The Greek Letter & Volatility Smiles, Part I

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0.6 × $1 =0.6, producing a loss of $1,200. On the other hand, if ST goes down by $1, producing a loss of $1,200 on the shares, but the option price goes down by (approx.) $0.6. In this example, the delta of the investor’s option position is 0.6×(−2,000) = −1,200. In other words, the investor loses 1,200δS on the short position when S goes up by δS. The delta of the stock is 1.0, and the long position of in 1,200 shares has delta +1,200. Therefore, the investor’s overall position is zero. The delta of the stock position offsets the delta of the option position. A position with delta of zero is called delta neutral.
I. An Example
A financial institution has sold for $300,000 a European call option on 100,000 shares of a non-dividend-paying stock, and we assume that:
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Alternatively, the institution can take a covered position by buying 100,000 shares as soon as the option is sold: Clearly if the option is exercised, the strategy will make profit. But if S T goes down to $40, the loss on the position is $9×100,000=$900,000. Neither a naked position nor a covered position provides a satisfactory hedge. It is important to note that if the assumptions underlying the Black-Scholes formula hold, the cost to the financial institution should always be $240,000 on average for both approaches. But on any individual occasion the cost is liable to range from 0 to over $1m.
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S0 = 49, K = 50, r = 0.05, σ = 0.20, T = 0.3846 (20 wks), µ = 0.13
The theoretical price of the option from Black-Scholes formula is about $240,000, so there is a $60,000 top-up. But how to hedge the risk involved in the option? One strategy open to the financial institution is to do nothing, which is referred to as taking a naked position: 1. Clearly, if ST < K, it will work well, as the option costs the financial institution nothing and it makes a profit of $300,000. 2. But if ST > K , and the option is exercised the institution may incur loss. For instance, if ST = $60, the option costs it $1,000,000.
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Q = max(S0 −K, 0) This is because all purchases and sales after the initial purchase at S0 are made at price K. In the absence of transaction cost, this hedging scheme would work perfectly well. On the other hand, we have shown in Lecture Note 3 (p.22) that the lower bound of European call c is -rT c ≥ max(S0 −Ke , 0) therefore Q is always less than its Black-Scholes price of the stock. An investor could earn riskless profits by writing option and hedging them by the stop-loss strategy. But there are two problems:
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III. Delta Hedging
Option price
Slope = ∆ = 1.2 Slope = ∆ = 0.6 B Stock price
Calculation of delta
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We have discussed the delta of an option, ∆ , which is given by
∆=
∂c ∂S
and it is the slope of the curve that relates the option price to the underlying asset price. Suppose in the above diagram S0 is $100, and c is $10. An investor has sold 20 call options − options to buy 2,000 share, his position can be hedged by buying 0.6 × 2,000 = 1,200 shares. The gain (loss) on the option position would be offset by the loss (gain) on the stock position. For instance, if ST goes up by $1, giving a gain of $1,200 on shares, but the option price will go up by (approx.)
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Because delta changes, the investor’s position remains delta hedged only for a relatively short period of time. The hedge has to be adjusted periodically, known as rebalancing. As the diagram indicates, an increase in the stock price leads to an increase in delta. Suppose delta rises from 0.6 to 0.65, an extra 0.05×2,000 = 100 shares are needed to maintain the hedge. The delta hedging is therefore referred to as dynamic hedging scheme. Delta of European Stock Options The Black-Scholes formula for European call price is
The Greek Letters and Volatility Smiles Part I
Here is the question: if a financial institution sells an option to a client in the over-the-counter market, how should the institution hedge its risk? In the following, we discuss the “Greek letters” or simply “Greeks”, which measure the risks in option positions.
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1. The cash flows to the hedger occur at different times and must be discounted;
St
t1
t2
t3t4T源自Time t 6Stop-loss strategy: buying at t1 and t3; selling at t2 and t4. 2. Purchases and sales can not be made at exactly the same price K . With non-zero interest rates, we can not assume that both purchases and sales are made at the same price. Also, the hedger can not predict for sure whether, when the stock price equals K, it will continue above or below K. If the path of the stock price crosses K many times, the scheme can be expensive when there are transaction costs and non-zero interest rates.