Continuous densities and distribution functions

What you’ll learn

• Normalise and integrate a probability density.
• Find a CDF, moments, median and quartiles.
• Distinguish density values from probabilities.

Understand the idea

A continuous density $f(x)$ is non-negative and has total area one. It can exceed one in height: only an area represents probability. A single point has zero probability, so endpoint inclusion does not change a continuous interval probability. Always include the support of the density.

The cumulative distribution function integrates the density from the left edge up to $x$. Outside the support it is zero or one, and inside it is continuous and non-decreasing. For a piecewise density, carry forward the area already accumulated from earlier pieces. Differentiate the CDF on a smooth piece to recover the density.

For the mean, integrate $xf(x)$; for the second moment, integrate $x^2f(x)$. Quantiles solve $F(q)=p$. The median cuts the area in half; the mode occurs where the density is largest, possibly at a boundary. These three measures need not coincide.

Key formulae

$$\int_{-\infty}^{\infty}f(x)\,dx=1$$

$$F(x)=\int_{-\infty}^{x}f(u)\,du$$

$$E(X)=\int_{-\infty}^{\infty}xf(x)\,dx$$

$$\operatorname{Var}(X)=\int_{-\infty}^{\infty}x^2f(x)\,dx-[E(X)]^2$$

Worked example

A density is $f(x)=kx$ for $0<x<2$ and zero elsewhere. Find $k$, the mean and the variance.

1. Normalisation gives $\int_0^2kx\,dx=2k=1$, so $k=1/2$.
2. $E(X)$ $=\int_0^2x(x/2)\,dx$ $=[x^3/6]_0^2$ $=4/3$.
3. $E(X^2)$ $=\int_0^2x^2(x/2)\,dx$ $=[x^4/8]_0^2$ $=2$.
4. The variance is $2-(4/3)^2=2/9$.

Answer: $k=1/2$, $E(X)=4/3$, $\operatorname{Var}(X)=2/9$.