Rayleigh Distribution Calculator
Written by Thierno Sadou Diallo, formula verified per our methodology • Last checked on 9/10/2026
The Rayleigh distribution has density f(x) = (x/σ²) × exp(−x²/(2σ²)) and cumulative distribution function F(x) = 1 − exp(−x²/(2σ²)), with σ the scale parameter and x≥0. It typically describes the magnitude of a vector with two independent Gaussian components, such as the horizontal wind speed or the amplitude of a radio signal subject to Gaussian noise.
Explanation
The Rayleigh distribution appears naturally whenever you are interested in the MAGNITUDE (the norm) of a two-dimensional vector whose each component independently follows a centered normal distribution with the same variance: this is exactly the case of the horizontal wind speed (the resultant of a North-South component and an East-West component, each approximately Gaussian), or the amplitude of a radio signal passing through a complex urban environment where multiple reflections produce Gaussian noise on its in-phase and quadrature components — a phenomenon known in telecommunications as "Rayleigh fading", which explains why the reception of a mobile signal fluctuates characteristically as you move. This distribution is in fact a special case of the Weibull distribution already on this site: by setting the Weibull shape parameter to exactly k=2 and its scale parameter to λ=σ√2, you recover exactly the Rayleigh distribution — a relationship verified numerically with perfect agreement between the two calculators. The Rayleigh parameter σ also corresponds to the MODE of the distribution (the most probable value), a property that distinguishes it from many other distributions where the scale parameter does not directly coincide with such an easily interpretable notable point.
Example: at the mode of the distribution
Inputs
Value x = 2. Scale parameter σ = 2 (so x=σ, exactly at the mode).
Calculation
f(2) = (2÷4) × exp(−4÷8) = 0.5 × exp(−0.5) ≈ 0.5 × 0.6065 ≈ 0.3033. F(2) = 1 − exp(−0.5) ≈ 1 − 0.6065 ≈ 0.3935.
Result
At the point x=σ, the density reaches its maximum value (≈0.3033) and about 39.3% of the probability mass is already below this value.
Frequently asked questions
Why does wind speed approximately follow a Rayleigh distribution?
Horizontal wind speed results from two independent components (generally modeled as approximately Gaussian and centered): a North-South component and an East-West component. The norm of this speed vector (what you actually measure with an anemometer, independent of direction) then naturally follows a Rayleigh distribution — a model widely used in meteorology and in assessing a site's wind potential, although a two-parameter Weibull distribution (more flexible) is sometimes preferred for a more precise fit to real data.
Why does the parameter σ correspond exactly to the mode of the distribution?
It is a remarkable mathematical property of this specific distribution, proved by finding where the derivative of the density vanishes: this search gives x=σ directly as the only positive solution. Concretely, this means that if you measure the σ parameter of a Rayleigh distribution, you immediately know the most frequently observed value, with no further calculation — a notable practical advantage over distributions where the link between parameters and notable points (mode, median, mean) is less direct.
What is the difference from the Weibull distribution, of which it is a special case?
The Weibull distribution has two independent parameters (shape k and scale λ), which makes it much more flexible for fitting real data; the Rayleigh distribution is the special case where k is fixed at exactly 2, leaving only a single scale parameter σ to fit. In practice, you use the Rayleigh distribution when the underlying theory (norm of two independent Gaussian components) precisely justifies this value k=2, and the full Weibull when you prefer to let the data freely determine the shape of the distribution.