de.md

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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 $\int_0^\infty e^{-\beta x},dx = e^{-\beta x}/(-\beta)\mid_0^\infty = 0 - 1/(-\beta) = 1/\beta$. 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 $\int_0^\infty e^{sx} e^{-\beta x},dx = 1/(\beta - s)$, $s &lt; \beta$.

Solution

$$ \begin{aligned} E[e^{sX}] &= \int_{-\infty}^\infty e^{sx} e^{-\beta |x|}\beta/2\,dx \\ E[e^{sX}] &= \int_{-\infty}^0 e^{sx} e^{\beta x}\beta/2\,dx + \int_0^\infty e^{sx} e^{-\beta x}\beta/2\,dx \\ &= \frac{\beta}{2}\bigl(\frac{1}{\beta + s} + \frac{1}{\beta - s}\bigr) \\ &= \frac{\beta}{2}\bigl(\frac{1}{\beta + s} + \frac{1}{\beta - s}\bigr) \\ &= \frac{\beta^2}{(\beta + s)(\beta - s)} \\ \end{aligned} $$

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 $\kappa'(s) = -1/(\beta + s) + 1/(\beta - s)$, $s < \beta$.

Solution

$$ \begin{aligned} \kappa'(s) &= -1/(\beta + s) + 1/(\beta -s) \\ &= -1/(\beta + s) + 1/(\beta - s) \\ \end{aligned} $$

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 $\beta = \sqrt{2}$.

Solution

$$ \begin{aligned} \kappa''(s) &= 1/(\beta + s)^2 + 1/(\beta - s)^2 \\ \end{aligned} $$

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 $x &lt; 0$.

Solution

$$ \begin{aligned} \Phi(x, s) &= E[e^{sX - \kappa(s)} 1(X \le x)] \\ &= e^{-\kappa(s)} e^{(s + \beta)x}(\beta/2)/(s + \beta) \\ &= \frac{(\beta + s)(\beta - s)}{\beta^2} e^{(s + \beta)x}(\beta/2)/(s + \beta) \\ &= \frac{(\beta - s)}{\beta} e^{(s + \beta)x}/2 \\ &= (1 - s/\beta) e^{(s + \beta)x}/2 \\ \end{aligned} $$

Note $\Phi(0,s) = (1 - s/\beta)/2$.

If $x &gt; 0$ 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 $\Phi(x, s) = 1 - (1 + s/\beta) e^{(s - \beta)x}/2$, $s < \beta$.

Solution

$$ \begin{aligned} \Phi(x, s) &= \Phi(0, s) + \int_0^x e^{su - \kappa(s)} e^{-\beta u}\beta/2\,du \\ &= \Phi(0, s) + e^{-\kappa(s)} \bigl(e^{(s - \beta)x}/(s - \beta) - 1/(s - \beta)\bigr)\beta/2 \\ &= \Phi(0, s) + \frac{(\beta + s)(\beta - s)}{\beta^2} \bigl(e^{(s - \beta)x}/(s - \beta) - 1/(s - \beta)\bigr)\beta/2 \\ &= \Phi(0, s) - \frac{(\beta + s)}{\beta} (e^{(s - \beta)x} - 1)/2 \\ &= \Phi(0, s) - (1 + s/\beta) e^{(s - \beta)x}/2 + (1 + s/\beta)/2 \\ &= (1 - s/\beta)/2 - (1 + s/\beta) e^{(s - \beta)x}/2 + (1 + s/\beta)/2 \\ &= 1 - (1 + s/\beta) e^{(s - \beta)x}/2 \\ \end{aligned} $$

Note $\Phi(0, s) = 1 - (1 + s/\beta)/2 = (1 - s/\beta)/2$ and $\lim_{x\to\infty} \Phi(x,s) = 1$.