Visual walkthrough — Discrete random variables — PMF, CDF

We use one running example the whole way: toss a fair coin 3 times, count the heads. Everything else is scenery.

Step 1 — Start with the raw outcomes (no numbers yet)

WHAT. Before any "random variable," we just have the list of things that can physically happen. Toss a coin 3 times: each toss is Heads (H) or Tails (T). There are $2\times 2\times 2=8$ equally likely sequences.

WHY. We must earn every symbol. The letter $\Omega$ (a hollow "O", read "omega") is just a name for the box holding all 8 sequences — the sample space. Nothing mathematical yet; it's a bag of tickets.

PICTURE. Each of the 8 tickets is drawn as a little card. They are equally likely, so each carries a probability of $\frac{1}{8}$.

Step 2 — The random variable $X$ turns each card into a number

WHAT. Define $X=$ "number of Heads on the card." This is our random variable. It reads a card and outputs a plain number.

$X:\Omega ⟶\{0,1,2,3\}$

• $X$ — the rule (a function, i.e. a machine: one input card, one output number).
• $\Omega$ — the input box from Step 1.
• $\{0,1,2,3\}$ — the only numbers that can come out (you can't get 4 heads in 3 tosses).

WHY this tool? We don't want to do algebra on the word "$HTH$" — we want a number so we can add, compare, and total probabilities. $X$ is the translator. Because its output list $\{0,1,2,3\}$ is finite (hence countable), we are allowed to put probability on each individual number and simply add — that is the whole meaning of discrete.

PICTURE. Arrows carry each of the 8 cards to its head-count. Notice several cards land on the same number — that piling-up is what creates unequal probabilities.

Step 3 — Count how many cards land on each number → the PMF

WHAT. For each output value $k$, gather the arrows that landed there and add up their weights. That total is the probability mass on $k$, written $p_X(k)$.

$p_X(k)=P(X=k)=\underset{\text{count}}{\underbrace{(\text{how many cards give }k\text{ heads})}}\times \underset{\text{weight per card}}{\underbrace{\frac{1}{8}}}$

Working the four buckets:

$p_X(0)=\frac{1}{8},p_X(1)=\frac{3}{8},p_X(2)=\frac{3}{8},p_X(3)=\frac{1}{8}$

• $p_X(1)=\frac{3}{8}$ because three cards ($HTT,THT,TTH$) have exactly one head, each worth $\frac{1}{8}$.
• The subscript $X$ just says "this PMF belongs to the variable $X$."
• $P(X=k)$ reads "the probability that the output equals $k$."

WHY. These are the heights of our bars. And they must obey two rules — not because a textbook says so, but because they are counts of the same 8 cards:

1. $p_X(k)\ge 0$ — a count of cards can't be negative. (Probability axiom: $P\ge 0$.)
2. ${\sum }_k{p}_X(k)=\frac{1}{8}+\frac{3}{8}+\frac{3}{8}+\frac{1}{8}=1$ — every card lands somewhere, so all 8 are accounted for exactly once. (Countable additivity over disjoint buckets → $P(\Omega )=1$.)

PICTURE. A bar (stick) at each value; its height is the mass. The four heights sum to a full 1 — the total ink is the same as the 8 cards we started with.

Step 4 — Walk left-to-right and accumulate → the CDF is born

WHAT. Now imagine dragging a marker from far left ($-\infty$) toward the right, and every time you step over a bar you scoop its mass into a bucket. The amount in the bucket when your marker sits at position $x$ is the CDF:

$F_X(x)=P(X\le x)={\sum }_{x_i\le x}{p}_X(x_i)$

• $F_X(x)$ — the running total collected up to and including $x$.
• $X\le x$ — "output is at most $x$"; the $\le$ (less-than-or-equal) matters, as Step 6 shows.
• ${\sum }_{x_i\le x}$ — add masses of every value at or to the left of $x$.

WHY sum, not integrate? Because the mass sits on isolated points, not spread over a region. There is nothing between the points to integrate. Adding sticks is the natural operation — that's the payoff of "discrete."

PICTURE. The marker moves; the bucket level rises in flat stretches broken by sudden lifts. Between values you scoop nothing (flat); crossing a value you scoop its whole stick (a jump).

Evaluating a few points:

$F_X(0)=\frac{1}{8},F_X(1)=\frac{1}{8}+\frac{3}{8}=\frac{4}{8},F_X(2)=\frac{4}{8}+\frac{3}{8}=\frac{7}{8},F_X(3)=\frac{7}{8}+\frac{1}{8}=1$

Step 5 — Why the graph is a staircase (read every property off it)

WHAT. Plot $F_X(x)$ for all real $x$, not just the four values. We get a step function: horizontal treads and vertical risers.

WHY each feature is forced:

• Non-decreasing. Moving right can only add non-negative sticks; the bucket never drains. So the staircase only ever climbs or stays level — never drops.
• Flat between values. Between, say, $1$ and $2$ there is no mass to scoop, so the total is unchanged — a horizontal tread at height $\frac{4}{8}$.
• Limits. Far left, before any bar, the bucket is empty: $F_X(-\infty )=0$. Far right, after the last bar, everything is collected: $F_X(+\infty )=1$. The top tread is exactly $1$ because the masses summed to $1$ in Step 3.

PICTURE. The full staircase over the number line, with each riser's height labelled by the mass it swallowed.

Step 6 — The degenerate corner: which endpoint owns the jump?

WHAT. Zoom into a single riser, at $x=1$. Question: at exactly $x=1$, is the bucket at the low tread $\frac{1}{8}$ or the high tread $\frac{4}{8}$? Answer: the high one. Because $F_X(x)=P(X\le x)$ uses $\le$, the point $x=1$ itself is included, so the mass $p_X(1)$ is already scooped the instant the marker reaches $1$.

WHY it matters. This is right-continuity: as you approach $1$ from the right you get $\frac{4}{8}$ and stay there — the value at $1$ agrees with the right side. But approaching from the left you get only $\frac{1}{8}$. We name that left-limit $F_X(1^{-})$ ("just left of 1"). The gap between the two sides is exactly the mass:

$p_X(x_i)=F_X(x_i)-F_X(x_i^{-})(\text{riser height}=\text{mass})$

• $F_X(x_i)$ — top of the riser (mass included, closed dot ●).
• $F_X(x_i^{-})$ — bottom, the value just to the left (mass not yet included, open dot ○).
• Their difference hands the PMF back to us — the CDF secretly stores every mass in its jumps.

PICTURE. One riser, drawn with a filled dot on top (value belongs here) and a hollow dot at the bottom (value does not belong to the left tread), and a bracket measuring the jump $=p_X(1)=\frac{3}{8}$.

Step 7 — Interval probabilities: reading a range off the staircase

WHAT. Suppose we want $P(0<X\le 2)$ — "more than 0, at most 2 heads." Read two tread heights and subtract:

$P(a<X\le b)=F_X(b)-F_X(a)$ $P(0<X\le 2)=F_X(2)-F_X(0)=\frac{7}{8}-\frac{1}{8}=\frac{6}{8}=\frac{3}{4}$

WHY this exact bracketing. $F_X(2)$ counts everything $\le 2$; $F_X(0)$ counts everything $\le 0$. Subtracting cancels the shared part and leaves the values $\{1,2\}$ — the left end $0$ is thrown out, the right end $2$ is kept. That is precisely the $(0,2]$ bracket: open on the left, closed on the right.

Check by masses: the values strictly above $0$ and at most $2$ are $1$ and $2$, so $p_X(1)+p_X(2)=\frac{3}{8}+\frac{3}{8}=\frac{6}{8}=\frac{3}{4}$. ✓

PICTURE. The staircase with the two treads $F_X(0)$ and $F_X(2)$ highlighted and their difference shaded — literally the vertical height captured between the two levels.

The one-picture summary

Everything on one canvas: the 8 cards fold into 4 sticks (the PMF); those sticks, scooped left-to-right, stack into the staircase (the CDF); each riser's height equals its stick; the top tread is $1$; and any interval is a difference of two tread heights.

Active recall

Connections

• Parent topic — PMF, CDF — this page is its visual derivation.
• Probability Axioms — "heights $\ge 0$" and "heights sum to 1" are the axioms drawn as pictures.
• Continuous random variables — PDF, CDF — replace sticks by a smooth density; the staircase becomes a smooth curve.
• Expectation and Variance of Discrete RVs — weight each value by its stick height to get $E[X]$.
• Binomial Distribution — these 3-coin-toss masses are $\text{Binomial}(3,\frac{1}{2})$.
• Poisson Distribution, Conditional Probability — same stick-and-staircase machinery, new masses.