In probability theory and statistics, the continuous uniform distributions or rectangular distributions are a family of symmetric probability distributions. Such a distribution describes an experiment where there is an arbitrary outcome that lies between certain bounds. The bounds are defined by the parameters, <math>a</math> and <math>b,</math> which are the minimum and maximum values. The interval can either be closed (i.e. <math>[a,b]</math>) or open (i.e. <math>(a,b)</math>). Therefore, the distribution is often abbreviated <math>U(a,b),</math> where <math>U</math> stands for uniform distribution.
Definitions
Probability density function
The probability density function of the continuous uniform distribution is
<math display="block">
f(x) = \begin{cases}
\dfrac{1}{b-a} & \text{for } a \le x \le b, \\[8pt]
0 & \text{for } x < a \ \text{ or } \ x > b.
\end{cases}
</math>
The values of <math>f(x)</math> at the two boundaries <math>a</math> and <math>b</math> are usually unimportant, because they do not alter the value of <math display="inline">\int_c^d f(x)dx</math> over any interval <math>[c,d],</math> nor of <math display="inline">\int_a^b x f(x) \, dx,</math> nor of any higher moment. Sometimes they are chosen to be zero, and sometimes chosen to be <math>\tfrac{1}{b-a} .</math> The latter is appropriate in the context of estimation by the method of maximum likelihood. In the context of Fourier analysis, one may take the value of <math>f(a)</math> or <math>f(b)</math> to be <math>\tfrac{1}{2(b-a)} ,</math> because then the inverse transform of many integral 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 zero measure. Also, it is consistent with the sign function, which has no such ambiguity.
Any probability density function integrates to <math>1,</math> so the probability density function of the continuous uniform distribution is graphically portrayed as a rectangle where is the base length and is the height. As the base length increases, the height (the density at any particular value within the distribution boundaries) decreases.
In terms of mean <math>\mu</math> and variance <math>\sigma ^2 ,</math> the probability density function of the continuous uniform distribution is
<math display="block">
f(x) = \begin{cases}
\dfrac{1}{2 \sigma \sqrt{3 & \text{for } - \sigma \sqrt{3} \le x - \mu \le \sigma \sqrt{3} , \\[2pt]
0 & \text{otherwise} .
\end{cases}
</math>
Cumulative distribution function
The cumulative distribution function of the continuous uniform distribution is:
<math display="block"> F(x) = \begin{cases}
0 & \text{for } x < a, \\[8pt]
\frac{x-a}{b-a} & \text{for } a \le x \le b, \\[8pt]
1 & \text{for } x > b.
\end{cases} </math>
Its inverse is:
<math display="block">F^{-1} (p) = a + p (b - a) \quad \text{ for } 0 < p < 1.</math>
In terms of mean <math>\mu</math> and variance <math>\sigma ^2 ,</math> the cumulative distribution function of the continuous uniform distribution is:
<math display="block">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} \le x - \mu < \sigma \sqrt{3} , \\
1 & \text{for } x - \mu \ge \sigma \sqrt{3} ;
\end{cases}</math>
its inverse is:
<math display="block">F^{-1} (p) = \sigma \sqrt{3} (2p-1) + \mu \quad \text{ for } 0 \le p \le 1.</math>
<u>Example 1.</u> Using the continuous uniform distribution function
For a random variable <math>X \sim U(0,23),</math> find <math>\Pr(2 < X < 18):</math>
<math display="block">\Pr(2 < X < 18) = (18-2) \cdot \frac{1}{23-0} = \frac{16}{23} .</math>
In a graphical representation of the continuous uniform distribution function <math>[f(x) \text{ vs } x],</math> the area under the curve within the specified bounds, displaying the probability, is a rectangle. For the specific example above, the base would be and the height would be
<u>Example 2.</u> Using the continuous uniform distribution function (conditional)
For a random variable <math>X \sim U(0,23),</math> find <math>\Pr(X > 12 \mid X > 8):</math>
<math display="block">\Pr(X > 12 \mid X > 8) = (23-12) \cdot \frac{1}{23-8} = \frac{11}{15}.</math>
The example above is a conditional probability case for the continuous uniform distribution: given that is true, what is the probability that Conditional probability changes the sample space, so a new interval length has to be calculated, where <math>b = 23</math> and <math>a' = 8.</math>
<math display="block">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)} ,</math>
from which we may calculate the raw moments <math>m_k :</math>
<math display="block">m_1 = \frac{a+b}{2} ,</math>
<math display="block">m_2 = \frac{a^2+ab+b^2}{3} ,</math>
<math display="block">m_k = \frac{ \sum_{i=0}^k a^i b^{k-i} }{k+1} .</math>
For a random variable following the continuous uniform distribution, the expected value is <math>m_1 = \tfrac{a+b}{2} ,</math> and the variance is <math>m_2 - m_1 ^2 = \tfrac{(b-a)^2}{12} .</math>
For the special case <math>a = -b,</math> the probability density function of the continuous uniform distribution is:
<math display="block">
f(x) = \begin{cases}
\frac{1}{2b} & \text{for } -b \le x \le b, \\[8pt]
0 & \text{otherwise} ;
\end{cases}
</math>
the moment-generating function reduces to the simple form:
<math display="block">M_X = \frac{ \sinh bt }{bt} .</math>
Cumulant-generating function
For the <math>n</math>-th cumulant of the continuous uniform distribution on the interval is <math>\tfrac{B_n}{n} ,</math> where <math>B_n</math> is the <math>n</math>-th Bernoulli number.
Standard uniform distribution
The continuous uniform distribution with parameters <math>a = 0</math> and <math>b = 1,</math> i.e. <math>U(0,1),</math> is called the standard uniform distribution.
One interesting property of the standard uniform distribution is that if <math>u_1</math> has a standard uniform distribution, then so does <math>1 - u_1 .</math> This property can be used for generating antithetic variates, among other things. In other words, this property is known as the inversion method where the continuous standard uniform distribution can be used to generate random numbers for any other continuous distribution..
Related distributions
- If X has a standard uniform distribution, then by the inverse transform sampling method, Y = − λ<sup>−1</sup> ln(X) has an exponential distribution with (rate) parameter λ.
- If X has a standard uniform distribution, then Y = X<sup>n</sup> has a beta distribution with parameters (1/n,1). As such,
- The standard uniform distribution is a special case of the beta distribution, with parameters (1,1).
- The Irwin–Hall distribution is the sum of n i.i.d. U(0,1) distributions.
- The Bates distribution is the average of n i.i.d. U(0,1) distributions.
- The sum of two independent uniform distributions U<sub>1</sub>(a,b)+U<sub>2</sub>(c,d) yields a trapezoidal distribution, symmetric about its mean, on the support [a+c,b+d]. The plateau has width equals to the absolute different of the width of U<sub>1</sub> and U<sub>2</sub>. 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 a triangular distribution, symmetric about its mean, on the support [a+c,a+c+2w].
- The sum of two independent, equally distributed, uniform distributions U<sub>1</sub>(a,b)+U<sub>2</sub>(a,b) yields a symmetric triangular distribution on the support [2a,2b].
- The distance between two i.i.d. uniform random variables |U<sub>1</sub>(a,b)−U<sub>2</sub>(a,b)| also has a triangular distribution, although not symmetric, on the support [0,b−a].
Statistical inference
Estimation of parameters
Estimation of maximum
Minimum-variance unbiased estimator
Given a uniform distribution on <math>[0,b]</math> with unknown <math>b,</math> the minimum-variance unbiased estimator (UMVUE) for the maximum is:
<math display="block">\hat{b} _\text{UMVU} = \frac{k+1}{k} m = m + \frac{m}{k} ,</math>
where <math>m</math> is the sample maximum and <math>k</math> is the sample size, sampling without replacement (though this distinction almost surely makes no difference for a continuous distribution). This follows for the same reasons as estimation for the discrete distribution, and can be seen as a very simple case of maximum spacing estimation. This problem is commonly known as the German tank problem, due to application of maximum estimation to estimates of German tank production during World War II.
Method of moments estimator
The method of moments estimator is:
<math display="block">\hat{b} _{MM} = 2 \bar{X} ,</math>
where <math>\bar{X}</math> is the sample mean.
Maximum likelihood estimator
The maximum likelihood estimator is:
<math display="block">\hat{b} _{ML} = m ,</math>
where <math>m</math> is the sample maximum, also denoted as <math>m = X_{(n)} ,</math> the maximum order statistic of the sample.
Estimation of minimum
Given a uniform distribution on <math>[a,b]</math> with unknown a, the maximum likelihood estimator for a is:
<math display="block">\hat a_{ML}=\min\{X_1,\dots,X_n\},</math>
the sample minimum.
Estimation of midpoint
The midpoint of the distribution, <math>\tfrac{a+b}{2} ,</math> is both the mean and the median of the uniform distribution. Although both the sample mean and the sample median are unbiased estimators of the midpoint, neither is as efficient as the sample mid-range, i.e. the arithmetic mean of the sample maximum and the sample minimum, which is the UMVU estimator of the midpoint (and also the maximum likelihood estimate).
Confidence interval
For the maximum
Let <math>X_1 , X_2 , X_3 , ..., X_n</math> be a sample from <math>U_{[0,L]} ,</math> where <math>L</math> is the maximum value in the population. Then <math>X_{(n)} = \max ( X_1 , X_2 , X_3 , ..., X_n )</math> has the Lebesgue–Borel density <math>f = \frac{ d \Pr _{X_{(n) }{ d \lambda } :</math>
<math display="block">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),</math> where <math>1 \! \! 1 _{[0,L]}</math> is the indicator function of <math>[0,L] .</math>
The confidence interval given before is mathematically incorrect, as
<math display="block">\Pr \big( [ \hat{\theta}, \hat{\theta} + \varepsilon ] \ni \theta \big) \ge 1 - \alpha</math>
cannot be solved for <math>\varepsilon</math> without knowledge of <math>\theta</math>. However, one can solve
<math display="block">\Pr \big( [ \hat{\theta}, \hat{\theta} (1 + \varepsilon) ] \ni \theta \big) \ge 1 - \alpha</math> for <math>\varepsilon \ge \alpha ^{-1/n} - 1</math> for any unknown but valid <math>\theta ;</math>
one then chooses the smallest <math>\varepsilon</math> possible satisfying the condition above. Note that the interval length depends upon the random variable <math>\hat{\theta} .</math>
Occurrence and applications
The probabilities for uniform distribution function are simple to calculate due to the simplicity of the function form. But according to Wanke (2008), in the particular case of investigating lead-time for inventory management at the beginning of the life cycle when a completely new product is being analyzed, the uniform distribution proves to be more useful.
See also
- Discrete uniform distribution
- Beta distribution
- Box–Muller transform
- Probability plot
- Q–Q plot
- Rectangular function
- Irwin–Hall distribution — In the degenerate case where n=1, the Irwin-Hall distribution generates a uniform distribution between 0 and 1.
- Bates distribution — Similar to the Irwin-Hall distribution, but rescaled for n. Like the Irwin-Hall distribution, in the degenerate case where n=1, the Bates distribution generates a uniform distribution between 0 and 1.
References
Further reading
External links
- Online calculator of Uniform distribution (continuous)
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