1. Random
2. 4. Special Distributions
3. The Lévy Distribution

The Lévy Distribution

The Lévy distribution, named for the French mathematician Paul Lévy, is important in the study of Brownian motion, and is one of only three stable distributions whose probability density function can be expressed in a simple, closed form.

The Standard Lévy Distribution

Distribution Functions

A random variable $$U$$ with values in $$(0, \infty)$$ has the standard Lévy distribution if it has probability density function $$g$$ given by $g(u) = \frac{1}{\sqrt{2 \pi}} \frac{1}{u^{3/2}} \exp\left(-\frac{1}{2 u}\right), \quad u \in (0, \infty)$ $$g$$ is a valid probability density function.

1. $$g$$ increases and then decreasing with mode at $$x = \frac{1}{3}$$.
2. $$g$$ is concave upward, then downward, then upward again, with inflection points at $$x = \frac{1}{3} - \frac{\sqrt{10}}{15} \approx 0.1225$$ and at $$x = \frac{1}{3} + \frac{\sqrt{10}}{15} \approx 0.5442$$.
Proof:

The substitution $$u = z^{-2}$$ gives $\int_0^\infty g(u) \, du = 2 \int_0^\infty \frac{1}{\sqrt{2 \pi}} e^{-z^2/2} dz$ We recognize the integrand of the new integral as the standard normal PDF, so the result is $$2 \frac{1}{2} = 1$$. Parts (a) and (b) follow by standard calculus. \begin{align} g^\prime(u) & = \frac{1}{2 \sqrt{2 \pi}} u^{-7/2} e^{-u^{-1}/2} (-3 u + 1) \\ g^{\prime\prime}(u) & = \frac{1}{4 \sqrt{2 \pi}} u^{-11/2} e^{-u^{-1}/2} (15 u^2 - 10 u + 1) \end{align}

Open the Special Distribtion Simulator and select the Lévy distribution. Keep the default parameter values. Note the shape of the probability density function. Run the simulation 1000 times and compare the empirical density function to the probability density function.

$$U$$ has distribution function $$G$$ given by $G(u) = 2\left[1 - \Phi\left(\frac{1}{\sqrt{u}}\right)\right], \quad u \in (0, \infty)$ where $$\Phi$$ is the standard normal distribution function.

Proof:

Using the same subsitution as before, $$t = z^{-2}$$ gives $G(u) = \int_0^u g(t) \, dt = 2 \int_{1/\sqrt{u}}^\infty \frac{1}{\sqrt{2 \pi}} e^{-z^2/2} dz = 2\left[1 - \Phi\left(\frac{1}{\sqrt{u}}\right)\right], \quad u \in (0, \infty)$

$$U$$ has quantile function $$G^{-1}$$ given by $G^{-1}(p) = \frac{1}{\left[\Phi^{-1}\left(1 - \frac{p}{2}\right)\right]^2}, \quad p \in [0, 1)$ where $$\Phi^{-1}$$ is the standard normal quantile function. The quartiles of $$U$$ are

1. $$q_1 = \left[\Phi^{-1}\left(\frac{7}{8}\right)\right]^{-2} \approx 0.7557$$, the first quartile.
2. $$q_2 = \left[\Phi^{-1}\left(\frac{3}{4}\right)\right]^{-2} \approx 2.1980$$, the median.
3. $$q_3 = \left[\Phi^{-1}\left(\frac{5}{8}\right)\right]^{-2} \approx 9.8516$$, the first quartile.
Proof:

The quantile function can be obtained from the distribution function by solving $$p = G(u)$$ for $$u = G^{-1}(p)$$.

Open the Special Distribution Calculator and select the Lévy distribution. Keep the default parameter values. Note the shape and location of the distribution function. Compute a few values of the distribution function and the quantile function.

Moments

After exploring the graphs of $$g$$ and $$G$$, you probably noticed that the Lévy distribution has a very heavy tail. The 99th percentile is about 6400, for example. The following result is not surprising.

$$\E(U) = \infty$$

Proof:

Note that $$u \mapsto e^{-1/2u}$$ is increasing. Hence $\E(U) = \int_0^\infty u \frac{1}{\sqrt{2 \pi} u^{3/2}} e^{-1/2u} du \gt \int_1^\infty \frac{1}{\sqrt{2 \pi}} u^{-1/2} e^{-1/2} du = \infty$

Of course, the higher-order moments are infinite as well, and the variance, skewness, and kurtosis do not exist. The moment generating function is infinite at every positive value, and so is of no use. On the other hand, the characteristic function of the standard Lévy distribution is very useful.

$$U$$ has characteristic function $$\chi_0$$ given by $\chi_0(t) = \E\left(e^{i t U}\right) = \exp\left(-\left|t\right|^{1/2}\left[1 + i \sgn(t)\right]\right), \quad t \in \R$ where $$\sgn$$ is the usual sign function: $$\sgn(t) = 1$$ for $$t \gt 0$$, $$\sgn(t) = - 1$$ for $$t \lt 0$$, and $$\sgn(0) = 0$$.

Related Distributions

A standard Lévy variable can be obtained from a standard normal variable by a simple transformation.

If $$Z$$ has the standard normal distribution then $$U = 1/Z^2$$ the standard Lévy distribution

Proof:

For $$u \gt 0$$, $\P\left(\frac{1}{Z^2} \le u \right) = \P\left(Z^2 \ge \frac{1}{u}\right) = \P\left(Z \ge \frac{1}{\sqrt{u}}\right) + \P\left(Z \le -\frac{1}{\sqrt{u}}\right) = 2\left[1 - \Phi\left(\frac{1}{\sqrt{u}}\right)\right]$ which is the CDF of the standard Lévy distribution.

The following result is bascially the converse.

If $$U$$ has the standard Lévy distribution, then $$V = 1/\sqrt{U}$$ has the standard half normal distribution.

Proof:

By the previous result, we can take $$U = 1 / Z^2$$ where $$Z$$ has the standard normal distribution. Then $$1/\sqrt{U} = \left|Z\right|$$, and $$\left|Z\right|$$ has the standard half normal distribution.

The General Lévy Distribution

Like so many other standard distributions, the standard Lévy distribution is generalized by adding location and scale parameters.

Definition

Suppose that $$U$$ has the standard Lévy distribution, and $$a \in \R$$ and $$b \in (0, \infty)$$. Then $$X = a + b U$$ has the Lévy distribution with location parameter $$a$$ and scale parameter $$b$$.

Distribution Functions

$$X$$ takes values in $$(a, \infty)$$ and has probability density function $$f$$ given by $f(x) = \sqrt{\frac{b}{2 \pi}} \frac{1}{(x - a)^{3/2}} \exp\left[-\frac{b}{2 (x - a)}\right], \quad x \in (a, \infty)$

1. $$f$$ increases and then decreases with mode at $$x = a + \frac{1}{3} b$$.
2. $$f$$ is concave upward, then downward, then upward again with inflection points at $$x = a + \left(\frac{1}{3} \pm \frac{\sqrt{10}}{15}\right) b$$.
Proof:

Recall that $$f(x) = \frac{1}{b} g\left(\frac{x - a}{b}\right)$$ where $$g$$ is the standard Lévy PDF, so the formula for $$f$$ follow from the definition of $$g$$ and simple algebra. Parts (a) and (b) follow from the corresponding results for $$g$$.

Open the Special Distribtion Simulator and select the Lévy distribution. Vary the parameters and note the shape and location of the probability density function. For various parameter values, run the simulation 1000 times and compare the empirical density function to the probability density function.

$$X$$ has distribution function $$G$$ given by $F(x) = 2\left[1 - \Phi\left(\sqrt{\frac{b}{(x - a)}}\right)\right], \quad x \in (a, \infty)$ where $$\Phi$$ is the standard normal distribution function.

Proof:

Recall that $$F(x) = G\left(\frac{x - a}{b}\right)$$ where $$G$$ is the standard Lévy CDF.

$$X$$ has quantile function $$F^{-1}$$ given by $F^{-1}(p) = a + \frac{b}{\left[\Phi^{-1}\left(1 - p/2\right)\right]^2}, \quad p \in [0, 1)$ where $$\Phi^{-1}$$ is the standard normal quantile function. The quartiles of $$U$$ are

1. $$q_1 = a + b \left[\Phi^{-1}\left(\frac{7}{8}\right)\right]^{-2}$$, the first quartile.
2. $$q_2 = a + b \left[\Phi^{-1}\left(\frac{3}{4}\right)\right]^{-2}$$, the median.
3. $$q_3 = a + b \left[\Phi^{-1}\left(\frac{5}{8}\right)\right]^{-2}$$, the first quartile.
Proof:

Recall that $$F^{-1}(p) = a + b G^{-1}(p)$$, where $$G^{-1}$$ is the standard Lévy quantile function.

Open the Special Distribution Calculator and select the Lévy distribution. Vary the parameter values and note the shape of the graph of the distribution function function. For various values of the parameters, compute a few values of the distribution function and the quantile function.

Moments

Of course, since $$U$$ has infinite mean, so does $$X$$. Also as before, the variance, skewness, and kurtosis are undefined.

$$\E(X) = \infty$$

$$X$$ has characteristic function $$\chi$$ given by $\chi(t) = \E\left(e^{i t X}\right) = \exp\left(i t a - b^{1/2} \left|t\right|^{1/2}[1 + i \sgn(t)]\right), \quad t \in \R$

Proof:

This follows from the standard characteristic function since $$\chi(t) = e^{i t a} \chi_0(b t)$$. Note that $$\sgn(b t) = \sgn(t)$$ since $$b \gt 0$$.

Related Distributions

Since the Lévy distribution is a location-scale family, it is trivially closed under location-scale transformations.

If $$X$$ has the Lévy distribution with location parameter $$a$$ and scale parameter $$b$$, and if $$c \in \R$$ and $$d \in (0, \infty)$$, then $$c + d X$$ has the Lévy distribution with location parameter $$c + a d$$ and scale parameter $$b d$$.

Of more interest is the fact that the Lévy distribution is closed under convolution (corresponding to sums of independent variables).

Suppose that $$X_1$$ and $$X_2$$ are independent, and that, $$X_k$$ has the Lévy distribution with location parameter $$a_k \in \R$$ and scale parameter $$b_k \in (0, \infty)$$ for $$k \in \{1, 2\}$$. Then $$X_1 + X_2$$ has the Lévy distribution with location parameter $$a_1 + a_2$$ and scale parameter $$(b_1^{1/2} + b_2^{1/2})^2$$.

Proof:

The characteristic function of $$X_k$$ is $\chi_k(t) = \exp\left(i t a_k - b_k^{1/2} \left| t \right|^{1/2}[1 + i \sgn(t)]\right), \quad t \in \R$ for $$k \in \{1, 2\}$$. Hence the characteristic function of $$X_1 + X_2$$ is \begin{align*} \chi(t) = \chi_1(t) \chi_2(t) & = \exp\left[i t (a_1 + a_2) - \left(b_1^{1/2} + b_2^{1/2}\right) \left| t \right|^{1/2}[1 + i \sgn(t)]\right]\\ & = \exp\left[i t A - B^{1/2} \left| t \right|^{1/2}[1 + i \sgn(t)]\right], \quad t \in \R \end{align*} where $$A = a_1 + a_2$$ is the location parameter and $$B = \left(b_1^{1/2} + b_2^{1/2}\right)^2$$ is the scale parameter.

As a corollary, the Lévy distribution is a stable distribution with index $$\alpha = \frac{1}{2}$$:

Suppose that $$n \in \N_+$$ and that $$(X_1, X_2, \ldots, X_n)$$ is a sequence of independent random variables, each having the Lévy distribution with location parameter $$a \in \R$$ and scale parameter $$b \in (0, \infty)$$. Then $$X_1 + X_2 + \cdots + X_n$$ has the Lévy distribution with location parameter $$n a$$ and scale parameter $$n^2 b$$.

Stability is one of the reasons for the importance of the Lévy distribution. From the characteristic function, it follows that the skewness parameter is $$\beta = 1$$.