Discrete random variables — PMF, CDF: Worked examples

The scenario matrix

Before solving anything, let's map the whole battlefield. A PMF/CDF problem can only ever land in one of these cells:

#	Case class	What is different / dangerous	Example
A	Finite uniform support	all masses equal — pure counting	Ex 1
B	Non-uniform finite support	masses differ; must track each	Ex 2
C	Endpoint bracketing ( $<$ vs $\leq$ )	points carry mass, so which bracket matters	Ex 3
D	Solve for a parameter	use $\sum p = 1$ to pin an unknown constant	Ex 4
E	Degenerate RV	all mass on one point; PMF/CDF collapse	Ex 5
F	Recover PMF from a CDF staircase	read jumps, not heights	Ex 6
G	Countably infinite support	infinite sum must still total 1 (geometric series)	Ex 7
H	Real-world word problem	translate English $\to$ RV $\to$ probability	Ex 8
I	Exam twist — $P (a \leq X \leq b)$ and left limits	the subtle inclusive-both-ends formula	Ex 9

We now hit every cell.

Example 1 — Finite uniform (cell A)

Figure — Discrete random variables — PMF, CDF

Look at the staircase: it is flat on $(2, 3)$ , so anywhere in that gap — including $2.7$ — gives the same height $\frac{1}{2}$ .

Example 2 — Non-uniform finite (cell B)

Example 3 — Endpoint bracketing (cell C)

Example 4 — Solve for a parameter (cell D)

Example 5 — Degenerate RV (cell E)

Worked example Statement

$X = 7$ always (a "constant random variable"). Write its PMF and CDF, and check the CDF properties.

Forecast: what does the staircase look like when there is only one step?

PMF. $p_{X} (7) = 1$ and $p_{X} (x) = 0$ for every other $x$ . Why this step? All the probability mass sits on the single reachable value.
CDF. $F_{X} (x) = 0$ for $x < 7$ and $F_{X} (x) = 1$ for $x \geq 7$ . Why this step? Walking rightward you collect nothing until $7$ , then jump straight to full.
Property audit. Non-decreasing ✓ (goes $0 \to 1$ ), right-continuous ✓ (the jump belongs to $x = 7$ since $\leq$ includes it), limits $0$ and $1$ ✓.

Verify: jump size $= F_{X} (7) - F_{X} (7^{-}) = 1 - 0 = 1 = p_{X} (7)$ ✓ — recovering the PMF from the single jump. This degenerate case is the seed of Expectation and Variance of Discrete RVs: $E [X] = 7$ , variance $= 0$ .

Example 6 — Recover PMF from a CDF (cell F)

Example 7 — Countably infinite support (cell G)

Worked example Statement

$X$ = number of coin flips until the first head, with a fair coin. So $X \in {1, 2, 3, \dots}$ (infinite!). Find the PMF, confirm it sums to $1$ , and compute $P (X \leq 3)$ .

Forecast: an infinite list of positive masses — can they really add to exactly $1$ ?

PMF. To get the first head on flip $k$ , you need $k - 1$ tails then a head: $p_{X} (k) = (\frac{1}{2})^{k - 1} \cdot \frac{1}{2} = (\frac{1}{2})^{k}$ . Why this step? Independent flips multiply; this is the geometric pattern.
Sum check via geometric series. $\sum_{k = 1}^{\infty} (\frac{1}{2})^{k} = \frac{\frac{1}{2}}{1 - \frac{1}{2}} = 1$ . Why this step? A geometric series $\sum a r^{k}$ with $∣ r ∣ < 1$ converges to $\frac{a}{1 - r}$ . This is the tool we need because we cannot add infinitely many terms by hand — the closed form does it. Countable additivity from Probability Axioms still applies to a countably infinite union.
CDF. $F_{X} (3) = P (X \leq 3) = \frac{1}{2} + \frac{1}{4} + \frac{1}{8} = \frac{7}{8}$ . Why this step? Sum the first three point masses.

Verify: $\frac{1}{2} + \frac{1}{4} + \frac{1}{8} = \frac{7}{8} = 0.875$ ✓, and the full infinite sum $= 1$ ✓. Related named PMF: the Poisson Distribution is another infinite-support model.

Example 8 — Word problem (cell H)

Worked example Statement

A shop sells $0$ , $1$ , or $2$ umbrellas per rainy hour with probabilities $0.5, 0.3, 0.2$ . Let $X$ = umbrellas sold in one hour. Find $F_{X} (1)$ and the probability that at least one umbrella is sold.

Forecast: which is larger, $F_{X} (1)$ or $P (X \geq 1)$ ?

Translate. The English gives the PMF directly: $p_{X} (0) = 0.5, p_{X} (1) = 0.3, p_{X} (2) = 0.2$ . Why this step? Naming the RV turns the story into numbers we can sum.
CDF at $1$ . $F_{X} (1) = p_{X} (0) + p_{X} (1) = 0.5 + 0.3 = 0.8$ . Why this step? " $\leq 1$ " collects the $0$ and $1$ masses.
At least one. $P (X \geq 1) = 1 - p_{X} (0) = 1 - 0.5 = 0.5$ . Why this step? Complement: everything except "sold none." Cheaper than adding $0.3 + 0.2$ , and matches it.

Verify: $p_{X} (1) + p_{X} (2) = 0.3 + 0.2 = 0.5$ ✓ equals the complement answer. And $F_{X} (1) = 0.8 > P (X \geq 1) = 0.5$ (forecast: $F_{X} (1)$ larger, correct). Units: probabilities, all in $[0, 1]$ ✓.

Example 9 — Exam twist: inclusive both ends & left limits (cell I)

Worked example Statement

Using the CDF from Example 6, compute (a) $P (1 < X \leq 3)$ , (b) $P (1 \leq X \leq 3)$ , and (c) $P (X = 3)$ from the CDF alone.

Forecast: will (a) and (b) differ, and by how much?

(a) strict left. $P (1 < X \leq 3) = F_{X} (3) - F_{X} (1) = 0.7 - 0.2 = 0.5$ . Why this step? Excludes the mass at $1$ , keeps mass at $3$ . This is exactly $p_{X} (3) = 0.5$ .
(b) inclusive both ends. $P (1 \leq X \leq 3) = F_{X} (3) - F_{X} (1^{-}) = 0.7 - 0 = 0.7$ . Why this step? To include the left endpoint we subtract the CDF just to the left of $1$ , i.e. $F_{X} (1^{-}) = 0$ , so we do not lose $p_{X} (1)$ .
(c) point mass. $P (X = 3) = F_{X} (3) - F_{X} (3^{-}) = 0.7 - 0.2 = 0.5$ . Why this step? A point's probability is its jump: $F$ at the value minus $F$ just left of it.

Verify: (b) $-$ (a) $= 0.7 - 0.5 = 0.2 = p_{X} (1)$ ✓ — the inclusive version added exactly the left-endpoint mass. (c) $= 0.5$ matches step 1's $p_{X} (3)$ ✓. See Conditional Probability for how these interval events feed conditional PMFs.

Active recall

Recall Which formula gives

$P (a \leq X \leq b)$ and why? $P (a \leq X \leq b) = F_{X} (b) - F_{X} (a^{-})$ ::: using $a^{-}$ (just left of $a$ ) keeps the mass $p_{X} (a)$ in, unlike $F_{X} (a)$ which would remove it.

Recall How do you find a point mass

$p_{X} (x_{i})$ from a CDF staircase? Measure the jump height ::: $p_{X} (x_{i}) = F_{X} (x_{i}) - F_{X} (x_{i}^{-})$ .

Recall For

$p_{X} (k) = (1/2)^{k}$ , $k \geq 1$ , why does the total equal 1? Geometric series ::: $\sum_{k \geq 1} (1/2)^{k} = \frac{1/2}{1 - 1/2} = 1$ .

Recall What is the variance of a degenerate RV

$X = 7$ ? Zero ::: all mass on one point means no spread, so variance $= 0$ .

Related core notes: Discrete random variables — PMF, CDF · Probability Axioms · Continuous random variables — PDF, CDF · Expectation and Variance of Discrete RVs.

Worked examples — Discrete random variables — PMF, CDF

The scenario matrix

Example 1 — Finite uniform (cell A)

Example 2 — Non-uniform finite (cell B)

Example 3 — Endpoint bracketing (cell C)

Example 4 — Solve for a parameter (cell D)

Example 5 — Degenerate RV (cell E)

Example 6 — Recover PMF from a CDF (cell F)

Example 7 — Countably infinite support (cell G)

Example 8 — Word problem (cell H)

Example 9 — Exam twist: inclusive both ends & left limits (cell I)

Active recall

Case map