Suppose the time t price St of a non-dividend paying stock follows geometric Brownian motion, i.e.,...

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Suppose the time t price St of a non-dividend paying stock follows geometric Brownian motion, i.e., dS = μSidt + o Sądz = where is the expected return of the stock, o is the volatility of the stock, and dz with ~ N(0, 1). As we saw in class, under some additional assumptions (which you may assume are satisfied) any derivative written on this stock must satisfy the following differential equation: where where t is time, ft is the price of the derivative at time t, and r is the risk-free rate. This is the Black-Scholes differential equation. and af af as Ət In particular, a European style call option written on this stock must satisfy the Black- Scholes differential equation. The formula for the time 0 < t < T price of a European style call option with strike price K and expiration T is given by C = S₂N (d₁) - e-r(T-t) KN (d₂) d₁ = +rSt + 1/0²520²5 = Tft მS2 In ( ) + (r + ) (T − t) - o√T-t E√dt d₂d₁-o√T-t 1 N(x) -1 √²+d= n(z)_dN (2) -dz, e = 2πT dx 88 1 √2π that is, N(x) and n(x) are the CDF and PDF, respectively, of a standard normal random variable. = In this exercise, you will verify that the formula for the price of a call option satisfies the Black-Scholes differential equation. In what follows, you may use the following fact (without proof): Sin(d₁) = e(Tt) Kn(d₂) 1. Calculate od₁/0S and ad₂/0S for t = 0. 2. For t=0 we have Co = SoN (d₁) - e-T KN (d₂) Using this, show that ac/0S = N(d₁). 3. Calculate ²c/as² for t = 0. 4. Show that for 0 < t < T Əc Ət which for t= 0 yields [Hint: 1 = Əc Ət Əd₁ Ət -Sin(d₁). = - O 2√T-t - Son(dı) 2√T - Əd₂ Ət rKer(T-t) N (d₂) -rKeN(d₂) (d₁d₂) Ət σ 2√T-t 5. Show that the formula for the price of a European style call option satisfies the Black- Scholes differential equation. Do this for t = 0. Suppose the time t price St of a non-dividend paying stock follows geometric Brownian motion, i.e., dS = μSidt + o Sądz = where is the expected return of the stock, o is the volatility of the stock, and dz with ~ N(0, 1). As we saw in class, under some additional assumptions (which you may assume are satisfied) any derivative written on this stock must satisfy the following differential equation: where where t is time, ft is the price of the derivative at time t, and r is the risk-free rate. This is the Black-Scholes differential equation. and af af as Ət In particular, a European style call option written on this stock must satisfy the Black- Scholes differential equation. The formula for the time 0 < t < T price of a European style call option with strike price K and expiration T is given by C = S₂N (d₁) - e-r(T-t) KN (d₂) d₁ = +rSt + 1/0²520²5 = Tft მS2 In ( ) + (r + ) (T − t) - o√T-t E√dt d₂d₁-o√T-t 1 N(x) -1 √²+d= n(z)_dN (2) -dz, e = 2πT dx 88 1 √2π that is, N(x) and n(x) are the CDF and PDF, respectively, of a standard normal random variable. = In this exercise, you will verify that the formula for the price of a call option satisfies the Black-Scholes differential equation. In what follows, you may use the following fact (without proof): Sin(d₁) = e(Tt) Kn(d₂) 1. Calculate od₁/0S and ad₂/0S for t = 0. 2. For t=0 we have Co = SoN (d₁) - e-T KN (d₂) Using this, show that ac/0S = N(d₁). 3. Calculate ²c/as² for t = 0. 4. Show that for 0 < t < T Əc Ət which for t= 0 yields [Hint: 1 = Əc Ət Əd₁ Ət -Sin(d₁). = - O 2√T-t - Son(dı) 2√T - Əd₂ Ət rKer(T-t) N (d₂) -rKeN(d₂) (d₁d₂) Ət σ 2√T-t 5. Show that the formula for the price of a European style call option satisfies the Black- Scholes differential equation. Do this for t = 0.

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