.
4. Finally, we could find the risk neutral probabilities first and
then apply them to the two possible outcomes and then
discount the total at the risk free rate.
GET THE PUT VALUE - PUT/CALL
PARITY FORMULA
Put Price = C - S + E/(1 + risk-free rate)t
For this case:
Put = 26.85 - 100 + 125/(1 + .08)1 = 42.6.
This model shows that, to get an option value, ones needs to know
the
•
•
•
•
•
current stock price
option’s exercise price
risk-free rate
option maturity
stock price volatility.
Proof of Put-Call Parity
Consider two Portfolios A and B:
A. Purchase the stock for P0, purchase a put for T0 with exercise
price E and borrow E/(1 + r) at the risk-free rate r (in one year at
maturity you will have to pay back E).
B. Purchase the call at C0 also with exercise E and one year
maturity.
Portfolio Action
A
E
Investment
Buy Stock
Buy Put
Borrow
TOTALS
B
0
Buy Call
P0
T0
Value at Expiration
P1 > E
P1
0
-E/(1 + r)
P0 + T0 - E/(1+r)
C0
P1 – E
P1  E
P1
E - P1
-E
-
0
P1
-
E
Both A and B have the same payoff whether P1 > E or P1  E,
therefore, arbitrage arguments imply that both must have the same
initial investment. That is,
C0  P0  T0 
E
E
C0  P0  T0 
(1  r ) , or for maturity t,
(1  r )t
r
C0  T0  P0
(1  r )
If we select E = P0 then
BLACK-SCHOLES MODEL - A
NEARLY EXACT OPTION PRICING
MODEL
C0 = P0N(d1) - E e-rt N(d2)
where
Price of Stock = P0
Exercise price = E
Risk free rate = r
Time until expiration in years = t
Normal distribution function = N( )
Exponential function (base of natural log) = e
where:
d1 
d2 
ln( P0 / E )  (r .5 2 )t
 t
ln( P0 / E )  (r .5 2 )t
 t
 d1   t
where Standard deviation of stocks return = 
Natural log function = ln
Intuitively, the Black Sholes model can be explained as the
discounted expected value of the cash flows at expiration of
the option. At expiration, assuming P > E, we pay the
exercise price E (a cash outflow) and receive the stock with
the value P (a cash inflow). Before expiration, we attach
probabilities to these events. The N(d) are probabilities.
This is also a statement of an arbitrage relationship
(replicating strategy). The first term, P0N(d1) is the cost of
the shares in a portfolio that closely tracks the option value
and the second term, E e-rt N(d2) is the amount borrowed at
the risk-free rate to partly finance the stock purchase. The
difference between the two terms is the value of a call
option.
N(d2) is the probability that the stock will exceed the option’s
exercise price at expiration and N(d1) is the present value of
the expected stock price at expiration conditional on P > E
times the probability that P > E.
The N(d) are cumulative probabilities of a normal
distribution. These probabilities are sometimes described as
“risk-adjusted” probabilities.
For each (d), we have the term ln(P0/E) which is the percent
by which the stock price exceeds the exercise price (i.e. is
in the money). Clearly, if the stock price exceeds the
exercise price by a large percentage, the more likely the call
option will be valuable (i.e. exercised) at expiration.
But note that ln(P0/E) is divided by  t so that the
probability is adjusted for the stock’s risk and the time to
expiration. A call on a risky stock (relatively large  ) that is
in the money by a given percent has less probability of
staying in the money.
Another way to look at N(d1) is as the partial derivative of call
price with respect to stock price.
C/P0 = N(d1)
This is the instantaneous hedge ratio comparable to the
inverse of m in the binomial model. This derivative is referred
to as the “Delta” for an option.
Tests of the Black-Scholes option pricing model are joint tests
of the model and market efficiency. Most tests show strong
support, with only deep out-of-the-money options showing any
pricing bias.
How Model Variables Impact Options
Prices
1.
Stock price – Positively related to call price.
Negatively related to put price.
2.
Exercise price – Negatively related to call price.
Positively related to put price.
3.
Return Stand. Deviation – Positively related to call price.
Positively related to put price.
4.
Maturity –
Positively related to call price.
Positively related to put price (usually).
5.
Interest rate – Positively related to call price.
Negatively related to put price.
All but the interest rate effects are clear. To explain the effect
of interest rate intuitively, consider the following.
From the call option formula, we have the discounted exercise
price enter the equation with a negative sign. This
reflects the fact that if we exercise we have to make a
cash payment of the exercise price. If interest rates
increase, the present value of that payment today falls,
so that the call price must rise.
From the put-call parity formula, we see the opposite occurs
since, for a put, we will be receiving the exercise price in
the future in exchange for the stock if we exercise.
TO GET THE VALUE OF THE CALL,
C0
EXAMPLE: ASSUME
Price of Stock
Exercise price
Risk free rate
time period 3 mo.
Std Dev of stock return
P0 = 36
E = 40
r = .05
t = .25
s = .50
•Substitute into d1 and d2.
ln(36 / 40)  [.05.5(.50) 2 ].25
d1 
 .25
.50 .25
d  .25  .50 .25  .50
2
•Substitute d1, d2 and other variables in the main equation
C0 = 36N(-.25) - 40e-.05(.25)N(-.50)
•Look up in the normal table for d to get N(d).
here N(d1) = N(-.25) = .4013
and N(d2) = N(-.50) =.3085
•Substitute in the main equation
C0  36(.4013)  40e .05(.25) (.3085)  2.26
USE PUT CALL PARITY FORMULA
TO GET PUT PRICE
T0 = PUT PRICE
T0  C0  P0  Ee rt
To see why this holds, look at the stock price distribution
and how the put gives you the left tail of the distribution.
Then see that shorting the stock and buying the call leaves
you with the same left tail. Or see that payoff at time t=0 is
equal on both sides no matter what price is.
EXAMPLE - use info above - you need the call price
T0  2.26  36  40e
.05(.25)
= 2.26 - 36 + 39.5 = 5.76
Put-call parity is acceptable for European puts and usually
for American puts of short maturity (< 1 year). However,
when the chance of early exercise is relatively large, for
example, for long maturity puts, put prices are derived using
computerized numerical methods. The main reason is that if
the underlying stock price were to fall to very low levels
(think internet stocks) then it pays to exercise early. Time
becomes a negative there because the most that you can
gain by holding is for the stock to fall to zero, but by waiting,
the stock still has unlimited upside. It is then best to
exercise an American put early.
NET PRESENT VALUE RULE FOR
PROJECT ACCEPTANCE MUST BE
ADJUSTED IF OPTIONS ARE
INVOLVED.
There are two types of options to consider for most projects
A. The call option to delay a project to the future when
the project may have a larger NPV. A project that can
be delayed effectively competes with itself in the future.
This call option is more valuable when a project can be
delayed for a longer time (t), when a project’s (returns)
are very risky (), and when interest rates (r) are high.
This could explain why it may be rational to delay a
positive NPV project; Managers have often been
criticized by governments for not investing in plant and
equipment during recessions. Managers are not being
indecisive or too risk-averse but simply evaluating
projects based upon their option values which may be
high during recessions.
The basic idea is that if you undertake a project now, you
can’t undertake it in the future when it may have a higher
NPV. The more likely a project could have a higher NPV
in the future, the larger its option’s time value.
If the project is accepted, its time value is lost.
Thus, time value must be considered in the project selection
criteria. Thus instead of
NPVproject > 0
we use
NPVproject > time value of the option to delay > 0
Hence we should accept a project only when it
has a relatively large NPV.
A large NPV in
options terms means that the market value or
present value of the project’s cash flows greatly
exceeds its exercise price (cost of the project). In
other words - when its option is sufficiently “in the
money” i.e., it has much intrinsic value.
B. When a project’s acceptance allows one to undertake
additional projects in the future then we must make another
adjustment to the NPV criteria above.
NPVproject + Value of option on extended projects >
time value of option to delay
For example, if we delay building a new pentium
chip-making plant it may be cheaper in the future,
all else equal. However, if not building the plant
means we may forfeit the opportunity to build the
next generation chip, then this extra option must
be considered.
Example: You have a project that requires a $20 million
investment. You expect the project to provide cash flows with
present value of $22 million. Assume the risk-free rate is 10%
and the return standard deviation is .60. If you can delay the
project for two years should you accept the project now or
wait? What if the project gives us the option to make future
investments where this option is worth $8 million? Assume
that the investment remains $20 million whenever it is made
and the present value of cash flows remains $22 million. Also
assume that if you delay then you lose the option to make
future investments. (You don’t have to present value $20
because it is discounted in the Options model).
P0 = 22
X = 20
 = 0.6
t=2
r = .10
Find Value of option to delay - Call option
d1 = [ ln(22/20) + (.10 + 0.5(0.36)2] / 0.6(2).5
= [0.095 + (.10 + 0.18)(2) ] / 0.849
= .77
N(d1) = N(.75) = 0.7734
d2 = .77 - 0.849 = -0.079
N(d2) = N(-.10) = 0.4602
Vc = P0 N(d1) - Xe- r t [N(d2)]
= 22 (0.7734) - [20e-(.10)(2)](0.4602)
= 17.015 - 16.37(0.4602)
= 9.5
Time value = 9.5 - (22 - 20) = 7.5
Since NPV = (22 -20) = 2 < 7.5 then wait.
If the project gives us the option to make future investments
but only if we invest now and this option is worth 8 then we
would have
NPV + Option on Future Project = 2 + 8 = 10 > 7.5
- so now we would go ahead with the project.
Measuring Implied Volatility
We can use options prices to get the market’s prediction of
the volatility of a company’s stock price over the life of the
option. But because  enters the Black-Scholes model
nonlinearly, the implied value of  derived from it will be
biased. To get a nearly unbiased estimate:
1. Choose an option with an exercise price equal to the
discounted stock price.
E = P0e-rt
2. It can be shown that using this and a linear approximation
to the Black-Sholes model that
C = T = P0(t).5/(2).5
3. Get the implied standard deviation ()
 = C(2).5/P0(t).5
where  = 3.1416.
A widely followed index of overall stock market volatility is the
VIX - the standard deviation of the S&P 100 implied by onemonth index option premiums. See www.cboe.com.
See the S&P 500 volatility term structure at www.cboe.com
Suppose we have the following set of 3-month call option
premiums for the S&P 100:
Exercise Price
770
775
780
Premium
43
39
34
Assume that the risk-free rate is 6 percent and the S&P 100
index is trading at 786. To find the implied volatility, select the
option that has an exercise price closest to E = P0e-rt which in
this case is the 775 strike (E = 786e-(.06)(.25) = 774.4). Then
= C(2).5/P0(t).5 = 39(2*3.1416).5/786(.25).5
= 98/393 = .25
Like interest rate term structure, you can get a term structure
of stock market volatilities from the implied volatilities of
options of different maturities. On December 3, 1990, implied
volatilities for S&P 100 options of various maturities was
Expiration
Dec 1990
Jan 1991
Feb 1991
Implied Volatility
.187
.205
.257
Incremental Volatility
--.218
.344
Question: Does this pattern make sense given that Iraq was
given until January 15, 1991 to leave Kuwait?