Rayleigh Distribution Calculator
Rayleigh distribution CDF and density for a scale σ and value x: the chance a magnitude built from two zero-mean normal components stays at or below x.
Rayleigh Distribution Calculator
Background.
Take an arrow whose horizontal and vertical parts are independent random errors, each normally distributed around zero with the same spread, and measure its length. That length follows a Rayleigh distribution, and the situation turns up more often than it sounds: a position fix scattered evenly around the true point, the envelope of narrowband radio noise, a signal reaching a receiver only by scattered paths with no direct line of sight. Each is a magnitude assembled from two Gaussian parts, and each is described, at least to a first approximation, by the same one-parameter curve.
That parameter, σ, is the standard deviation of each underlying component, and everything else follows from it. The most likely value is σ itself, the median is about 1.1774σ and the mean about 1.2533σ. The right tail is long but thins quickly: the chance of exceeding 3σ is 1.11%.
Two numbers come back for a given σ and value x. The cumulative probability F(x) = 1 − exp(−x²/2σ²) is the share of outcomes at or below x — the probability that a position error stays inside a circle of radius x, say. The density f(x) is the height of the curve at x. It is useful for plotting, or for comparing how likely values near different points are, but it is not itself a probability.
The calculation takes the Rayleigh model as given. It does not fit σ to data, test whether your measurements follow the distribution, or allow for a steady underlying offset. When the two components have a nonzero mean, as with a strong direct signal path or a biased sensor, the magnitude follows a Rice distribution instead; when the tail shape needs to be fitted freely, the Weibull distribution, of which Rayleigh is the special case with shape parameter 2, is the usual next step.
What is rayleigh distribution calculator?
The Rayleigh distribution is a continuous probability distribution for non-negative values with a single scale parameter σ > 0. If X and Y are independent normal variables with mean zero and standard deviation σ, the distance √(X² + Y²) is Rayleigh-distributed. Its cumulative distribution function is F(x) = 1 − exp(−x²/(2σ²)) and its probability density is f(x) = (x/σ²)·exp(−x²/(2σ²)), both defined for x ≥ 0.
How to use this calculator.
- Enter the scale σ: the standard deviation of each of the two underlying components, not of the magnitude itself. If you know only the mean magnitude, σ = mean × √(2/π), about 0.7979 × the mean.
- Enter the value x in the same unit as σ. Zero is allowed and gives a cumulative probability of 0.
- Read the cumulative probability as P(magnitude ≤ x). Subtract it from 1 for the chance of exceeding x.
- Read the density in reciprocal units of x, per metre if x is in metres, and use it for the shape of the curve rather than as a probability.
The formula.
F(x) = 1 − exp(−x² / (2σ²)) and f(x) = (x / σ²)·exp(−x² / (2σ²)). Both depend on x only through the ratio x/σ, so the scale stretches the curve without changing its shape: doubling σ and x together leaves F unchanged. Because the exponent is squared in x, the probability of exceeding x, exp(−x²/2σ²), falls away much faster than an exponential tail. Setting F equal to a probability p and solving gives the quantile x = σ√(−2 ln(1 − p)). With p = 0.5 that is the median, σ√(2 ln 2) ≈ 1.1774σ, which navigation calls the circular error probable. The density peaks at x = σ, where it equals e^(−1/2)/σ.
A worked example.
A receiver's horizontal position error has independent east and north components, each normal with mean zero and σ = 2 m. How often does a fix land within 3 m of the true point? Square the radius: 3² = 9. Double the variance: 2 × 2² = 8. Their ratio is 9 ÷ 8 = 1.125, and exp(−1.125) = 0.3247 is the chance of landing outside the circle. So F(3) = 1 − 0.3247 = 0.6753, or 0.6753 to the engine's twelve significant figures: about two fixes in three fall within 3 m. The density there is 3 ÷ 2² = 0.75, times 0.3247, giving 0.2435 per metre. Widening the circle to 4 m raises the share to 1 − exp(−2) = 0.8647, while the radius that captures exactly half the fixes is 2 × 1.1774 = 2.3548 m.
Frequently asked questions.
Is σ the standard deviation of the Rayleigh values themselves?
How can I estimate σ from a set of measured magnitudes?
When should I use a Rice or Weibull distribution instead?
If the density is not a probability, what does 0.2435 per metre tell me?
What is the circular error probable, and how does it relate to σ?
References& sources.
- [1]NIST/SEMATECH e-Handbook, Rayleigh Distribution.
- [2]Papoulis and Pillai, Probability, Random Variables, and Stochastic Processes, 4th ed. (PRINT).
- [3]BIPM, The International System of Units (SI Brochure), 9th ed., version 3.01, coherent derived units and quantity equations.
How this page was produced
- Published by
- Quanta Calculator
- Overseen by
- Daniel, Quanta Calculator Researcher
- Primary sources
- 3 cited below
- Method
- F(x)=1-exp[-x²/(2σ²)]
- Published
- Last verified
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