Inverse Gaussian distribution

Parameters Probability density function $\lambda > 0$$\mu > 0$ $x \in (0,\infty)$ $\left[\frac{\lambda}{2 \pi x^3}\right]^{1/2} \exp{\frac{-\lambda (x-\mu)^2}{2 \mu^2 x}}$ $\Phi\left(\sqrt{\frac{\lambda}{x}} \left(\frac{x}{\mu}-1 \right)\right)$$+\exp\left(\frac{2 \lambda}{\mu}\right) \Phi\left(-\sqrt{\frac{\lambda}{x}}\left(\frac{x}{\mu}+1 \right)\right)$where $\Phi \left(\right)$ is the standard normal (standard Gaussian) distribution c.d.f. $\mu$ $\mu\left[\left(1+\frac{9 \mu^2}{4 \lambda^2}\right)^\frac{1}{2}-\frac{3 \mu}{2 \lambda}\right]$ $\frac{\mu^3}{\lambda}$ $3\left(\frac{\mu}{\lambda}\right)^{1/2}$ $\frac{15 \mu}{\lambda}$ $e^{\left(\frac{\lambda}{\mu}\right)\left[1-\sqrt{1-\frac{2\mu^2t}{\lambda}}\right]}$ $e^{\left(\frac{\lambda}{\mu}\right)\left[1-\sqrt{1-\frac{2\mu^2\mathrm{i}t}{\lambda}}\right]}$

In probability theory, the inverse Gaussian distribution (also known as the Wald distribution) is a two-parameter family of continuous probability distributions with support on (0,∞).

Its probability density function is given by

$f(x;\mu,\lambda) = \left[\frac{\lambda}{2 \pi x^3}\right]^{1/2} \exp{\frac{-\lambda (x-\mu)^2}{2 \mu^2 x}}$

for x > 0, where $\mu > 0$ is the mean and $\lambda > 0$ is the shape parameter.

As λ tends to infinity, the inverse Gaussian distribution becomes more like a normal (Gaussian) distribution. The inverse Gaussian distribution has several properties analogous to a Gaussian distribution. The name can be misleading: it is an "inverse" only in that, while the Gaussian describes a Brownian Motion's level at a fixed time, the inverse Gaussian describes the distribution of the time a Brownian Motion with positive drift takes to reach a fixed positive level.

Its cumulant generating function (logarithm of the characteristic function) is the inverse of the cumulant generating function of a Gaussian random variable.

To indicate that a random variable X is inverse Gaussian-distributed with mean μ and shape parameter λ we write

$X \sim IG(\mu, \lambda).\,\!$

Properties

Summation

If Xi has a IG(μ0wiλ0wi2) distribution for i = 1, 2, ..., n and all Xi are independent, then

$S=\sum_{i=1}^n X_i \sim IG \left( \mu_0 \sum w_i, \lambda_0 \left(\sum w_i \right)^2 \right).$

Note that

$\frac{\textrm{Var}(X_i)}{\textrm{E}(X_i)}= \frac{\mu_0^2 w_i^2 }{\lambda_0 w_i^2 }=\frac{\mu_0^2}{\lambda_0}$

is constant for all i. This is a necessary condition for the summation. Otherwise S would not be inverse Gaussian.

Scaling

For any t > 0 it holds that

$X \sim IG(\mu,\lambda) \,\,\,\,\,\, \Rightarrow \,\,\,\,\,\, tX \sim IG(t\mu,t\lambda).$

Exponential family

The inverse Gaussian distribution is a two-parameter exponential family with natural parameters -λ/(2μ²) and -λ/2, and natural statistics X and 1/X.

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Relationship with Brownian motion

The stochastic process Xt given by

$X_0 = 0\quad$
$X_t = \nu t + \sigma W_t\quad\quad\quad\quad$

(where Wt is a standard Brownian motion and $\nu > 0$) is a Brownian motion with drift ν.

Then, the first passage time for a fixed level $\alpha > 0$ by Xt is distributed according to an inverse-gaussian:

$T_\alpha = \inf\{ 0 < t \mid X_t=\alpha \} \sim IG(\tfrac\alpha\nu, \tfrac {\alpha^2} {\sigma^2}).\,$

When drift is zero

A common special case of the above arises when the Brownian motion has no drift. In that case, parameter μ tends to infinity, and the first passage time for fixed level α has probability density function

$f \left( x; 0, \left(\frac{\alpha}{\sigma}\right)^2 \right) = \frac{\alpha}{\sigma \sqrt{2 \pi x^3}} \exp\left(-\frac{\alpha^2 }{2 x \sigma^2}\right).$

This is a Lévy distribution with parameters $c=\frac{\alpha^2}{\sigma^2}$ and $\mu=0$.

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Maximum likelihood

The model where

$X_i \sim IG(\mu,\lambda w_i), \,\,\,\,\,\, i=1,2,\ldots,n$

with all wi known, (μλ) unknown and all Xiindependent has the following likelihood function

$L(\mu, \lambda)= \left( \frac{\lambda}{2\pi} \right)^\frac n 2 \left( \prod^n_{i=1} \frac{w_i}{X_i^3} \right)^{\frac{1}{2}} \exp\left(\frac{\lambda}{\mu}\sum_{i=1}^n w_i -\frac{\lambda}{2\mu^2}\sum_{i=1}^n w_i X_i - \frac\lambda 2 \sum_{i=1}^n w_i \frac1{X_i} \right).$

Solving the likelihood equation yields the following maximum likelihood estimates

$\hat{\mu}= \frac{\sum_{i=1}^n w_i X_i}{\sum_{i=1}^n w_i}, \,\,\,\,\,\,\,\, \frac{1}{\hat{\lambda}}= \frac{1}{n} \sum_{i=1}^n w_i \left( \frac{1}{X_i}-\frac{1}{\hat{\mu}} \right).$

$\hat{\mu}$ and $\hat{\lambda}$ are independent and

$\hat{\mu} \sim IG \left(\mu, \lambda \sum_{i=1}^n w_i \right) \,\,\,\,\,\,\,\, \frac{n}{\hat{\lambda}} \sim \frac{1}{\lambda} \chi^2_{n-1}.$
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Generating random variates from an inverse-Gaussian distribution

The following algorithm may be used.[1]

Generate a random variate from a normal distribution with a mean of 0 and 1 standard deviation

$\displaystyle \nu = N(0,1).$

Square the value

$\displaystyle y = \nu^2$

and use this relation

$x = \mu + \frac{\mu^2 y}{2\lambda} - \frac{\mu}{2\lambda}\sqrt{4\mu \lambda y + \mu^2 y^2}.$

Generate another random variate, this time sampled from a uniformed distribution between 0 and 1

$\displaystyle z = U(0,1).$

If

$z \le \frac{\mu}{\mu+x}$

then return

$\displaystyle x$

else return

$\frac{\mu^2}{x}.$

Sample code in Java:

public double inverseGaussian(double mu, double lambda) {
Random rand = new Random();
double v = rand.nextGaussian();   // sample from a normal distribution with a mean of 0 and 1 standard deviation
double y = v*v;
double x = mu + (mu*mu*y)/(2*lambda) - (mu/(2*lambda)) * Math.sqrt(4*mu*lambda*y + mu*mu*y*y);
double test = rand.nextDouble();  // sample from a uniform distribution between 0 and 1
if (test <= (mu)/(mu + x))
return x;
else
return (mu*mu)/x;
}

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Related distributions

• If $X \sim \textrm{IG}(\mu,\lambda)\,$ then $k X \sim \textrm{IG}(k \mu,k \lambda)\,$
• If $X_i \sim \textrm{IG}(\mu,\lambda)\,$ then $\sum_{i=1}^{n} X_i \sim \textrm{IG}(n \mu,n^2 \lambda)\,$
• If $X_i \sim \textrm{IG}(\mu,\lambda)\,$ for $i=1,\ldots,n\,$ then $\bar{X} \sim \textrm{IG}(\mu,n \lambda)\,$
• If $X_i \sim \textrm{IG}(\mu_i,2 \mu^2_i)\,$ then $\sum_{i=1}^{n} X_i \sim \textrm{IG}\left(\sum_{i=1}^n \mu_i, 2 {\left( \sum_{i=1}^{n} \mu_i \right)}^2\right)\,$

The convolution of a Wald distribution and an exponential (the ex-Wald distribution) is used as a model for response times in psychology.[2]

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History

This distribution appears to have been first derived by Schrödinger in 1915 as the time to first passage of a Brownian motion.[3] The name inverse Gaussian was proposed by Tweedie in 1945.[4] Wald re-derived this distribution in 1947 as the limiting form of a sample in a sequential probability ratio test. Tweedie investigated this distribution in 1957 and established some of its statistical properties.

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Software

The R programming language has software for this distribution. [1]

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Notes

1. ^ Generating Random Variates Using Transformations with Multiple Roots by John R. Michael, William R. Schucany and Roy W. Haas, American Statistician, Vol. 30, No. 2 (May, 1976), pp. 88–90
2. ^ Schwarz W (2001) The ex-Wald distribution as a descriptive model of response times. Behav Res Methods Instrum Comput 33(4):457-469
3. ^ Schrodinger E (1915) Zur Theorie der Fall—und Steigversuche an Teilchenn mit Brownscher Bewegung. Physikalische Zeitschrift 16, 289-295
4. ^ Folks JL & Chhikara RS (1978) The inverse Gaussian and its statistical application - a review. J Roy Stat Soc 40(3) 263-289
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References

• The inverse gaussian distribution: theory, methodology, and applications by Raj Chhikara and Leroy Folks, 1989 ISBN 0-8247-7997-5
• System Reliability Theory by Marvin Rausand and Arnljot Høyland
• The Inverse Gaussian Distribution by Dr. V. Seshadri, Oxford Univ Press, 1993
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Last modified on 7 May 2013, at 15:02