Double Exponential Model

Keith A. Lewis

April 25, 2024

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 < \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 \operatorname{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 < 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 > 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.