Pricing Cont’d &

Beginning Greeks

Assumptions of the Black-Scholes Model

European exercise style Markets are efficient No transaction costs The stock pays no dividends during

the option’s life Interest rates and volatility remain

constant

The Stock Pays no Dividends During the Option’s Life

If you apply the BSOPM to two securities, one with no dividends and the other with a dividend yield, the model will predict the same call premium

Robert Merton developed a simple extension to the BSOPM to account for the payment of dividends

The Stock Pays Dividends During the Option’s Life (cont’d)

Adjust the BSOPM by following (=continuous dividend yield):

Tdd

T

TRXS

d

dNXedSNeC RTT

*1

*2

2

*1

*2

*1

*

and

2ln

where

)()(

Interest Rates and Volatility Remain Constant

There is no real “riskfree” interest rate

Often use the closest T-bill rate to expiry

Volatility expectations change constantly. That’s why option prices can change when everything else remains constant!

Calculating Black-Scholes Prices

from Historical Data: S, R, T that just was, and as standard

deviation of historical returns from some arbitrary past period

from Actual Data: S, R, T that just was, and implied from pricing of nearest “at-the-money” option (termed “implied volatility).

Implied Volatility Introduction Calculating implied volatility Volatility smiles

Introduction Instead of solving for the call

premium, assume the market-determined call premium is correct

Then solve for the volatility that makes the equation hold

This value is called the implied volatility

Calculating Implied Volatility Setup spreadsheet for pricing “at-the-

money” call option. Input actual price. Run SOLVER to equate actual and

calculated price by varying .

Volatility Smiles Volatility smiles are in contradiction

to the BSOPM, which assumes constant volatility across all strike prices

When you plot implied volatility against striking prices, the resulting graph often looks like a smile

Volatility Smiles (cont’d)Volatility Smile

Microsoft August 2000

0

10

20

30

40

50

60

40 45 50 55 60 65 70 75 80 85 90 95 100 105

Striking Price

Imp

lie

d V

ola

tili

ty (

%)

Current Stock Price

Problems Using the Black-Scholes Model

Does not work well with options that are deep-in-the-money or substantially out-of-the-money

Produces biased values for very low or very high volatility stocks Increases as the time until expiration

increases

May yield unreasonable values when an option has only a few days of life remaining

Beginning Greeks & Hedging Hedge Ratios Greeks (Option Price Sensitivities)

delta, gamma (Stock Price) rho (riskless rate) theta (time to expiration)vega (volatility)

Delta Hedging

Hedge Ratios Number of units of hedging security to

moderate value change in exposed position If trading options: Number of units of

underlying to hedge options portfolio If trading underlying: Number of options to

hedge underlying portfolio For now: we will act like trading European

Call Stock Options with no dividends on underlying stock.

Delta, Gamma Sensitivity of Call Option Price to

Stock Price change (Delta):

= N(d1) We calculated this to get option price. Gamma is change in Delta measure as

Stock Price changes….we’ll get to this later!

Delta Hedging If an option were on 1 share of stock,

then to delta hedge an option, we would take the overall position: +C - S = 0 (change)

This means whatever your position is in the option, take an opposite position in the stock (+ = bought option sell stock) (+ = sold option buy stock)

Recall the Pricing Example IBM is trading for $75. Historically, the volatility is 20% (A

call is available with an exercise of $70, an expiry of 6 months, and the risk free rate is 4%.

ln(75/70) + (.04 + (.2)2/2)(6/12)d1 = -------------------------------------------- = .70, N(d1) =.7580

.2 * (6/12)1/2

d2 = .70 - [ .2 * (6/12)1/2 ] = .56, N(d2) = .7123

C = $75 (.7580) - 70 e -.04(6/12) (.7123) = $7.98Intrinsic Value = $5, Time Value = $2.98

Hedge the IBM Option Say we bought (+) a one share IBM

option and want to hedge it:

+ C - S means

1 call option hedged with shares of IBM stock sold short (-).

= N(d1) = .758 shares sold short.

Overall position value:Call Option cost = -$

7.98 Stock (short) gave = +$ 56.85 (S = .758*75 = 56.85)

Overall account value: +$ 48.87

Hedge the IBM Option

Why a Hedge? Suppose IBM goes to $74.

ln(74/70) + (.04 + (.2)2/2)(6/12)d1 = -------------------------------------------- = 0.61, N(d1) =.7291

.2 * (6/12)1/2

d2 = 0.61 - [ .2 * (6/12)1/2 ] = 0.47, N(d2) = .6808

C = $74 (.7291) - 70 e -.04(6/12) (.6808) = $7.24

Results Call Option changed:

(7.24 - 7.98)/7.98 = -9.3% Stock Price changed:

(74 - 75)/75 = -1.3% Hedged Portfolio changed:

(Value now –7.24 + (.758*74) = $48.85) (48.85 - 48.87)/48.87 = -0.04%!

Now that’s a hedge!

Hedging Reality #1 Options are for 100 shares, not 1 share. You will rarely have one option to

hedge. Both these issues are just multiples!

+ C - S becomes + 100 C - 100 S for 1 actual option, or + X*100 C - X*100 S for X actual options

Hedging Reality #2 Hedging Stock more likely:

+ C - S = 0 becomes algebraically - (1/) C + S

So to hedge 100 shares of long stock (+), you would sell (-) 1/ options

For example, (1/.758) = 1.32 options

Hedging Reality #3 Convention does not hedge long stock

by selling call options (covered call). Convention hedges long stock with

bought put options (protective put).

Instead of - (1/) C + S- (1/P) P + S

Hedging Reality #3 cont’d P = [N(d1) - 1], so if

N(d1) < 1 (always), thenP < 0

This means- (1/P) P + S

actually has the same positions in stock and puts ( -(-) = + ).

This is what is expected, protective put is long put and long stock.

Reality #3 Example Remember IBM pricing:

ln(75/70) + (.04 + (.2)2/2)(6/12)d1 = -------------------------------------------- = .70, N(d1) =.7580

.2 * (6/12)1/2

d2 = .70 - [ .2 * (6/12)1/2 ] = .56, N(d2) = .7123C = $75 (.7580) - 70 e -.04(6/12) (.7123) = $7.98Put Price = Call Price + X e-rT - S

Put = $7.98 + 70 e -.04(6/12) - 75

= $1.59

Hedge 100 Shares of IBM - (1/P) P + S =

- 100 * (1/P) P + 100 * S P = N(d1) – 1 = .758 – 1 = -.242

- (1/P) = - (1/ -.242) = + 4.13 options Thus if “ + “ of + S means bought

stock, then “ + “ of +4.13 means bought put options!

That’s a protective put!

Hedge Setup Position in Stock: $75 * 100 = +$7500 Position in Put Options:

$1.59 * +4.13 * 100 = +$656.67 Total Initial Position =+$8156.67

IBM drops to $74 Remember call now worth $7.24

Puts now worth $1.85 * 4.13 * 100 = $ 764.05

Total Position = $7400 + 764.05 = $8164.05

Put Price = Call Price + X e-rT - S

Put = $7.24 + 70 e -.04(6/12) - 74

= $1.85

Results Stock Price changed:

(74 - 75)/75 = -1.3%

Portfolio changed: (8164.05 – 8156.67) / 8156.67 = +0.09%!!!!

Now that’s a hedge!