FIN 685: Risk Management Topic 3: Non-Linear Hedging Larry Schrenk, Instructor.

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FIN 685: Risk Management Topic 3: Non-Linear Hedging Larry Schrenk, Instructor
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Transcript of FIN 685: Risk Management Topic 3: Non-Linear Hedging Larry Schrenk, Instructor.

Page 1: FIN 685: Risk Management Topic 3: Non-Linear Hedging Larry Schrenk, Instructor.

FIN 685: Risk Management

Topic 3: Non-Linear Hedging

Larry Schrenk, Instructor

Page 2: FIN 685: Risk Management Topic 3: Non-Linear Hedging Larry Schrenk, Instructor.

TOPICS

Black-Scholes Model

Greeks

Hedging

An Extended Example

Page 3: FIN 685: Risk Management Topic 3: Non-Linear Hedging Larry Schrenk, Instructor.

The Black-Scholes Model

Page 4: FIN 685: Risk Management Topic 3: Non-Linear Hedging Larry Schrenk, Instructor.

BLACK-SCHOLES FORMULA

1 2

2 1

1

2 1

( ) ( )

( ) ( )

2ln( / ) ( / 2)where

2ln( / ) ( / 2)

rt

rt

c S N d Xe N d

p Xe N d S N d

S X r td

t

S X r td d t

t

Page 5: FIN 685: Risk Management Topic 3: Non-Linear Hedging Larry Schrenk, Instructor.

VARIABLES

S = Spot Price X = Exercise Price r = Risk Free Rate s = Standard Deviation of

Underlying t = Time to Maturity c = Price of a Call Option p = Value of a Put Option

Page 6: FIN 685: Risk Management Topic 3: Non-Linear Hedging Larry Schrenk, Instructor.

EXAMPLE

Base Case Data– S = 100– K = 100– r = 5% – s = 30%– T = 3 months

Page 7: FIN 685: Risk Management Topic 3: Non-Linear Hedging Larry Schrenk, Instructor.

BLACK-SCHOLES FORMULA

2

1

2

0.05 0.25

ln(100 / 100) 0.05 0.3 / 2 0.250.1583

0.3 0.25

0.1583 0.3 0.

$6.58

25 0.0083

100 (0.1583) 100 (0.0083) = 1 3

d

d

c N e N

Page 8: FIN 685: Risk Management Topic 3: Non-Linear Hedging Larry Schrenk, Instructor.

The Greeks

Page 9: FIN 685: Risk Management Topic 3: Non-Linear Hedging Larry Schrenk, Instructor.

THE GREEKS

Sensitivity Analysis

5 First Derivatives; 1 Second Derivative

Linear Approximations (except Gamma)

Page 10: FIN 685: Risk Management Topic 3: Non-Linear Hedging Larry Schrenk, Instructor.

THE GREEKS

Delta (D) = Price Sensitivity Gamma (G) = 2nd Order Price

Sensitivity Rho (R) = Interest Rate

Sensitivity Theta ( )Q = Time Sensitivity Vega ( )n = Volatility Sensitivity

Page 11: FIN 685: Risk Management Topic 3: Non-Linear Hedging Larry Schrenk, Instructor.

DELTA HEDGING

• The hedge position must be frequently rebalanced

• Delta hedging a written option involves a “buy high, sell low” trading rule

Page 12: FIN 685: Risk Management Topic 3: Non-Linear Hedging Larry Schrenk, Instructor.

EUROPEAN CALL DELTA

8 8.5 9 9.5 10 10.5 11 11.5 120

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

S

Del

ta

(S=10, K=10, T=0.2, r=0.05, s=0.2)

Page 13: FIN 685: Risk Management Topic 3: Non-Linear Hedging Larry Schrenk, Instructor.

GAMMA

• 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.

Page 14: FIN 685: Risk Management Topic 3: Non-Linear Hedging Larry Schrenk, Instructor.

EUROPEAN CALL GAMMA

8 8.5 9 9.5 10 10.5 11 11.5 120

0.05

0.1

0.15

0.2

0.25

0.3

0.35

0.4

0.45

0.5

S

Gam

ma

(S=10, K=10, T=0.2, r=0.05, s=0.2)

Page 15: FIN 685: Risk Management Topic 3: Non-Linear Hedging Larry Schrenk, Instructor.

INTERPRETATION OF GAMMA• For a delta neutral portfolio,

DP » Q Dt + ½GDS 2

DP

DS

Negative Gamma

DP

DS

Positive Gamma

Page 16: FIN 685: Risk Management Topic 3: Non-Linear Hedging Larry Schrenk, Instructor.

THETA

• 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

Page 17: FIN 685: Risk Management Topic 3: Non-Linear Hedging Larry Schrenk, Instructor.

EUROPEAN CALL THETA

(S=10, K=10, T=0.2, r=0.05, s=0.2)

8 8.5 9 9.5 10 10.5 11 11.5 12-1.4

-1.2

-1

-0.8

-0.6

-0.4

-0.2

0

S

The

ta

Page 18: FIN 685: Risk Management Topic 3: Non-Linear Hedging Larry Schrenk, Instructor.

VEGA

• 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.

Page 19: FIN 685: Risk Management Topic 3: Non-Linear Hedging Larry Schrenk, Instructor.

EUROPEAN CALL VEGA

(S=10, K=10, T=0.2, r=0.05, s=0.2)

8 8.5 9 9.5 10 10.5 11 11.5 120

0.2

0.4

0.6

0.8

1

1.2

1.4

1.6

1.8

S

Veg

a

Page 20: FIN 685: Risk Management Topic 3: Non-Linear Hedging Larry Schrenk, Instructor.

MANAGING DELTA, GAMMA, & VEGA

· 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

Page 21: FIN 685: Risk Management Topic 3: Non-Linear Hedging Larry Schrenk, Instructor.

RHO

• 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

Page 22: FIN 685: Risk Management Topic 3: Non-Linear Hedging Larry Schrenk, Instructor.

EUROPEAN CALL RHO

(S=10, K=10, T=0.2, r=0.05, s=0.2)

8 8.5 9 9.5 10 10.5 11 11.5 120

0.2

0.4

0.6

0.8

1

1.2

1.4

1.6

1.8

2

Rho

S

Page 23: FIN 685: Risk Management Topic 3: Non-Linear Hedging Larry Schrenk, Instructor.

FORMULAE

)N(drXe t22

eσS- Theta Call 2

rt2

d21

2

etS Vega Call

2

d-

21

)N(dtXeRho Call 2rt

t2Sσ

eGamma Call

2

d21

1Delta Call dN

Page 24: FIN 685: Risk Management Topic 3: Non-Linear Hedging Larry Schrenk, Instructor.

Hedging

Page 25: FIN 685: Risk Management Topic 3: Non-Linear Hedging Larry Schrenk, Instructor.

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.

Page 26: FIN 685: Risk Management Topic 3: Non-Linear Hedging Larry Schrenk, Instructor.

15.26

HEDGING IN PRACTICE

• 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

Page 27: FIN 685: Risk Management Topic 3: Non-Linear Hedging Larry Schrenk, Instructor.

DELTA

0 5 10 15-6

-4

-2

0

2

4

6

Stock Price

Call

Option P

rice

Delta Hedge

calldelta

Hedged Call Option Parameters: (K=10, T=0.2, s=0.3)

Current Price: S = 10, Risk Free Rate: r = 0.05

Position in Bonds

Slope =D

Page 28: FIN 685: Risk Management Topic 3: Non-Linear Hedging Larry Schrenk, Instructor.

DELTA + GAMMA

0 5 10 15-6

-4

-2

0

2

4

6

Stock Price

Cal

l Opt

ion

Pric

e

Delta vs. Delta-Gamma Hedge

calldeltadelta-gamma

Hedged Call Option Parameters: (K=10, T=0.2, s=0.3)

Current Price: S = 10, Risk Free Rate: r = 0.05

2nd Call Option Parameters: (K=8, T=0.4, s=0.25)

Page 29: FIN 685: Risk Management Topic 3: Non-Linear Hedging Larry Schrenk, Instructor.

EXAMPLES

Base Case Data– S = 100– K = 100– r = 5% – s = 30%– T = 3 months

Call Value for Black-Scholes: c = $6.5831

Page 30: FIN 685: Risk Management Topic 3: Non-Linear Hedging Larry Schrenk, Instructor.

EXAMPLE 1: DELTA + GAMMA Using both delta and gamma for

the estimate, what is the change in value (c’) if the stock price increases by 5%?

221 0.1583d

2 2e eCall Gamma 0.0263

Sσ 2 t 100 0.3 2 (0.25)

c 100 0.05 0.5629 0.0263 100 0. $3.471105

1Call Delta 0.158 0.5629N d N

Page 31: FIN 685: Risk Management Topic 3: Non-Linear Hedging Larry Schrenk, Instructor.

EXAMPLE 2: RHO

What is what is the change in value (c’) of the option if the interest rate increases by 5%?

rt2

0.05 0.25

Call Rho tXe N(d )

= 0.25 100e N(0.0083) =12.4268

c 0.05 0.05 12.4268 $0.0311

Page 32: FIN 685: Risk Management Topic 3: Non-Linear Hedging Larry Schrenk, Instructor.

EXAMPLE 3: VEGA

What is what is the change in value (c’) of the option if the standard deviation increases by 5%?

21

2

d-

2

0.1583-

2

S teCall Vega

2

100 0.25e 19.6986

2

c 0.3 0.05 19.6986 $0.2955

Page 33: FIN 685: Risk Management Topic 3: Non-Linear Hedging Larry Schrenk, Instructor.

EXAMPLE 4: THETA

What is what is the change in value (c’) of the above option if the time decreases by 5%?

σ

21

2

d

2rt

2

0.1583

20.05 0.25

S eCall Theta - rXe N(d )

2 2 t

100 0.3 e - 0.05 100 e N(0.0083) = -14.3045

2 2 0.25

c 0.25 0.05 14.304 $0.17885

Page 34: FIN 685: Risk Management Topic 3: Non-Linear Hedging Larry Schrenk, Instructor.

An Extended Example

Page 35: FIN 685: Risk Management Topic 3: Non-Linear Hedging Larry Schrenk, Instructor.

MANAGING THE RISK OF OPTIONS

• 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.

Page 36: FIN 685: Risk Management Topic 3: Non-Linear Hedging Larry Schrenk, Instructor.

MANAGING THE RISK OF OPTIONS

• 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.

Page 37: FIN 685: Risk Management Topic 3: Non-Linear Hedging Larry Schrenk, Instructor.

MANAGING THE RISK OF OPTIONS

• 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.

Page 38: FIN 685: Risk Management Topic 3: Non-Linear Hedging Larry Schrenk, Instructor.

MANAGING THE RISK OF OPTIONS

• 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 1, 2, 1 and 2. We use hS shares of stock and hC of the June 130 calls.

Page 39: FIN 685: Risk Management Topic 3: Non-Linear Hedging Larry Schrenk, Instructor.

MANAGING THE RISK OF OPTIONS

• The delta hedge condition is– hS(1) - 1,0001 + hC 2 = 0

• The gamma hedge condition is– -1,0001 + hC 2 = 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.

Page 40: FIN 685: Risk Management Topic 3: Non-Linear Hedging Larry Schrenk, Instructor.

MANAGING THE RISK OF OPTIONS

• 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.

Page 41: FIN 685: Risk Management Topic 3: Non-Linear Hedging Larry Schrenk, Instructor.

MANAGING THE RISK OF OPTIONS• 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.