April 25, 2024
The forward value of a European put option having strike k > 0 on underlying F is E[\max\{k - F, 0\}] = E[(k - F)1(F \le k)] = k P(F\le k) - E[F] P^s(F\le k), where P^s is share measure defined by dP^s/dP = F/f.
Every positive random variable can be written F = f e^{sX -\kappa(s)} where X has mean 0, variance 1 and \kappa(s) = \log E[e^{s X}] is the cumulant of X.
If X is standard normal and s = \sigma\sqrt{t}, then this is the Black put formula. Recall E[\exp(N)] = \exp(E[N] + \mathop{\rm{Var}}(N)/2) if N is normal, and E[\exp(N) f(M)] = E[\exp(N)] E[f(M + \mathop{\rm{Cov}}(N,M))] if N and M are jointly normal. We have \kappa(s) = \log E[\exp(sX)] = s^2/2 and E^s[f(X)] = E[\exp(sX-s^2/2) f(X)] = E[f(X + s)]. In the usual notation, m = -d_2 = (\log (k/f) + \sigma^2t/2)/\sigma\sqrt{t} and P^s(X \le m) = P(X + \sigma\sqrt{t} \le -d_2) = P(X \le -d_1) where d_1 = d_2 + \sigma\sqrt{t}.
The cumulant generating function of the random variable X is \kappa(s) = \log E[e^{sX}]. When they exist, the cumulants (\kappa_n) are defined by \kappa(s) = \sum_{n=1}^\infty \kappa_n \frac{s^n}{n!} Since the n-th derivative evaluated at 0 satisfies \kappa^{(n)}(s)|_{s = 0} = \kappa_n it is easy to work out that \kappa_1 = E[X] and \kappa_2 = \mathop{\rm{Var}}(X). Higher order cumulants are less intuitive but the third and fourth are related to skew and kurtosis.
The cumulants of a random variable plus a constant are the same except the first cumulant is increased by the constant. More generally, the cumulants of the sum of two independent random variables are the sums of their cumulants. They scale homogeneously, the n-th cumulant of a constant times a random variable is \kappa_n(cX) = c^n\kappa_n(X).
The exponential of the cumulant in terms of powers of s is E[e^{sX}] = \exp(\sum_{n=1}^\infty \kappa_n \frac{s^n}{n!}) = \sum_{n=0}^\infty B_n(\kappa_1,\dots,\kappa_n) \frac{s^n}{n!} where B_n(\kappa_1,\dots,\kappa_n) is the n-th complete Bell polynomial. This is just a special case of the Fa`a di Bruno formula first proved by Louis Franois Antoine Arborgast in 1800. Bell polynomials satisfy the recurrence B_0 = 1 and B_{n+1}(x_1,\dots,x_{n+1}) = \sum_{k=0}^n \binom{n}{k} B_{n - k}(x_1,\dots, x_{n - k}) x_{k+1}.
Let \psi be the probability density of a random variable, X, with expected value 0 and mean 1. The the Fourier transform is
\begin{aligned} E[e^{-iuX}] &= e^{\sum_{n=1}^\infty \kappa_n (-iu)^n/n!}\\ &= e^{-u^2/2} e^{\sum_{n=3}^\infty \kappa_n (-iu)^n/n!}\\ &= e^{-u^2/2} \sum_{n=0}^\infty B_n(0, 0, \kappa_3,...,\kappa_n)(-iu)^n/n!.\\ &= e^{-u^2/2} (1 + \sum_{n=3}^\infty B_n(0, 0, \kappa_3,...,\kappa_n)(-iu)^n/n!.\\ \end{aligned}
where (\kappa_n) are the cumulants of X.
The Fourier transform, \hat{f}(u) = E[e^{-iuX}], of f' is iu \hat f(u) so the Fourier transform of the n-th derivative f^{(n)} is (iu)^n\hat f(u) hence \begin{aligned} \hat{\psi}(u) &= \hat{\phi}(u)(1 + \sum_{n=3}^\infty B_n(0, 0,\kappa_3,...,\kappa_n) (-iu)^n/n!\\ &= \hat{\phi}(u) + \sum_{n=3}^\infty (-1)^n B_n(0, 0, \kappa_3,...,\kappa_n) \widehat{\phi^{(n)}}(u)/n!\\ \end{aligned} where \phi(x) = e^{-x^2/2}. Note \hat{\phi}(u) = e^{-u^2/2}. Taking inverse Fourier transforms and integrating both sides yields \Psi(x) = \Phi(x) + \sum_{n=3}^\infty (-1)^n B_n(0,0,\kappa_3,...,\kappa_n) \Phi^{(n)}(x)/n!.
The derivatives of the standard normal density can be computed using Hermite polynomials pp. 793–801. One definition is H_n(x) = (-1)^n e^{x^2/2}\frac{d^n}{dx^n}e^{-x^2/2}. They satisfy the recurrence H_0(x) = 1, H_1(x) = x and H_{n+1}(x) = xH_n(x) - n H_{n-1}(x). Note some authors use He_n(x) instead of H_n(x). This shows \phi^{(n)}(x) = (-1)^n\phi(x) H_n(x) so \Phi^{(n)} = (-1)^{n-1} H_{n-1}(x) for n > 0.
We now have an explicit formula for the cumulative distibution function of X: \Psi(x) = \Phi(x) - \phi(x)\sum_{n=1}^\infty B_n(\kappa_1,\dots,\kappa_n) H_{n-1}(x)/n!
Given any random variable, X, and number s, define X^s by P(X^s\le x) = P^s(X\le x) where dP^s/dP = e^{-\kappa(s) + sX}. The cumulant of X^s is \kappa^s(u) = \kappa(u + s) - \kappa(s), where \kappa is the cumulant of X. This follows from E^s[e^{uX}] = E[e^{sX - \kappa(s)} e^{uX}] = e^{\kappa(s + u) - \kappa(s)}.
It follows that the n-th derivative
satisfies \kappa^{s(n)}(u) = \kappa^{(n)}(u +
s), for n > 0,
and setting u=0, \kappa^s_n = \kappa^{s(n)}(0) =
\kappa^{(n)}(s).
The last expression can be expressed as \kappa^{(n)}(s) = \sum_{k=0}^\infty \kappa_{n - k} s^k/k!, but for many random variables, we can use the closed form solution on the left-hand side instead of the infinite sum.
A normally distributed random variable, X, has density function f(x) = e^{-(x - \mu)^2/2\sigma^2}/\sigma\sqrt{2\pi}, -\infty<x<\infty, with mean \mu and variance \sigma^2. The cumulant is \kappa(s) = \mu s + \sigma^2s^2/2 so \kappa_1 = \mu and \kappa_2 = \sigma^2 are the only non-zero cumulants.
Since \kappa^s(u) = \kappa(u + s) - \kappa(s) = (\mu + \sigma^2s)u + \sigma^2u/2 we see that \kappa^s_1 = \kappa_1 + \sigma^2 and \kappa^s_2 = \kappa_2. The Esscher transform of a normally distributed random variable is normal with mean \mu^s = \mu + \sigma^2 and variance \sigma^{s2} = \sigma^2.
If the cumulants of a random variable vanish after some some point, then it must be normal (Theorem 7.3.5).
A gamma distributed random variable, X, has density function f(x) = x^{\alpha - 1} e^{-x/\beta}/\beta^\alpha\Gamma(\alpha), x > 0, with mean \alpha\beta and variance \alpha\beta^2. The cumulant is \kappa(s) = -\alpha\log(1 - \beta s) so \kappa_n = \kappa^{(n)}(0) = (n-1)!\alpha\beta^n since \kappa^{(n)}(s) = (n-1)!\alpha\beta^n/(1 - \beta s)^n.
Since \kappa^s(u) = \kappa(u + s) - \kappa(s) = -\alpha\log (1 - \beta u/(1 - \beta s), the Esscher transform of a gamma distributed random variable is gamma with \alpha^* = \alpha and \beta^* = \beta/(1 - \beta s).
Note that an exponentially distributed random variable is the special case when \alpha = 1.
If X is Poisson with mean \mu then \kappa(s) = \mu(e^s - 1) so \kappa_n = \mu for all n and \kappa_n^* = \mu e^s.
The Variance Gamma model is the difference of independent Gamma distributions so \kappa_n = (n-1)!(\alpha\beta^n - \alpha'\beta'^n). In order for this to be a perturbation of a standard normal distribution we need 0 = \alpha\beta - \alpha'\beta' and 1 = \alpha\beta^2 - \alpha'\beta'^2. Using mean and standard deviation \mu = \mu' and \sigma^2 = 1 + \sigma'^2. For convergence we need 1 < \sigma^2 \lll \mu.
Assume Y_i are independent and identically distributed. If N is Poisson with parameter \lambda then X is {} if X = \sum_{i=0}^N Y_i. The exponential of the cumulant of X is \begin{aligned} Ee^{sX} &= Ee^{s\sum_{i=0}^N Y_i}\\ &= \sum_{k=0}^\infty \frac{\lambda^k}{k!}e^{-\lambda}(Ee^{Y_i})^k\\ &= \sum_{k=0}^\infty \frac{\lambda^k}{k!}e^{-\lambda}(e^{\kappa^Y(s)})^k\\ &= \sum_{k=0}^\infty \frac{(\lambda e^{\kappa^Y(s)})^k}{k!}e^{-\lambda}\\ &= e^{\lambda(e^{\kappa(s) - 1})} \end{aligned}
Merton’s jump diffusion model assumes X = \alpha Z + \beta\sum_{k=0}^N Y_k where N is Poisson and Y_k are independent normal.
f(y) = p\eta_1 e^{-\eta_1 y}1(y > 0) + (1 - p)\eta_2 e^{\eta_2 y}1(y < 0) \begin{aligned} Ee^{sY} &= \int_0^\infty e^{sx} pe^{-\eta_1 y}\,dy + \int_{-\infty}^0 e^{sy} (1-p)e^{\eta_2 y}\,dy\\ &= \frac{p}{1-s/\eta_1} + \frac{1-p}{1 + s/\eta_2} \end{aligned}
\kappa^Y_n = n!(\frac{p}{\eta_1^n} + \frac{1-p}{(-\eta_2)^n})
Kolmogorov’s precursor to the Lévy-Khintchine theorem states that if a random variable X is infinitely divisible and has finite variance there exists a number \gamma and a non-decreasing function G defined on the real line such that \kappa(s) = \log Ee^{sX} = \gamma s + \int_{-\infty}^\infty K_s(x)\,dG(x), where K_s(x) = (e^{sx} - 1 - sx)/x^2 = \sum_{n=2}^\infty s^nx^{n-2}/n!. Note the first cumulant of X is \gamma and for n\ge 2, \kappa_n = \int_{-\infty}^\infty x^{n-2}\,dG(x).
The Hamburger moment problem[cite?] provides the answer to what the allowable cumulants are: the Hankel matrix [\kappa_{i+j}]_{i,j\ge 2} must be positive definite.
Since K_s(0) = s^2/2 is the cumulant of the standard normal distribution and a^2K_s(a) + as is the cumulant of a Poisson distribution having mean a, infinitely divisible random variables can be approximated by a normal plus a linear combination of independent Poisson distributions.
The Gram-Charlier A series expands the quotients of cumulative distribution functions G/F using Hermite polynomials, but does not have asymptotic convergence, whereas the Edgeworth expansion involves the quotient of characteristic functions \hat G/\hat F in terms of cumulants and does have asymptotic convergence, ignoring some dainty facts .
If (X_t) is a Lévy process then X_1 is infinitely divisible and \log Ee^{sX_t} = t\kappa(s). A consequence is that the volatility smile at a single maturity determines the entire volatility surface, a fact that may indicate Lévy processes are not appropriate for modeling stock prices.