Probability Density Function (p. d. f.)

Let $X$ be a continuous random variable on the interval $S=(a,b)$. A non-negative, integrable function $f(x)$ is called the probability density function (p.d.f.) of $X$ if it satisfies three conditions: (1) $f(x) \ge 0$ for every $x \in S$; (2) the total area under $y=f(x)$ over $S$ is 1, i.e. $\int_S f(x)\,dx = 1$; (3) for any interval $A$ inside $S$, $P[X \in A] = \int_A f(x)\,dx$.

Worked examples.

Example 1. Let $X$ have p.d.f. $f(x) = 3x^2$ for $0 < x < 1$. Note carefully that $f(x)$ is not $P[X=x]$: for instance $f(0.9) = 3(0.9)^2 = 2.43 > 1$, which could never be a probability. In the continuous case $f(x)$ is only the height of the density curve at $x$; it is the area under the curve that gives probability.

Checking the two p.d.f. conditions: $f(x)=3x^2 \ge 0$ on $[0,1]$, and $\int_0^1 3x^2\,dx = [x^3]_0^1 = 1$, so $f$ is a valid p.d.f. For $0 \le c < d \le 1$, $P[c<X<d] = \int_c^d 3x^2\,dx = d^3-c^3$. Taking $c=1/2, d=1$: $P[1/2<X<1] = 1^3-(1/2)^3 = 1-1/8 = 7/8$. And $P[X=1/2] = \int_{1/2}^{1/2} f(x)\,dx = 0$ — geometrically, the 'rectangle' over a single point has zero width and hence zero area. This is a general fact: for a continuous random variable, $P[X=x]=0$ for every $x$ in the support.

Example 2. Let $X$ have p.d.f. $f(x) = x^3/4$ on $0<x<c$. Find $c$ so that $f$ is a valid p.d.f.

Since $\int_0^c \dfrac{x^3}{4}\,dx = \left[\dfrac{x^4}{16}\right]_0^c = \dfrac{c^4}{16}$ must equal 1, $c^4 = 16$, and since $c>0$, $c=2$.

Example 6. Verify whether the following are p.d.f.s of a continuous r.v. $X$: (i) $f(x)=e^{-x}$ for $0<x<\infty$, $=0$ otherwise; (ii) $f(x) = x/2$ for $-2<x<2$, $=0$ otherwise.

(i) $e^{-x} > 0$ for every $x>0$, and $\int_0^\infty e^{-x}\,dx = \left[-e^{-x}\right]_0^\infty = -(0-1) = 1$; both conditions hold, so $f$ is a valid p.d.f. (ii) $f(x)=x/2$ is negative for $-2<x<0$ (e.g. $f(-1)=-1/2<0$), violating $f(x)\ge0$; so $f$ is not a valid p.d.f.

Example 7. Find $k$ so that $f(x) = kx^2(1-x)$ for $0<x<1$ (and 0 otherwise) is a p.d.f.

$\int_0^1 kx^2(1-x)\,dx = k\int_0^1 (x^2-x^3)\,dx = k\left[\dfrac{x^3}{3}-\dfrac{x^4}{4}\right]_0^1 = k\left(\dfrac13-\dfrac14\right) = \dfrac{k}{12}$. Setting this equal to 1 gives $k=12$.

Example 8. For each p.d.f., find (a) $P(X<1)$ and (b) $P(|X|<1)$: (i) $f(x)=x^2/18$ for $-3<x<3$; (ii) $f(x)=(x+2)/18$ for $-2<x<4$.

(i)(a) $P(X<1) = \int_{-3}^1 \dfrac{x^2}{18}\,dx = \dfrac{1}{54}\left[x^3\right]_{-3}^1 = \dfrac{1}{54}(1-(-27)) = \dfrac{28}{54} = \dfrac{14}{27}$. …