PowerPoint Presentation
Topic 5 – Options continued
Black – Scholes Model and Put-Call parity
The Black-Scholes Model
1973
Theoretical equation for pricing European options
Uses risk-neutral argument (same as the Binomial model)
Widely used in options trading
The Black-Scholes Model
The fair price depends on the following factors : S, X, T, r and 2.
All these factors are readily observable except for the variance, which has to be estimated.
There are two ways of estimating 2.
First method uses historical data on security’s price movements and calculates variance based on log price relatives
Second method of estimating is to use the Black and Scholes equation in reverse.
Volatility
Probability (discrete & continuous)
50%
50%
5
The normal distribution function
The cumulative probability (z) gives the area under the curve
This is analagous to the percentage of red balls that fall to the right of a particular point
5
Market Analogy
105
104
103
102
101
100
99
98
97
96
95
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92
91
Time (t)
Volatility (s)
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Market profile
6
Normal probabilities
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Using the tables for N(x) in Hull p868-9 calculate the:
– probability of P > 0.55z?
– probability of P < 0.60z?
The ‘x’ that corresponds to a cumulative probability of 0.90?
Answers
0.2912
0.7257
≈ +1.28
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Black Scholes (European Options)
8
N() is the cumulative normal distribution function
K is the option strike
σ is the annualised lognormal volatility
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Black Scholes (European Options)
9
Example
S = 100, K = 105, σ is 10%, r = 5%
What is the value of a 1 year call option?
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Black Scholes (European Options)
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Make sure that you use the correct part of the probability distribution to calculate N(d)
Solution:
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Equation – interpretation
As becomes very large, the call price tends to and the put price tends to zero, i.e. it is very likely that the call option will be exercised (and not the put), and today’s value is given by the difference between the current stock price and the present value of the strike.
As becomes very small, the call price tends to zero and the put price tends to , i.e. it is very likely that the put option will be exercised (and not the call) and today’s value is given by the difference between the present value of the strike and the current stock price.
https://www.khanacademy.org/economics-finance-domain/core-finance/derivative-securities/black-scholes/v/introduction-to-the-black-scholes-formula
Implied vs historical volatility
Historical volatility can be observed from history of stock price but is backward looking
Future volatility is implied by the option prices observed in the market
Premiums and levels of volatility vary together all other things being equal
Traders generally quote % volatility rather than prices – implied volatility is less variable than the option price
Hypothetical call option price changes
S X T r %volatility Call price
100 100 30 0.02 30 6
100 100 30 0.02 15 4
100 100 30 0.02 40 7
102 100 30 0.02 30 7.5
100 100 20 0.02 30 4.5
100 100 5 0.02 30 2
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Put-Call Parity: (No Dividends)
Consider the following 2 portfolios:
Portfolio I: European put on the stock + the stock
Portfolio II: European call on a stock + zero-coupon bond that pays K at time T
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P 263-264 Hull
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© 2017 ICMA Centre, Henley Business School
Put Call Parity
See p279 12.1 Hull the Bond pays K at time T
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The Put-Call Parity Result (Equation 11.6, page 264)
Both are worth max(ST , K ) at the maturity of the options
They must therefore be worth the same today. This means that c + Ke -rT = p + S0
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Example of Put-Call Parity
The price of a non-dividend paying stock is $19 and the price of a three-month European call option on the stock with a strike price of $20 is $1. The risk-free rate is 4% per annum. What is the price of a three-month European put option with a strike price of $20?
So C=1, T= 0.25, S = 19, K=20, r= 0.04
P= 1+20e^(-0.04*0.25) -19
= 1.80
So the European Put price is $1.80
C0 = S0N (d1)− e
−rtKN (d2 )
P0 = e
−rtKN (−d2 )− S0N (−d1)
d1 =
ln(
S0
K
)+ rt +σ
2t
2
σ t
d2 = d1 −σ t
C
0
=S
0
N(d
1
)-e
-rt
KN(d
2
)
P
0
=e
-rt
KN(-d
2
)-S
0
N(-d
1
)
d
1
=
ln(
S
0
K
)+rt+
s
2
t
2
st
d
2
=d
1
-st
C0 = S0N (d1)− e
−rtKN (d2 )
P0 = e
−rtKN (−d2 )− S0N (−d1)
d1 =
ln(
S0
K
)+ rt +σ
2t
2
σ t
d2 = d1 −σ t
C
0
=S
0
N(d
1
)-e
-rt
KN(d
2
)
P
0
=e
-rt
KN(-d
2
)-S
0
N(-d
1
)
d
1
=
ln(
S
0
K
)+rt+
s
2
t
2
st
d
2
=d
1
-st
d1 =
ln(100
105
)+0.05+ 0.10
2 *1
2
0.10*1
= 0.0621
d2 = 0.0621−0.10 = −0.0379
N (d1) = 0.525
N (d2 ) = 0.485
e−rt = 0.9512
d
1
=
ln(
100
105
)+0.05+
0.10
2
*1
2
0.10*1
=0.0621
d
2
=0.0621-0.10=-0.0379
N(d
1
)=0.525
N(d
2
)=0.485
e
-rt
=0.9512
C0 =100*0.525−0.9512*105*0.485= 4.06
C
0
=100*0.525-0.9512*105*0.485=4.06
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