Continuous Random Variables & Probability Density Functions

A discrete random variable piles its probability onto isolated values through a mass function $p(k)=\mathbb{P}(X=k)$. A continuous random variable spreads probability smoothly instead: Definition 12.5 says $X$ is continuous if there is a function $f:\mathbb{R}\to\mathbb{R}$, the probability density function (pdf), with $\mathbb{P}(X\le b)=\int_{-\infty}^{b}f(x)\,dx\qquad\text{for every }b\in\mathbb{R}.$ Geometrically, $\mathbb{P}(X\le b)$ is the area under the graph of $f$ from $-\infty$ up to $b$. The same picture delivers every probability: for any subset $B\subseteq\mathbb{R}$ for which the integral makes sense, $\mathbb{P}(X\in B)=\int_B f(x)\,dx,$ so $\mathbb{P}(a\le X\le b)=\int_a^b f(x)\,dx$ and $\mathbb{P}(X>a)=\int_a^{\infty}f(x)\,dx$. The crucial mental shift is that $f(x)$ is not a probability — it is a density, a rate of probability per unit length, and may even exceed $1$ — and only its integral over a region returns an actual probability.

Not every function can serve as a density. Remark 12.1 gives the two conditions, and they are exactly the ones that make areas-under-$f$ behave like probabilities: $f(x)\ge 0\ \text{for all }x\in\mathbb{R},\qquad \int_{-\infty}^{\infty}f(x)\,dx=1.$ Non-negativity keeps every area $\int_B f\ge 0$, so no event gets a negative probability; and total area $1$ encodes $\mathbb{P}(X\in\mathbb{R})=1$, the certainty that $X$ lands somewhere. Conversely, any function meeting these two conditions defines a valid continuous random variable. In practice a density often arrives with an unknown normalising constant: given a shape like $f(x)=c$ on $[a,b]$ or $f(x)=cx$ on $[0,L]$ (and $0$ elsewhere), you find $c$ by forcing the total area to $1$. For the constant, $\int_a^b c\,dx=c(b-a)=1$ gives $c=\tfrac{1}{b-a}$; for the ramp, $\int_0^{L}cx\,dx=c\,\tfrac{L^2}{2}=1$ gives $c=\tfrac{2}{L^2}$.

Because probability is area under $f$, the chance of landing on any one exact value is the area over a single point — a region of zero width. Proposition 12.1 makes this precise: for any real number $c$, $\mathbb{P}(X=c)=\int_c^c f(x)\,dx=0.$ Two consequences follow. First, a continuous random variable is never discrete: no value carries positive probability, so the discrete mass-function picture cannot apply. Second, because each endpoint contributes nothing, including or excluding endpoints changes no probability: $\mathbb{P}(a\le X\le b)=\mathbb{P}(a<X<b)=\int_a^b f(x)\,dx.$ This is a sharp break from the discrete case, where $\mathbb{P}(X=k)$ is typically positive and the difference between $\le$ and $<$ matters. For continuous $X$ you may move endpoints freely — only the interval, i.e. the region of integration, counts.

The simplest continuous model is the uniform distribution on $[0,1]$ (Example 12.5): pick a real number at random from $[0,1]$ with every location equally likely. Its density is the flat function $f(x)=1$ for $0\le x\le 1$ and $f(x)=0$ otherwise, whose total area is that of a $1\times 1$ square, namely $1$. Probabilities are then simply lengths: for $0\le a\le b\le 1$, $\mathbb{P}(a\le X\le b)=\int_a^b 1\,dx=b-a.$ So $\mathbb{P}(0.25\le X\le 0.6)=0.35$, while $\mathbb{P}(X=0.5)=0$ as the proposition demands. The uniform on a general interval $[\alpha,\beta]$ has the constant density $\tfrac{1}{\beta-\alpha}$. Densities need not be flat, though: Example 12.6 takes $f(x)=3e^{-3x}$ for $x>0$ (and $0$ otherwise), a decaying exponential density. It is non-negative and $\int_0^{\infty}3e^{-3x}\,dx=1$, so it too is a legitimate density — one whose areas are read off by calculus rather than by a rectangle.