Black-Scholes Option Pricing Model

Abstract

The Black-Scholes formula is one of the most recognizable formulae in quantitative finance.

Modelul Black-Scholes de Evaluare a OptiunilorUlciorul nu merge de multe ori la apă.The pitcher goes so often to the well that it comes home broken at last.

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The Black-Scholes formula is one of the most recognizable formulae in quantitative finance. The formula for the price C(S, τ) of a European call option is given by:

$$C(S,\tau ) =\exp \{ (b - r)\tau \}S\Phi (y + \sigma \sqrt{\tau }) -\exp (-r\tau )K\Phi (y),$$

(6.1)

where we use y as an abbreviation for

$$y = \frac{\log \left (S/K\right ) + \left \{b - {\sigma }^{2}/2\right \}\tau } {\sigma \sqrt{\tau }}$$

(6.2)

and b − r denotes the cost of carry b subtracted by the interest rate r.

The corresponding Black–Scholes formula for the price P(S, τ) of a European put option can be found, for example, by using the put–call parity as in Franke et al. [2011]:

$$P(S,\tau ) = C(S,\tau ) - S\exp \{(b - r)\tau \} + K\exp (-r\tau ).$$

From this and Eq. (6.1) we obtain

$$P(S,\tau ) =\exp (-r\tau )K\Phi (-y) -\exp \{ (b - r)\tau \}S\Phi (-y - \sigma \sqrt{\tau }).$$

(6.3)

The stop–loss strategy is a strategy to decrease the risk associated with a long call as an expensive hedging strategy, i.e. the bank selling the option takes an uncovered position as long as the stock price is below the exercise price, S _t  < K, and sets up a covered position as soon as the call is in–the–money, S _t  > K. 

FormalPara Exercise 6.1 (European Call Option).

Consider the European call detailed in Table  6.1 and the corresponding stock price movement given in Table  6.2 . We assume, that there are no transaction costs nor opportunity costs arising from binding capital. However bear in mind that in this example it’s only possible to sell or buy stocks at K ± δ.

• (a) Calculate the BS price of the call and the cost of hedging.

• (b) Modify your quantlet for $S_{0}\,=\,51$ , i.e. the call is in the money.

Table 6.1 Parameters for the European call

Table 6.2 Stock price movement

• (a) The solution is provided in Table 6.3, where a dummy variable indicates the need of a hedge strategy and the resulting consequences, e.g. the number of shares purchased, the costs of buying shares, the revenues due to selling shares and the cumulative costs.

  Table 6.3 Solution for Exercise 6.1 (a)

  When the stock price crosses K = 50 at time point 2, the open position is hedged at a price of S _t  = 52, resulting in costs of 5,200,000. In the next period, as S _t  ≤ K, we close the position at a price of $S_{t}\,=\,50$ and receive 5,000,000. These transactions results in a loss of 2 per share, i.e. a loss of 200,000. We continue the stop- loss strategy, checking every period whether $S_{t} \geq K$ and $S_{t} \leq K$ respectively.

  At the exercise point T = 20, if $S_{t} \geq K$ we will hold 100,000 shares bought at S _t  ≥ K and will be able to serve the call option for which we receive K units per share. If $S_{t} \leq K$ the option will not be exercised and we hold no shares, as we will have sold them at some $S_{t} \leq K$.

  

  SFSstoploss

• (b) The solution is provided in Table 6.4. Note the difference in the final costs compared to Table 6.3

SFSstoploss

FormalPara Exercise 6.2 (Stop-Loss Strategy).

Check the performance measure of a Stop-Loss strategy for an increasing hedging frequency Δt:

• (a) Consider the stock outlined in Table  6.1 and simulate m = 500 stock paths starting from $S_{0}\,=\,49.00$ .

• (b) Calculate, via a quantlet, the costs Λ _m for applying the Stop-Loss strategy over all m. Thereby set Δt = 5.

• (c) Calculate the variance $\upsilon _{\Lambda }^{2}$ of these costs.

• (d) Calculate the following performance measure

  $$L\,=\, \frac{\sqrt{\upsilon _{\Lambda }^{2}}} {C(S_{0},T)}$$

• (e) Modify your quantlet to calculate the performance measures L for  $\varDelta t\,=\,\left \{4,2,1,0.5,0.25\right \}$ .

Table 6.4 Solution for Exercise 6.1 (b)

Compare these differing values of L, as it has exemplary been done in Table  6.5.

FormalPara Exercise 6.3 (Delta Ratio).

Calculate the value of $\varDelta \,=\,\frac{\partial C} {\partial S}$ , the ratio of change of the option price with respect to the underlying stock price, for a European call with strike price K and maturity T.

The BS formula for a European call may be written as follows:

$$C(S,\tau )\,=\,S\Phi (d_{1}) - K\exp (-r\tau )\Phi (d_{2}),$$

where we use the abbreviations

$$\begin{array}{rcl} d_{1}& =& y + \sigma \sqrt{\tau }, \\ d_{2}& =& y \\ y& =& \frac{\log S/K + \left \{r - {\sigma }^{2}/2\right \}\tau } {\sigma \sqrt{\tau }} .\end{array}$$

A simple derivative of the formula yields

$$\frac{\partial C(S,\tau )} {\partial S} \,=\,\Phi (d_{1}) + S\frac{\partial \Phi (d_{1})} {\partial S} - K\exp (-r\tau )\frac{\partial \Phi (d_{2})} {\partial S} ,$$

where we again assume b = r to clarify the calculations.

Table 6.5 Exemplary values for Exercise 6.2.   SFShullhedgeratio

By inserting the following intermediate results

$$\begin{array}{rcl} \frac{\partial \Phi (d_{1})} {\partial S} & =& \varphi (d_{1})\frac{\partial d_{1}} {\partial S} \\ \frac{\partial \Phi (d_{2})} {\partial S} & =& \varphi (d_{2})\frac{\partial d_{2}} {\partial S} \\ \frac{\partial d_{1}} {\partial S} = \frac{\partial d_{2}} {\partial S} & =&{ \left (S\sigma \sqrt{\tau }\right )}^{-1} \\ \varphi (d_{2})& =& \varphi (d_{1})\exp (r\tau ) \frac{S} {K} \\ \end{array}$$

we obtain the following delta ratio:

$$\begin{array}{rcl} \frac{\partial C(S,\tau )} {\partial S} & =& \Phi (d_{1}) + S\varphi (d_{1}){\left (S\sigma \sqrt{\tau }\right )}^{-1} \\ & & \ \ \ -K\exp (-r\tau )\varphi (d_{1})\exp (r\tau ) \frac{S} {K}{\left (S\sigma \sqrt{\tau }\right )}^{-1} \\ & & = \Phi (d_{1}).\end{array}$$

FormalPara Exercise 6.4 (Delta Hedging).

Calculate the delta ratio $\varDelta \,=\,\frac{\partial C(S,\tau )} {\partial S}$ for the European call stated in Exercise 6.1. Furthermore think of the consequences resulting from the delta hedge, e.g. the amount of shares to be held and the cumulative costs for holding these stocks across t.

To calculate the delta ratio, which changes across time t due to the stock price S _t and the declining time to maturity τ, we use the result from Exercise 6.3, namely

$$\frac{\partial C(S,\tau )} {\partial S} \,=\,\Phi \left \{\frac{\log S/K + \left (r - {\sigma }^{2}/2\right )\tau } {\sigma \sqrt{\tau }} + \sigma \sqrt{\tau }\right \}.$$

With this ratio, we can calculate the number of shares we need to hold at any t and the resulting costs for holding these shares across time. The solution is provided in Table 6.6.

Table 6.6 Solution to Exercise 6.4.  SFSdeltahedging

FormalPara Exercise 6.5 (Delta Neutral Position).

A bank sold 3,000 Calls and 2,500 Puts with the same maturity and exercise price on the same underlying stock. Δ Call is 0.42 and Δ Put is −0.38. How many stocks should the company sell or buy in order to have a delta neutral position?

Let’s call the number of stocks needed for a delta neutral position x. The Δ of the stock is 1. From Franke et al. [2011, p. 114] we will have

$$\begin{array}{rcl} 0.42 \cdot (-3000) - 0.38 \cdot (-2500) + x \cdot 1& \,=\,& 0 \\ x& =& 310\end{array}$$

So, the bank will have a delta neutral position when it buys 310 stocks.

FormalPara Exercise 6.6 (Gamma and Delta Hedging).

Mrs. Ying Chen already has a portfolio with $\varDelta \,=\, - 300$ and Γ = 250. She wants to make her portfolio Δ and Γ neutral. In the market, she can buy/sell stocks and call options to achieve this. She calculated the Δ and Γ of the call option as 0.55 and 1.25 respectively. What should Mrs. Chen do in order to realize her goal?

Let us call the additional number of calls and stocks that Mrs. Chen will need as x and y respectively. For a Δ and Γ hedged portfolio, we can write the following equations from the corresponding Gamma and Delta Hedging equations:

$$\begin{array}{rcl} 1.25 \cdot x + 250& =& 0 \\ 0.55 \cdot x + y \cdot 1 - 300& =& 0\end{array}$$

After solving this set of equations, we see, that Mrs. Chen should buy y = 410 stocks and short sell x = 200 call options in order to achieve a Δ and Γ neutral position.

FormalPara Exercise 6.7 (Option Pricing with Black-Scholes Model).

Consider a European call option on a stock when there are ex-dividend dates in 3 and 6 months. The dividend on each ex-dividend date is expected to be 1, the current price is 80, the exercise price is 80, the volatility is 25 % per annum, the annually risk free rate is 7 %, the time to maturity is 1 year, calculate the option price using the Black-Scholes model.

The present value of the expected dividends can be calculated as

$$\exp (-0.25 \cdot 0.07) +\exp (-0.5 \cdot 0.07)\,=\,1.9483.$$

We deduct this present value of the dividends from the stock price at t = 0 to arrive at the purely random component of the stock price:

$$\begin{array}{rcl} S_{0}& =& 80 - 1.9483 \\ & =& 78.0517.\end{array}$$

Afterwards, as a first step, we calculate Φ(y) and $\Phi (y + \sigma \sqrt{\tau })$ according to the formulas given in Exercise 6.3:

$$\begin{array}{rcl} \Phi (y)& =& \Phi \left [\frac{\log (\frac{78.0517} {80} ) + \left \{0.07 - {(0.25)}^{2}/2\right \} \cdot 1} {0.25 \cdot \sqrt{1}} \right ] \\ & =& \Phi (0.0564) \\ & =& 0.5225 \\ \Phi (y + \sigma \sqrt{\tau })& =& \Phi \left (0.0564 + 0.25 \cdot \sqrt{1}\right ) \\ & =& \Phi (0.3064) \\ & =& 0.6203\end{array}$$

We now possess all information we need to calculate the price of the call:

$$\begin{array}{rcl} C(78.0517,1)& =& 78.0517 \cdot 0.6203 - 80 \cdot \exp (-0.07 \cdot 1) \cdot 0.5225 \\ & =& 9.4462\end{array}$$

SFSBSCopt1

FormalPara Exercise 6.8 (Delta, Gamma and Theta of Portfolio).

An investor longs a 3 months maturity call option with K = 220, and shorts a half year call with K = 220. Currently, S ₀  = 220, risk free rate r = 0.06 (continuously compounded), and σ = 0.25. The stock pays no dividends.

• (a) Calculate the Δ of the portfolio.

• (b) Calculate the Γ of the portfolio.

• (c) Calculate the Θ of the portfolio.

• (a) We calculate $\varDelta \,=\,\Phi (y + \sigma \sqrt{\tau })$ for both calls separately and sum up the results to obtain the overall delta of the portfolio:

  • Call to be longed:

    $$\varDelta _{1}\,=\,0.6018$$

  • Call to be shorted:

    $$\varDelta _{2}\,=\, - 0.5724$$

  The delta of the portfolio is therefore

  $$\begin{array}{rcl} \varDelta _{1} + \varDelta _{2}& \,=\,& 0.6018 + (-0.5724) \\ & \,=\,& 0.0294\end{array}$$

• (b) To obtain the portfolio’s gamma we calculate

  $$\Gamma \,=\, \frac{1} {\sigma S\sqrt{\tau }}\varphi \left (y + \sigma \sqrt{\tau }\right )$$

  for each call:

  • Call to be longed:

    $$\Gamma _{1}\,=\,0.0099$$

  • Call to be shorted:

    $$\Gamma _{2}\,=\, - 0.0142$$

  This leads to the following overall gamma:

  $$\begin{array}{rcl} \Gamma _{1} + \Gamma _{2}& \,=\,& 0.0099 + (-0.0142) \\ & \,=\,& -0.0043\end{array}$$

• (c) By the derivative of the BS formula with respect to t

$$\begin{array}{rcl} \Theta & =& \frac{\partial C(S,\tau )} {\partial t} \\ & =& - \frac{\sigma S} {2\sqrt{\tau }}\varphi \left (y + \sigma \sqrt{\tau }\right ) - rK\exp (-r\tau )\Phi (y) \\ \end{array}$$

we can calculate the theta for each call:

• Call to be longed:

  $$\Theta _{1}\,=\, - 21.827$$

• Call to be shorted:

  $$\Theta _{2}\,=\, - 28.279$$

The theta of the portfolio may then be calculated as:

$$\begin{array}{rcl} \Theta _{1} - \Theta _{2}& =& -21.827 + (-28.279) \\ & =& -50.106\end{array}$$

SFSgreeks

FormalPara Exercise 6.9 (Collar Portfolio).

Mr. Wang constructed a collar portfolio, which was established by buying a share of a stock for 15 JPY, buying a 1-year put option with exercise price 12.5 JPY, and short selling a 1-year call option with exercise price 17.5 JPY. If the BS model holds, based on the volatility of the stock, Mr. Wang calculated that for a strike price of 12.5 JPY and maturity of 1 year, $\Phi (y + \sigma \sqrt{\tau })\,=\,0.63$ , whereas for the exercise price of 17.5 JPY, $\Phi (y + \sigma \sqrt{\tau })\,=\,0.32$ where we use y as an abbreviation for

$$y = \frac{\log S/K + \left (b - {\sigma }^{2}/2\right )\tau } {\sigma \sqrt{\tau }} .$$

(6.4)

• (a) Draw a payoff graphic of this collar at the option expiration date.

• (b) If the stock price increases by 1 JPY, what will Mr. Wang gain or lose from this portfolio?

• (c) What happens to the delta of the portfolio if the stock price becomes very high or very low?

• (a) The payoff can be drawn as in the Fig. 6.1.

• (b) The resulting consequences for Mr. Wang’s portfolio may be calculated via the delta of the portfolio, which indicates any losses or gains. In detail, the value of the stock will raise by 1 JPY, the loss on the long put and on the short call will amount to 0.37 and 0.32 JPY respectively. For the whole portfolio, this leads to a gain of 0.31 JPY, as stated in Table 6.7.

• (c) For very large stock prices: the delta of the collar approaches zero, because both $\pm \Phi (y + \sigma \sqrt{\tau })$ approach 1. The value of the portfolio is simply the present value of the exercise price of the call, and is unaffected by small changes in the stock price.

For very small stock prices: as stock price approaches zero, the delta also approaches zero, because both $\pm \Phi (y + \sigma \sqrt{\tau })$ terms approach 0. The value of the portfolio is simply the exercise price of the put, and is unaffected by small changes in the stock price.

Fig. 6.1

Payoff of a collar. SFSpayoffcollar

Table 6.7 Delta of the collar

FormalPara Exercise 6.10 (Gamma and Delta Neutral Positions).

A financial institution has the portfolio given in Table  6.8 of OTC options on a particular stock.

Table 6.8 Portfolio on a particular stock

A traded call option is available which has a delta Δ = 0.3 and a gamma Γ = 1.8. What position in that traded option and underlying stock would make the portfolio both gamma neutral and delta neutral?

We can calculate the current position of portfolio:

$$\begin{array}{rcl} \varDelta & =& -2000 \cdot 0.5 + 1500 \cdot 0.7 - 4000 \cdot (-0.4) \\ & =& -1000 + 1050 + 1600 \\ & =& 1650 \\ \Gamma & =& -2000 \cdot 2.5 + 1500 \cdot 0.7 - 4000 \cdot 1.1 \\ & =& -5000 + 1,050 - 4,400 \\ & =& -8350\end{array}$$

In order to become Gamma-neutral according to Franke et al. [2011, Sect. 6.4], one should buy traded options:

$$\frac{8350} {1.8} \,=\,4638.89.$$

This will make the delta of the position equal to

$$1650 + 4638.89 \cdot 0.3 = 3041.67.$$

Therefore, to become delta neutral we need to sell 3,041.67 shares.

FormalPara Exercise 6.11 (Hypothetical Call Option).

Knowing that the current price of oil is 100 EUR per barrel, a petrochemical firm PetroCC plans to buy a call option on oil with strike price = 100 EUR. The volatility of oil prices is 10 % per month, and the risk-free rate is 3 % per month.

• (a) What is the value of the 4-month call option?

• (b) Suppose that instead of buying 4-month call options on 100,000 barrels of oil, the firm will synthetically replicate a call position by buying oil directly now, and delta-matching the hypothetical call option. How many barrels of oil should it buy?

• (c) If oil prices increase by 1 % after the first day of trading, how many barrels of oil should it buy or sell?

• (a) We can calculate the call option price using BS formula directly:

  $$C(S,\tau )\,=\,S\Phi (y + \sigma \sqrt{\tau }) -\exp (-r\tau )K\Phi (y)$$

  where $S\,=\,100,K\,=\,100,\sigma \,=\,0.1,b\,=\,r\,=\,0.03$ and τ = 4.Then,

  $$\begin{array}{rcl} y + \sigma \sqrt{\tau }& =& \frac{\log S/K + (b + {\sigma }^{2}/2)\tau } {\sigma \sqrt{\tau }} \\ & =& \frac{0 + (0.03 + 0.1{0}^{2}/2) \cdot 4} {0.10 \cdot \sqrt{4}} \\ & =& 0.7000 \\ y& =& 0.7000 - \sigma \sqrt{\tau } \\ & =& 0.5000 \\ \Phi (y + \sigma \sqrt{\tau })& =& 0.7580 \\ \Phi (y)& =& 0.6915 \\ C(S,\tau )& =& S\Phi (y + \sigma \sqrt{\tau }) -\exp (-r\tau )K\Phi (y) \\ & =& 100 \cdot 0.7580 -\exp (-0.03 \cdot 4) \cdot 100 \cdot 0.6915 \\ & =& 14.4695\end{array}$$

  So the price of 4-month call option should be 14.4695 EUR.

• (b) Since the delta of the call option is,

  $$\varDelta \,=\,\frac{\partial C} {\partial S}\,=\,\Phi (y + \sigma \sqrt{\tau })\,=\,0.7580$$

  and the delta of the stock is 1.So, we should buy 75,800 barrels of oil, which can provide the same delta value as the call options.

• (c) If the oil price increases by 1 %, then the $y + \sigma \sqrt{\tau }$ also increases, so as the delta.

$$\begin{array}{rcl} y + \sigma \sqrt{\tau }& =& \frac{\log S/K + (b + {\sigma }^{2}/2)\tau } {\sigma \sqrt{\tau }} \\ & =& \frac{0.01 + (0.03 + 0.1{0}^{2}/2) \cdot 4} {0.10 \cdot \sqrt{4}} \\ & =& 0.8000 \\ \Phi (0.800)& =& 0.7881\end{array}$$

The delta increases $0.7881 - 0.7580\,=\,0.0301$, so one should buy an additional 3,010 barrels of oil.

FormalPara Exercise 6.12 (Implied Volatility and Delta Neutrality).

There are two calls on the same stock with the same time to maturity (1 year) but different strike price. Option A has a strike price K ₁  = 10 USD, while option B has a strike price K ₂  = 9.5 USD. The current stock price is S _t  = 10 USD. The stock does not pay dividend. Risk free rate is 3 %. One applies the Black-Scholes equation as the option pricing model. Yet despite that fact that one is confident that the appropriate volatility of the stock is 0.15 p.a. One observes option A selling for 0.8 USD and option B selling for 1 USD.

• (a) Is the implied volatility of Option A more or less than 15 %? What about that of option B?

• (b) Determine a delta-neutral position in the two calls that will exploit their apparent mispricing. Use σ = 0.15 to compute Delta.

• (a) We can calculate the call option price using BS formula directly:For option A:

  $$\begin{array}{rcl} y + \sigma \sqrt{\tau }& =& \frac{\log S/K + (b + {\sigma }^{2}/2)\tau } {\sigma \sqrt{\tau }} \\ & =& \frac{0 + (0.03 + 0.1{5}^{2}/2) \cdot 1} {0.15 \cdot \sqrt{1}} \\ & =& 0.28 \\ y& =& 0.28 - \sigma \sqrt{\tau } \\ & =& 0.13 \\ \Phi (y + \sigma \sqrt{\tau })& =& 0.6103 \\ \Phi (y)& =& 0.5517 \\ C_{A}(S,\tau )& =& S\Phi (y + \sigma \sqrt{\tau }) -\exp (-r\tau )K\Phi (y) \\ & =& 10 \cdot 0.6103 -\exp (-0.03 \cdot 1) \cdot 10 \cdot 0.5517 \\ & =& 0.75 < 0.8\end{array}$$

  For option B:

  $$\begin{array}{rcl} y + \sigma \sqrt{\tau }& =& \frac{\log S/K + (b + {\sigma }^{2}/2)\tau } {\sigma \sqrt{\tau }} \\ & =& \frac{0.051 + (0.03 + 0.1{5}^{2}/2) \cdot 1} {0.15 \cdot \sqrt{1}} \\ & =& 0.62 \\ y& =& 0.62 - \sigma \sqrt{\tau } \\ & =& 0.47 \\ \Phi (y + \sigma \sqrt{\tau })& =& 0.7324 \\ \Phi (y)& =& 0.6808 \\ C_{B}(S,\tau )& =& S\Phi (y + \sigma \sqrt{\tau }) -\exp (-r\tau )K\Phi (y) \\ & =& 10 \cdot 0.7324 -\exp (-0.03 \cdot 1) \cdot 9.5 \cdot 0.6808 \\ & =& 1.05 > 1\end{array}$$

  For Option A, the real price is higher than the Black-Scholes value when using 0.15 as the volatility. For option B, the price is lower. Thus, for option A, the implied volatility should be higher than 0.15, and the implied volatility of option B is lower than 0.15.

• (b) From task (a), it is clear that option A is overvalued, and option B is undervalued. Thus, we can long option B and short option A to do arbitrage. For every option A we short, we should long $\varDelta _{A}/\varDelta _{B}\,=\,0.6103/0.7324 = 0.8333$ option B.

FormalPara Exercise 6.13 (Implied Volatility).

E-Tech Corp. stock sells for 80 EUR and pays no dividends. A 6-month call option with exercise price 80 EUR is priced at 7.23 EUR, while a 6-month call option with exercise price 90 EUR is priced at 5.38. The risk-free interest rate is 8 % per year.

• (a) What is the implied volatility of these two options?

• (b) Based on the above information, construct a profitable trading strategy.

• (a) From Black-Scholes formula, we have:

$$\begin{array}{rcl} C(S,\tau )& =& S\Phi (y + \sigma \sqrt{\tau }) -\exp (-r\tau )K\Phi (y)\end{array}$$

(6.5)

$$\begin{array}{rcl} y + \sigma \sqrt{\tau }& =& \frac{\log S/K + (b + {\sigma }^{2}/2)\tau } {\sigma \sqrt{\tau }} \end{array}$$

(6.6)

After substituting data in this example, solve the equations w.r.t. σ, we have the implied volatility of the call option with exercise price 80 EUR is 25 %; the implied volatility of the call option with exercise price 90 EUR is 35 %. (The calculations in this question can also been done in the DerivaGem Options Calculator software of John Hull.)

• (b) Based on implied volatilities, the call option with exercise price 90 EUR is overpriced. Thus we can exploit this by constructing a delta-neutral position: buy the call with exercise price of 80 EUR and write the call with exercise price of 90 EUR.

To calculate delta, we need to know σ. One strategy is to use the middle point of 25 and 35 %, i.e., σ = 35 %. When σ = 35 %, the delta of the call with exercise price of 80 EUR is 0.616, and the delta of the call with exercise price of 90 EUR is 0.397. Thus the delta-neutral proportion is 0.616/0.397 = 1.552. So we buy one call with exercise price of 80 EUR, and write 1.552 calls with exercise price of 90 EUR. The net position is delta neutral, but since $7.32 - 1.552 \times 5.38\,=\, - 1.03$, there is an initial cash inflow (profit) of 1.03 EUR.

FormalPara Exercise 6.14 (Greeks).

The Black-Scholes price of a call option with strike price K, maturity T is defined as follows in t ∈ [0,T) at a stock price x:

$$\nu (x,T - t) = x\Phi \{d_{+}(x,T - t)\} - K\exp \{ - r(T - t)\}\Phi \{d_{-}(x,T - t)\}$$

$r \in \mathbb{R}$ denotes the riskless interest rate and σ > 0 the volatility of the stock. The function d ₊ and d ₋ are given by

$$d_{\pm }(x,r) =\{\log x/K + (r \pm {\sigma }^{2}/2)\tau \}/\sigma \sqrt{\tau }$$

Calculate the “Greeks”

• (a) $\varDelta \,=\, \frac{\partial } {\partial x}\nu (x,T - t)$

• (b) $\Gamma \,=\, \frac{{\partial }^{2}} {\partial {x}^{2}} \nu (x,T - t)$

• (c) $\Theta \,=\, \frac{\partial } {\partial t}\nu (x,T - t)$

and verify that ν solves the partial differential equation

$$\left (\frac{{\sigma }^{2}} {2} {x}^{2} \frac{{\partial }^{2}} {\partial {x}^{2}} + rx \frac{\partial } {\partial x} - \frac{\partial } {\partial t}\right )\nu (x,t)\,=\,r\nu (x,t)\text{ on }(0,\infty ) \times (0,\infty )$$

and satisfies the boundary condition

$$\nu (x,T - t) \rightarrow {(x - K)}^{+}\text{ for }t \rightarrow T$$

From the Black-Scholes formula,

$$\nu (x,T - t) = x\Phi \{d_{+}(x,T - t)\} - K\exp \{ - r(T - t)\}\Phi \{d_{-}(x,T - t)\}$$

where Φ is the distribution function of the standard normal distribution and $(T\,-\,t)\,=\,\tau$ the time to maturity.

Recall (Exercise 6.3), the ratio of change of the option price with respect to the underlying stock price (Delta, $\varDelta \,=\,\frac{\partial \nu } {\partial x}$) can be expressed as:

• (a)
  $$\begin{array}{rcl} \frac{\partial } {\partial x}\nu (x,T - t)& =& \Phi \{d_{+}(x,T - t)\} + x\varphi \{d_{+}(x,T - t)\} \frac{\partial } {\partial x}d_{+}(x,T - t) - \\ & & \varphi \{d_{-}(x,T - t)\} \frac{\partial } {\partial x}d_{-}(x,T - t) \end{array}$$

  (6.7)

  Note that,

  $$\begin{array}{rcl} d_{+}(x,\tau ) - d_{-}(x,\tau )& =& \left \{\left (r + \frac{{\sigma }^{2}} {2} \right )\tau -\left (r -\frac{{\sigma }^{2}} {2} \right )\tau \right \}/\sigma \sqrt{\tau } = \sigma \sqrt{\tau } \\ \frac{\partial } {\partial x}d_{\pm }(x,\tau )& =& k/(\sigma \sqrt{\tau }xk) = 1/x\sigma \sqrt{\tau } \\ \varphi (d_{-}(x,\tau ))& =& \exp \{-{d}^{2}/2 - (x,\tau )\}/2\pi \\ & =& \exp \{-{(d_{+}(x,\tau ) - \sigma \sqrt{\tau })}^{2}/2\}/\sqrt{2\pi } \\ & =& \exp \{-({d}^{2} + (x,\tau ) - 2d + (x,\tau )\sigma \sqrt{\tau } + {\sigma }^{2}\tau )/2\}/\sqrt{2\pi } \\ & =& \varphi \{d_{+}(x,\tau )\}\exp \{\sigma \sqrt{\tau }d_{+}(x,\tau ) - {\sigma }^{2}\tau /2\} \\ & =& \varphi \{d_{+}(x,\tau )\}\exp (\log x/K + r\tau ) \\ & =& \varphi \{d_{+}(x,\tau )\}x/K\exp (r\tau ) \\ \end{array}$$

  By substitution in Eq. 6.7,

  $$\begin{array}{rcl} \text{ Delta: }\varDelta = \frac{\partial \nu } {\partial x}& =& \Phi \{d_{+}(x,T - t)\} + x\varphi \{d_{+}(x,T - t)\} \frac{\partial } {\partial x}d_{+}(x,T - t) - \\ & & \varphi \{d_{-}(x,T - t)\} \frac{\partial } {\partial x}d_{-}(x,T - t) = \Phi \{d_{+}(x,T - t)\} \\ \end{array}$$

• (b) The ratio of change of the option Δ with respect to the underlying stock price (Gamma, $\Gamma \,=\, \frac{\partial \nu } {\partial \varDelta }$) can be expressed as:

  $$\begin{array}{rcl} \Gamma = \frac{\partial \nu } {\partial \varDelta } = \frac{\partial } {\partial {x}^{2}}\nu (x,T - t)& =& \frac{\partial } {\partial x}\Phi \{d_{+}(x,T - t)\} \\ & =& \varphi \{d_{+}(x,T - t)\} \frac{\partial } {\partial x}d_{+}(x,T - t) \\ & =& \frac{\varphi \{d_{+}(x,T - t)\}} {x\sigma \sqrt{T - t}} \\ \end{array}$$

• (c) The ratio of change of the price of the underlying with respect to time (Theta, $\Theta \,=\,\frac{\partial \nu } {\partial t}$) can be expressed as:

$$\begin{array}{rcl} \Theta = \frac{\partial } {\partial t}\nu (x,T - t)& =& x\varphi \{d_{+}(x,T - t)\} \frac{\partial } {\partial t}d_{+}(x,T - t) \\ & & -kr\exp \{ - r(T - t)\}\Phi \{d_{-}(x,T - t)\} \\ & & +\exp \{ - r(T - t)\}\varphi \{d_{-}(x,T - t)\} \frac{\partial } {\partial t}d_{-}(x,T - t) \\ & =& x/k\varphi \{d_{+}(x,T - t)\} \\ & =& x\varphi \{d_{+}(x,T - t)\} \frac{\partial } {\partial t}(d_{+}d_{-})(x,T - t) - \\ & &-kr\exp \left \{-r(T - t)\right \}\Phi \{d_{-}(x,T - t)\} \\ & =& - \frac{x\sigma } {d\sqrt{(}T - t)}\varphi \{d_{+}(x,T - t)\} \\ & & -kr\exp \{ - r(T - t)\}\Phi \{d_{-}(x,T - t)\} \\ \end{array}$$

To verify that ν solves the partial differential equation, we need to show that the Black-Scholes option price model gives the same price as a model free no-arbitrage approach. Applying It$\hat{\text{ o}}$’s lemma:

$$d\nu (x,t) = \sigma x\frac{\partial \nu } {\partial x}dW_{t} + \left (\mu x\frac{\partial \nu } {\partial x} + \frac{{\sigma }^{2}} {2} {x}^{2}\frac{{\partial }^{2}\nu } {\partial {x}^{2}} + \frac{\partial \nu } {\partial t} \right )\mathit{dt}.$$

Consider a portfolio Π containing an option and − Δ units of the underlying stocks:

$$\Pi = \nu (x,t) - \varDelta x$$

$$d\Pi = d\nu (x,t) - \varDelta dx$$

$$d\Pi = d\nu (x,t) - \varDelta (\mu x\mathit{dt} + \sigma xdW_{t})$$

For $\varDelta \,=\,\frac{\partial \nu (x,t)} {\partial x}$

$$d\Pi = \left (\frac{{\sigma }^{2}} {2} {x}^{2}\frac{{\partial }^{2}\nu } {\partial {x}^{2}} + \frac{\partial \nu } {\partial t} \right )\mathit{dt}$$

Now if Π was invested in riskless assets it would see a growth of $r\Pi dt$ in the interval of length dt. Then for a fair price we should have $d\Pi \,=\,r\Pi dt.$

$$r\Pi \mathit{dt} = \left \{\frac{{\sigma }^{2}} {2} {x}^{2}\frac{{\partial }^{2}\nu (x,t)} {\partial {x}^{2}} + \frac{\partial \nu (x,t)} {\partial t} \right \}dt$$

Hence,

$$r\left \{\nu -\frac{\partial \nu (x,t)} {\partial x} x\right \} = \frac{{\sigma }^{2}} {2} {x}^{2}\frac{{\partial }^{2}\nu (x,t)} {\partial {x}^{2}} + \frac{\partial \nu (x,t)} {\partial t} .$$

Re-arranging gives

$$\frac{{\sigma }^{2}} {2} {x}^{2}\frac{{\partial }^{2}\nu (x,t)} {\partial {x}^{2}} + rx\frac{\partial \nu (x,t)} {\partial x} + \frac{\partial \nu (x,t)} {\partial t} - r\nu (x,t) = 0,$$

satisfying $\nu (x,\tau ) \rightarrow {(x - K)}^{+}\text{ for }t \rightarrow T.$

FormalPara Exercise 6.15 (Black-Scholes Price of a Call Option and Vega).

• (a) Show that for x,y > 0 with x≠t, the following holds:

  $$\vert \nu (x,t) - \nu (y,t)\vert < \vert x - y\vert ,$$
  $$\frac{\nu (x,t) - \nu (y,t)} {\nu (y,t)} > \frac{x - y} {y} \text{ for }x > y,$$

  and

  $$\frac{\nu (x,t) - \nu (y,t)} {\nu (y,t)} < \frac{x - y} {y} \text{ for }x < y$$

  for the special cases, where ν(x,t) is the Black-Scholes price of a call option with stock price x, time to expiration of the option as t.

• (b) Show that the “Vega” $\,=\, \frac{\partial } {\partial \sigma }\nu (x,t)$ is always positive and calculate the value x where “Vega” is maximal.

• (a) For ν(x, T − t) (see, question Exercise 6.14), let time to maturity $\tau \,=\,T - t.$ It holds that

  $$\frac{\partial } {\partial {x}^{2}}\nu (x,\tau ) = \varphi \{d_{+}(x,\tau )\}/x\sigma \sqrt{\tau } > 0,$$

  hence

  $$\nu (x,\tau )\text{ is strictly convex on }(0,\infty )$$

  Following Lipschitz condition, $\vert \nu (x,\tau ) - \nu (y,\tau )\vert < \vert x - y\vert$ for $x,y\,>\,0,x\neq y$ $\vert \nu (x,\tau ) - \nu (y,\tau )\vert \,=\,\vert \nu _{x}(z,\tau )\vert \vert x - y\vert$

  It holds ∀z ∈ (0, y):

  $$\frac{\nu (x,\tau ) - \nu (y,\tau )} {x - y} > \frac{\nu (y,\tau ) - \nu (z,\tau )} {y - z} ,$$

  therefore

  $$\frac{\nu (x,\tau ) - \nu (x,\tau )} {x - y} > \frac{\nu (y,\tau )} {y}$$

  Also for x < y, 

  $$\frac{\nu (x,\tau ) - \nu (y,\tau )} {\nu (y,\tau )} < \frac{x - y} {y} ,$$

  if and only if

  $$\frac{y - x} {x} < \frac{\nu (y,\tau ) - \nu (x,\tau )} {\nu (x,\tau )}$$

• (b) The “Vega” is the rate of change the option with respect to the volatility. For a European call option on a non-dividend-paying stock,

$$\begin{array}{rcl} \text{ Vega: }\frac{\partial \nu (x,\tau )} {\partial \sigma } & =& \varphi \{d_{+}(x,\tau )\}\sqrt{\tau } \\ & =& \frac{x\sqrt{\tau }} {\sqrt{2\pi }}\exp (-d_{1}^{2}/2), \\ \end{array}$$

where

$$d_{1} = \frac{\log x/K + \left (r - {\sigma }^{2}/2\right )\tau } {\sigma \sqrt{\tau }} ,$$

which is always positive. Let V = Vega, then

$${V }^{{\prime}} = \frac{\sqrt{\tau }} {\sqrt{2}\pi }\exp (-d_{1}^{2}/2)\left (1 - \frac{d_{1}} {\sigma \sqrt{\tau }}\right ).$$

Solving for x in ${V }^{{\prime\prime}}$, the value for which the Vega reaches maximum is obtained:

$${V }^{{\prime\prime}} = \frac{1} {\sqrt{2}\pi }\exp (-d_{1}^{2}/2)\left \{-d_{ 1} + \frac{(d_{1}^{2} - 1)} {\sigma \sqrt{\tau }} \right \},$$

with maximum Vega at $x\,=\,k\exp \{({\sigma }^{2} - r)\tau /2\}$

FormalPara Exercise 6.16 (Price of Risk, Stochastic Process and Girsanov Transformation).

In the Black-Scholes model, the stock price is modelled by

$$dS(t) = S(t)\left \{\mu \mathit{dt} + \sigma dW(t)\right \},$$

where μ denotes the drift, σ the volatility and Wt, the standard Brownian motion.

• (a) The market price of risk is defined as excess return. How is this defined in the Black-Scholes framework?

• (b) Give the explicit form of the stochastic process S(t)

• (c) Suppose now that a class of parametrized class of equivalent probabilities Q are introduced via the Girsanov transformation:

  $${W}^{\theta }(t)\,=\,W(t) -\int \nolimits _{0}^{t}\theta (u)\mathit{du}$$

  where θ is a real valued, bounded continuous function. By using the Girsanov Theorem there exists an equivalent probability measure denoted ${Q}^{\theta }$ so that ${W}^{\theta }(t)$ is a Brownian motion for t. Show the dynamics of S(t) under ${Q}^{\theta }$

• (a) The market price of risk is the rate of extra return above r per unit risk. In the Black-Scholes model, the stock price is modelled by

  $$dS(t)\,=\,S(t)\left \{\mu \mathit{dt} + \sigma dW(t)\right \}.$$

  Under an equivalent risk neutral measure the model can be expressed as

  $$\frac{dS(t)} {S(t)} \,=\,(\mu - r)\mathit{dt} + \sigma dW(t).$$

  By setting $X_{t}\,=\,\frac{\mu -r} {\sigma }$, the market price of risk:

  $$\frac{dS(t)} {S(t)} \,=\,\sigma \{X_{t}\mathit{dt} + dW(t)\},$$

  where r is the constant risk-free interest rate.

• (b) By It$\hat{\text{ o}}$’s formula,

  $$d\log S(t)\,=\,\left (\mu - {\sigma }^{2}/2\right )\mathit{dt} + \sigma dW(t),$$

  so that S(t) satisfies the Black-Scholes model, if and only if

  $$S(t)\,=\,S(0)\exp \left \{\left (\mu - {\sigma }^{2}/2\right )t + \sigma W(t)\right \}$$

• (c) We construct under ${Q}^{\theta }$ a martingale price process:

$$dS(t) = S(t)\{(\mu - r)\mathit{dt} + \sigma dW(t)\},$$

where (by Girsanov transformation)

$${W}^{\theta }(t) = W(t) -\int \nolimits _{0}^{t}\theta (u)du$$

is a ${Q}^{\theta }-$Brownian motion. Applying the Radon- Nicodym derivative, $\zeta _{t}\,=\, \frac{dQ} {d{Q}^{\theta }}$, gives:

$$\zeta _{t} =\exp \left \{-\int \nolimits _{0}^{t}\theta (u)dW(u) -\frac{1} {2}\int \nolimits _{0}^{t}{\theta }^{2}(u)\mathit{du}\right \}.$$

From $dS(t)\,=\,S(t)\{(\mu - r)\mathit{dt} + \sigma dW(t)\}$ and by definition of ${W}^{\theta }(t)$, it holds that

$$dS(t) = S(t)\sigma d{W}^{\theta }(t).$$

Applying It$\hat{\text{ o}}$’s lemma the dynamics of the stochastic process is expressed as

$$S(t) = S(0)\exp \left \{-\int \nolimits _{0}^{t}\sigma d{W}^{\theta }(u) -\frac{1} {2}\int \nolimits _{0}^{t}{\sigma }^{2}d(u)\right \}.$$