1. Pricing Cont’d & Beginning Greeks

2. 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

3. 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

4. The Stock Pays Dividends During the Option’s Life (cont’d) Adjust the BSOPM by following (  =continuous dividend yield): Tdd T T  R X S d dNXedSNeC RT TT                       * 1 * 2 2 * 1 * 2 * 1 * and 2 ln where )()(

5. 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!

6. 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).

7. Implied Volatility  Introduction  Calculating implied volatility  Volatility smiles

8. 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

9. 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 .

10. 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

11. Volatility Smiles (cont’d)

12. 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

13. Beginning Greeks & Hedging  Hedge Ratios  Greeks (Option Price Sensitivities)  delta, gamma (Stock Price)  rho (riskless rate)  theta (time to expiration)  vega (volatility)  Delta Hedging

14. 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.

15. Delta, Gamma  Sensitivity of Call Option Price to Stock Price change (Delta):  = N(d 1 )  We calculated this to get option price.  Gamma is change in Delta measure as Stock Price changes…. we’ll get to this later!

16. 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)

17. 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) d 1 = -------------------------------------------- =.70, N(d 1 ) =.7580.2 * (6/12) 1/2 d 2 =.70 - [.2 * (6/12) 1/2 ] =.56, N(d 2 ) =.7123 C = $75 (.7580) - 70 e -.04(6/12) (.7123) = $7.98 Intrinsic Value = $5, Time Value = $2.98

18. 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(d 1 ) =.758 shares sold short.

19.  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

20. Why a Hedge?  Suppose IBM goes to $74. ln(74/70) + (.04 + (.2) 2 /2)(6/12) d 1 = -------------------------------------------- = 0.61, N(d 1 ) =.7291.2 * (6/12) 1/2 d 2 = 0.61 - [.2 * (6/12) 1/2 ] = 0.47, N(d 2 ) =.6808 C = $74 (.7291) - 70 e -.04(6/12) (.6808) = $7.24

21. 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!

22. 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

23. 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

24. 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

25. Hedging Reality #3 cont’d   P = [N(d 1 ) - 1], so if N(d 1 ) < 1 (always), then  P < 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.

26. Reality #3 Example  Remember IBM pricing: ln(75/70) + (.04 + (.2) 2 /2)(6/12) d 1 = -------------------------------------------- =.70, N(d 1 ) =.7580.2 * (6/12) 1/2 d 2 =.70 - [.2 * (6/12) 1/2 ] =.56, N(d 2 ) =.7123 C = $75 (.7580) - 70 e -.04(6/12) (.7123) = $7.98 Put Price = Call Price + X e -rT - S Put = $7.98 + 70 e -.04(6/12) - 75 =$1.59

27. Hedge 100 Shares of IBM  - (1/  P ) P + S = - 100 * (1/  P ) P + 100 * S   P = N(d 1 ) – 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!

28. Hedge Setup  Position in Stock: $75 * 100 = +$7500  Position in Put Options: $1.59 * +4.13 * 100 = +$656.67  Total Initial Position =+$8156.67

29. 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

30. Results  Stock Price changed: (74 - 75)/75 = -1.3%  Portfolio changed: (8164.05 – 8156.67) / 8156.67 = +0.09%!!!!  Now that’s a hedge!