FIN 685: Risk Management
Topic 3: Non-Linear Hedging
Larry Schrenk, Instructor

Black-Scholes Model

Greeks

Hedging

An Extended Example
The Black-Scholes Model
c  S N (d1 )  Xe  rt N (d 2 )
p  Xe
 rt
N ( d 2 )  S N ( d1 )
where d1 
d2 
ln(S / X )  (r   2 / 2)t
 t
2
ln(S / X )  (r   / 2)t
 t
 d1   t
S = Spot Price
 X = Exercise Price
 r = Risk Free Rate
  = Standard Deviation of Underlying
 t = Time to Maturity
 c = Price of a Call Option
 p = Value of a Put Option


Base Case Data
– S = 100
– K = 100
– r = 5%
–  = 30%
– T = 3 months
d1 


ln(100 / 100)  0.05   0.3  / 2 0.25
2
0.3 0.25
 0.1583
d 2  0.1583  0.3 0.25  0.0083
c  100N (0.1583)  100e
0.05 0.25 
N(0.0083) = $6.5831
The Greeks

Sensitivity Analysis

5 First Derivatives; 1 Second Derivative

Linear Approximations (except Gamma)
Delta (D) = Price Sensitivity
 Gamma (G) = 2nd Order Price Sensitivity
 Rho (R) = Interest Rate Sensitivity
 Theta (Q = Time Sensitivity
 Vega (n = Volatility Sensitivity


The hedge position must be frequently
rebalanced

Delta hedging a written option involves a
“buy high, sell low” trading rule
1
0.9
0.8
0.7
Delta
0.6
0.5
0.4
0.3
0.2
0.1
0
8
8.5
9
9.5
10
S
10.5
11
11.5
(S=10, K=10, T=0.2, r=0.05, =0.2)
12
Gamma (G) is the rate of change of delta
(D) with respect to the price of the
underlying asset
 Gamma is greatest for options that are
close to the money.

0.5
0.45
0.4
0.35
Gamma
0.3
0.25
0.2
0.15
0.1
0.05
0
8
8.5
9
9.5
10
S
10.5
11
11.5
(S=10, K=10, T=0.2, r=0.05, =0.2)
12

For a delta neutral portfolio,
DP  Q Dt + ½GDS 2
DP
DP
DS
DS
Positive Gamma
Negative Gamma


Theta (Q) of a derivative (or portfolio of
derivatives) is the rate of change of the value
with respect to the passage of time
The theta of a call or put is usually negative.
This means that, if time passes with the price of
the underlying asset and its volatility remaining
the same, the value of the option declines
0
-0.2
-0.4
Theta
-0.6
-0.8
-1
-1.2
-1.4
8
8.5
9
9.5
10
S
10.5
11
11.5
(S=10, K=10, T=0.2, r=0.05, =0.2)
12
Vega (n) is the rate of change of the value
of a derivatives portfolio with respect to
volatility
 Vega tends to be greatest for options that
are close to the money.

1.8
1.6
1.4
Vega
1.2
1
0.8
0.6
0.4
0.2
0
8
8.5
9
9.5
10
S
10.5
11
11.5
(S=10, K=10, T=0.2, r=0.05, =0.2)
12


D can be changed by taking a position in the
underlying
To adjust G & n it is necessary to take a
position in an option or other derivative

Rho is the rate of change of the value
of a derivative with respect to the
interest rate

For currency options there are 2 rhos
2
1.8
1.6
1.4
Rho
1.2
1
0.8
0.6
0.4
0.2
0
8
8.5
9
9.5
10
S
10.5
11
11.5
(S=10, K=10, T=0.2, r=0.05, =0.2)
12
Call Delta  N d1 
d12

2
e
Call Gamma 
Sσ 2t
Call Theta  Call Vega 
Sσ e
d12

2
2 2t
d12
2
S te
2
Call Rho  tXe  rt N(d 2 )
 rXe  rt N(d 2 )
Hedging
Hedging is about the reduction of risk.
 We will consider dynamic hedging in
which a portfolio is dynamically traded in
order to reduce risk.
 Simply put, a portfolio is hedged against
a certain risk if the portfolio value is not
sensitive to that risk.

Traders usually ensure that their
portfolios are delta-neutral at least once
a day
 Whenever the opportunity arises, they
improve gamma and vega
 As portfolio becomes larger hedging
becomes less expensive

15.26
Current Price: S = 10, Risk Free Rate: r = 0.05
Delta Hedge
6
call
delta
4
Call Option Price
2
0
Slope =D
-2
-4
Position in
Bonds
-6
0
5
10
15
Stock Price
Hedged Call Option Parameters: (K=10, T=0.2, =0.3)
Current Price: S = 10, Risk Free Rate: r = 0.05
Delta vs. Delta-Gamma Hedge
6
call
delta
delta-gamma
4
Call Option Price
2
0
-2
-4
-6
0
5
10
15
Stock Price
Hedged Call Option Parameters: (K=10, T=0.2, =0.3)
2nd Call Option Parameters: (K=8, T=0.4, =0.25)

Base Case Data
– S = 100
– K = 100
– r = 5%
–  = 30%
– T = 3 months

Call Value for Black-Scholes: c = $6.5831

Using both delta and gamma for the
estimate, what is the change in value (c’)
if the stock price increases by 5%?
Call Delta  N  d1   N  0.158  0.5629
Call Gamma 
e
d12

2
Sσ 2 t

e
2
0.1583 


2
100  0.3  2 (0.25)
 0.0263
c  100  0.05  0.5629  0.0263 100  0.05    $3.4711

What is what is the change in value (c’) of
the option if the interest rate increases by
5%?
Call Rho  tXe rtN(d2 )
=  0.25  100e
0.05 0.25 
N(0.0083) =12.4268
c  0.05  0.05   12.4268  $0.0311

What is what is the change in value (c’) of
the option if the standard deviation
increases by 5%? - d
2
1
Call Vega 

2
S te
2
100 0.25e
2
2
0.1583 

-
2
 19.6986
c  0.3  0.05   19.6986  $0.2955

What is what is the change in value (c’) of
the above option if the time decreases by
5%?
Call Theta  -
 -
100  0.3  e
Sσ e

d12
2
2 2 t

 rXe rtN(d2 )
 0.1583 2
2 2 0.25
2
 0.05 100  e
0.05 0.25 
N(0.0083) = -14.3045
c  0.25  0.05    14.3045  $0.1788
An Extended Example


Here we talk about how option dealers hedge
the risk of option positions they take.
Assume a dealer sells 1,000 AOL June 125 calls
at the Black-Scholes price of 13.5512 with a
delta of .5692. Dealer will buy 569 shares and
adjust the hedge daily.
– To buy 569 shares at $125.9375 and sell 1,000 calls
at $13.5512 will require $58,107.
– We simulate the daily stock prices for 35 days, at
which time the call expires.


The second day, the stock price is 120.5442. There are now 34 days
left. Using bsbin2.xls, we get a call price of 10.4781 and delta of
.4999. We have
– Stock worth 569($120.5442) = $68,590
– Options worth -1,000($10.4781) = -$10,478
– Total of $58,112
– Had we invested $58,107 in bonds, we would have had
$58,107e.0446(1/365) = $58,114.
Table 5.9, p. 202 shows the remaining outcomes. We must adjust to
the new delta of .4999. We need 500 shares so sell 69 and invest the
money ($8,318) in bonds.


At the end of the second day, the stock goes to 106.9722 and the call
to 4.7757. The bonds accrue to a value of $8,319. We have
– Stock worth 500($106.9722) = $53,486
– Options worth -1,000($4.7757) = -$4,776
– Bonds worth $8,319 (includes one days’ interest)
– Total of $57,029
– Had we invested the original amount in bonds, we would have
had $58,107e.0446(2/365) = $58,121. We are now short by over
$1,000.
At the end we have $56,540, a shortage of $1,816.



What we have seen is the second order or gamma effect. Large price
changes, combined with an inability to trade continuously result in
imperfections in the delta hedge.
To deal with this problem, we must gamma hedge, i.e., reduce the
gamma to zero. We can do this only by adding another option. Let
us use the June 130 call, selling at 11.3772 with a delta of .5086 and
gamma of .0123. Our original June 125 call has a gamma of .0121.
The stock gamma is zero.
We shall use the symbols D1, D2, G1 and G2. We use hS shares of stock
and hC of the June 130 calls.




The delta hedge condition is
– hS(1) - 1,000D1 + hC D 2 = 0
The gamma hedge condition is
– -1,000G1 + hC G2 = 0
We can solve the second equation and get hC and then substitute
back into the first to get hS. Solving for hC and hS, we obtain
– hC = 1,000(.0121/.0123) = 984
– hS = 1,000(.5692 - (.0121/.0123).5086) = 68
So buy 68 shares, sell 1,000 June 125s, buy 985 June 130s.

The initial outlay will be
– 68($125.9375) - 1,000($13.5512) + 985($11.3772) =
$6,219

At the end of day one, the stock is at
120.5442, the 125 call is at 10.4781, the 130
call is at 8.6344. The portfolio is worth
– 68($120.5442) - 1,000($10.4781) + 985($8.6344) =
$6,224


It should be worth $6,219e.0446(1/365) = $6,220.
The new deltas are .4999 and .4384 and the
new gammas are .0131 and .0129.




The new values are 1,012 130 calls so we buy 27. The new number of
shares is 56 so we sell 12. Overall, this generates $1,214, which we
invest in bonds.
The next day, the stock is at $106.9722, the 125 call is at $4.7757 and
the 130 call is at $3.7364. The bonds are worth $1,214. The portfolio
is worth
– 56($106.9722) - 1,000($4.7757) + 1,012($3.7364) + $1,214 = $6,210.
The portfolio should be worth $6,219e.0446(2/365) = $6,221.
Continuing this, we end up at $6,589 and should have $6,246, a
difference of $343. We are much closer than when only delta
hedging.