Random variables: PMF, PDF and CDF

Objectives and prerequisites

Translate outcomes into numerical random variables; distinguish mass from density; validate a PMF or PDF; construct and read a cumulative distribution function; and handle endpoints correctly. Prerequisites: event probability, sums and elementary integration. Random variables are explicitly listed in GATE 2026 CS Engineering Mathematics. The lesson prepares for the named distributions; mixed distributions and multivariate densities are outside its scope.

1. A random variable is a function

A real-valued random variable X assigns a number X(ω) to each outcome ω. The function is fixed; randomness comes from the experiment's outcome. Several outcomes may map to the same number. The event {X=x} is the set of original outcomes mapped to x, and its probability adds their weights.

Example: on two independent fair coin tosses, define X as the number of heads. HH maps to 2, HT and TH to 1, and TT to 0. Thus P(X=0)=1/4, P(X=1)=1/2, P(X=2)=1/4. The support consists of attainable values with positive mass here: {0,1,2}. A random variable's values are not automatically equiprobable, even when the original elementary outcomes are.

A discrete variable has finite or countably infinite support. A continuous variable with a density is described by area rather than point mass. A count of requests is discrete; an idealized waiting time often uses a continuous model. Real measurements may be rounded, but the assumed mathematical model determines how to calculate.

2. Probability mass function

For discrete X, its PMF is p_X(x)=P(X=x). It must satisfy p_X(x)≥0 and Σₓp_X(x)=1 over the support. For any set S, P(X∈S)=Σₓ∈S p_X(x). Values outside the support have mass zero.

Worked example: suppose p_X(x)=kx for x=1,2,3, and zero elsewhere. Normalization gives k(1+2+3)=6k=1, so k=1/6. The masses are 1/6,1/3,1/2. P(X≥2)=1/3+1/2=5/6. A proposed k=1/3 would make total mass two, so it is invalid even though some individual entries are in [0,1]. Both nonnegativity and total mass matter.

For Y=X² with this positive support, Y takes 1,4,9 with the same respective masses. When a transformation is many-to-one, add masses: if X has probabilities 1/4,1/2,1/4 at −1,0,1, then X² has probability 1/2 at zero and 1/2 at one. Do not list the value one twice as if those were different values of Y.

3. The cumulative distribution function works for every type

F_X(t)=P(X≤t). Every CDF is nondecreasing, right-continuous, lies between zero and one, and has limits zero as t→−∞ and one as t→∞. It need not be strictly increasing or continuous. If F is constant on [a,b], then P(a<X≤b)=0; mass at the left endpoint a may still be present.

For the head-count variable, F(t)=0 for t<0; 1/4 for 0≤t<1; 3/4 for 1≤t<2; and 1 for t≥2. At t=1 the value includes X=1, so F(1)=3/4. The jump at a point equals its mass: P(X=a)=F(a)−F(a−), where F(a−)=lim_{t↑a}F(t)=P(X<a).

For a<b, P(a<X≤b)=F(b)−F(a). For an inclusive left endpoint, P(a≤X≤b)=F(b)−F(a−). For the head count, P(1≤X≤2)=1−1/4=3/4, while P(1<X≤2)=1−3/4=1/4. Writing the endpoint inequalities before substituting prevents a common discrete-case error.

4. Density is not point probability

For an absolutely continuous variable X with PDF f_X(x), probabilities are integrals: P(a≤X≤b)=∫ₐᵇ f_X(x)dx. A valid PDF is nonnegative and has integral one over the entire real line. Its CDF is F_X(t)=∫_{−∞}ᵗ f_X(x)dx, and f_X(t)=F'_X(t) wherever the derivative relation applies.

The PDF's height can exceed one. Uniform density f(x)=2 on [0,1/2], zero elsewhere, is valid because area is 2×1/2=1. The height two is not a probability. For a variable with a density, P(X=a)=0 at every individual a, even when f(a)>0. Probability comes from interval area. Consequently open or closed endpoints do not change its interval probabilities.

A probability-zero point is still an allowed value in a continuous model. Do not conclude that X cannot take any value merely because each singleton has probability zero: countable additivity does not say one can add uncountably many singleton probabilities to obtain the whole interval's probability.

5. Worked continuous model

Let f(x)=cx for 0≤x≤2 and zero elsewhere. Nonnegativity requires c≥0; normalization gives ∫₀²cx dx=2c=1, hence c=1/2. The CDF is F(t)=0 for t<0, t²/4 for 0≤t≤2, and 1 for t>2. It joins continuously at both endpoints.

P(1<X≤2)=F(2)−F(1)=1−1/4=3/4. Also P(X=1)=0, although f(1)=1/2. To find a median m, solve F(m)=1/2: m²/4=1/2 gives m=√2 in the support. The negative root is outside the support and is not a median. Medians are developed further in the next lesson.

6. Choosing the right representation

A PMF is summed; a PDF is integrated; a CDF is differenced with attention to endpoints. A CDF may represent a discrete, continuous or mixed law, so never assume its derivative alone captures all mass: jumps represent point masses. We do not develop mixed laws here, but this warning explains why a step CDF's ordinary derivative being zero between jumps does not imply zero total probability.

Checklist: identify support first; normalize using the correct operation; reject negative masses or density; write the target inequalities; use sums, areas or CDF differences; check the result is in [0,1]. When deriving a CDF, include all pieces outside support and check right-continuity and limiting values. These tests often catch a correct-looking but incomplete formula.

Original practice: six single-correct MCQs

These are original questions, not previous-year questions. Choose exactly one option.

RV01. p(x)=kx for x=1,2,3 and zero elsewhere is a PMF. Find k.

A. 1/3 B. 1/6 C. 1/2 D. 1

Answer: B

A: This gives total mass two.

B: The total mass is k(1+2+3)=6k=1.

C: This gives total mass three.

D: The values must be probabilities summing to one, not the raw labels.

RV02. X counts heads in two independent fair tosses. What is F_X(1)?

A. 1/4 B. 1/2 C. 3/4 D. 1

Answer: C

A: This is P(X<1), which excludes mass at one.

B: This is P(X=1), not the cumulative probability.

C: Include X=0 and X=1: 1/4+1/2=3/4.

D: X=2 has probability 1/4 and is excluded.

RV03. For any CDF F and a<b, which expression equals P(a<X≤b)?

A. F(b)−F(a) B. F(b)−F(a−) C. F(b)+F(a) D. F(a)−F(b)

Answer: A

A: Subtract P(X≤a) from P(X≤b).

B: This includes mass at a and gives P(a≤X≤b).

C: Cumulative events overlap; adding does not isolate the interval.

D: A CDF is nondecreasing, so this is nonpositive.

RV04. f(x)=2 on [0,1/2] and zero elsewhere. Which statement is correct?

A. It is invalid because 2>1 B. P(X=1/4)=2 C. It is a valid PDF and P(X=1/4)=0 D. Its total probability is 2

Answer: C

A: PDF height may exceed one; total area must be one.

B: Density height is not point probability.

C: Area is 2×1/2=1, and a singleton has zero probability.

D: Total area is one, not the height two.

RV05. f(x)=x/2 for 0≤x≤2 and zero elsewhere. Find P(1<X≤2).

A. 1/4 B. 1/2 C. 1 D. 3/4

Answer: D

A: This is the area from zero to one.

B: This is the density at one, not the interval area.

C: This is the area over the entire support.

D: Integrate x/2 from one to two: [x²/4]₁²=3/4.

RV06. X has masses 1/4,1/2,1/4 at −1,0,1. If Y=X², find P(Y=1).

A. 1/4 B. 1/2 C. 1 D. 0

Answer: B

A: This counts only one of X=−1 and X=1.

B: Both −1 and 1 map to one, so add 1/4+1/4.

C: The mass at X=0 maps to Y=0 and is excluded.

D: Values ±1 both produce Y=1 with positive probability.

Source and scope

Random variables are explicitly part of Section 1 Probability and Statistics, official GATE 2026 CS syllabus. PMF, PDF and CDF are the representations used to teach this topic. Original examples and questions; named distributions and summary measures continue in separate lessons.

Official GATE 2026 CS syllabus (verified 2026-10-14).