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 $\text{max}(X_1, X_2, \cdots, X_n)$.

---

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 $$q(x) = \left\{ \begin{array}{lr} n x^{n-1} & : x \in [0,1]\\ 0 & : x \notin [0,1] \end{array} \right.$$