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

2.  Black-Scholes Model  Greeks  Hedging  An Extended Example

3. The Black-Scholes Model

5.  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

6.  Base Case Data – S = 100 – K = 100 – r = 5% –  = 30% – T = 3 months

8. The Greeks

9.  Sensitivity Analysis  5 First Derivatives; 1 Second Derivative  Linear Approximations (except Gamma)

10.  Delta (  ) = Price Sensitivity  Gamma (  ) = 2 nd Order Price Sensitivity  Rho (  ) = Interest Rate Sensitivity  Theta (  = Time Sensitivity  Vega (  = Volatility Sensitivity

11.  The hedge position must be frequently rebalanced  Delta hedging a written option involves a “buy high, sell low” trading rule

12. (S=10, K=10, T=0.2, r=0.05,  =0.2)

13.  Gamma (  ) is the rate of change of delta (  ) with respect to the price of the underlying asset  Gamma is greatest for options that are close to the money.

14. (S=10, K=10, T=0.2, r=0.05,  =0.2)

15.  For a delta neutral portfolio,     t + ½  S 2  SS Negative Gamma  SS Positive Gamma

16.  Theta (  ) 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

17. (S=10, K=10, T=0.2, r=0.05,  =0.2)

18.  Vega ( ) 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.

19. (S=10, K=10, T=0.2, r=0.05,  =0.2)

20.   can be changed by taking a position in the underlying  To adjust  & it is necessary to take a position in an option or other derivative

21.  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

22. (S=10, K=10, T=0.2, r=0.05,  =0.2)

24. Hedging

25.  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.

26. 15.26  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

27. Hedged Call Option Parameters: (K=10, T=0.2,  =0.3) Current Price: S = 10, Risk Free Rate: r = 0.05 Position in Bonds Slope = 

28. Hedged Call Option Parameters: (K=10, T=0.2,  =0.3) Current Price: S = 10, Risk Free Rate: r = 0.05 2 nd Call Option Parameters: (K=8, T=0.4,  =0.25)

29.  Base Case Data – S = 100 – K = 100 – r = 5% –  = 30% – T = 3 months  Call Value for Black-Scholes: c = $6.5831

30.  Using both delta and gamma for the estimate, what is the change in value (c’) if the stock price increases by 5%?

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

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

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

34. An Extended Example

35.  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.

36.  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.

37.  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.

38.  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 h S shares of stock and h C of the June 130 calls.

39.  The delta hedge condition is – h S (1) - 1,000  1 + h C  2 = 0  The gamma hedge condition is – -1,000  1 + h C  2 = 0  We can solve the second equation and get h C and then substitute back into the first to get h S. Solving for h C and h S, we obtain – h C = 1,000(.0121/.0123) = 984 – h S = 1,000(.5692 - (.0121/.0123).5086) = 68  So buy 68 shares, sell 1,000 June 125s, buy 985 June 130s.

40.  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.

41.  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.