April 25, 2024
A European option pays some function of the underlying price at expiration. If F is the price at expiration and \nu is the payoff function then its forward value is {v = E[\nu(F)]}.
If F is positive and \log F has finite mean and variance then F = fe^{sX - \kappa(s)} for some X with mean zero and variance one where \kappa(s) = \log E[\exp(sX)] is the cumulant of X.
Exercise. Show E[F] = f and \operatorname{Var}(\log F) = s^2.
We call f the forward and s the vol.
Exercise. Prove {X = (\log F/f + \kappa(s))/s} has mean 0 and variance 1 if s > 0.
If an option pays off in shares of F then E[F\nu(F)] = fE[ e^{sX - \kappa(s)}\nu(F)] = fE^s[\nu(F)] where E^s is the share measure. Since E[e^{sX - \kappa(s)}] = 1 share measure is a probability measure that we denote by P^s.
The delta of an option is the derivative of its value with respect to the forward, \partial_f v = E[\nu'(F)\partial_f F] = E[\nu'(F)e^{sX - \kappa(s)}] = E^s[\nu'(F)]. The gamma of an option is the second derivative of its value with respect to the forward, The vega of an option is the derivative of its value with respect to the vol, \partial_s v = E[\nu'(F)\partial_s F] = E[\nu'(F) F (X - \kappa'(s))] = f E^s[\nu'(F) (X - \kappa'(s))].
A put with strike k pays \nu(F) = {\max\{k - F, 0\}} = {(k - F)^+} = {(k - F)1(F\ge k)} at expiration. Note F\le k is equivalent to X\le x where {x = x(k) = x(k;f,s) = (\log k/f + \kappa(s))/s} is the moneyness of a put with strike k.
The forward value of a put option is
\begin{aligned} p &= E[(k - F)^+] \\ &= E[(k - F)1(F\le k)] \\ &= kE[1(F \le k)] - E[F1(F\le k)] \\ &= kP(X \le x) - fP^s(X\le x). \\ \end{aligned}
Let \Psi(x) = P(X\le x) be the cumulative distribution function of X. The cumulative share distribution is {\Psi(x, s) = E[e^{sX - \kappa(s)} 1(X\le x)] = P^s(X\le x)}. Note {\Psi(x, 0) = \Psi(x)}.
We can write the put value as p = k\Psi(x) - f\Psi(x, s).
Since \partial_f (k - f)^+ = -1(f \le k) the put delta is \begin{aligned} \partial_f p &= E[-1(F \le k)\partial_f F] \\ &= -E[1(F \le k)e^{sX - \kappa(s)}] \\ &= -P^s(F\le k) \\ &= -\Psi(x, s). \end{aligned}
The second derivative with respect to forward is \begin{aligned} \partial_f^2 p &= \partial_f (-\Psi(x, s)) \\ &= -\partial_x\Psi(x, s)\partial_f x \\ &= \partial_x\Psi(x, s)/fs \\ \end{aligned}
Exercise. Show \partial_f x = -1/fs.
Hint: Use x = (\log k/f + \kappa(s))/s.
The derivative with respect to vol is \partial_s E[(k - F)^+] = {-E[1(F\le k)F(X - \kappa'(s))]}. Since \partial_s \Psi(x, s) = {E[e^{sX - \kappa(s)}(X - \kappa'(s))1(X\le x)]} we have \partial_s p = -f\partial_s\Psi(x, s).
Exercise. Show \partial_s p = -k\partial_s\Psi(x, 0).
A call with strike k pays {\nu(F) = (F - k)^+} at expiration and has forward value {c = E[(F - k)^+]}. Since {(F - k)^+ - (k - F)^+ = F - k} we have put-call parity {c - p = f - k}.
Let \overline{\Psi}(x) = 1 - \Psi(x) and {\overline{\Psi}(x, s) = 1 - \Psi(x, s)} be the complementary distributions. If P^s does not have a point mass at k then {P^s(F\ge k) = P^s(X\ge x) = \overline{\Psi}(x, s)}.
Exercise. Show the forward call value is c = f\overline{\Psi}(x, s) - k \overline{\Psi}(x).
Exercise. Show call delta \partial_f c = \overline{\Psi}(x, s).
Hint: Use put-call parity to show \partial_f c = \partial_ p + 1.
Exercise. Show call gamma equals put gamma.
Exercise. Show call vega equals put vega.
The Fischer Black model takes f = f, s = \sigma\sqrt{t} and X = B_t/\sqrt{t} where \sigma is the volatility, t is time in years to expiration, and B_t is Brownian motion at time t. We only use the fact B_t is normally distributed with mean 0 and variance t.
Exercise. Show B_t/\sqrt{t} has mean 0 and variance 1.
Exercise Show E[e^{\sigma B_t}] = e^{\sigma^2t/2} for any \sigma\in 𝑹.
Hint. Use E[\exp(N)] = \exp(E[N] + \operatorname{Var}(N)/2) if N is normally distributed.1
Exercise. Show F = fe^{\sigma B_t - \sigma^2t/2} in the Black model.
Exercise. If X is standard normal then E[e^{sX - s^2/2}g(X)] = E[g(X + s)].
Hint: Use E[\exp(N)g(N)] = E[\exp(N)] E[g(N + \operatorname{Var}(N))] if N is normal.2
Exercise. Show \Psi(x, s) = E[e^{sX - s^2/2}1(X\le x)] = P(X + s\le x) = \Psi(x - s).
Exercise. Show x = \log(k/f)/\sigma\sqrt{t} + \sigma\sqrt{t}/2.
Black defined {d_2 = \log(f/k)/\sigma\sqrt{t} - \sigma\sqrt{t}/2} and {d_1 = \log(f/k)/\sigma\sqrt{t} + \sigma\sqrt{t}/2}.
Exercise Show x = -d_2 and x - \sigma\sqrt{t} = -d_1.
The forward value of the Black put is p = k\Psi(x) - f\Psi(x - \sigma\sqrt{t}), delta is \partial_f p = -\Psi(x - \sigma\sqrt{t}), and gamma is \partial_f^2 p = \psi(x - \sigma\sqrt{t})/f\sigma\sqrt{t}, where {x = (\log k/f)/\sigma\sqrt{t} + \sigma\sqrt{t}/2} and \psi(x) = \Psi'(x) is the standard normal density.
Exercise. Show Black vega is \partial_\sigma p = f\psi(x - \sigma\sqrt{t})\sqrt{t}.
Hint: Use \partial_\sigma s = \sqrt{t}.
The B-S/M model assumes the risk-neutral price of a stock is {S_t = S_0e^{rt + \sigma B_t - \sigma^2t/2}} where r is the continuously compounded risk-free rate. The current price S_0 is called the spot price of the stock. The present value of an option paying \nu(S_t) at time t is {v = e^{-rt}E[\nu(S_t)]}. Note {S_t = F} where F is the forward price at expiration. This gives the relationship between the spot and forward f = e^{rt}S_0 and is called the cost of carry.
The Black-Scholes/Merton model takes f = e^{rt}S_0 and discounts the Black forward value by e^{-rt} to get the present option value v = e^{-rt}E[\nu(S_t)]. The B-S/M put delta is \begin{aligned} \partial_{S_0}v &= e^{-rt}E[\nu'(S_t)\partial_{S_0}S_t] \\ &= e^{-rt}E[\nu'(S_t)e^{rt + \sigma B_t - \sigma^2t/2}] \\ &= E^{\sigma\sqrt{t}}[\nu'(S_t)] \\ \end{aligned} This shows the B-S/M delta is the same as the Black forward delta for any option.
Let X be standard normal and N = \mu + \sigma X. Since 1 = \int_{-\infty}^\infty e^{-x^2/2}\,dx/\sqrt{2\pi} we have \begin{aligned} E[e^N] &= \int_{-\infty}^\infty e^{\mu + \sigma x} e^{-x^2/2}\,dx/\sqrt{2\pi} \\ &= e^{\mu + \sigma^2/2} \int_{-\infty}^\infty e^{-(x - \sigma)^2/2}\,dx/\sqrt{2\pi} \\ &= e^{\mu + \sigma^2/2} \\ &= e^{E[N] + \operatorname{Var}(N)/2} \\ \end{aligned} ↩︎
We have \begin{aligned} E[\exp(\sigma X) g(X)] &= \int_{-\infty}^\infty e^{\sigma x} g(x) e^{-x^2/2}\,dx/\sqrt{2\pi} \\ &= e^{\sigma^2/2}\int_{-\infty}^\infty g(x) e^{-(x - \sigma)^2/2}\,dx/\sqrt{2\pi} \\ &= e^{\sigma^2/2}\int_{-\infty}^\infty g(x + \sigma) e^{-x^2/2}\,dx/\sqrt{2\pi} \\ &= E[e^{\sigma X}]E[g(X + \sigma)]. \\ \end{aligned} ↩︎