The cumulative distribution function
It is awkward to specify a measure on every Borel set B, so we
encode the whole law in one function. The cumulative distribution function
(CDF) accumulates probability up to a point:
F(x) \;=\; \mathbb{P}(X \le x) \;=\; \mathbb{P}_X\big((-\infty,\,x]\big).
Three properties of F follow directly from the
probability axioms.
Let us derive each one, naming the event behind every step.
Interval probabilities by subtraction. Fix
a < b. The outcomes with X \le b split
cleanly into those already at most a and those strictly between,
and the two pieces are disjoint:
\{X \le b\} = \{X \le a\} \;\sqcup\; \{a < X \le b\}.
Apply \mathbb{P}; additivity over the disjoint union turns it
into a sum:
\mathbb{P}(X \le b) = \mathbb{P}(X \le a) + \mathbb{P}(a < X \le b).
Rewrite the two cumulative terms as values of F and solve for the
middle piece:
F(b) = F(a) + \mathbb{P}(a < X \le b) \;\Longrightarrow\; \mathbb{P}(a < X \le b) = F(b) - F(a).
F is non-decreasing. Take any
a \le b. The events nest, since being at most
a certainly implies being at most b:
\{X \le a\} \subseteq \{X \le b\}.
Monotonicity of \mathbb{P} on nested events then gives the
ordering of the values directly:
a \le b \;\Longrightarrow\; F(a) = \mathbb{P}(X \le a) \le \mathbb{P}(X \le b) = F(b).
(The same fact reads off the previous derivation:
F(b) - F(a) = \mathbb{P}(a < X \le b) \ge 0.)
The end limits are 0 and
1. As x \to +\infty the events
\{X \le x\} grow to fill the whole space,
\{X \le x\} \uparrow \Omega, and continuity of probability along
increasing limits gives
\lim_{x \to +\infty} F(x) = \mathbb{P}(\Omega) = 1.
As x \to -\infty the events shrink to nothing,
\{X \le x\} \downarrow \emptyset, and the decreasing version
gives
\lim_{x \to -\infty} F(x) = \mathbb{P}(\emptyset) = 0.
There is a fourth property — right-continuity,
F(x) = \lim_{t \downarrow x} F(t) — which comes from the same
continuity argument applied to (-\infty, x] = \bigcap_n (-\infty, x + \tfrac1n];
the vignette below explains why it is right-continuity, not left, that holds.
Why does the CDF use \le rather than
<? The choice is what makes F
right-continuous. The half-lines telescope from the right,
(-\infty, x] = \bigcap_{n=1}^{\infty} \left(-\infty,\; x + \tfrac1n\right],
so by continuity along the decreasing events
\{X \le x + \tfrac1n\} \downarrow \{X \le x\},
\lim_{n \to \infty} F\!\left(x + \tfrac1n\right) = F(x).
Approaching from the left, however,
\{X \le x - \tfrac1n\} \uparrow \{X < x\}, so the left limit
is F(x^{-}) = \mathbb{P}(X < x), and the gap is exactly the
probability piled on the point itself,
F(x) - F(x^{-}) = \mathbb{P}(X = x).
Such a point of positive mass is an atom: the CDF makes a vertical
jump of height \mathbb{P}(X = x) there, and the
closed top of the jump (from \le) is where the value
F(x) sits. A purely discrete law is all jumps (the dice
staircase below); a purely continuous law has none. A mixed law has both —
for example an insurance payout that is 0 with probability
\tfrac12 (an atom at 0) and otherwise
spread continuously over positive amounts. Its CDF jumps at
0 and then rises smoothly: neither a staircase nor a smooth
curve, but a blend of the two.
Two flavours: discrete and continuous
When X lands on a countable set of values the law is described
by a probability mass function (PMF)
p(x) = \mathbb{P}(X = x), and the CDF is a staircase that jumps by
p(x) at each value:
F(x) = \sum_{x_k \le x} p(x_k), \qquad \sum_k p(x_k) = 1.
When F is instead smooth, the law has a
probability density function (PDF) f \ge 0 with
F(x) = \int_{-\infty}^{x} f(t)\,dt, \qquad \int_{-\infty}^{\infty} f(t)\,dt = 1.
Here a single point carries no probability (\mathbb{P}(X = x) = 0),
so {<} and {\le} coincide.
A worked staircase: the sum of two dice
Roll two fair dice and let X be their sum. The PMF is the
familiar triangle (the sum 7 is most likely, with
p(7) = \tfrac{6}{36}), so the CDF is a staircase:
flat between integers, jumping by p(k) at each
k, rising from 0 up to
1. Read off any interval probability by subtraction, e.g.
\mathbb{P}(4 < X \le 7) = F(7) - F(4).
Let us compute a few CDF values from the masses (counts out of
36). The lowest sums are
p(2) = \tfrac{1}{36} (only
(1,1)), p(3) = \tfrac{2}{36}
((1,2), (2,1)), and
p(4) = \tfrac{3}{36}
((1,3), (2,2), (3,1)). Accumulating these:
F(2) = p(2) = \tfrac{1}{36}, \qquad F(3) = p(2) + p(3) = \tfrac{1}{36} + \tfrac{2}{36} = \tfrac{3}{36},
F(4) = \tfrac{1}{36} + \tfrac{2}{36} + \tfrac{3}{36} = \tfrac{6}{36} = \tfrac{1}{6}.
Because F is flat between integers, a fractional argument just
reads the last integer reached:
F(3.5) = F(3) = \tfrac{3}{36}, while
F(4) = \tfrac{6}{36} already includes the jump at
4 (right-continuity — the value sits at the top of the
step). Now an interval probability falls straight out by subtraction, e.g.
\mathbb{P}(3 < X \le 4) = F(4) - F(3) = \tfrac{6}{36} - \tfrac{3}{36} = \tfrac{3}{36} = p(4),
which is exactly the single mass at 4, as it should be.