Unless otherwise noted, LibreTexts content is licensed by CC BY-NC-SA 3.0. distribution function and cumulative distributions function for this discrete uniform Note that $$X$$ takes values in $S = \{a, a + h, a + 2 h, \ldots, a + (n - 1) h\}$ so that $$S$$ has $$n$$ elements, starting at $$a$$, with step size $$h$$, a discrete interval. (probability density function) given by: P (X = x) = 1/ (k+1) for all values of x = 0,... k Note that the mean is the average of the endpoints (and so is the midpoint of the interval $$[a, b]$$) while the variance depends only on the number of points and the step size. The expected value of discrete uniform random variable is E(X)=N+12. Open the Special Distribution Simulator and select the discrete uniform distribution. Recall that $$f(x) = g\left(\frac{x - a}{h}\right)$$ for $$x \in S$$, where $$g$$ is the PDF of $$Z$$. Walk through homework problems step-by-step from beginning to end. We specialize further to the case where the finite subset of $$\R$$ is a discrete interval, that is, the points are uniformly spaced. For , 2, ..., the first few values are For $$k \in \N$$ $\E\left(X^k\right) = \frac{1}{n} \sum_{i=1}^n x_i^k$. Vary the number of points, but keep the default values for the other parameters. and kurtosis excess are, The mean deviation for a uniform distribution on elements is given by. Then $H(X) = \E\{-\ln[f(X)]\} = \sum_{x \in S} -\ln\left(\frac{1}{n}\right) \frac{1}{n} = -\ln\left(\frac{1}{n}\right) = \ln(n)$. Hints help you try the next step on your own. Note that the last point is $$b = a + (n - 1) h$$, so we can clearly also parameterize the distribution by the endpoints $$a$$ and $$b$$, and the step size $$h$$. This follows from the definition of the distribution function: $$F(x) = \P(X \le x)$$ for $$x \in \R$$. Run the simulation 1000 times and compare the empirical density function to the probability density function. Vary the parameters and note the graph of the distribution function. The distribution function $$F$$ of $$X$$ is given by. Then $$X = a + h Z$$ has the uniform distribution on $$n$$ points with location parameter $$a$$ and scale parameter $$h$$. Of course, the fact that $$\skw(Z) = 0$$ also follows from the symmetry of the distribution. This follows from the definition of the (discrete) probability density function: $$\P(X \in A) = \sum_{x \in A} f(x)$$ for $$A \subseteq S$$. The discrete uniform distribution is implemented in the Wolfram We will assume that the points are indexed in order, so that $$x_1 \lt x_2 \lt \cdots \lt x_n$$. Recall that $$\E(X) = a + h \E(Z)$$ and $$\var(X) = h^2 \var(Z)$$, so the results follow from the corresponding results for the standard distribution. Note that $$\skw(Z) \to \frac{9}{5}$$ as $$n \to \infty$$. The probability density function $$f$$ of $$X$$ is given by $f(x) = \frac{1}{\#(S)}, \quad x \in S$. Recall that \begin{align} \sum_{k=1}^{n-1} k^3 & = \frac{1}{4}(n - 1)^2 n^2 \\ \sum_{k=1}^{n-1} k^4 & = \frac{1}{30} (n - 1) (2 n - 1)(3 n^2 - 3 n - 1) \end{align} Hence $$\E(Z^3) = \frac{1}{4}(n - 1)^2 n$$ and $$\E(Z^4) = \frac{1}{30}(n - 1)(2 n - 1)(3 n^2 - 3 n - 1)$$. Most classical, combinatorial probability models are … Discrete Uniform Distribution The discrete uniform distribution is also known as the "equally likely outcomes" distribution. Suppose that $$Z$$ has the standard discrete uniform distribution on $$n \in \N_+$$ points, and that $$a \in \R$$ and $$h \in (0, \infty)$$. With this parametrization, the number of points is $$n = 1 + (b - a) / h$$. of Integer Sequences. Compute a few values of the distribution function and the quantile function. The uniform distribution on a discrete interval converges to the continuous uniform distribution on the interval with the same endpoints, as the step size decreases to 0. $$G^{-1}(1/4) = \lceil n/4 \rceil - 1$$ is the first quartile. Without some additional structure, not much more can be said about discrete uniform distributions. From MathWorld--A Wolfram Web Resource. A random variable with p.d.f. Vary the parameters and note the graph of the probability density function. The simplest is the uniform distribution. Suppose that $$X$$ has the uniform distribution on $$S$$. $$G^{-1}(1/2) = \lceil n / 2 \rceil - 1$$ is the median. For odd. https://mathworld.wolfram.com/DiscreteUniformDistribution.html. Then $$Y = c + w X = (c + w a) + (w h) Z$$. The quantile function $$G^{-1}$$ of $$Z$$ is given by $$G^{-1}(p) = \lceil n p \rceil - 1$$ for $$p \in (0, 1]$$. For the remainder of this discussion, we assume that $$X$$ has the distribution in the definiiton. Recall that \begin{align} \sum_{k=0}^{n-1} k & = \frac{1}{2}n (n - 1) \\ \sum_{k=0}^{n-1} k^2 & = \frac{1}{6} n (n - 1) (2 n - 1) \end{align} Hence $$\E(Z) = \frac{1}{2}(n - 1)$$ and $$\E(Z^2) = \frac{1}{6}(n - 1)(2 n - 1)$$. Of course, the results in the previous subsection apply with $$x_i = i - 1$$ and $$i \in \{1, 2, \ldots, n\}$$. [ "article:topic", "showtoc:no", "license:ccby", "authorname:ksiegrist" ], $$\newcommand{\R}{\mathbb{R}}$$ $$\newcommand{\N}{\mathbb{N}}$$ $$\newcommand{\Z}{\mathbb{Z}}$$ $$\newcommand{\E}{\mathbb{E}}$$ $$\newcommand{\P}{\mathbb{P}}$$ $$\newcommand{\var}{\text{var}}$$ $$\newcommand{\sd}{\text{sd}}$$ $$\newcommand{\cov}{\text{cov}}$$ $$\newcommand{\cor}{\text{cor}}$$ $$\newcommand{\skw}{\text{skew}}$$ $$\newcommand{\kur}{\text{kurt}}$$, 5.21: The Uniform Distribution on an Interval, Uniform Distributions on Finite Subsets of $$\R$$, Uniform Distributions on Discrete Intervals, probability generating function of $$Z$$, $$F(x) = \frac{k}{n}$$ for $$x_k \le x \lt x_{k+1}$$ and $$k \in \{1, 2, \ldots n - 1 \}$$, $$\sigma^2 = \frac{1}{n} \sum_{i=1}^n (x_i - \mu)^2$$. Note the graph of the distribution function. of them having the same probability, then, Restricting the set to the set of positive If $$c \in \R$$ and $$w \in (0, \infty)$$ then $$Y = c + w X$$ has the discrete uniform distribution on $$n$$ points with location parameter $$c + w a$$ and scale parameter $$w h$$.