Continuous Random Variables

A discrete random variable has a probability at each value. A continuous one does not: $P(X = 2)$ is zero, because a single point has no width. Probability lives in areas instead, and every formula in this topic is an integral where the discrete version was a sum.

Probability density functions

A function $f(x)$ is a probability density function for $X$ if

$$f(x) \ge 0 \quad\text{for all } x \qquad\text{and}\qquad \int_{-\infty}^{\infty} f(x)\,dx = 1$$

Probability is then area under the curve:

$$P(a < X < b) = \int_a^b f(x)\,dx$$

Finding an unknown constant

Almost every question begins here: the total area must be 1.

Given $f(x) = kx(4-x)$ for $0 \le x \le 4$ and zero elsewhere, find $k$.

$$\int_0^4 (4x - x^2)\,dx = \left[2x^2 - \frac{x^3}{3}\right]_0^4 = 32 - \frac{64}{3} = \frac{32}{3}$$

$$k \times \frac{32}{3} = 1 \quad\Longrightarrow\quad k = \frac{3}{32}$$

Then, for instance,

$$P(1 < X < 3) = \frac{3}{32}\int_1^3 (4x - x^2)\,dx = \frac{11}{16} = 0.6875$$

Expectation and variance

$$E(X) = \int x\,f(x)\,dx \qquad E(X^2) = \int x^2 f(x)\,dx$$

$$\operatorname{Var}(X) = E(X^2) - \big[E(X)\big]^2$$

Continuing with $f(x) = \tfrac{3}{32}x(4-x)$:

$$E(X) = \frac{3}{32}\int_0^4 (4x^2 - x^3)\,dx = \frac{3}{32}\left(\frac{256}{3} - 64\right) = 2$$

$$E(X^2) = \frac{3}{32}\int_0^4 (4x^3 - x^4)\,dx = \frac{3}{32}\left(256 - \frac{1024}{5}\right) = \frac{24}{5} = 4.8$$

$$\operatorname{Var}(X) = 4.8 - 2^2 = 0.8$$

The cumulative distribution function

$$F(x) = P(X \le x) = \int_{-\infty}^{x} f(t)\,dt$$

For the same density,

$$F(x) = \frac{3}{32}\left(2x^2 - \frac{x^3}{3}\right) = \frac{3x^2}{16} - \frac{x^3}{32}, \qquad 0 \le x \le 4$$

with $F(x) = 0$ below 0 and $F(x) = 1$ above 4. Check: $F(4) = 3 - 2 = 1$ ✓

Differentiating recovers the density, so $f(x) = F'(x)$ — the two are always one integration apart.

Median and quartiles

The median $m$ splits the area in half:

$$F(m) = 0.5$$

and the quartiles solve $F(Q_1) = 0.25$ and $F(Q_3) = 0.75$.

Here symmetry gives $m = 2$ immediately, and the algebra agrees: $F(2) = \tfrac{12}{16} - \tfrac{8}{32} = 0.5$ ✓

For a density without symmetry the equation must be solved. With $f(x) = \tfrac19 x^2$ on $0 \le x \le 3$:

$$F(x) = \frac{x^3}{27} = 0.5 \quad\Longrightarrow\quad m = \sqrt[3]{13.5} = 2.38$$

The continuous uniform distribution

If $X$ is equally likely to take any value between $a$ and $b$, written $X \sim U(a,b)$:

$$f(x) = \frac{1}{b-a} \quad (a \le x \le b), \qquad E(X) = \frac{a+b}{2}, \qquad \operatorname{Var}(X) = \frac{(b-a)^2}{12}$$

A bus arrives at a time uniformly distributed between 2 and 10 minutes past the hour.

$$f(x) = \tfrac18 \qquad E(X) = 6 \qquad \operatorname{Var}(X) = \frac{64}{12} = 5.33$$

$$P(X > 7) = \frac{10 - 7}{8} = \frac38$$

The exponential distribution

Waiting times between random events follow

$$f(x) = \lambda e^{-\lambda x} \quad (x \ge 0), \qquad F(x) = 1 - e^{-\lambda x}, \qquad E(X) = \frac1\lambda, \qquad \operatorname{Var}(X) = \frac{1}{\lambda^2}$$

A component has a mean lifetime of 5 hours, so $\lambda = 0.2$.

$$P(X > 8) = e^{-0.2 \times 8} = e^{-1.6} = 0.2019$$

$$P(X < 3) = 1 - e^{-0.6} = 0.4512$$

The median solves $1 - e^{-\lambda m} = 0.5$, giving

$$m = \frac{\ln 2}{\lambda} = \frac{0.6931}{0.2} = 3.47 \text{ hours}$$