A discrete random variable takes values you can list, each with a probability. Two descriptions carry the same information: the probability mass function lists $P(X = x)$, and the cumulative distribution function adds them up. Moving between them, with care at the endpoints, is the skill here.
Idea clave
The pmf is $p_X(x) = P(X = x)$: never negative, adding to 1 over all values. The CDF is $F_X(x) = P(X \le x)$: it climbs in steps from 0 to 1, jumping by $P(X = a)$ at each value $a$. For $a < b$: $$P(a < X \le b) = F(b) - F(a).$$
Por qué funciona
The events $\{X \le a\}$ and $\{a < X \le b\}$ are disjoint, and together they make $\{X \le b\}$. By the addition axiom, $F(b) = F(a) + P(a < X \le b)$. The size of a jump is a special case: $P(X = a) = F(a) - P(X < a)$, the CDF at $a$ minus its value just before $a$. Between two values of $X$ nothing happens, so $F$ is flat there: if $X$ is a whole number, $F(2.5) = F(2)$.
On the CDF a filled dot sits at the top of each jump and an open dot at the bottom: at $x = a$ the CDF already includes $P(X = a)$.
Worked examples
Ejemplo resuelto: Reading the CDF from the pmf
$X$ is the number of heads in three fair flips: $p(0) = \frac{1}{8}$, $p(1) = \frac{3}{8}$, $p(2) = \frac{3}{8}$, $p(3) = \frac{1}{8}$. Find $F(1)$, $F(1.5)$ and $P(1 \le X \le 2)$.
- $F(1) = p(0) + p(1) = \frac{1}{2}$, and $F(1.5) = \frac{1}{2}$ too: no value lies between 1 and 1.5.
- $P(1 \le X \le 2) = p(1) + p(2) = \frac{3}{4}$.
- With the CDF: $F(2) - F(0) = \frac{7}{8} - \frac{1}{8} = \frac{3}{4}$. Subtract $F(0)$, not $F(1)$, so that $X = 1$ stays in.
Respuesta: $\frac{1}{2}$, $\frac{1}{2}$ and $\frac{3}{4}$
Ejemplo resuelto: From a CDF table
A whole-number $X$ has $F(0) = 0.1$, $F(1) = 0.3$, $F(2) = 0.6$, $F(3) = 0.85$, $F(4) = 1$. Find $P(X = 2)$, $P(1 \le X \le 3)$ and $P(X \ge 2)$.
- $P(X = 2)$ is the jump at 2: $0.6 - 0.3 = 0.3$.
- $$P(1 \le X \le 3) = F(3) - F(0) = 0.85 - 0.1 = 0.75$$
- $$P(X \ge 2) = 1 - P(X \le 1) = 1 - 0.3 = 0.7$$
Respuesta: $0.3$, $0.75$ and $0.7$
Error común
Using $F(b) - F(a)$ for $P(a \le X \le b)$ when $X$ is discrete. That difference is $P(a < X \le b)$ and leaves out $P(X = a)$, which can be large. Check by listing the values that should count and adding their probabilities directly. (For continuous variables single points have probability 0, so there the endpoints do not matter.)