Transcript Options and Risk Measurement

Options and risk measurement
Definition of a call option
 A call
option is the right but not the
obligation to buy 100 shares of the
stock at a stated exercise price on or
before a stated expiration date.
 The price of the option is not the
exercise price.
Example
 A share
of IBM sells for 75.
 The call has an exercise price of 76.
 The value of the call seems to be zero.
 In fact, it is positive and in one example
equal to 2.
t=0
t=1
S = 80, call = 4
S = 75
S = 70, call = 0
Value of call = .5 x 4 = 2
Definition of a put option
 A put
option is the right but not the
obligation to sell 100 shares of the
stock at a stated exercise price on or
before a stated expiration date.
 The price of the option is not the
exercise price.
Example
 A share
of IBM sells for 75.
 The put has an exercise price of 76.
 The value of the put seems to be 1.
 In fact, it is more than 1 and in our
example equal to 3.
t=0
t=1
S = 80, put = 0
S = 75
S = 70, put = 6
Value of put = .5 x 6 = 3
Put-call parity
S
+ P = X*exp(-r(T-t)) + C at any time t.
 s + p = X + c at expiration
 In the previous examples, interest was
zero or T-t was negligible.
 Thus S + P=X+C
 75+3=76+2
 If not true, there is a money pump.
Puts and calls as random
variables
 The
exercise price is always X.
 s, p, c, are cash values of stock, put,
and call, all at expiration.
 p = max(X-s,0)
 c = max(s-X,0)
 They are random variables as viewed
from a time t before expiration T.
 X is a trivial random variable.
Puts and calls before expiration
 S,
P, and C are the market values at
time t before expiration T.
 Xe-r(T-t) is the market value at time t of
the exercise money to be paid at T
 Traders tend to ignore r(T-t) because it
is small relative to the bid-ask spreads.
Put call parity at expiration
 Equivalence
at expiration (time T)
s+p=X+c
 Values at time t in caps:
S + P = Xe-r(T-t) + C
No arbitrage pricing implies
put call parity in market prices
 Put
call parity holds at expiration.
 It also holds before expiration.
 Otherwise, a risk-free arbitrage is
available.
Money pump one
S + P = Xe-r(T-t) + C + e
 S and P are overpriced.
 Sell short the stock.
 Sell the put.
 Buy the call.
 “Buy” the bond. For instance deposit Xe-r(T-t)
in the bank.
 The remaining e is profit.
 The position is riskless because at expiration
s + p = X + c. i.e.,
 If
Money pump two
S + P + e = Xe-r(T-t) + C
 S and P are underpriced.
 “Sell” the bond. That is, borrow Xe-r(T-t).
 Sell the call.
 Buy the stock and the put.
 You have + e in immediate arbitrage
profit.
 The position is riskless because at
expiration s + p = X + c. i.e.,
 If
Money pump either way
 If
the prices persist, do the same thing
over and over – a MONEY PUMP.
 The existence of the e violates noarbitrage pricing.
Measuring risk
Rocket science
Rate of return =
 (price
increase + dividend)/purchase
price.
Pt 1  Pt  divt 1
Rj 
Pt
Sample average
Year
Rate of return
on common stocks
1926
1927
1928
1929
11.62
37.49
43.61
-8.42
Sample average
11.62  37.49  43.61  8.42
R
4
 21.075
Sample versus population
 A sample
is a series of random draws
from a population.
 Sample is inferential. For instance the
sample average.
 Population: model: For instance the
probabilities in the problem set.
Population mean
 The
value to which the sample average
tends in a very long time.
 Each sample average is an estimate,
more or less accurate, of the population
mean.
Abstraction of finance
 Theory
works for the expected values.
 In practice one uses sample means.
Deviations
Rate of return
on common stocks 11.62
37.49
43.61
sample average
21.075
21.075
21.075
deviation
-9.455
16.415
22.535
deviation squared
89.39703 269.4522 507.8262
sample variance
578.8768
standard deviation 24.05986
-8.42
21.075
-29.495
869.955
Explanation
 Square
deviations to measure both
types of risk.
 Take square root of variance to get
comparable units.
 Its still an estimate of true population
risk.
Why divide by 3 not 4?
 Sample
deviations are probably too
small …
 because the sample average minimizes
them.
 Correction needed.
 Divide by T-1 instead of T.
Derivation of sample average
as an estimate of population
mean.
Select m to min imize
(11.62  m) 2  (37.49  m) 2  (43.62  m) 2  (8.420  m) 2
Solution
 2(11.62  m)  2(37.49  m)  2(43.62  m)
 2( 8.420  m)  0
4m  11.62  37.49  43.62  8.42
11.62  37.49  43.62  8.42
m
4
Rough interpretation of
standard deviation
 The
usual amount by which returns
miss the population mean.
 Sample standard deviation is an
estimate of that amount.
 About 2/3 of observations are within
one standard deviation of the mean.
 About 95% are within two S.D.’s.
Estimated risk and return
1926-1999
Sample average Sample sigma Sample Premium
T-Bills
3.8
3.2
0
Common stocks
13.3
20.1
9.5
Small cap stocks
17.6
33.6
13.8
LT Corp bonds
5.9
8.7
2.1
Inflation
3.2
4.5
-0.6
Review question
 What
is the difference between the
population mean and the sample
average?
Answer
 Take
a sample of T observations drawn
from the population
 The sample average is
(sum of the rates)/T
 The sample average tends to the
population mean as the number of
observations T becomes large.