title: Double Exponential Model
author: Keith A. Lewis
institute: KALX, LLC
classoption: fleqn
fleqn: true
...
\newcommand{\Var}{\operatorname{Var}}
\newcommand{\RR}{𝑹}
The exponential random variable with parameter $\beta > 0$ has
density $f(x) = \beta e^{-\beta x}$, $x \ge 0$.
Since the indefinite integral of $e^{\gamma x}$ is $e^{\gamma x}/\gamma$
we have $\int_0^\infty e^{-\beta x},dx = e^{-\beta x}/(-\beta)|_0^\infty = 0 - 1/(-\beta) = 1/\beta$.
This shows $\int_0^\infty f(x),dx = 1$ so it is a density function.
Let $f(x) = \alpha e^{-\beta|x|}$, $-\infty < x < \infty$, where $\beta > 0$.
Note $|x| = x$ if $x \ge 0$ and $|x| = -x$ if $x\le 0$.
Exercise. Show $\int{-\infty}^\infty f(x),dx = 1$ implies $\alpha = \beta/2$_.
Hint: $\int_{-\infty}^0 e^{\beta x},dx = \int_0^\infty e^{-\beta x},dx$.
This shows $f(x) = e^{-\beta|x|}\beta/2$, $-\infty < x < \infty$,
is a density function for the double exponential random variable $X$.
The moment generating function of $X$ is $E[e^{sX}]$.
Exercise. Show $E[e^{sX}] = \beta^2/(\beta + s)(\beta - s)$, $s < \beta$.
Hint: Use
$\int_{-\infty}^0 e^{sx} e^{\beta x},dx = 1/(\beta + s)$.
and
$\int_0^\infty e^{sx} e^{-\beta x},dx = 1/(\beta - s)$, $s < \beta$.
The cumulant of $X$ is the log of the moment generating function
$$
\kappa(s) = \log E[e^{sX}] = 2\log\beta - \log (\beta + s) - \log(\beta - s).
$$
Note $\kappa(0) = 2\log\beta - \log\beta - \log\beta = 0$, as it should.
Exercise. Show $\kappa'(s) = -1/(\beta + s) + 1/(\beta - s)$, $s < \beta$.
The mean of $X$ is $\kappa'(0) = -1/\beta + 1/\beta = 0$.
The variance of $X$ is $\kappa''(0)$.
Exercise. Show $\Var(X) = 2/\beta^2$, $s < \beta$.
This shows $X$ has variance 1 if $\beta = \sqrt{2}$.
Valuing options and their greeks with underlying $F = fe^{sX - \kappa(s)}$,
where $X$ has mean 0 and variance 1, boils down to computing
$$
\Phi(x,s) = E[e^{sX - \kappa(s)} 1(X \le x)]
$$
and its derivatives with respect to $x$ and $s$.
Exercise. If $x \le 0$ then $\Phi(x, s) = (1 - s/\beta) e^{(s + \beta)x}/2$.
Hint: Use $\int_{-\infty}^x e^{su} e^{\beta u},du = e^{(s + \beta)x}/(s + \beta)$ if $x < 0$.
Note $\Phi(0,s) = (1 - s/\beta)/2$.
If $x > 0$ then $\int_0^x e^{su} e^{-\beta u},du = e^{(s - \beta)x}/(s - \beta) - 1/(s - \beta)$.
Exercise. If $x \ge 0$ then $\Phi(x, s) = 1 - (1 + s/\beta) e^{(s - \beta)x}/2$, $s < \beta$.
Note $\Phi(0, s) = 1 - (1 + s/\beta)/2 = (1 - s/\beta)/2$ and $\lim_{x\to\infty} \Phi(x,s) = 1$.