Continuous Distributions: The Uniform and the Standard Normal

A uniform variable $X\sim\mathrm{Unif}(a,b)$ models a point dropped completely at random on $[a,b]$, with no part of the interval favoured over any other (Definition 12.11). Because a continuous variable takes uncountably many values, we cannot give each a positive probability; instead probability is described by a density $f$, and the probability of an event is the area under $f$ over that event. 'Equally likely across $[a,b]$' forces $f$ to be constant there, and for the total area to equal $1$ that constant must be the reciprocal of the width: $f(x)=\begin{cases}\dfrac{1}{b-a},& a<x<b,\\[2pt] 0,&\text{otherwise.}\end{cases}$ The graph is a rectangle of width $b-a$ and height $\frac{1}{b-a}$, whose area is exactly $1$. This generalises the familiar $\mathrm{Unif}(0,1)$ (height $1$ on $[0,1]$) to any interval. Note that the density value $\frac{1}{b-a}$ is not a probability and may even exceed $1$ on a short interval; only areas are probabilities.

For a uniform variable the area over a subinterval $[c,d]\subseteq[a,b]$ is a rectangle of height $\frac{1}{b-a}$ and width $d-c$, so $P\{c\le X\le d\}=\frac{d-c}{b-a},$ the fraction of the interval's length that $[c,d]$ occupies. The c.d.f. accumulates this area from the left: $F(x)=P\{X\le x\}=0$ for $x<a$, rises linearly as $F(x)=\frac{x-a}{b-a}$ for $a\le x<b$, and equals $1$ for $x\ge b$; indeed $P\{c\le X\le d\}=F(d)-F(c)=\frac{d-c}{b-a}$. A defining feature of every continuous distribution appears here: a single point has zero width, so $P\{X=c\}=0.$ Consequently the endpoints never matter, $P\{c\le X\le d\}=P\{c<X<d\}$, a convenience we exploit throughout.

The standard normal (or Gaussian) variable $Z\sim N(0,1)$ (Definition 12.12) has the bell-shaped density $\varphi(x)=\frac{1}{\sqrt{2\pi}}\,e^{-x^2/2},\qquad x\in\mathbb{R}.$ Several features are visible in the formula. It is symmetric about $0$, since $\varphi(-x)=\varphi(x)$ (the exponent depends only on $x^2$); it peaks at $x=0$ with height $\frac{1}{\sqrt{2\pi}}\approx0.399$; and it decays extremely fast in both tails because of the $e^{-x^2/2}$ factor, yet stays strictly positive for every real $x$. The constant $\frac{1}{\sqrt{2\pi}}$ is the normalising factor that makes the total area equal $1$. Unlike the uniform, the normal spreads probability over the whole real line, concentrating it near $0$. The meaning of the parameters $0$ and $1$ (the mean and variance) is taken up in the next chapter; for now, $N(0,1)$ is simply the standard bell curve.

Probabilities for $Z$ come from its cumulative distribution function $\Phi(x)=P\{Z\le x\}=\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{x}e^{-s^2/2}\,ds,$ the area under the bell curve to the left of $x$. This integral has no closed-form antiderivative, so values of $\Phi$ are obtained from a standard-normal table or from technology, never by elementary integration. Three facts make the table go a long way. First, any interval probability is a difference, $P\{a\le Z\le b\}=\Phi(b)-\Phi(a)$. Second, by symmetry of $\varphi$ about $0$, $\Phi(-x)=1-\Phi(x),$ so a table listing only positive $x$ covers negative arguments too; in particular $\Phi(0)=\tfrac12$, i.e. $P\{Z\le0\}=0.5$. Third, $\Phi(x)\to1$ as $x\to+\infty$ and $\Phi(x)\to0$ as $x\to-\infty$. Together these yield the landmark values $P(-1\le Z\le1)\approx0.683$ and $P(-2\le Z\le2)\approx0.954$.