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Continuous uniform distribution

From Wikipedia, the free encyclopedia
Uniform distribution on an interval
Continuous uniform
Probability density function
PDF of the uniform probability distribution using the maximum convention at the transition points.
Usingmaximum convention
Cumulative distribution function
CDF of the uniform probability distribution.
NotationU[a,b]{\displaystyle {\mathcal {U}}_{[a,b]}}
Parameters−∞<a<b<∞{\displaystyle -\infty <a<b<\infty }
Support[a,b]{\displaystyle [a,b]}
PDF{1b−afor x∈[a,b]0otherwise{\displaystyle {\begin{cases}{\frac {1}{b-a}}&{\text{for }}x\in [a,b]\\0&{\text{otherwise}}\end{cases}}}
CDF{0for x<ax−ab−afor x∈[a,b]1for x>b{\displaystyle {\begin{cases}0&{\text{for }}x<a\\{\frac {x-a}{b-a}}&{\text{for }}x\in [a,b]\\1&{\text{for }}x>b\end{cases}}}
Mean12(a+b){\displaystyle {\tfrac {1}{2}}(a+b)}
Median12(a+b){\displaystyle {\tfrac {1}{2}}(a+b)}
Modeany value in (a,b){\displaystyle {\text{any value in }}(a,b)}
Variance112(b−a)2{\displaystyle {\tfrac {1}{12}}(b-a)^{2}}
MAD14(b−a){\displaystyle {\tfrac {1}{4}}(b-a)}
Skewness0{\displaystyle 0}
Excess kurtosis−65{\displaystyle -{\tfrac {6}{5}}}
Entropylog⁡(b−a){\displaystyle \log(b-a)}
MGF{etb−etat(b−a)for t≠01for t=0{\displaystyle {\begin{cases}{\frac {\mathrm {e} ^{tb}-\mathrm {e} ^{ta}}{t(b-a)}}&{\text{for }}t\neq 0\\1&{\text{for }}t=0\end{cases}}}
CF{eitb−eitait(b−a)for t≠01for t=0{\displaystyle {\begin{cases}{\frac {\mathrm {e} ^{\mathrm {i} tb}-\mathrm {e} ^{\mathrm {i} ta}}{\mathrm {i} t(b-a)}}&{\text{for }}t\neq 0\\1&{\text{for }}t=0\end{cases}}}

Inprobability theory andstatistics, thecontinuous uniform distributions orrectangular distributions are a family ofsymmetricprobability distributions. Such a distribution describes an experiment where there is an arbitrary outcome that lies between certain bounds.[1] The bounds are defined by the parameters,a{\displaystyle a} andb,{\displaystyle b,} which are the minimum and maximum values. The interval can either beclosed (i.e.[a,b]{\displaystyle [a,b]}) oropen (i.e.(a,b){\displaystyle (a,b)}).[2] Therefore, the distribution is often abbreviatedU(a,b),{\displaystyle U(a,b),} whereU{\displaystyle U} stands for uniform distribution.[1] The difference between the bounds defines the interval length; allintervals of the same length on the distribution'ssupport are equally probable. It is themaximum entropy probability distribution for arandom variableX{\displaystyle X} under no constraint other than that it is contained in the distribution's support.[3]

Definitions

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Probability density function

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Theprobability density function of the continuous uniform distribution isf(x)={1b−afor a≤x≤b,0for x<a  or  x>b.{\displaystyle f(x)={\begin{cases}{\dfrac {1}{b-a}}&{\text{for }}a\leq x\leq b,\\[8pt]0&{\text{for }}x<a\ {\text{ or }}\ x>b.\end{cases}}}

The values off(x){\displaystyle f(x)} at the two boundariesa{\displaystyle a} andb{\displaystyle b} are usually unimportant, because they do not alter the value of∫cdf(x)dx{\textstyle \int _{c}^{d}f(x)dx} over any interval[c,d],{\displaystyle [c,d],} nor of∫abxf(x)dx,{\textstyle \int _{a}^{b}xf(x)\,dx,} nor of any higher moment. Sometimes they are chosen to be zero, and sometimes chosen to be1b−a.{\displaystyle {\tfrac {1}{b-a}}.} The latter is appropriate in the context of estimation by the method ofmaximum likelihood. In the context ofFourier analysis, one may take the value off(a){\displaystyle f(a)} orf(b){\displaystyle f(b)} to be12(b−a),{\displaystyle {\tfrac {1}{2(b-a)}},} because then the inverse transform of manyintegral transforms of this uniform function will yield back the function itself, rather than a function which is equal "almost everywhere", i.e. except on a set of points with zeromeasure. Also, it is consistent with thesign function, which has no such ambiguity.

Any probability density function integrates to1,{\displaystyle 1,} so the probability density function of the continuous uniform distribution is graphically portrayed as a rectangle where⁠b−a{\displaystyle b-a}⁠ is the base length and⁠1b−a{\displaystyle {\tfrac {1}{b-a}}}⁠ is the height. As the base length increases, the height (the density at any particular value within the distribution boundaries) decreases.[4]

In terms of meanμ{\displaystyle \mu } and varianceσ2,{\displaystyle \sigma ^{2},} the probability density function of the continuous uniform distribution isf(x)={12σ3for −σ3≤x−μ≤σ3,0otherwise.{\displaystyle f(x)={\begin{cases}{\dfrac {1}{2\sigma {\sqrt {3}}}}&{\text{for }}-\sigma {\sqrt {3}}\leq x-\mu \leq \sigma {\sqrt {3}},\\[2pt]0&{\text{otherwise}}.\end{cases}}}

Cumulative distribution function

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Thecumulative distribution function of the continuous uniform distribution is:F(x)={0for x<a,x−ab−afor a≤x≤b,1for x>b.{\displaystyle F(x)={\begin{cases}0&{\text{for }}x<a,\\[8pt]{\frac {x-a}{b-a}}&{\text{for }}a\leq x\leq b,\\[8pt]1&{\text{for }}x>b.\end{cases}}}

Its inverse is:F−1(p)=a+p(b−a) for 0<p<1.{\displaystyle F^{-1}(p)=a+p(b-a)\quad {\text{ for }}0<p<1.}

In terms of meanμ{\displaystyle \mu } and varianceσ2,{\displaystyle \sigma ^{2},} the cumulative distribution function of the continuous uniform distribution is:F(x)={0for x−μ<−σ3,12(x−μσ3+1)for −σ3≤x−μ<σ3,1for x−μ≥σ3;{\displaystyle F(x)={\begin{cases}0&{\text{for }}x-\mu <-\sigma {\sqrt {3}},\\{\frac {1}{2}}\left({\frac {x-\mu }{\sigma {\sqrt {3}}}}+1\right)&{\text{for }}-\sigma {\sqrt {3}}\leq x-\mu <\sigma {\sqrt {3}},\\1&{\text{for }}x-\mu \geq \sigma {\sqrt {3}};\end{cases}}}

its inverse is:F−1(p)=σ3(2p−1)+μ for 0≤p≤1.{\displaystyle F^{-1}(p)=\sigma {\sqrt {3}}(2p-1)+\mu \quad {\text{ for }}0\leq p\leq 1.}

Example 1. Using the continuous uniform distribution function

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For arandom variableX∼U(0,23),{\displaystyle X\sim U(0,23),} findPr(2<X<18):{\displaystyle \Pr(2<X<18):}Pr(2<X<18)=(18−2)⋅123−0=1623.{\displaystyle \Pr(2<X<18)=(18-2)\cdot {\frac {1}{23-0}}={\frac {16}{23}}.}

In a graphical representation of the continuous uniform distribution function[f(x) vs x],{\displaystyle [f(x){\text{ vs }}x],} the area under the curve within the specified bounds, displaying the probability, is a rectangle. For the specific example above, the base would be⁠16,{\displaystyle 16,}⁠ and the height would be⁠123.{\displaystyle {\tfrac {1}{23}}.}⁠[5]

Example 2. Using the continuous uniform distribution function (conditional)

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For a random variableX∼U(0,23),{\displaystyle X\sim U(0,23),} findPr(X>12∣X>8):{\displaystyle \Pr(X>12\mid X>8):}Pr(X>12∣X>8)=(23−12)⋅123−8=1115.{\displaystyle \Pr(X>12\mid X>8)=(23-12)\cdot {\frac {1}{23-8}}={\frac {11}{15}}.}

The example above is aconditional probability case for the continuous uniform distribution: given that⁠X>8{\displaystyle X>8}⁠ is true, what is the probability that⁠X>12?{\displaystyle X>12?}⁠ Conditional probability changes the sample space, so a new interval length⁠b−a′{\displaystyle b-a'}⁠ has to be calculated, whereb=23{\displaystyle b=23} anda′=8.{\displaystyle a'=8.}[5] The graphical representation would still follow Example 1, where the area under the curve within the specified bounds displays the probability; the base of the rectangle would be⁠11,{\displaystyle 11,}⁠ and the height would be⁠115.{\displaystyle {\tfrac {1}{15}}.}⁠[5]

Generating functions

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Moment-generating function

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Themoment-generating function of the continuous uniform distribution is:[6]MX=E⁡[etX]=∫abetxdxb−a=etb−etat(b−a)=Bt−Att(b−a),{\displaystyle M_{X}=\operatorname {E} \left[e^{tX}\right]=\int _{a}^{b}e^{tx}{\frac {dx}{b-a}}={\frac {e^{tb}-e^{ta}}{t(b-a)}}={\frac {B^{t}-A^{t}}{t(b-a)}},}from which we may calculate theraw momentsmk:{\displaystyle m_{k}:}m1=a+b2,{\displaystyle m_{1}={\frac {a+b}{2}},}m2=a2+ab+b23,{\displaystyle m_{2}={\frac {a^{2}+ab+b^{2}}{3}},}mk=∑i=0kaibk−ik+1.{\displaystyle m_{k}={\frac {\sum _{i=0}^{k}a^{i}b^{k-i}}{k+1}}.}

For a random variable following the continuous uniform distribution, theexpected value ism1=a+b2,{\displaystyle m_{1}={\tfrac {a+b}{2}},} and thevariance ism2−m12=(b−a)212.{\displaystyle m_{2}-m_{1}^{2}={\tfrac {(b-a)^{2}}{12}}.}

For the special casea=−b,{\displaystyle a=-b,} the probability density function of the continuous uniform distribution is:f(x)={12bfor −b≤x≤b,0otherwise;{\displaystyle f(x)={\begin{cases}{\frac {1}{2b}}&{\text{for }}-b\leq x\leq b,\\[8pt]0&{\text{otherwise}};\end{cases}}}

the moment-generating function reduces to the simple form:MX=sinh⁡btbt.{\displaystyle M_{X}={\frac {\sinh bt}{bt}}.}

Cumulant-generating function

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For⁠n≥2,{\displaystyle n\geq 2,}⁠ then{\displaystyle n}-thcumulant of the continuous uniform distribution on the interval⁠[−12,12]{\displaystyle [-{\tfrac {1}{2}},{\tfrac {1}{2}}]}⁠ isBnn,{\displaystyle {\tfrac {B_{n}}{n}},} whereBn{\displaystyle B_{n}} is then{\displaystyle n}-thBernoulli number.[7]

Standard uniform distribution

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The continuous uniform distribution with parametersa=0{\displaystyle a=0} andb=1,{\displaystyle b=1,} i.e.U(0,1),{\displaystyle U(0,1),} is called thestandard uniform distribution.

One interesting property of the standard uniform distribution is that ifu1{\displaystyle u_{1}} has a standard uniform distribution, then so does1−u1.{\displaystyle 1-u_{1}.} This property can be used for generatingantithetic variates, among other things. In other words, this property is known as theinversion method where the continuous standard uniform distribution can be used to generaterandom numbers for any other continuous distribution.[4] Ifu1{\displaystyle u_{1}} is a uniform random number with standard uniform distribution, i.e. withU(0,1),{\displaystyle U(0,1),} thenx=F−1(u1){\displaystyle x=F^{-1}(u_{1})} generates a random numberx{\displaystyle x} from any continuous distribution with the specifiedcumulative distribution functionF.{\displaystyle F.}[4]

Relationship to other functions

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As long as the same conventions are followed at the transition points, the probability density function of the continuous uniform distribution may also be expressed in terms of theHeaviside step function as:f(x)=H⁡(x−a)−H⁡(x−b)b−a,{\displaystyle f(x)={\frac {\operatorname {H} (x-a)-\operatorname {H} (x-b)}{b-a}},}

or in terms of therectangle function as:f(x)=1b−a rect⁡(x−a+b2b−a).{\displaystyle f(x)={\frac {1}{b-a}}\ \operatorname {rect} \left({\frac {x-{\frac {a+b}{2}}}{b-a}}\right).}

There is no ambiguity at the transition point of thesign function. Using the half-maximum convention at the transition points, the continuous uniform distribution may be expressed in terms of the sign function as:f(x)=sgn⁡(x−a)−sgn⁡(x−b)2(b−a).{\displaystyle f(x)={\frac {\operatorname {sgn} {(x-a)}-\operatorname {sgn} {(x-b)}}{2(b-a)}}.}

Properties

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Moments

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The mean (first rawmoment) of the continuous uniform distribution is:E⁡[X]=∫abxdxb−a=b2−a22(b−a)=b+a2.{\displaystyle \operatorname {E} [X]=\int _{a}^{b}x{\frac {dx}{b-a}}={\frac {b^{2}-a^{2}}{2(b-a)}}={\frac {b+a}{2}}.}

The second raw moment of this distribution is:E⁡[X2]=∫abx2dxb−a=b3−a33(b−a).{\displaystyle \operatorname {E} \left[X^{2}\right]=\int _{a}^{b}x^{2}{\frac {dx}{b-a}}={\frac {b^{3}-a^{3}}{3(b-a)}}.}

In general, then{\displaystyle n}-th raw moment of this distribution is:E⁡[Xn]=∫abxndxb−a=bn+1−an+1(n+1)(b−a).{\displaystyle \operatorname {E} \left[X^{n}\right]=\int _{a}^{b}x^{n}{\frac {dx}{b-a}}={\frac {b^{n+1}-a^{n+1}}{(n+1)(b-a)}}.}

The variance (secondcentral moment) of this distribution is:Var⁡[X]=E⁡[(X−E⁡[X])2]=∫ab(x−a+b2)2dxb−a=(b−a)212.{\displaystyle \operatorname {Var} [X]=\operatorname {E} \left[{\left(X-\operatorname {E} [X]\right)}^{2}\right]=\int _{a}^{b}\left(x-{\frac {a+b}{2}}\right)^{2}{\frac {dx}{b-a}}={\frac {(b-a)^{2}}{12}}.}

Order statistics

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LetX1,...,Xn{\displaystyle X_{1},...,X_{n}} be ani.i.d. sample fromU(0,1),{\displaystyle U(0,1),} and letX(k){\displaystyle X_{(k)}} be thek{\displaystyle k}-thorder statistic from this sample.

X(k){\displaystyle X_{(k)}} has abeta distribution, with parametersk{\displaystyle k} and⁠n−k+1.{\displaystyle n-k+1.}⁠

The expected value is:E⁡[X(k)]=kn+1.{\displaystyle \operatorname {E} \left[X_{(k)}\right]={\frac {k}{n+1}}.}

This fact is useful when makingQ–Q plots.

The variance is:Var⁡[X(k)]=k(n−k+1)(n+1)2(n+2).{\displaystyle \operatorname {Var} \left[X_{(k)}\right]={\frac {k(n-k+1)}{(n+1)^{2}(n+2)}}.}

See also:Order statistic § Probability distributions of order statistics

Uniformity

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The probability that a continuously uniformly distributed random variable falls within any interval of fixed length is independent of the location of the interval itself (but it is dependent on the interval size(ℓ){\displaystyle (\ell )}), so long as the interval is contained in the distribution's support.

Indeed, ifX∼U(a,b){\displaystyle X\sim U(a,b)} and if[x,x+ℓ]{\displaystyle [x,x+\ell ]} is a subinterval of[a,b]{\displaystyle [a,b]} with fixedℓ>0,{\displaystyle \ell >0,} then:Pr(X∈[x,x+ℓ])=∫xx+ℓdyb−a=ℓb−a,{\displaystyle \Pr {\big (}X\in [x,x+\ell ]{\big )}=\int _{x}^{x+\ell }{\frac {dy}{b-a}}={\frac {\ell }{b-a}},}which is independent ofx.{\displaystyle x.} This fact motivates the distribution's name.

Uniform distribution on more general sets

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The uniform distribution can be generalized to sets more general than intervals.

Formally, letS{\displaystyle S} be aBorel set of positive, finiteLebesgue measureλ(S),{\displaystyle \lambda (S),} i.e.0<λ(S)<+∞.{\displaystyle 0<\lambda (S)<+\infty .} The uniform distribution onS{\displaystyle S} can be specified by defining the probability density function to be zero outsideS{\displaystyle S} and constantly equal to1λ(S){\displaystyle {\tfrac {1}{\lambda (S)}}} onS.{\displaystyle S.}

An interesting special case is when the set S is asimplex. It is possible to obtain a uniform distribution on the standardn-vertex simplex in the following way.[8]: Thm.4.1 taken independent random variables with the sameexponential distribution; denote them by X1,...,Xn; and let Yi := Xi / (sumi Xi). Then, the vector Y1,...,Yn is uniformly distributed on the simplex.

Related distributions

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  • IfX has a standard uniform distribution, then by theinverse transform sampling method,Y = −λ−1 ln(X) has anexponential distribution with (rate) parameterλ.
  • IfX has a standard uniform distribution, thenY =Xn has abeta distribution with parameters (1/n,1). As such,
  • The standard uniform distribution is a special case of thebeta distribution, with parameters (1,1).
  • TheIrwin–Hall distribution is the sum ofni.i.d.U(0,1) distributions.
  • TheBates distribution is the average ofni.i.d.U(0,1) distributions.
  • The sum of two independent uniform distributionsU1(a,b)+U2(c,d) yields atrapezoidal distribution, symmetric about its mean, on the support [a+c,b+d]. The plateau has width equals to the absolute different of the width ofU1 andU2. The width of the sloped parts corresponds to the width of the narrowest uniform distribution.
    • If the uniform distributions have the same width w, the result is atriangular distribution, symmetric about its mean, on the support [a+c,a+c+2w].
    • The sum of two independent, equally distributed, uniform distributionsU1(a,b)+U2(a,b) yields a symmetrictriangular distribution on the support [2a,2b].
  • The distance between twoi.i.d. uniform random variables |U1(a,b)-U2(a,b)| also has atriangular distribution, although not symmetric, on the support [0,b-a].

Statistical inference

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Estimation of parameters

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Estimation of maximum

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Minimum-variance unbiased estimator
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Main article:German tank problem

Given a uniform distribution on[0,b]{\displaystyle [0,b]} with unknownb,{\displaystyle b,} theminimum-variance unbiased estimator (UMVUE) for the maximum is:b^UMVU=k+1km=m+mk,{\displaystyle {\hat {b}}_{\text{UMVU}}={\frac {k+1}{k}}m=m+{\frac {m}{k}},}wherem{\displaystyle m} is thesample maximum andk{\displaystyle k} is thesample size, sampling without replacement (though this distinction almost surely makes no difference for a continuous distribution). This follows for the same reasons asestimation for the discrete distribution, and can be seen as a very simple case ofmaximum spacing estimation. This problem is commonly known as theGerman tank problem, due to application of maximum estimation to estimates of German tank production duringWorld War II.

Method of moments estimator
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Themethod of moments estimator is:b^MM=2X¯,{\displaystyle {\hat {b}}_{MM}=2{\bar {X}},}whereX¯{\displaystyle {\bar {X}}} is the sample mean.

Maximum likelihood estimator
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Themaximum likelihood estimator is:b^ML=m,{\displaystyle {\hat {b}}_{ML}=m,}wherem{\displaystyle m} is thesample maximum, also denoted asm=X(n),{\displaystyle m=X_{(n)},} the maximumorder statistic of the sample.

Estimation of minimum

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Given a uniform distribution on[a,b]{\displaystyle [a,b]} with unknowna, the maximum likelihood estimator fora is:a^ML=min{X1,…,Xn}{\displaystyle {\hat {a}}_{ML}=\min\{X_{1},\dots ,X_{n}\}},thesample minimum.[9]

Estimation of midpoint

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The midpoint of the distribution,a+b2,{\displaystyle {\tfrac {a+b}{2}},} is both the mean and the median of the uniform distribution. Although both the sample mean and the sample median areunbiased estimators of the midpoint, neither is asefficient as the samplemid-range, i.e. the arithmetic mean of the sample maximum and the sample minimum, which is theUMVU estimator of the midpoint (and also themaximum likelihood estimate).

Confidence interval

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For the maximum

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LetX1,X2,X3,...,Xn{\displaystyle X_{1},X_{2},X_{3},...,X_{n}} be a sample fromU[0,L],{\displaystyle U_{[0,L]},} whereL{\displaystyle L} is the maximum value in the population. ThenX(n)=max(X1,X2,X3,...,Xn){\displaystyle X_{(n)}=\max(X_{1},X_{2},X_{3},...,X_{n})} has the Lebesgue–Borel densityf=dPrX(n)dλ:{\displaystyle f={\frac {d\Pr _{X_{(n)}}}{d\lambda }}:}[10]

f(t)=n1L(tL)n−1=ntn−1Ln11[0,L](t),{\displaystyle f(t)=n{\frac {1}{L}}\left({\frac {t}{L}}\right)^{n-1}\!=n{\frac {t^{n-1}}{L^{n}}}1\!\!1_{[0,L]}(t),} where11[0,L]{\displaystyle 1\!\!1_{[0,L]}} is theindicator function of[0,L].{\displaystyle [0,L].}

The confidence interval given before is mathematically incorrect, asPr([θ^,θ^+ε]∋θ)≥1−α{\displaystyle \Pr {\big (}[{\hat {\theta }},{\hat {\theta }}+\varepsilon ]\ni \theta {\big )}\geq 1-\alpha }cannot be solved forε{\displaystyle \varepsilon } without knowledge ofθ{\displaystyle \theta }. However, one can solvePr([θ^,θ^(1+ε)]∋θ)≥1−α{\displaystyle \Pr {\big (}[{\hat {\theta }},{\hat {\theta }}(1+\varepsilon )]\ni \theta {\big )}\geq 1-\alpha } forε≥α−1/n−1{\displaystyle \varepsilon \geq \alpha ^{-1/n}-1} for any unknown but validθ;{\displaystyle \theta ;}one then chooses the smallestε{\displaystyle \varepsilon } possible satisfying the condition above. Note that the interval length depends upon the random variableθ^.{\displaystyle {\hat {\theta }}.}

Occurrence and applications

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The probabilities for uniform distribution function are simple to calculate due to the simplicity of the function form.[2] Therefore, there are various applications that this distribution can be used for as shown below: hypothesis testing situations, random sampling cases, finance, etc. Furthermore, generally, experiments of physical origin follow a uniform distribution (e.g. emission of radioactiveparticles).[1] However, it is important to note that in any application, there is the unchanging assumption that the probability of falling in an interval of fixed length is constant.[2]

Economics example for uniform distribution

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In the field of economics, usuallydemand andreplenishment may not follow the expected normal distribution. As a result, other distribution models are used to better predict probabilities and trends such asBernoulli process.[11] But according to Wanke (2008), in the particular case of investigatinglead-time for inventory management at the beginning of thelife cycle when a completely new product is being analyzed, the uniform distribution proves to be more useful.[11] In this situation, other distribution may not be viable since there is no existing data on the new product or that the demand history is unavailable so there isn't really an appropriate or known distribution.[11] The uniform distribution would be ideal in this situation since the random variable of lead-time (related to demand) is unknown for the new product but the results are likely to range between a plausible range of two values.[11] Thelead-time would thus represent the random variable. From the uniform distribution model, other factors related tolead-time were able to be calculated such ascycle service level andshortage per cycle. It was also noted that the uniform distribution was also used due to the simplicity of the calculations.[11]

Sampling from an arbitrary distribution

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Main article:Inverse transform sampling

The uniform distribution is useful for sampling from arbitrary distributions. A general method is the inverse transform sampling method, which uses thecumulative distribution function (CDF) of the target random variable. This method is very useful in theoretical work. Since simulations using this method require inverting the CDF of the target variable, alternative methods have been devised for the cases where the CDF is not known in closed form. One such method isrejection sampling.

Thenormal distribution is an important example where the inverse transform method is not efficient. However, there is an exact method, theBox–Muller transformation, which uses the inverse transform to convert two independent uniformrandom variables into two independentnormally distributed random variables.

Quantization error

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Main article:Quantization error

In analog-to-digital conversion, a quantization error occurs. This error is either due to rounding or truncation. When the original signal is much larger than oneleast significant bit (LSB), the quantization error is not significantly correlated with the signal, and has an approximately uniform distribution. TheRMS error therefore follows from the variance of this distribution.

Random variate generation

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There are many applications in which it is useful to run simulation experiments. Manyprogramming languages come with implementations to generatepseudo-random numbers which are effectively distributed according to the standard uniform distribution.

On the other hand, the uniformly distributed numbers are often used as the basis fornon-uniform random variate generation.

Ifu{\displaystyle u} is a value sampled from the standard uniform distribution, then the valuea+(b−a)u{\displaystyle a+(b-a)u} follows the uniform distribution parameterized bya{\displaystyle a} andb,{\displaystyle b,} as described above.

History

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While the historical origins in the conception of uniform distribution are inconclusive, it is speculated that the term "uniform" arose from the concept ofequiprobability in dice games (note that the dice games would havediscrete and not continuous uniform sample space).Equiprobability was mentioned inGerolamo Cardano'sLiber de Ludo Aleae, a manual written in 16th century and detailed on advanced probability calculus in relation to dice.[12]

See also

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References

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  1. ^abcDekking, Michel (2005).A modern introduction to probability and statistics : understanding why and how. London, UK: Springer. pp. 60–61.ISBN 978-1-85233-896-1.
  2. ^abcWalpole, Ronald; Myers, Raymond H.; Myers, Sharon L.; Ye, Keying (2012).Probability & Statistics for Engineers and Scientists. Boston, USA: Prentice Hall. pp. 171–172.ISBN 978-0-321-62911-1.
  3. ^Park, Sung Y.; Bera, Anil K. (2009). "Maximum entropy autoregressive conditional heteroskedasticity model".Journal of Econometrics.150 (2):219–230.CiteSeerX 10.1.1.511.9750.doi:10.1016/j.jeconom.2008.12.014.
  4. ^abc"Uniform Distribution (Continuous)".MathWorks. 2019. RetrievedNovember 22, 2019.
  5. ^abcIllowsky, Barbara; et al. (2013).Introductory Statistics. Rice University, Houston, Texas, USA: OpenStax College. pp. 296–304.ISBN 978-1-938168-20-8.
  6. ^Casella & Berger 2001, p. 626
  7. ^Wichura, Michael J. (January 11, 2001)."Cumulants"(PDF).Stat 304 Handouts. University of Chicago.
  8. ^Non-Uniform Random Variate Generation.doi:10.1007/978-1-4613-8643-8.
  9. ^Ln(a,b)=∏i=1nf(Xi)=1(b−a)n1[a,b](X1,…,Xn){\displaystyle L_{n}(a,b)=\prod _{i=1}^{n}f(X_{i})={\frac {1}{(b-a)^{n}}}\mathbf {1} _{[a,b]}(X_{1},\dots ,X_{n})}
    =1(b−a)n1{a≤min{X1,…,Xn}}1{max{X1,…,Xn}≤b}{\displaystyle ={\frac {1}{(b-a)^{n}}}\mathbf {1} _{\{a\leq \min\{X_{1},\dots ,X_{n}\}\}}\mathbf {1} _{\{\max\{X_{1},\dots ,X_{n}\}\leq b\}}}.
    Since we haven≥1{\displaystyle n\geq 1} the factor1(b−a)n{\displaystyle {\frac {1}{(b-a)^{n}}}} is maximized by biggest possiblea, which is limited inLn(a,b){\displaystyle L_{n}(a,b)} bymin{X1,…,Xn}{\displaystyle \min\{X_{1},\dots ,X_{n}\}}. Thereforea=min{X1,…,Xn}{\displaystyle a=\min\{X_{1},\dots ,X_{n}\}} is the maximum ofLn(a,b){\displaystyle L_{n}(a,b)}.
  10. ^Nechval KN, Nechval NA, Vasermanis EK, Makeev VY (2002)Constructing shortest-length confidence intervals. Transport and Telecommunication 3 (1) 95-103
  11. ^abcdeWanke, Peter (2008)."The uniform distribution as a first practical approach to new product inventory management".International Journal of Production Economics.114 (2):811–819.doi:10.1016/j.ijpe.2008.04.004 – via Research Gate.
  12. ^Bellhouse, David (May 2005)."Decoding Cardano's Liber de Ludo".Historia Mathematica.32:180–202.doi:10.1016/j.hm.2004.04.001.

Further reading

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External links

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