Rayleigh distribution

Not to be confused with Rayleigh mixture distribution.

Rayleigh

Probability density function
Cumulative distribution function
Parameters	scale: $\sigma >0\,$$\sigma >0\,$
Support	$x\in [0,+\infty )$$x\in [0,+\infty )$
PDF	${\frac {x}{\sigma ^{2}}}e^{-x^{2}/\left(2\sigma ^{2}\right)}$${\frac {x}{\sigma ^{2}}}e^{-x^{2}/\left(2\sigma ^{2}\right)}$
CDF	$1-e^{-x^{2}/\left(2\sigma ^{2}\right)}$$1-e^{-x^{2}/\left(2\sigma ^{2}\right)}$
Quantile	$Q(F;\sigma )=\sigma {\sqrt {-\ln[(1-F)^{2}]}}$$Q(F;\sigma )=\sigma {\sqrt {-\ln[(1-F)^{2}]}}$
Mean	$\sigma {\sqrt {\frac {\pi }{2}}}$$\sigma {\sqrt {\frac {\pi }{2}}}$
Median	$\sigma {\sqrt {2\ln(2)}}\,$$\sigma {\sqrt {2\ln(2)}}\,$
Mode	$\sigma \,$$\sigma \,$
Variance	${\frac {4-\pi }{2}}\sigma ^{2}$${\frac {4-\pi }{2}}\sigma ^{2}$
Skewness	${\frac {2{\sqrt {\pi }}(\pi -3)}{(4-\pi )^{3/2}}}$${\frac {2{\sqrt {\pi }}(\pi -3)}{(4-\pi )^{3/2}}}$
Ex. kurtosis	$-{\frac {6\pi ^{2}-24\pi +16}{(4-\pi )^{2}}}$$-{\frac {6\pi ^{2}-24\pi +16}{(4-\pi )^{2}}}$
Entropy	$1+\ln \left({\frac {\sigma }{\sqrt {2}}}\right)+{\frac {\gamma }{2}}$$1+\ln \left({\frac {\sigma }{\sqrt {2}}}\right)+{\frac {\gamma }{2}}$
MGF	$1+\sigma t\,e^{\sigma ^{2}t^{2}/2}{\sqrt {\frac {\pi }{2}}}\left({\textrm {erf}}\left({\frac {\sigma t}{\sqrt {2}}}\right)\!+\!1\right)$$1+\sigma t\,e^{\sigma ^{2}t^{2}/2}{\sqrt {\frac {\pi }{2}}}\left({\textrm {erf}}\left({\frac {\sigma t}{\sqrt {2}}}\right)\!+\!1\right)$
CF	$1\!-\!\sigma te^{-\sigma ^{2}t^{2}/2}{\sqrt {\frac {\pi }{2}}}\!\left({\textrm {erfi}}\!\left({\frac {\sigma t}{\sqrt {2}}}\right)\!-\!i\right)$$1\!-\!\sigma te^{-\sigma ^{2}t^{2}/2}{\sqrt {\frac {\pi }{2}}}\!\left({\textrm {erfi}}\!\left({\frac {\sigma t}{\sqrt {2}}}\right)\!-\!i\right)$

In probability theory and statistics, the Rayleigh distribution /ˈreɪli/ is a continuous probability distribution for positive-valued random variables.

A Rayleigh distribution is often observed when the overall magnitude of a vector is related to its directional components. One example where the Rayleigh distribution naturally arises is when wind velocity is analyzed into its orthogonal 2-dimensional vector components. Assuming that each component is uncorrelated, normally distributed with equal variance, and zero mean, then the overall wind speed (vector magnitude) will be characterized by a Rayleigh distribution. A second example of the distribution arises in the case of random complex numbers whose real and imaginary components are independently and identically distributed Gaussian with equal variance and zero mean. In that case, the absolute value of the complex number is Rayleigh-distributed.

The distribution is named after Lord Rayleigh

Definition

The probability density function of the Rayleigh distribution is

$f(x;\sigma )={\frac {x}{\sigma ^{2}}}e^{-x^{2}/(2\sigma ^{2})},\quad x\geq 0,$$f(x;\sigma )={\frac {x}{\sigma ^{2}}}e^{-x^{2}/(2\sigma ^{2})},\quad x\geq 0,$

where $\sigma$$\sigma$ is the scale parameter of the distribution. The cumulative distribution function is

$F(x;\sigma )=1-e^{-x^{2}/(2\sigma ^{2})}$$F(x;\sigma )=1-e^{-x^{2}/(2\sigma ^{2})}$

for $x\in [0,\infty ).$$x\in [0,\infty ).$

Relation to random vector length

Consider the two-dimensional vector $Y=(U,V)$$Y=(U,V)$ which has components that are Gaussian-distributed, centered at zero, and independent. Then $f_{U}(u;\sigma )={\frac {e^{-u^{2}/\left(2\sigma ^{2}\right)}}{\sqrt {2\pi \sigma ^{2}}}}$$f_{U}(u;\sigma )={\frac {e^{-u^{2}/\left(2\sigma ^{2}\right)}}{\sqrt {2\pi \sigma ^{2}}}}$, and similarly for $f_{V}(v;\sigma )$$f_{V}(v;\sigma )$.

Let $x$$x$ be the length of $Y$$Y$. It is distributed as

$f(x;\sigma )={\frac {1}{2\pi \sigma ^{2}}}\int _{-\infty }^{\infty }du\,\int _{-\infty }^{\infty }dv\,e^{-u^{2}/\left(2\sigma ^{2}\right)}e^{-v^{2}/\left(2\sigma ^{2}\right)}\delta (x-{\sqrt {u^{2}+v^{2}}}).$$f(x;\sigma )={\frac {1}{2\pi \sigma ^{2}}}\int _{-\infty }^{\infty }du\,\int _{-\infty }^{\infty }dv\,e^{-u^{2}/\left(2\sigma ^{2}\right)}e^{-v^{2}/\left(2\sigma ^{2}\right)}\delta (x-{\sqrt {u^{2}+v^{2}}}).$

By transforming to the polar coordinate system one has

$f(x;\sigma )={\frac {1}{2\pi \sigma ^{2}}}\int _{0}^{2\pi }\,d\phi \int _{0}^{\infty }dr\,\delta (r-x)re^{-r^{2}/\left(2\sigma ^{2}\right)}={\frac {x}{\sigma ^{2}}}e^{-x^{2}/\left(2\sigma ^{2}\right)},$$f(x;\sigma )={\frac {1}{2\pi \sigma ^{2}}}\int _{0}^{2\pi }\,d\phi \int _{0}^{\infty }dr\,\delta (r-x)re^{-r^{2}/\left(2\sigma ^{2}\right)}={\frac {x}{\sigma ^{2}}}e^{-x^{2}/\left(2\sigma ^{2}\right)},$

which is the Rayleigh distribution. It is straightforward to generalize to vectors of dimension other than 2. There are also generalizations when the components have unequal variance or correlations.

Properties

The raw moments are given by:

$\mu _{k}=\sigma ^{k}2^{\frac {k}{2}}\,\Gamma \left(1+{\frac {k}{2}}\right)$$\mu _{k}=\sigma ^{k}2^{\frac {k}{2}}\,\Gamma \left(1+{\frac {k}{2}}\right)$

where $\Gamma (z)$$\Gamma (z)$ is the Gamma function.

The mean and variance of a Rayleigh random variable may be expressed as:

$\mu (X)=\sigma {\sqrt {\frac {\pi }{2}}}\ \approx 1.253\sigma$$\mu (X)=\sigma {\sqrt {\frac {\pi }{2}}}\ \approx 1.253\sigma$

and

${\textrm {var}}(X)={\frac {4-\pi }{2}}\sigma ^{2}\approx 0.429\sigma ^{2}$${\textrm {var}}(X)={\frac {4-\pi }{2}}\sigma ^{2}\approx 0.429\sigma ^{2}$

The mode is $\sigma$$\sigma$ and the maximum pdf is

$f_{\text{max}}=f(\sigma ;\sigma )={\frac {1}{\sigma }}e^{-{\frac {1}{2}}}\approx {\frac {1}{\sigma }}0.606$$f_{\text{max}}=f(\sigma ;\sigma )={\frac {1}{\sigma }}e^{-{\frac {1}{2}}}\approx {\frac {1}{\sigma }}0.606$

The skewness is given by:

$\gamma _{1}={\frac {2{\sqrt {\pi }}(\pi -3)}{(4-\pi )^{\frac {3}{2}}}}\approx 0.631$$\gamma _{1}={\frac {2{\sqrt {\pi }}(\pi -3)}{(4-\pi )^{\frac {3}{2}}}}\approx 0.631$

The excess kurtosis is given by:

$\gamma _{2}=-{\frac {6\pi ^{2}-24\pi +16}{(4-\pi )^{2}}}\approx 0.245$$\gamma _{2}=-{\frac {6\pi ^{2}-24\pi +16}{(4-\pi )^{2}}}\approx 0.245$

The characteristic function is given by:

$\varphi (t)=1-\sigma te^{-{\frac {1}{2}}\sigma ^{2}t^{2}}{\sqrt {\frac {\pi }{2}}}\left[{\textrm {erfi}}\left({\frac {\sigma t}{\sqrt {2}}}\right)-i\right]$$\varphi (t)=1-\sigma te^{-{\frac {1}{2}}\sigma ^{2}t^{2}}{\sqrt {\frac {\pi }{2}}}\left[{\textrm {erfi}}\left({\frac {\sigma t}{\sqrt {2}}}\right)-i\right]$

where $\operatorname {erfi} (z)$$\operatorname {erfi} (z)$ is the imaginary error function. The moment generating function is given by

$M(t)=1+\sigma t\,e^{{\frac {1}{2}}\sigma ^{2}t^{2}}{\sqrt {\frac {\pi }{2}}}\left[{\textrm {erf}}\left({\frac {\sigma t}{\sqrt {2}}}\right)+1\right]$$M(t)=1+\sigma t\,e^{{\frac {1}{2}}\sigma ^{2}t^{2}}{\sqrt {\frac {\pi }{2}}}\left[{\textrm {erf}}\left({\frac {\sigma t}{\sqrt {2}}}\right)+1\right]$

where $\operatorname {erf} (z)$$\operatorname {erf} (z)$ is the error function.

Differential entropy

The differential entropy is given by

$H=1+\ln \left({\frac {\sigma }{\sqrt {2}}}\right)+{\frac {\gamma }{2}}$$H=1+\ln \left({\frac {\sigma }{\sqrt {2}}}\right)+{\frac {\gamma }{2}}$

where $\gamma$$\gamma$ is the Euler–Mascheroni constant.

Differential equation

The pdf of the Rayleigh distribution is a solution of the following differential equation:

$\left\{{\begin{array}{l}\sigma ^{2}xf'(x)+f(x)\left(x^{2}-\sigma ^{2}\right)=0\\[10pt]f(1)={\frac {\exp \left(-{\frac {1}{2\sigma ^{2}}}\right)}{\sigma ^{2}}}\end{array}}\right\}$$\left\{{\begin{array}{l}\sigma ^{2}xf'(x)+f(x)\left(x^{2}-\sigma ^{2}\right)=0\\[10pt]f(1)={\frac {\exp \left(-{\frac {1}{2\sigma ^{2}}}\right)}{\sigma ^{2}}}\end{array}}\right\}$

Parameter estimation

Given a sample of N independent and identically distributed Rayleigh random variables $x_{i}$$x_{i}$ with parameter $\sigma$$\sigma$,

${\widehat {\sigma ^{2}}}\approx \!\,{\frac {1}{2N}}\sum _{i=1}^{N}x_{i}^{2}$${\widehat {\sigma ^{2}}}\approx \!\,{\frac {1}{2N}}\sum _{i=1}^{N}x_{i}^{2}$ is an unbiased maximum likelihood estimate.

${\hat {\sigma }}\approx \!\,{\sqrt {{\frac {1}{2N}}\sum _{i=1}^{N}x_{i}^{2}}}$${\hat {\sigma }}\approx \!\,{\sqrt {{\frac {1}{2N}}\sum _{i=1}^{N}x_{i}^{2}}}$ is a biased estimator that can be corrected via the formula

$\sigma ={\hat {\sigma }}{\frac {\Gamma (N){\sqrt {N}}}{\Gamma (N+{\frac {1}{2}})}}={\hat {\sigma }}{\frac {4^{N}N!(N-1)!{\sqrt {N}}}{(2N)!{\sqrt {\pi }}}}$$\sigma ={\hat {\sigma }}{\frac {\Gamma (N){\sqrt {N}}}{\Gamma (N+{\frac {1}{2}})}}={\hat {\sigma }}{\frac {4^{N}N!(N-1)!{\sqrt {N}}}{(2N)!{\sqrt {\pi }}}}$

Confidence intervals

To find the (1 − α) confidence interval, first find the two numbers $\chi _{1}^{2},\ \chi _{2}^{2}$$\chi _{1}^{2},\ \chi _{2}^{2}$ where:

  $Pr(\chi ^{2}(2N)\leq \chi _{1}^{2})=\alpha /2,\quad Pr(\chi ^{2}(2N)\leq \chi _{2}^{2})=1-\alpha /2$$Pr(\chi ^{2}(2N)\leq \chi _{1}^{2})=\alpha /2,\quad Pr(\chi ^{2}(2N)\leq \chi _{2}^{2})=1-\alpha /2$

then

  ${\frac {N{\overline {x^{2}}}}{\chi _{2}^{2}}}\leq {\widehat {\sigma }}^{2}\leq {\frac {N{\overline {x^{2}}}}{\chi _{1}^{2}}}$${\frac {N{\overline {x^{2}}}}{\chi _{2}^{2}}}\leq {\widehat {\sigma }}^{2}\leq {\frac {N{\overline {x^{2}}}}{\chi _{1}^{2}}}$

Generating random variates

Given a random variate U drawn from the uniform distribution in the interval (0, 1), then the variate

$X=\sigma {\sqrt {-2\ln(1-U)}}\,$$X=\sigma {\sqrt {-2\ln(1-U)}}\,$

has a Rayleigh distribution with parameter $\sigma$$\sigma$. This is obtained by applying the inverse transform sampling-method.

Related distributions

• $R\sim \mathrm {Rayleigh} (\sigma )$$R\sim \mathrm {Rayleigh} (\sigma )$ is Rayleigh distributed if $R={\sqrt {X^{2}+Y^{2}}}$$R={\sqrt {X^{2}+Y^{2}}}$, where $X\sim N(0,\sigma ^{2})$$X\sim N(0,\sigma ^{2})$ and $Y\sim N(0,\sigma ^{2})$$Y\sim N(0,\sigma ^{2})$ are independent normal random variables. (This gives motivation to the use of the symbol "sigma" in the above parametrization of the Rayleigh density.)
• The chi distribution with v = 2 is equivalent to the Rayleigh Distribution with σ = 1. I.e., if $R\sim \mathrm {Rayleigh} (1)$$R\sim \mathrm {Rayleigh} (1)$, then $R^{2}$$R^{2}$ has a chi-squared distribution with parameter $N$$N$, degrees of freedom, equal to two (N = 2)

$[Q=R^{2}]\sim \chi ^{2}(N)\ .$$[Q=R^{2}]\sim \chi ^{2}(N)\ .$

• If $R\sim \mathrm {Rayleigh} (\sigma )$$R\sim \mathrm {Rayleigh} (\sigma )$, then $\sum _{i=1}^{N}R_{i}^{2}$$\sum _{i=1}^{N}R_{i}^{2}$ has a gamma distribution with parameters $N$$N$ and $2\sigma ^{2}$$2\sigma ^{2}$

$\left[Y=\sum _{i=1}^{N}R_{i}^{2}\right]\sim \Gamma (N,2\sigma ^{2}).$$\left[Y=\sum _{i=1}^{N}R_{i}^{2}\right]\sim \Gamma (N,2\sigma ^{2}).$

• The Rice distribution is a generalization of the Rayleigh distribution.
• The Weibull distribution is a generalization of the Rayleigh distribution. In this instance, parameter $\sigma$$\sigma$ is related to the Weibull scale parameter $\lambda$$\lambda$: $\lambda =\sigma {\sqrt {2}}.$$\lambda =\sigma {\sqrt {2}}.$
• The Maxwell–Boltzmann distribution describes the magnitude of a normal vector in three dimensions.
• If $X$$X$ has an exponential distribution $X\sim \mathrm {Exponential} (\lambda )$$X\sim \mathrm {Exponential} (\lambda )$, then $Y={\sqrt {X}}\sim \mathrm {Rayleigh} (1/{\sqrt {2\lambda }}).$$Y={\sqrt {X}}\sim \mathrm {Rayleigh} (1/{\sqrt {2\lambda }}).$

Applications

An application of the estimation of σ can be found in magnetic resonance imaging (MRI). As MRI images are recorded as complex images but most often viewed as magnitude images, the background data is Rayleigh distributed. Hence, the above formula can be used to estimate the noise variance in an MRI image from background data.

Proof of correctness – Unequal variances

Click [show] to expand

We start with

$f(x;\sigma )={\frac {1}{2\pi \sigma _{1}\sigma _{2}}}\int _{-\infty }^{\infty }du\,\int _{-\infty }^{\infty }dv\,e^{-u^{2}/\left(2\sigma _{1}^{2}\right)}e^{-v^{2}/\left(2\sigma _{2}^{2}\right)}\delta (x-{\sqrt {u^{2}+v^{2}}}).$$f(x;\sigma )={\frac {1}{2\pi \sigma _{1}\sigma _{2}}}\int _{-\infty }^{\infty }du\,\int _{-\infty }^{\infty }dv\,e^{-u^{2}/\left(2\sigma _{1}^{2}\right)}e^{-v^{2}/\left(2\sigma _{2}^{2}\right)}\delta (x-{\sqrt {u^{2}+v^{2}}}).$

as above, except with $\sigma _{1}$$\sigma _{1}$ and $\sigma _{2}$$\sigma _{2}$ distinct.

Let $a=u\sigma _{2}/\sigma _{1}$$a=u\sigma _{2}/\sigma _{1}$ so that $a/\sigma _{2}=u/\sigma _{1}$$a/\sigma _{2}=u/\sigma _{1}$, and differentiating we have:

${\frac {\sigma _{1}}{\sigma _{2}}}\mathrm {d} a=\mathrm {d} u$${\frac {\sigma _{1}}{\sigma _{2}}}\mathrm {d} a=\mathrm {d} u$

Substituting,

$f(x;\sigma )={\frac {\sigma _{1}}{\sigma _{2}}}{\frac {1}{2\pi \sigma _{1}\sigma _{2}}}\int _{-\infty }^{\infty }da\,\int _{-\infty }^{\infty }dv\,e^{-a^{2}/\left(2\sigma _{2}^{2}\right)}e^{-v^{2}/\left(2\sigma _{2}^{2}\right)}\delta \left(x-{\sqrt {\left({\frac {\sigma _{1}}{\sigma _{2}}}a\right)^{2}+v^{2}}}\right)$$f(x;\sigma )={\frac {\sigma _{1}}{\sigma _{2}}}{\frac {1}{2\pi \sigma _{1}\sigma _{2}}}\int _{-\infty }^{\infty }da\,\int _{-\infty }^{\infty }dv\,e^{-a^{2}/\left(2\sigma _{2}^{2}\right)}e^{-v^{2}/\left(2\sigma _{2}^{2}\right)}\delta \left(x-{\sqrt {\left({\frac {\sigma _{1}}{\sigma _{2}}}a\right)^{2}+v^{2}}}\right)$

As before, we perform a polar coordinate transformation:

$\left\{{\begin{array}{l}a=r{\textrm {cos}}(\phi )\\v=r{\textrm {sin}}(\phi )\\{\textrm {d}}a{\textrm {d}}v=r{\textrm {d}}r{\textrm {d}}\phi \end{array}}\right.$$\left\{{\begin{array}{l}a=r{\textrm {cos}}(\phi )\\v=r{\textrm {sin}}(\phi )\\{\textrm {d}}a{\textrm {d}}v=r{\textrm {d}}r{\textrm {d}}\phi \end{array}}\right.$

Substituting,

$f(x;\sigma )={\frac {\sigma _{1}}{\sigma _{2}}}{\frac {1}{2\pi \sigma _{1}\sigma _{2}}}\int _{0}^{2\pi }{\textrm {d}}\phi \,\int _{0}^{\infty }r{\textrm {d}}r\,e^{-r^{2}/\left(2\sigma _{2}^{2}\right)}\delta \left(x-{\sqrt {\left({\frac {\sigma _{1}}{\sigma _{2}}}a\right)^{2}+v^{2}}}\right).$$f(x;\sigma )={\frac {\sigma _{1}}{\sigma _{2}}}{\frac {1}{2\pi \sigma _{1}\sigma _{2}}}\int _{0}^{2\pi }{\textrm {d}}\phi \,\int _{0}^{\infty }r{\textrm {d}}r\,e^{-r^{2}/\left(2\sigma _{2}^{2}\right)}\delta \left(x-{\sqrt {\left({\frac {\sigma _{1}}{\sigma _{2}}}a\right)^{2}+v^{2}}}\right).$

Simplifying,

$f(x;\sigma )={\frac {1}{2\pi \sigma _{2}^{2}}}\int _{0}^{2\pi }{\textrm {d}}\phi \,\int _{0}^{\infty }r{\textrm {d}}r\,e^{-r^{2}/\left(2\sigma _{2}^{2}\right)}\delta \left(x-{\sqrt {\left({\frac {\sigma _{1}}{\sigma _{2}}}a\right)^{2}+v^{2}}}\right).$$f(x;\sigma )={\frac {1}{2\pi \sigma _{2}^{2}}}\int _{0}^{2\pi }{\textrm {d}}\phi \,\int _{0}^{\infty }r{\textrm {d}}r\,e^{-r^{2}/\left(2\sigma _{2}^{2}\right)}\delta \left(x-{\sqrt {\left({\frac {\sigma _{1}}{\sigma _{2}}}a\right)^{2}+v^{2}}}\right).$

See Hoyt distribution for more information.

See also

• Normal distribution
• Rayleigh fading
• Rayleigh mixture distribution
• Circular error probable