Probability mass and density functions: Worked examples

Before we start, three plain-word reminders so no symbol sneaks in undefined:

The scenario matrix

Every problem this topic can throw at you falls into one of these cells. The right column names the example that kills it.

#	Case class	What makes it tricky	Killed by
A	Discrete, finite, uniform	just count	Ex 1
B	Discrete, non-uniform (weighted)	outcomes have different weights	Ex 2
C	Discrete, find missing mass so $\sum = 1$	normalization on a PMF	Ex 3
D	Continuous, find the constant $c$	normalization on a PDF	Ex 4
E	Continuous, probability of an interval	integrate the density	Ex 5
F	Degenerate: single point of a continuous variable	$P (X = c) = 0$	Ex 5 (part b)
G	Continuous, go PDF ↔ CDF	differentiate / integrate	Ex 6
H	Limiting behaviour: tail probability, $x \to \infty$	improper integral	Ex 7
I	Density that exceeds 1	density ≠ probability	Ex 8
J	Real-world word problem	translate words → math	Ex 9
K	Exam twist: mixed / piecewise	glue two pieces, check continuity	Ex 10

Figure — Probability mass and density functions

Continuous cases (PDF)

Worked example Example 5 — Cells E & F: interval probability & a single point

Statement. Using $f_{X} (x) = \frac{1}{2} x$ on $[0, 2]$ from Ex 4: (a) find $P (1 \leq X \leq 2)$ ; (b) find $P (X = 1.4)$ .

Forecast: part (a) — more or less than half? Part (b) — guess it exactly.

(a) Set up the area: $P (1 \leq X \leq 2) = \int_{1}^{2} \frac{1}{2} x d x$ . Why this step? For a PDF, probability of an interval = area under the curve over that interval.
Integrate: $\frac{1}{2} [\frac{x ^{2}}{2}]_{1}^{2} = \frac{1}{2} (\frac{4}{2} - \frac{1}{2}) = \frac{1}{2} \cdot \frac{3}{2} = \frac{3}{4}$ .
(b) A single point has zero width: $P (X = 1.4) = \int_{1.4}^{1.4} f_{X} d x = 0$ . Why this step? This is the degenerate case F: an integral over zero width is zero area, so any exact value has probability $0$ — even though its density $f_{X} (1.4) = 0.7$ is nonzero.

Verify: $\frac{3}{4} = 0.75 \in [0, 1]$ ✓. And since $P (X = 1) = 0$ , the boundary doesn't matter: $P (1 \leq X \leq 2) = P (1 < X < 2)$ ✓.

Worked example Example 6 — Cell G: PDF ↔ CDF both directions

Statement. $X$ is uniform on $[0, 1]$ so $f_{X} (x) = 1$ there. (a) Build $F_{X} (x)$ . (b) Recover $f_{X}$ by differentiating.

Forecast: what shape is $F_{X}$ — flat, straight, or curved?

(a) Integrate the density from the left: $F_{X} (x) = \int_{0}^{x} 1 d t = x$ for $x \in [0, 1]$ . Why this step? The CDF is "area accumulated up to $x$ ." Integrating a flat height-1 slab of width $x$ gives area $x$ — a straight ramp.
Handle the ends: $F_{X} (x) = 0$ for $x < 0$ (no area yet) and $F_{X} (x) = 1$ for $x > 1$ (all area collected). Why this step? Cover all cases — a CDF must run from $0$ up to $1$ .
(b) Differentiate: $\frac{d}{d x} (x) = 1 = f_{X} (x)$ on $(0, 1)$ . Why this step? By the Fundamental Theorem of Calculus, the density is the slope of the CDF; a straight ramp of slope 1 gives constant density 1.

Verify: $F_{X} (1) - F_{X} (0) = 1 - 0 = 1$ = total probability ✓, and slope $= 1$ matches $f_{X} = 1$ ✓.

Word problem & exam twist

Worked example Example 9 — Cell J: real-world translation

Statement. A support-line's call durations (minutes) follow $f_{X} (x) = \frac{1}{3} e^{- x /3}$ for $x \geq 0$ (exponential with mean $3$ min). What fraction of calls last between 2 and 5 minutes?

Forecast: roughly a quarter, a half, or three-quarters?

Translate words → math: "between 2 and 5 minutes" $= P (2 \leq X \leq 5) = \int_{2}^{5} \frac{1}{3} e^{- x /3} d x$ . Why this step? Interval question ⇒ area under the density. Units: $x$ in minutes, $f_{X}$ in "probability per minute", so $f_{X} d x$ is dimensionless probability ✓.
Integrate: antiderivative of $\frac{1}{3} e^{- x /3}$ is $- e^{- x /3}$ . Evaluate $[- e^{- x /3}]_{2}^{5}$ .
Compute: $- e^{- 5/3} + e^{- 2/3} = e^{- 2/3} - e^{- 5/3} \approx 0.5134 - 0.1889 = 0.3245$ .

Verify: result $\approx 0.325 \in [0, 1]$ ✓. Cross-check with the CDF $F_{X} (x) = 1 - e^{- x /3}$ : $F_{X} (5) - F_{X} (2) = (1 - e^{- 5/3}) - (1 - e^{- 2/3}) = e^{- 2/3} - e^{- 5/3}$ ✓ same number.

Recall

Recall One-line trap: density vs probability

Can a valid PDF have $f_{X} (x) = 4$ ? ::: Yes — density can exceed 1; only its area is capped at 1 (Ex 8).

Recall Degenerate continuous case

For continuous $X$ , what is $P (X = 1.4)$ ? ::: Exactly $0$ — zero width means zero area (Ex 5b).

Recall Which rule normalizes?

PMF vs PDF total-probability condition? ::: $\sum_{x} p_{X} (x) = 1$ (discrete) vs $\int_{- \infty}^{\infty} f_{X} (x) d x = 1$ (continuous).

Connections. Normalizing constants here reappear when you fit models by 1.4.02-Maximum-likelihood-estimation; the discrete PMF over classes is exactly what 2.1.05-Softmax-activation produces; the Gaussian density underlies 3.2.04-Gaussian-processes. Expected values built from these functions live in 1.3.08-Expected-value-and-variance, and the underlying objects in 1.3.01-Random-variables-and-distributions.

Worked examples — Probability mass and density functions