Inverse Gaussian distribution

The probability density function (pdf) of the inverse Gaussian distribution has a single parameter form given by

:$f(x;\backslash mu,\backslash mu^2)$

= \frac{\mu}{\sqrt{2 \pi x^3}} \exp\biggl(-\frac{(x-\mu)^2}{2x}\biggr).

In this form, the mean and variance of the distribution are equal, $\backslash mathbb\{E\}[X]\; =\; \backslash text\{Var\}(X).$

Also, the cumulative distribution function (cdf) of the single parameter inverse Gaussian distribution is related to the standard normal distribution by

:$$

\begin{align}

\Pr(X < x) &= \Phi(-z_1) + e^{2 \mu} \Phi(-z_2),

\end{align}

where $z\_1\; =\; \backslash frac\{\backslash mu\}\{x^\{1/2\}\}\; -\; x^\{1/2\}$, $z\_2\; =\; \backslash frac\{\backslash mu\}\{x^\{1/2\}\}\; +\; x^\{1/2\},$ and the $\backslash Phi$ is the cdf of standard normal distribution. The variables $z\_1$ and $z\_2$ are related to each other by the identity $z\_2^2\; =\; z\_1^2\; +\; 4\backslash mu.$

In the single parameter form, the MGF simplifies to

:$M(t)\; =\; \backslash exp[\backslash mu(1-\backslash sqrt\{1-2\; t\})].$

An inverse Gaussian distribution in double parameter form $f(x;\backslash mu,\backslash lambda)$ can be transformed into a single parameter form $f(y;\backslash mu\_0,\backslash mu\_0^2)$ by appropriate scaling $y\; =\; \backslash frac\{\backslash mu^2\; x\}\{\backslash lambda\},$ where $\backslash mu\_0\; =\; \backslash mu^3/\backslash lambda.$

The above paragraph can be re-written as: if $Y=\backslash lambda\; X/\backslash mu^2$, then $Y\; \backslash sim\; \backslash operatorname\{IG\}(\backslash lambda/\backslash mu,(\backslash lambda/\backslash mu)^2)${{r|Folks1978}}. This approach is better in the sense that it clearly shows dimensionless nature of the single parameter form (note that $\backslash dim\; \backslash lambda\; =\; \backslash dim\; \backslash mu\; =\; \backslash dim\; x$). This property follows from a more general fact: if $a>0$ and $Y=a\; X$, then $Y\; \backslash sim\; \backslash operatorname\{IG\}(a\backslash mu,a\backslash lambda)${{r|Tweedie1957a}}.

The standard form of inverse Gaussian distribution is

:$f(x;1,1)$

= \frac{1}{\sqrt{2 \pi x^3}} \exp\biggl(-\frac{(x-1)^2}{2x}\biggr).

If X_i has an $\backslash operatorname\{IG\}(\backslash mu\_0\; w\_i,\; \backslash lambda\_0\; w\_i^2\; )\backslash ,\backslash !$ distribution for i = 1, 2, ..., n

and all X_i are independent, then

:$$

S=\sum_{i=1}^n X_i

\sim

\operatorname{IG}\left( \mu_0 \sum w_i, \lambda_0 \left(\sum w_i \right)^2 \right).

Note that

:$$

\frac{\operatorname{Var}(X_i)}{\operatorname{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 distributed.

For any t > 0 it holds that

:$$

X \sim \operatorname{IG}(\mu,\lambda) \,\,\,\,\,\, \Rightarrow \,\,\,\,\,\, tX \sim \operatorname{IG}(t\mu,t\lambda).

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.

For $\backslash lambda>0$ fixed, it is also a single-parameter natural exponential family distribution{{r|Seshadri1999}} where the base distribution has density

: $$

h(x) = \sqrt{ \frac{\lambda}{2\pi x^3} } \exp\left( - \frac{\lambda}{2x} \right) \mathbb{1}_{[0,\infty)}(x)\,.

Indeed, with $\backslash theta\backslash le\; 0$,

:$$

p(x;\theta) = \frac{\exp(\theta x) h(x)}

{\int \exp(\theta y) h(y) dy}

is a density over the reals. Evaluating the integral, we get

:$$

p(x;\theta) =

\sqrt{ \frac{\lambda}{2\pi x^3} } \exp\left( - \frac{\lambda}{2x} +\theta x - \sqrt{-2\lambda\theta} \right) \mathbb{1}_{[0,\infty)}(x)\,.

Substituting $\backslash theta\; =\; -\backslash lambda/(2\backslash mu^2)$ makes the above expression equal to $f(x;\backslash mu,\backslash lambda)$.

File:Inverse gaussian as stopping time of random walk.png

Let the stochastic process X_t be given by

:$X\_0\; =\; 0\backslash quad$

:$X\_t\; =\; \backslash nu\; t\; +\; \backslash sigma\; W\_t\backslash quad\backslash quad\backslash quad\backslash quad$

where W_t is a standard Brownian motion. That is, X_t is a Brownian motion with drift $\backslash nu\; >\; 0$.

Then the first passage time for a fixed level $\backslash alpha\; >\; 0$ by X_t is distributed according to an inverse-Gaussian:

:$$

T_\alpha = \inf\{ t > 0 \mid X_t=\alpha \} \sim \operatorname{IG} \left(\frac\alpha\nu, \left(\frac \alpha \sigma \right)^2 \right)

= \frac{\alpha}{\sigma\sqrt{2 \pi x^3}} \exp\biggl(-\frac{(\alpha-\nu x)^2}{2 \sigma^2 x}\biggr)

i.e

:$$

P(T_{\alpha} \in (T, T + dT)) =

\frac{\alpha}{\sigma\sqrt{2 \pi T^3}} \exp\biggl(-\frac{(\alpha-\nu T)^2}{2 \sigma^2 T}\biggr)dT

(cf. Schrödinger{{r|Schrödinger1915}} equation 19, Smoluchowski{{r|Smoluchowski1915}}, equation 8, and Folks{{r|Folks1978}}, equation 1).

{{Collapse top|title=Derivation of the first passage time distribution}}

Suppose that we have a Brownian motion $X\_\{t\}$ with drift $\backslash nu$ defined by:

:$X\_\{t\}\; =\; \backslash nu\; t\; +\; \backslash sigma\; W\_\{t\},\; \backslash quad\; X(0)\; =\; x\_\{0\}$

And suppose that we wish to find the probability density function for the time when the process first hits some barrier $\backslash alpha\; >\; x\_\{0\}$ - known as the first passage time. The Fokker-Planck equation describing the evolution of the probability distribution $p(t,x)$ is:

:

where $\backslash delta(\backslash cdot)$ is the Dirac delta function. This is a boundary value problem (BVP) with a single absorbing boundary condition $p(t,\backslash alpha)=0$, which may be solved using the method of images. Based on the initial condition, the fundamental solution to the Fokker-Planck equation, denoted by $\backslash varphi(t,x)$, is:

:$\backslash varphi(t,x)\; =\; \{1\backslash over\{\backslash sqrt\{2\backslash pi\; \backslash sigma^\{2\}t\}\}\}\backslash exp\backslash left[\; -\; \{(x-x\_\{0\}-\backslash nu\; t)^\{2\}\backslash over\{2\backslash sigma^\{2\}t\}\}\; \backslash right]$

Define a point $m$, such that $m>\backslash alpha$. This will allow the original and mirror solutions to cancel out exactly at the barrier at each instant in time. This implies that the initial condition should be augmented to become:

:$p(0,x)\; =\; \backslash delta(x-x\_\{0\})\; -\; A\backslash delta(x-m)$

where $A$ is a constant. Due to the linearity of the BVP, the solution to the Fokker-Planck equation with this initial condition is:

:$p(t,x)\; =\; \{1\backslash over\{\backslash sqrt\{2\backslash pi\backslash sigma^\{2\}t\}\}\}\backslash left\backslash \{\; \backslash exp\backslash left[\; -\; \{(x-x\_\{0\}-\backslash nu\; t)^\{2\}\backslash over\{2\backslash sigma^\{2\}t\}\}\; \backslash right\; ]\; -\; A\backslash exp\backslash left[\; -\{(x-m-\backslash nu\; t)^\{2\}\backslash over\{2\backslash sigma^\{2\}t\}\}\; \backslash right\; ]\; \backslash right\backslash \}$

Now we must determine the value of $A$. The fully absorbing boundary condition implies that:

:$(\backslash alpha-x\_\{0\}-\backslash nu\; t)^\{2\}\; =\; -2\backslash sigma^\{2\}t\; \backslash log\; A\; +\; (\backslash alpha\; -\; m\; -\; \backslash nu\; t)^\{2\}$

At $p(0,\backslash alpha)$, we have that $(\backslash alpha-x\_\{0\})^\{2\}\; =\; (\backslash alpha-m)^\{2\}\; \backslash implies\; m\; =\; 2\backslash alpha\; -\; x\_\{0\}$. Substituting this back into the above equation, we find that:

:$A\; =\; e^\{2\backslash nu(\backslash alpha\; -\; x\_\{0\})/\backslash sigma^\{2\}\}$

Therefore, the full solution to the BVP is:

:$p(t,x)\; =\; \{1\backslash over\{\backslash sqrt\{2\backslash pi\backslash sigma^\{2\}t\}\}\}\backslash left\backslash \{\; \backslash exp\backslash left[\; -\; \{(x-x\_\{0\}-\backslash nu\; t)^\{2\}\backslash over\{2\backslash sigma^\{2\}t\}\}\; \backslash right\; ]\; -\; e^\{2\backslash nu(\backslash alpha-x\_\{0\})/\backslash sigma^\{2\}\}\backslash exp\backslash left[\; -\{(x+x\_\{0\}-2\backslash alpha-\backslash nu\; t)^\{2\}\backslash over\{2\backslash sigma^\{2\}t\}\}\; \backslash right\; ]\; \backslash right\backslash \}$

Now that we have the full probability density function, we are ready to find the first passage time distribution $f(t)$. The simplest route is to first compute the survival function $S(t)$, which is defined as:

:

where $\backslash Phi(\backslash cdot)$ is the cumulative distribution function of the standard normal distribution. The survival function gives us the probability that the Brownian motion process has not crossed the barrier $\backslash alpha$ at some time $t$. Finally, the first passage time distribution $f(t)$ is obtained from the identity:

:

Assuming that $x\_\{0\}\; =\; 0$, the first passage time follows an inverse Gaussian distribution:

:$f(t)\; =\; \{\backslash alpha\backslash over\{\backslash sqrt\{2\backslash pi\; \backslash sigma^\{2\}t^\{3\}\}\}\}\; e^\{-(\backslash alpha-\backslash nu\; t)^\{2\}/2\backslash sigma^\{2\}t\}\; \backslash sim\; \backslash text\{IG\}\backslash left[\; \{\backslash alpha\backslash over\{\backslash nu\}\},\backslash left(\; \{\backslash alpha\backslash over\{\backslash sigma\}\}\; \backslash right)^\{2\}\; \backslash right]$

{{Collapse bottom}}

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

: $f\; \backslash left(\; x;\; 0,\; \backslash left(\backslash frac\; \backslash alpha\; \backslash sigma\; \backslash right)^2\; \backslash right)$

= \frac \alpha {\sigma \sqrt{2 \pi x^3}} \exp\left(-\frac{\alpha^2 }{2 \sigma^2 x}\right)

(see also Bachelier{{r|Bachelier1900a|p=74}}{{r|Bachelier1900b|p=39}}). This is a Lévy distribution with parameters $c=\backslash left(\backslash frac\; \backslash alpha\; \backslash sigma\; \backslash right)^2$ and $\backslash mu=0$.

The model where

:$$

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

with all w_i known, (μ, λ) unknown and all X_i independent 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

:$$

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

$\backslash widehat\{\backslash mu\}$ and $\backslash widehat\{\backslash lambda\}$ are independent and

:$$

\widehat{\mu} \sim \operatorname{IG} \left(\mu, \lambda \sum_{i=1}^n w_i \right), \qquad \frac{n}{\widehat{\lambda}} \sim \frac{1}{\lambda} \chi^2_{n-1}.

The following algorithm may be used.{{r|Michael1976}}

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

:$$

\displaystyle \nu \sim N(0,1).

Square the value

:$$

\displaystyle y = \nu^2

and use the 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 uniform distribution between 0 and 1

:$$

\displaystyle z \sim 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;

}

File:Wald Distribution matplotlib.jpg

And to plot Wald distribution in Python using matplotlib and NumPy:

import matplotlib.pyplot as plt

import numpy as np

h = plt.hist(np.random.wald(3, 2, 100000), bins=200, density=True)

plt.show()

• If $X\; \backslash sim\; \backslash operatorname\{IG\}(\backslash mu,\backslash lambda)$, then $k\; X\; \backslash sim\; \backslash operatorname\{IG\}(k\; \backslash mu,k\; \backslash lambda)$ for any number $k\; >\; 0.${{r|Chhikara1989}}
• If $X\_i\; \backslash sim\; \backslash operatorname\{IG\}(\backslash mu,\backslash lambda)\backslash ,$ then $\backslash sum\_\{i=1\}^n\; X\_i\; \backslash sim\; \backslash operatorname\{IG\}(n\; \backslash mu,n^2\; \backslash lambda)\backslash ,$
• If $X\_i\; \backslash sim\; \backslash operatorname\{IG\}(\backslash mu,\backslash lambda)\backslash ,$ for $i=1,\backslash ldots,n\backslash ,$ then $\backslash bar\{X\}\; \backslash sim\; \backslash operatorname\{IG\}(\backslash mu,n\; \backslash lambda)\backslash ,$
• If $X\_i\; \backslash sim\; \backslash operatorname\{IG\}(\backslash mu\_i,2\; \backslash mu^2\_i)\backslash ,$ then $\backslash sum\_\{i=1\}^n\; X\_i\; \backslash sim\; \backslash operatorname\{IG\}\backslash left(\backslash sum\_\{i=1\}^n\; \backslash mu\_i,\; 2\; \backslash left(\; \backslash sum\_\{i=1\}^n\; \backslash mu\_i\; \backslash right)^2\backslash right)\backslash ,$
• If $X\; \backslash sim\; \backslash operatorname\{IG\}(\backslash mu,\backslash lambda)$, then $\backslash lambda\; (X-\backslash mu)^2/\backslash mu^2X\; \backslash sim\; \backslash chi^2(1)$.{{Cite journal | title = On the inverse Gaussian distribution function | journal = Journal of the American Statistical Association | volume = 63 | issue = 4 | pages = 1514–1516 | last1 = Shuster | first1 = J. | year = 1968| doi = 10.1080/01621459.1968.10480942 }}

The convolution of an inverse Gaussian distribution (a Wald distribution) and an exponential (an ex-Wald distribution) is used as a model for response times in psychology,{{r|Schwarz2001}} with visual search as one example.{{Cite journal | doi = 10.1037/a0020747| pmid = 21090905| pmc = 3062635| title = What are the shapes of response time distributions in visual search?| journal = Journal of Experimental Psychology: Human Perception and Performance| volume = 37| issue = 1| pages = 58–71| year = 2011| last1 = Palmer | first1 = E. M. | last2 = Horowitz | first2 = T. S. | last3 = Torralba | first3 = A. | last4 = Wolfe | first4 = J. M. }}

This distribution appears to have been first derived in 1900 by Louis Bachelier{{r|Bachelier1900a|Bachelier1900b}} as the time a stock reaches a certain price for the first time. In 1915 it was used independently by Erwin Schrödinger{{r|Schrödinger1915}} and Marian v. Smoluchowski{{r|Smoluchowski1915}} as the time to first passage of a Brownian motion. In the field of reproduction modeling it is known as the Hadwiger function, after Hugo Hadwiger who described it in 1940.{{cite journal |last=Hadwiger |first=H. |year=1940 |title=Eine analytische Reproduktionsfunktion für biologische Gesamtheiten |journal=Skandinavisk Aktuarietidskrijt |volume=7 |issue=3–4 |pages=101–113 |doi=10.1080/03461238.1940.10404802}} Abraham Wald re-derived this distribution in 1944{{r|Wald1944}} as the limiting form of a sample in a sequential probability ratio test. The name inverse Gaussian was proposed by Maurice Tweedie in 1945.{{Cite journal |last=Tweedie | first=M. C. K. |year=1945 | title=Inverse Statistical Variates |journal=Nature | volume=155 | issue=3937 | pages=453 |doi=10.1038/155453a0 | bibcode=1945Natur.155..453T | s2cid=4113244 | doi-access=free }} Tweedie investigated this distribution in 1956{{cite journal |last=Tweedie |first=M. C. K. |year=1956 |title=Some Statistical Properties of Inverse Gaussian Distributions |journal=Virginia Journal of Science |series=New Series |volume=7 |issue=3 |pages=160–165}} and 1957{{Cite journal |last=Tweedie | first=M. C. K. |year=1957 | title=Statistical Properties of Inverse Gaussian Distributions I |journal=Annals of Mathematical Statistics | volume=28 | issue=2 | pages=362–377 | doi=10.1214/aoms/1177706964 | jstor=2237158 | doi-access=free }}{{Cite journal |last=Tweedie | first=M. C. K. |year=1957 | title=Statistical Properties of Inverse Gaussian Distributions II |journal=Annals of Mathematical Statistics | volume=28 | issue=3 | pages=696–705 | doi=10.1214/aoms/1177706881 | jstor=2237229 | url=http://projecteuclid.org/euclid.aoms/1177706881 | doi-access=free | url-access=subscription }} and established some of its statistical properties. The distribution was extensively reviewed by Folks and Chhikara in 1978.{{r|Folks1978}}

Assuming that the time intervals between occurrences of a random phenomenon follow an inverse Gaussian distribution, the probability distribution for the number of occurrences of this event within a specified time window is referred to as rated inverse Gaussian.Capacity per unit cost-achieving input distribution of rated-inverse gaussian biological neuron M Nasiraee, HM Kordy, J Kazemitabar IEEE Transactions on Communications 70 (6), 3788-3803 While, first and second moment of this distribution are calculated, the derivation of the moment generating function remains an open problem.

Despite the simple formula for the probability density function, numerical probability calculations for the inverse Gaussian distribution nevertheless require special care to achieve full machine accuracy in floating point arithmetic for all parameter values.{{cite journal | last1=Giner | first1=Göknur | last2=Smyth | first2=Gordon |title=statmod: Probability Calculations for the Inverse Gaussian Distribution | journal=The R Journal| volume=8|issue=1|pages=339–351| date=August 2016|url=https://journal.r-project.org/archive/2016-1| doi=10.32614/RJ-2016-024 | arxiv=1603.06687 |doi-access=free}} Functions for the inverse Gaussian distribution are provided for the R programming language by several packages including rmutil,{{cite web |last1=Lindsey|first1=James |title=rmutil: Utilities for Nonlinear Regression and Repeated Measurements Models |url=http://www.commanster.eu/rcode.html |date=2013-09-09 }}{{cite web |last1=Swihart |first1=Bruce |last2=Lindsey|first2=James |title=rmutil: Utilities for Nonlinear Regression and Repeated Measurements Models |url=https://cran.r-project.org/package=rmutil |date=2019-03-04 }} SuppDists,{{cite web |last1=Wheeler |first1=Robert |title=SuppDists: Supplementary Distributions |url=https://cran.r-project.org/package=SuppDists |date=2016-09-23}} STAR,{{cite web |last1=Pouzat |first1=Christophe |title=STAR: Spike Train Analysis with R |url=https://cran.r-project.org/package=STAR |date=2015-02-19}} invGauss,{{cite web |last1=Gjessing |first1=Hakon K. |title=Threshold regression that fits the (randomized drift) inverse Gaussian distribution to survival data |url=https://cran.r-project.org/package=invGauss |date=2014-03-29}} LaplacesDemon,{{cite web |last1=Hall |first1=Byron |last2=Hall |first2=Martina |last3=Statisticat |first3=LLC |last4=Brown |first4=Eric |last5=Hermanson |first5=Richard |last6=Charpentier |first6=Emmanuel |last7=Heck |first7=Daniel |last8=Laurent |first8=Stephane |last9=Gronau |first9=Quentin F. |last10=Singmann |first10=Henrik |title=LaplacesDemon: Complete Environment for Bayesian Inference |url=https://cran.r-project.org/package=LaplacesDemon |date=2014-03-29}} and statmod.{{cite web |last1=Giner |first1=Göknur |last2=Smyth |first2=Gordon |title=statmod: Statistical Modeling |url=https://cran.r-project.org/web/packages/statmod/index.html|date=2017-06-18}}

• Generalized inverse Gaussian distribution
• Tweedie distributions—The inverse Gaussian distribution is a member of the family of Tweedie exponential dispersion models
• Stopping time

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{{ProbDistributions|continuous-semi-infinite}}

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