Continuous probability distribution

A real probability distribution with a continuous cumulative distribution function assigns probability zero to each individual value. Here this is the broad meaning of continuous probability distribution. Many elementary examples are more specifically absolutely continuous probability distributions, described by a probability density function with respect to Lebesgue measure. Some texts use "continuous" for that narrower class. The Cantor distribution shows why the distinction matters: its cumulative distribution function is continuous, yet no probability density function with respect to Lebesgue measure represents its whole law. ^[ref].

For scale $a>0$ and shape $\kappa >0$, the log-logistic distribution has the displayed cumulative distribution function for $x\ge 0$, with density $(\kappa /a)(x/a{)}^{\kappa -1}/[1+(x/a{)}^{\kappa }{]}^2$ for $x>0$. Its log-odds are affine in $\log x$, giving the name and motivating a log-logistic probability plot. In the exam's parameterization, $a=\alpha$ and $\kappa =\beta -1$.

For $F(x)=1-[1+(x/\alpha {)}^{\beta -1}{]}^{-1}$, plotting the logits of $p_i=i/(n+1)$ against sorted log observations gives an approximately straight line under the model. The probability integral transform makes $F(X_{(i)})$ a uniform order statistic with expectation $i/(n+1)$, motivating these plotting positions. With those axes the slope is $\beta -1$ and the intercept is $-(\beta -1)\log \alpha$.

The unit-radius Wigner semicircle distribution has the displayed probability density function. Its graph is a rescaled upper semicircle. The area of the unit semicircle is $\pi /2$, so the factor $2/\pi$ normalizes it. Symmetry gives mean zero, and the substitution $x=\sin \theta$ gives variance $1/4$.

A scaled radius-$R$ version has density $2\sqrt{R^2-x^2}/(\pi R^2)$ on $(-R,R)$ and variance $R^2/4$. Rejection sampling from a uniform proposal on $(-1,1)$ accepts a proposed $x$ with probability $\sqrt{1-x^2}$; its average acceptance probability is $\pi /4$. Thus the number of proposals for one output has geometric distribution with expectation $4/\pi$. This law also describes the limiting empirical eigenvalue density in standard ensembles of large symmetric random matrices.

A probability distribution is singular continuous if it has no point masses and assigns probability one to a set of zero Lebesgue measure. Its cumulative distribution function is continuous, but the distribution has no probability density function with respect to Lebesgue measure. This distinguishes continuity of the cumulative distribution function from an absolutely continuous probability distribution, whose probabilities are integrals of a probability density function.

For $0<p<1$, $\beta >0$ and $x>0$, this probability density function has cumulative distribution function $F(x)=1-\log (1-(1-p)e^{-\beta x})/\log p$. Differentiating gives the density, while the endpoint limits are zero and one. Its quantile function is $F^{-1}(u)={\beta }^{-1}[\log (1-p)-\log (1-p^{1-u})]$ for $0<u<1$, enabling inverse transform sampling.

Expanding the denominator as a geometric series givesTermwise integration, justified by the monotone convergence theorem, and the gamma function integral yield raw moments $\mathbb{E}[X^j]=-j!{\mathrm{Li}}_{j+1}(1-p)/({\beta }^j\log p)$. Thus the polylogarithm expresses its mean, variance and higher moments. Here $p$ is the complement of the series parameter $1-p$; parameter conventions must be checked when comparing formulas.

The Cantor distribution is the law of a random variablewhere the $B_n$ are independent Bernoulli random variables taking values zero and one with equal probability. Its values lie in the Cantor set, and its cumulative distribution function is the Cantor function. Every individual value has probability zero, but the Cantor set has probability one despite having zero Lebesgue measure. A probability density function on that set would integrate to zero, so it cannot describe this law. The example motivates the distinction between a continuous probability distribution and an absolutely continuous probability distribution.

For two independent random variables taking values in a three-dimensional ball with constant volume probability density, the density of their distance is the displayed expression for $0\le r\le 2R$, and zero otherwise. Both independence and the common radius $R>0$ are essential. To derive it, put $u=|{\mathbf{r}}_1|$, $v=|{\mathbf{r}}_2|$ and integrate the product volume differential form over orientations. The law of cosines gives the angular Jacobian determinant $v/(ur)$, leaving $f_R(r)=9r\int uvdudv/(2R^6)$ over $0\le u,v\le R$, $|u-v|\le r\le u+v$. With $x=u+v$, $y=u-v$, the integrand becomes $(x^2-y^2)dxdy/8$. Direct integration gives $\int uvdudv=2R^{3}r/3-R^2{r}^2/2+r^4/24$. An independent geometric derivation uses the overlap volume of two radius-$R$ balls displaced by $r$, namely $\pi (4R+r)(2R-r{)}^2/12$, multiplied by $4\pi r^2$ and divided by the square of $4\pi R^3/3$. The density integrates to one and has mean $36R/35$.

For $a,b>0$, this probability density function is obtained from a beta distribution variable $U$ by $X=U/(1-U)$. Indeed $u=x/(1+x)$ and $du/dx=(1+x{)}^{-2}$, and the change of variables formula gives the displayed density. In particular, the ratio of two independent equal-rate exponential distributions has $a=b=1$, density $(1+x{)}^{-2}$ and cumulative distribution function $x/(1+x)$.

A triangular probability density function rises linearly from zero at $a$ to its mode $c$, then falls linearly to zero at $b$. Its peak is $2/(b-a)$. Its inverse transform sampling rule is $a+\sqrt{u(b-a)(c-a)}$ for $u\le (c-a)/(b-a)$, and $b-\sqrt{(1-u)(b-a)(b-c)}$ otherwise.

If $U,V$ are independent random variables, each with the uniform distribution on $[0,a]$ for $a>0$, then $T=U-V$ has the symmetric triangular distributionThe convolution of probability densities gives $q(t)=a^{-2}\int 1_{[0,a]}(s)1_{[0,a]}(s+t)ds$. The overlap of these two intervals has length $a-|t|$ for $|t|<a$ and is empty otherwise, proving the formula. By linearity of expectation and the variance of a sum,For $0\le b\le a$, its upper-tail probability is $(a-b{)}^2/(2a^2)$.

A useful model for elapsed time measured from two independently truncated clock readings uses $a$ as the display resolution and $T$ as the endpoint-offset difference. Independence is essential: uniform marginal probability distributions alone do not imply this triangular distribution. For example $U=V$ gives $T=0$ with certainty. In general, with the same uniform marginals, $\mathrm{Var}(U-V)=a^2/6-2\mathrm{Cov}(U,V)$.

The standard arcsine distribution is a continuous probability distribution on $[0,1]$ with probability density functionIt is the beta distribution with both parameters equal to $1/2$. The density grows without bound toward both endpoints, yet neither endpoint has any probability mass. Its name comes from the inverse sine in its cumulative distribution function.^[ref]

A useful way to see the shape is to choose a point uniformly around a circle and project it onto a diameter. Equal angular intervals do not give equal horizontal distances: the projection moves slowly near either end. An ideal harmonic oscillator sampled at a uniformly chosen time in one complete period has the same endpoint concentration after rescaling its position. The distribution translates a simple geometric slowdown into a precise probability calculation.

For an original comparison, take equal-length regions: the middle half $[1/4,3/4]$, and the two outside quarters together. ThenA uniform distribution on $[0,1]$ instead assigns each region probability $1/2$. Both distributions have mean $1/2$; the mean alone hides their different concentration.

Choose $\Theta$ with uniform distribution on $(0,\pi /2)$ and set $X={\sin }^2\Theta$. Since the sine is increasing on this interval,Differentiation gives $f(x)=1/(\pi \sqrt{x(1-x)})$, and $F(0)=0$, $F(1)=1$ prove normalization. This also evaluates the density's improper integral without treating its divergent endpoint values as infinite probability.

For the circle construction, instead choose $\Phi$ with uniform distribution on $[0,2\pi )$ and put $Y=\cos \Phi$. Each $-1<y<1$ has two preimages in one revolution. The change of variables formula adds both contributions:The affine change of variables $X=(1+Y)/2$ gives the standard arcsine distribution. A single monotone-branch calculation without the second preimage would give only half the total probability.

The same geometry explains the ideal harmonic oscillator $y(t)=A\cos (\omega t+\varphi )$, with amplitude $A>0$ and angular frequency $\omega >0$. Choose $T$ uniformly in any complete period of duration $2\pi /\omega$. Its oscillation phase modulo $2\pi$ is uniform, and $Y=y(T)$ has densityNear a turning point, $|dy/dt|=\omega \sqrt{A^2-y^2}$ is small, so a fixed position interval takes more time to cross. For the chosen amplitude $A=2\mathrm{cm}$, exactly one third of the period is spent in $|y|\le 1\mathrm{cm}$, and two thirds outside that central interval. These are calculations for an ideal sinusoid, rather than measured oscillator data.

The sampling assumption matters. Taking one observation every full period returns the same position each time. It does not produce this probability distribution. Uniformly random observation times and a fixed regularly spaced sampling scheme are different models; aliasing can conceal the motion altogether.

Paul Lévy used a related geometric construction in his work on stochastic processes: project a uniformly chosen point on a circle onto its diameter to divide a segment. The projection description appears explicitly in §5 of his paper, following its arcsine density calculation.^[ref]

An unbounded probability density function does not imply an atom. HereThe last equivalence follows from $\arcsin u/u\to 1$. The endpoint interval has much more probability than the same interval under a uniform distribution, whose value is $\epsilon$, but its probability still tends to zero. This is an integrable singularity.

For example, the first 1% of the unit interval contains about $6.377\%$ of the probability; the last 1% contains the same amount because the arcsine distribution is a symmetric probability distribution. Each exact endpoint still has zero probability. Density and interval probability answer different questions.

A numerical integration routine must respect these endpoint integrable singularities. The substitution $x={\sin }^2\theta$ givesIt turns a difficult-looking density integral into a constant measure in angle. Integrating finite bins is also a faithful way to draw the distribution near its endpoints; sampling $f(0)$ to make a bar height is not.

Inverting $F(x)=p$ gives the quantile functionConsequently, inverse transform sampling uses $X={\sin }^2(\pi U/2)$ with $U$ uniform on $(0,1)$. The quartiles are $(2-\sqrt{2})/4$, $1/2$ and $(2+\sqrt{2})/4$.^[ref]

The first and third quartiles are about $0.1464$ and $0.8536$, so the central half of the probability spans about $0.7071$ of the unit interval. Under a uniform distribution, its width would be $0.5$.

The symmetric probability distribution gives $\mathbb{E}[X]=1/2$. The angle representation gives a short calculation of the second moment:Use ${\sin }^4\theta =(3-4\cos 2\theta +\cos 4\theta )/8$ to evaluate the integral. A uniform distribution on $[0,1]$ has variance $1/12$; the larger variance $1/8$ quantifies the greater spread toward the ends.

More generally, for a nonnegative integer $n$, the beta function yieldsFor $n=0$ this is one; for $n=1,2$ it reproduces the values above. The central binomial coefficient thus appears in an exact moment calculation for a continuous probability distribution.

Specify whether two parameters mean endpoints or location and width. With endpoints $a<b$, define $Y=a+(b-a)X$ for a standard arcsine distribution variable $X$. ThenThis is a location-scale family with location $a$ and width $b-a$. Its mean is $(a+b)/2$ and its variance is $(b-a{)}^2/8$. A convention using parameters $a,w$ for location and width instead has right endpoint $a+w$, so it must not be read as having endpoints $a,w$.^[ref]

For example, a chosen interval $[-2,6]$ gives mean $2$ and variance $8$, while its density still concentrates toward both endpoints. It is a bounded probability distribution; its endpoint singularities do not make its support infinite.

For $0<A<B<\infty$, this density makes $\log X$ uniform between $\log A$ and $\log B$. It is a proper finite-interval version of the scale-invariant prior. The chosen logarithm base changes only the endpoint coordinates.

The Pareto distribution describes a positive random variable whose tail probability falls as a power law. In the shape–minimum convention used here, $X\sim \mathrm{Pareto}(\alpha ,\beta )$ has shape parameter $\alpha >0$ and minimum $\beta >0$, with probability support $[\beta ,\infty )$. Its probability density function isThe minimum $\beta$ has the same units as $X$; the shape parameter $\alpha$ is dimensionless. At fixed $\alpha$, changing $\beta$ gives a scale family, since $X/\beta \sim \mathrm{Pareto}(\alpha ,1)$. This convention is also used in uniform-Pareto conjugacy.^[ref]

The survival function and cumulative distribution function areFor $0<p<1$, solving $F(x)=p$ gives the quantile functionIts right-hand limit at $p=0$ is $\beta$; it tends to infinity as $p\uparrow 1$.^[ref] There is no probability mass at $\beta$: changing a probability density function's value at that single point changes no probabilities.

Choose $\alpha =2$ and $\beta =10$ as an original theoretical example. Then $\mathbb{P}(X>20)=1/4$, $\mathbb{P}(X>100)=1/100$, the median is $10\sqrt{2}\simeq 14.142$, and the 0.99 quantile is $Q(0.99)=100$. These chosen values and all figures below describe mathematical probability distributions, not actual wealth observations.

On this log-log plot, write $r=x/\beta$ to obtain $\log \overline{F}(\beta r)=-\alpha \log r$. The slope is $-\alpha$, whereas the probability density function has power exponent $\alpha +1$. Multiplying a threshold by ten divides its tail probability by $10^{\alpha }$. A larger $\alpha$ therefore concentrates the Pareto distribution nearer $\beta$; a smaller $\alpha$ makes extreme values more plausible. This exact scaling is useful when modelling large positive losses or sizes over several orders of magnitude.

The Pareto distribution is a heavy-tailed distribution: for every $t>0$, the exponential moment $\mathbb{E}[e^{tX}]$ diverges because $e^{tx}{x}^{-\alpha -1}$ eventually grows without bound. In contrast, the Laplace distribution has exponentially decreasing tails and some finite positive exponential moments. The Student's t-distribution also has power tails, but on both sides of the real line, with a smooth central region rather than a positive minimum.

If $X$ has a Pareto distribution with positive scale $x_m$ and positive shape $\alpha$, then for $z\ge 0$,Thus the rescaled logarithm has an exponential distribution with rate $\alpha$. The tail probability calculation also identifies the support as $z\ge 0$.

Let $X$ have survival function $\mathbb{P}(X\ge x)=x^{-s}$ for $x\ge 1$, with $s>1$. Its expected value is $s/(s-1)$. The tail-sum formula for expectation givesThe absolutely continuous probability distribution gives zero mass to the integers, so $\lceil X\rceil =\lfloor X\rfloor +1$ almost surely and $\mathbb{E}[\lceil X\rceil ]=\zeta (s)+1$. Taking expectations of $\lfloor X\rfloor \le X\le \lceil X\rceil$ gives $\zeta (s)\le s/(s-1)\le \zeta (s)+1$, which rearranges to the displayed bounds on the Riemann zeta function. Finiteness is justified by $\lfloor X\rfloor \le X$ and $\lceil X\rceil \le X+1$.

For the Pareto distribution of shape $a>0$ and lower bound $b>0$, the density $a{b}^a{x}^{-a-1}$ is strictly decreasing on $[b,\infty )$. Its highest-density interval of probability $1-\alpha$, $0<\alpha <1$, must begin at $b$. The tail probability $\mathbb{P}(X>u)=(b/u{)}^a$ makes the upper endpoint $u=b{\alpha }^{-1/a}$. If this is a posterior distribution, the result is its highest posterior density interval. In particular, Uniform-Pareto conjugacy supplies posterior shape $a+n$ and lower bound $\max (b,{\max }_i{y}_i)$.

Let $S_n={\sum }_{i=1}^n{X}_i$, where the summands are independent and identically distributed with Pareto distribution $\mathrm{Pareto}(\alpha ,\beta )$. Their positivity implies $\{X_1\ge na\}\subseteq \{S_n\ge na\}$, so for fixed $a>0$ and $na\ge \beta$,Both outer expressions tend to zero. A single large summand already prevents a strictly negative exponential rate. No positive exponential moment exists, so the exponential-moment form of the Cramér theorem cannot be applied. The bound proves this logarithmic rate and does not assert an exact tail asymptotic. In particular, it remains true when $a$ lies below a finite mean, where the event need not even be rare. The stronger fixed-summand-count asymptotic in a heavy-tailed subexponential distribution is a separate statement.

A Pareto distribution is a proper probability distribution for every $\alpha ,\beta >0$, even when its mean is infinite. To see both facts, integrate the probability density function against $x^q$. For real $q\ge 0$,The improper integral converges exactly when $q-\alpha -1<-1$. At $q=\alpha$ it diverges logarithmically; above that it diverges as a positive power of the upper endpoint. Setting $q=0$ gives one, proving normalization regardless of whether higher moments exist.

Taking $q=1,2$ and using $\mathrm{Var}(X)=\mathbb{E}[X^2]-(\mathbb{E}[X]{)}^2$ when both terms are finite givesFor $0<\alpha \le 1$, the nonnegative random variable has expected value $+\infty$; a centered variance about a finite mean is unavailable. For $1<\alpha \le 2$, the mean is finite and the variance is infinite.^[ref] A finite total probability and a finite expected value answer different questions.

At the chosen minimum $\beta =10$, $\alpha =3$ gives $\mathbb{E}[X]=15$, $\mathbb{E}[X^2]=300$ and variance $300-15^2=75$, but an infinite third moment. At $\alpha =3/2$, the mean is 30 and the variance is infinite; at $\alpha =3/4$, even the mean is infinite, while the median remains the finite number $102^{4/3}\simeq 25.198$. At the boundary $\alpha =2$, the earlier example has mean 20 but infinite variance. A finite sample contains finite values almost surely; that fact cannot establish that the underlying mean or variance is finite.

Conditioning a Pareto distribution on exceeding a threshold changes its minimum while preserving its shape parameter. For $u\ge \beta$ and $x\ge u$, the definition of conditional probability givesEquivalently, for $r\ge 1$,The ratio to the threshold has the same Pareto distribution at every admissible threshold.^[ref] For the chosen $\alpha =2$, $\beta =10$, only $1/4$ of outcomes exceed 20, but among those outcomes the probability of exceeding 40 is again $1/4$. When $\alpha >1$, the same identity gives conditional expectation $\mathbb{E}[X\mid X\ge u]=\alpha u/(\alpha -1)$.

This multiplicative scaling differs from the additive memoryless property of an exponential distribution. Indeed $Y=\log (X/\beta )$ has exponential distribution of rate $\alpha$, because $\mathbb{P}(Y>y)=e^{-\alpha y}$ for $y\ge 0$. Thus logarithms turn multiplication into addition and explain the scaling property. Conversely, if $U$ has uniform distribution on $(0,1)$, inverse transform sampling gives $X=\beta (1-U{)}^{-1/\alpha }$.

The excess $Z=X-u$ conditional on $X\ge u$ has survival function $(1+z/u{)}^{-\alpha }$ for $z\ge 0$. In the generalized Pareto distribution convention $\mathbb{P}(Z>z)=(1+\xi z/\sigma {)}^{-1/\xi }$, this means $\xi =1/\alpha$ and $\sigma =u/\alpha$. Here the excess starts at zero; the original Pareto distribution starts at $\beta >0$.

For a nonnegative random variable with finite positive mean, the share of its total amount carried by the largest fraction $q$ of outcomes can be defined by dividing their contribution to the expected value by the whole expected value. For a Pareto distribution, let $0<q\le 1$ and choose $x_q=\beta q^{-1/\alpha }$, so that $\mathbb{P}(X\ge x_q)=q$. If $\alpha >1$, direct integration givesThis is a share of amount, distinct from the fraction of outcomes. It depends on $\alpha$, but not on the unit of measurement $\beta$.^[ref]

For the chosen $\alpha =2$, the top 20% starts at $x_{0.2}=10\sqrt{5}\simeq 22.361$ and carries $T(0.2)=\sqrt{0.2}\simeq 44.72\%$ of the amount. To make the top 20% carry exactly 80%, solveThe resulting Pareto distribution has a finite mean and infinite variance. The 80/20 rule selects one shape parameter; it is not a universal law, or even a property of every Pareto distribution. For $\alpha \le 1$, the population-share definition above is unavailable because its denominator is infinite; substituting such an $\alpha$ into the finite-mean formula is invalid.

The left panel compares three quantiles in units of the minimum. The right panel shows $T(0.2)$ only where the mean is finite, and marks the chosen 80/20 value. The thresholds $\alpha =1$ and $\alpha =2$ separate the moment regimes; they do not separate valid and invalid probability distributions.

A straight-looking empirical log-log plot does not by itself establish a Pareto distribution or a reliable tail exponent; estimating an exponent and assessing goodness of fit are separate tasks.^[ref] If a Pareto distribution is used only above a fitted lower threshold, $T(q)$ describes that conditional tail population, not automatically the whole population. A genuine finite upper cutoff gives a truncated distribution with finite nonnegative moments; its survival function, quantiles and top shares must be recalculated. Truncation is a modelling change, distinct from merely observing a finite sample. None of the chosen numbers here estimates an actual wealth distribution.

The Rayleigh distribution is the probability distribution of the Euclidean norm of a two-dimensional random vector with independent zero-mean normal random variables of equal variance as components. If $X,Y\sim N(0,{\sigma }^2)$ are independent, thenHere $\sigma$ is the standard deviation of each component, not of the radius $R$. At $\sigma =1$ this gives the unit-scale probability density function $r{e}^{-r^2/2}$.^[ref]

This geometry turns two signed errors into one nonnegative distance. For example, suppose horizontal and vertical positioning errors are independent $N(0,4{\mathrm{cm}}^2)$ random variables. Their means both vanish, but the distance from the intended point is usually positive. The Rayleigh distribution with $\sigma =2\mathrm{cm}$ gives the probability of landing inside any centered disk. These are chosen model assumptions, rather than measurements of a particular instrument.

An apparently paradoxical feature is that the most likely location per unit area is the origin, while the radial mode is $\sigma$. A narrow annulus farther out has more area. Near zero that growing circumference outweighs the falling planar density; farther out the exponential function wins. Distinguishing densities measured against different elements of area or length resolves the paradox.

The joint probability density of $X,Y$ isIn polar coordinates, $x=r\cos \theta$, $y=r\sin \theta$, and the absolute Jacobian determinant is $r$. Thus the joint density of radius and angle factors asIntegrating over angle gives the Rayleigh distribution. The same factorization shows that the angle has uniform distribution and is independent of the radius. This is a short change of variables argument: an annulus of width $dr$ has area $2\pi rdr$ to first order.

The assumptions are substantive. Unequal component variances give elliptical rather than circular density contours; a nonzero mean displaces the center. Even zero covariance and equal normal distributions do not suffice without independence. For an original counterexample, let $X\sim N(0,{\sigma }^2)$ and let $S$ be an independent Rademacher random variable. Put $Y=SX$. Then $Y$ has the same normal distribution as $X$, and $\mathrm{Cov}(X,Y)=\mathbb{E}[S{X}^2]=0$. Nevertheless, all points lie on $y=x$ or $y=-x$, and $R=\sqrt{2}|X|$ has densitywhich is positive at zero, unlike the Rayleigh distribution. This pair is not a Gaussian random vector.

For $r\ge 0$, integration gives the survival function and cumulative distribution functionSolving $F_R(r_p)=p$ gives the quantile function^[ref]

For the chosen positioning model $\sigma =2\mathrm{cm}$,The disk of radius one component standard deviation therefore contains about 39.35% of outcomes; the familiar one-dimensional normal interval percentage does not apply to a two-dimensional disk. The 95% and 99% radial thresholds areA circle defined by $R=r_p$ itself still has probability zero; its enclosed disk carries the stated probability.

The closed-form quantile function also gives inverse transform sampling: if $U$ has uniform distribution on $(0,1)$, then $\sigma \sqrt{-2\log U}$ has Rayleigh distribution. Using $1-U$ instead gives the same law. Endpoints are excluded so that the logarithm stays finite.

Squaring a Rayleigh distribution variable produces an exponential distribution. For $w\ge 0$,where the exponential distribution parameter is a rate. Equivalently, $R^2/{\sigma }^2$ has chi-squared distribution with two degrees of freedom (statistics). An amplitude and its square therefore have different probability density functions, even though one determines the other.

This makes the Rayleigh-normal scale mixture transparent: at unit scale, $R^2$ is exponential, allowing a random normal scale to produce a Laplace distribution. Another equivalent parametrization uses a Weibull distribution with shape two and scale $\sqrt{2}\sigma$. Conventions that call this latter scale the Rayleigh scale differ by a factor of $\sqrt{2}$; check the probability density function before comparing numerical parameter values.^[ref]

Using $u=r^2/(2{\sigma }^2)$ in the defining expected value integral givesThe last integral is the gamma function. Near zero, the integrand before substitution behaves as $r^{q+1}$, explaining the threshold $q>-2$; at infinity the Gaussian tail controls every power. In particular,^[ref]

For the chosen $\sigma =2\mathrm{cm}$, the mean radius is about $2.5066\mathrm{cm}$, its median is $2\sqrt{2\log 2}\simeq 2.3548\mathrm{cm}$, and its mode is $2\mathrm{cm}$. Its root mean square is $2\sqrt{2}\simeq 2.8284\mathrm{cm}$. Different summaries answer different questions; the component scale $\sigma$ is neither the radial mean nor the radial standard deviation.

In an 1880 paper, John William Strutt, known as Lord Rayleigh, studied the resultant of many oscillations with a common frequency and arbitrary oscillation phases. His large-number calculation distinguished a mean intensity from a definite resultant amplitude and used planar coordinates followed by a polar integral. The resulting amplitude density is the Rayleigh distribution in a different scale convention. This is a large-number description, not an exact law for every finite collection of oscillations.^[ref]

To see the connection in a chosen model, take independent oscillation phases ${\Theta }_j$ with uniform distribution on $[0,2\pi )$ and equal component amplitude $a$. The phasor sum has coordinatesBoth have mean zero, variance $n{a}^2/2$ and zero covariance. The multivariate central limit theorem gives an approximately circular Gaussian random vector for large $n$, with $\sigma =a\sqrt{n/2}$. Its amplitude $\sqrt{X_n^2+Y_n^2}$ is therefore approximately Rayleigh. For every finite $n$, that amplitude is bounded by $na$, whereas a Rayleigh distribution has an unbounded tail. At $n=1$ it is exactly $a$. This simple endpoint check prevents the approximation from becoming a false exact identity.

A hypoexponential distribution is the distribution of a sum of independent variables with exponential distributions. For two positive unequal rates $\lambda ,\mu$, convolution of independent random variables gives the probability density functionIt is zero for negative $s$. Its equal-rate limit is the shape-two gamma distribution with rate $\lambda$.

Two sequential independent memoryless reaction waits have a hypoexponential distribution. Convolution gives $p(t)=k_1{k}_2(e^{-k_{1}t}-e^{-k_{2}t})/(k_2-k_1)$. Equal rates give the Erlang distribution $k^{2}t{e}^{-kt}$. A nonzero mode distinguishes this ideal two-event model from a single exponential wait, but does not chemically identify the events.

If the independent exponential random variables $T_i$ have rates $i\lambda$, thenwhere the $E_i$ are independent exponential random variables of rate $\lambda$. To see this, the first of $N$ independent exponential clocks rings at rate $N\lambda$; after it rings, the memoryless property leaves $N-1$ independent rate-$\lambda$ clocks. The successive order statistic gaps are therefore independent with rates $N\lambda ,(N-1)\lambda ,\dots ,\lambda$. Their sum has the same distribution as $\sum T_i$. In particular, for $x>0$, the tail at $Nx$ has logarithm $-\lambda Nx+o(N)$, since $1-(1-e^{-\lambda Nx}{)}^N\sim N{e}^{-\lambda Nx}$. The sum divided by $N$ has a large deviation principle with speed $N$ and rate function $\lambda x$ on $x\ge 0$, infinity otherwise.

For independent exponential distributions with rates ${\lambda }_1>{\lambda }_2>0$, let $T=S_1+S_2$. Dividing the joint density of $(S_1,T)$ by the hypoexponential distribution density of $T$ gives a truncated exponential distribution on $(0,t)$ with rate ${\lambda }_1-{\lambda }_2$. Its displayed conditional expectation supplies an E-step when only pair totals are observed. At equal rates the conditional split is uniform, with mean $t/2$.

// Target: probability-and-statistics.bigb

The Student's t-distribution describes a standard normal random variable divided by an independent estimate of its scale. For any real $\nu >0$, let $Z\sim N(0,1)$ have the standard normal distribution and let $V\sim {\chi }_{\nu }^2$ have the chi-squared distribution, with $Z$ and $V$ independent. ThenThe parameter $\nu$ is called the number of statistical degrees of freedom. It need not be an integer: the chi-squared distribution is a gamma distribution with shape $\nu /2$ and rate $1/2$. The independence assumption is essential; specifying only the two marginal distributions does not determine the probability distribution of their ratio.

Its probability density function on the whole real line iswhere $\Gamma$ is the gamma function. Every member is a symmetric distribution about zero. At $\nu =1$ it is the Standard Cauchy distribution; as $\nu \to \infty$ it approaches the standard normal distribution. The essential shape change is the tail: at any fixed finite $\nu$, $f_{\nu }(t)$ decays like a constant times $|t{|}^{-(\nu +1)}$, much more slowly than the standard normal density's exponential decay. Thus it is a heavy-tailed distribution. ^[ref]