Random Variables, PMF & CDF

Toss three coins. Count the heads. You’ve just done the thing a random variable is named for — turning an outcome into a number, so we can chart it, average it, do arithmetic on it. The number is “discrete” when it sits on a countable list like 0, 1, 2, 3 heads, and “continuous” when it can land anywhere in a range (a waiting time, a measurement). This lesson is the discrete case.

Two functions tell you everything: the PMF (the list of probabilities, one per value) and the CDF (the running total). This is not just exam vocabulary: a classifier’s output layer (a softmax) is a PMF over the possible classes, and the CDF is what lets a computer draw random samples from any distribution. Most GATE questions on this topic boil down to “given a small PMF, read off some probability” — and they almost always pivot on one CDF property the exam loves to test.

PMF and CDF — list versus running total

The probability mass function p(x) = P(X = x) gives the probability of each individual value. Two rules make it valid: every mass is non-negative, and the masses sum to 1.

p(x) ≥ 0   for every x      and      Σ p(x) = 1

The cumulative distribution function F(x) = P(X ≤ x) is the running total of the masses up to and including x. Because it accumulates non-negative masses, the CDF can only climb or stay flat — and for a discrete RV it climbs in jumps, one jump at each value, the jump height equal to that value’s mass. Between values it is flat, so the graph is a step function.

The CDF properties GATE tests

A 2025 MSQ asked precisely which of these hold for every CDF. Memorise the list:

• Non-decreasing — F(x) never goes down as x increases.
• Right-continuous — at a jump the CDF takes the value after the jump, not before.
• Limits at the ends — F(−∞) = 0 and F(+∞) = 1.
• Step function for a discrete RV — flat stretches joined by jumps of height p(x).

From the running-total view, one identity does most of the interval work:

P(a < X ≤ b) = F(b) − F(a)

— the mass accumulated up to b, minus the mass already counted up to a.

How GATE asks this

Two shapes. The first is an MSQ on the bullet list above: a grid of statements (“the CDF is non-decreasing,” “the CDF is left-continuous,” “F(+∞) = 1,” “p(x) can exceed 1”) where you select all the true ones — exactly the 2025 question. The second is a NAT: a small PMF table is given and you must compute a single probability such as P(X ≤ 2) or P(1 < X ≤ 3), either by summing masses or by differencing the CDF.

Worked example — build a CDF, read off probabilities

A discrete random variable X takes values 0, 1, 2, 3 with PMF p = 0.1, 0.3, 0.4, 0.2. Find the CDF, then P(X ≤ 2) and P(1 < X ≤ 3).

First check it is a valid PMF: all masses are non-negative and 0.1 + 0.3 + 0.4 + 0.2 = 1. Now accumulate to get the CDF:

x        0      1      2      3
p(x)    0.1    0.3    0.4    0.2
F(x)    0.1    0.4    0.8    1.0     ← running total

P(X ≤ 2)      = F(2)              = 0.8
P(1 < X ≤ 3)  = F(3) − F(1)       = 1.0 − 0.4 = 0.6

P(X ≤ 2) is just the CDF value at 2 (the running total 0.1 + 0.3 + 0.4). The interval P(1 < X ≤ 3) excludes X = 1 but includes X = 3, so it is the masses at 2 and 3, 0.4 + 0.2 = 0.6 — which is exactly F(3) − F(1).

Quick check