Continuous Probability Distributions - Uniform Distribution

A real-valued continuous random variable \(X\) is uniformly distributed if the probability that \(X\) lands in an interval is proportional to the length of that interval.

Let \(S\) be a finite set. A uniformly distributed random variable \(X\) on \(S\) should be equally likely to land at any element of \(S\). Thus, for any \(x \in S\), the probability \(P(X = x) = 1/|S|\), where \(|S|\) denotes the cardinality of \(S\).

But if \(S\) is infinite, say, a subinterval of \(\mathbb{R}\), then \(1/|S| = 1/\infty = 0\), so defining \(X\) by giving the probabilities that \(X\) equals a certain element of \(S\) will not work. Instead, one defines \(X\) by assigning probabilities to subsets of \(S\). These probabilities are assigned by weighting subsets based on their measure. For example, when \(S = [0,1]\), for any \([a,b] \subset [0,1]\), one has the probability \(P(X \in [a,b]) = b-a\).

The uniform distribution on \([a,b]\) is denoted \(\mathcal{U}[a,b]\), and has PDF \[p(x) = \left\{ \begin{array}{lr} 1/(b-a) & : x \in [a,b]\\ 0 & : x \notin [a,b] \end{array} \right.\] Integrating this function, one observes the cumulative density function for \(\mathcal{U}[a,b]\) is \[f(x) = \int_{-\infty}^{x} p(y) \, dy = \left\{ \begin{array}{lr} 0 & : x \in (-\infty, a) \\ (x-a)/(b-a) & : x \in [a,b]\\ 1 & : x \in (b, \infty) \end{array} \right.\]

Consider the following example computation, using this information:

Let \(X_1, X_2, \cdots, X_n\) be independent, and identically distributed \(\mathcal{U}[0,1]\). Compute the probability distribution function for the random variable \(\text{max}(X_1, X_2, \cdots, X_n)\).

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Each \(X_i\) has PDF \(p(x) = 1\) for \(x\in [0,1]\) and \(p(x) = 0\) elsewhere. Let \(q(x)\) denote the CDF for \(Y:= \text{max}(X_1, X_2, \cdots, X_n)\). It follows \[g(x) = P(Y\le x) = P(X_i \le x \, : \, 1\le i \le n) = P(X_1 \le x)^n = \left(\int_{-\infty}^{x} p(y) \, dy \right)^n \] \[ = \left\{ \begin{array}{lr} 0 & : x \in (-\infty, 0) \\ x^n & : x \in [0,1]\\ 1 & : x \in (1, \infty) \end{array} \right.\] Differentiating this gives the PDF \[q(x) = \left\{ \begin{array}{lr} n x^{n-1} & : x \in [0,1]\\ 0 & : x \notin [0,1] \end{array} \right. \]