Random Variables

Master discrete and continuous random variables for JEE - From dice rolls to probability density functions

13 min read

Random Variables

The Lottery Earnings Mystery

In a lottery, you can win ₹10,000 with probability 0.001, ₹100 with probability 0.01, or nothing (probability 0.989). Is it worth paying ₹50 for a ticket?

To answer this, we need to find the expected value of your winnings. This requires treating the prize money as a random variable—a function that assigns numerical values to random outcomes!

Random variables are the foundation of probability theory and essential for JEE, statistics, finance, and physics!

What is a Random Variable?

A random variable is a function that assigns a real number to each outcome in a sample space.

Random Variable Definition

Random Variable: A function $X: S \to \mathbb{R}$

Maps: Sample Space → Real Numbers

Notation: Usually denoted by capital letters X, Y, Z

Why “Random Variable”?

  • “Variable”: Takes different numerical values
  • “Random”: The value depends on the outcome of a random experiment
  • Actually: It’s a function, not a variable!

Example: Tossing Two Coins

Sample Space: S = {HH, HT, TH, TT}

Random Variable X = “number of heads”

OutcomeX (heads)
HH2
HT1
TH1
TT0

X maps each outcome to a number: HH → 2, HT → 1, TH → 1, TT → 0

Types of Random Variables

Two Main Types

1. Discrete Random Variable (DRV)

Takes countable values (finite or countably infinite)

Examples:

  • Number of heads in coin tosses: {0, 1, 2, 3, …}
  • Score on a die: {1, 2, 3, 4, 5, 6}
  • Number of defective items in a batch
  • Number of students passing an exam

2. Continuous Random Variable (CRV)

Takes uncountable values (any value in an interval)

Examples:

  • Height of students: any value in [100, 250] cm
  • Time taken to finish a race: any value in [0, ∞)
  • Temperature: any real number
  • Measurement errors

Key Difference:

  • Discrete: Count (0, 1, 2, 3, …)
  • Continuous: Measure (175.3 cm, 23.78°C, …)

Discrete Random Variables (DRV)

Probability Mass Function (PMF)

For a discrete random variable X, the probability mass function gives the probability of each value.

Probability Mass Function (PMF)
$$P(X = x_i) = p_i$$

where $x_i$ are the possible values and $p_i$ are their probabilities.

Properties:

  1. $0 \leq p_i \leq 1$ for all $i$
  2. $\sum_{i} p_i = 1$ (total probability = 1)

Notation: Sometimes written as $p(x)$ or $f(x)$

Example: PMF of Two Coins

X = number of heads when tossing 2 coins

X012
P(X)1/41/21/4

Verification: 1/4 + 1/2 + 1/4 = 1 ✓

Cumulative Distribution Function (CDF)

The CDF gives the probability that X is at most x.

Cumulative Distribution Function (CDF)
$$F(x) = P(X \leq x) = \sum_{x_i \leq x} P(X = x_i)$$

Properties:

  1. $0 \leq F(x) \leq 1$
  2. $F$ is non-decreasing (never goes down)
  3. $\lim_{x \to -\infty} F(x) = 0$
  4. $\lim_{x \to \infty} F(x) = 1$
  5. $P(a < X \leq b) = F(b) - F(a)$

Example: CDF of Two Coins

For X = number of heads:

$$F(x) = \begin{cases} 0 & \text{if } x < 0 \\ 1/4 & \text{if } 0 \leq x < 1 \\ 3/4 & \text{if } 1 \leq x < 2 \\ 1 & \text{if } x \geq 2 \end{cases}$$

Graph: Step function (jumps at 0, 1, 2)

Expectation (Mean) of Discrete RV

The expected value or mean is the long-run average value.

Expectation (Expected Value)
$$E(X) = \mu = \sum_{i} x_i \cdot P(X = x_i)$$

Alternative notation: $\mathbb{E}[X]$, $\mu_X$

Interpretation: Weighted average, where weights are probabilities

Properties of Expectation

Expectation Properties

  1. Linearity: $E(aX + b) = aE(X) + b$

  2. Sum: $E(X + Y) = E(X) + E(Y)$ (always, even if dependent!)

  3. Constant: $E(c) = c$

  4. Product (if independent): $E(XY) = E(X) \cdot E(Y)$

  5. Non-negative: If $X \geq 0$, then $E(X) \geq 0$

Example: Expected Dice Roll

Fair die: X ∈ {1, 2, 3, 4, 5, 6}, each with P = 1/6

$$E(X) = 1 \cdot \frac{1}{6} + 2 \cdot \frac{1}{6} + 3 \cdot \frac{1}{6} + 4 \cdot \frac{1}{6} + 5 \cdot \frac{1}{6} + 6 \cdot \frac{1}{6}$$ $$= \frac{1+2+3+4+5+6}{6} = \frac{21}{6} = 3.5$$

Interpretation: On average, rolling 3.5 (even though you can never actually roll 3.5!)

Variance and Standard Deviation of DRV

Variance measures the spread of X around its mean.

Variance of Random Variable
$$\text{Var}(X) = \sigma^2 = E[(X - \mu)^2] = \sum_i (x_i - \mu)^2 \cdot P(X = x_i)$$

Alternative formula (easier to compute):

$$\text{Var}(X) = E(X^2) - [E(X)]^2$$

Standard Deviation:

$$\sigma = \text{SD}(X) = \sqrt{\text{Var}(X)}$$

Properties of Variance

Variance Properties

  1. Constant shift: $\text{Var}(X + b) = \text{Var}(X)$

  2. Scale: $\text{Var}(aX) = a^2 \text{Var}(X)$

  3. Constant: $\text{Var}(c) = 0$

  4. Non-negative: $\text{Var}(X) \geq 0$

  5. Sum (if independent): $\text{Var}(X + Y) = \text{Var}(X) + \text{Var}(Y)$

Example: Variance of Fair Die

$$E(X) = 3.5$$ $$E(X^2) = 1^2 \cdot \frac{1}{6} + 2^2 \cdot \frac{1}{6} + ... + 6^2 \cdot \frac{1}{6} = \frac{1+4+9+16+25+36}{6} = \frac{91}{6}$$ $$\text{Var}(X) = \frac{91}{6} - (3.5)^2 = \frac{91}{6} - \frac{49}{4} = \frac{182 - 147}{12} = \frac{35}{12} \approx 2.92$$

Continuous Random Variables (CRV)

For continuous RVs, we can’t list all values (uncountably many!), so we use density instead of mass.

Probability Density Function (PDF)

Probability Density Function (PDF)

A function $f(x)$ is a PDF if:

  1. Non-negative: $f(x) \geq 0$ for all $x$

  2. Total probability: $\int_{-\infty}^{\infty} f(x) \, dx = 1$

Probability of interval:

$$P(a \leq X \leq b) = \int_a^b f(x) \, dx$$

Important: For continuous RV, $P(X = a) = 0$ (probability at a single point is 0!)

Example: Uniform Distribution on [0, 1]

$$f(x) = \begin{cases} 1 & \text{if } 0 \leq x \leq 1 \\ 0 & \text{otherwise} \end{cases}$$

Check: $\int_0^1 1 \, dx = 1$ ✓

Probability: $P(0.2 \leq X \leq 0.5) = \int_{0.2}^{0.5} 1 \, dx = 0.3$

CDF for Continuous RV

CDF for Continuous RV
$$F(x) = P(X \leq x) = \int_{-\infty}^{x} f(t) \, dt$$

Relationship between PDF and CDF:

$$f(x) = \frac{d}{dx} F(x)$$

(PDF is the derivative of CDF)

Expectation and Variance for Continuous RV

Expectation for Continuous RV
$$E(X) = \int_{-\infty}^{\infty} x \cdot f(x) \, dx$$ $$E(g(X)) = \int_{-\infty}^{\infty} g(x) \cdot f(x) \, dx$$

Variance:

$$\text{Var}(X) = E(X^2) - [E(X)]^2 = \int_{-\infty}^{\infty} x^2 f(x) \, dx - \mu^2$$

Example: Uniform [0, 1] Expectation

$$E(X) = \int_0^1 x \cdot 1 \, dx = \left[\frac{x^2}{2}\right]_0^1 = \frac{1}{2}$$ $$E(X^2) = \int_0^1 x^2 \cdot 1 \, dx = \left[\frac{x^3}{3}\right]_0^1 = \frac{1}{3}$$ $$\text{Var}(X) = \frac{1}{3} - \left(\frac{1}{2}\right)^2 = \frac{1}{3} - \frac{1}{4} = \frac{1}{12}$$

Interactive Visualization

Memory Tricks

Memory Tricks

  1. “Discrete = Distinct, Continuous = Connected”

    • Discrete has Distinct separate values
    • Continuous is Connected (interval)

  2. PMF vs PDF:

    • PMF = Probability Mass (discrete, like point masses)
    • PDF = Probability Density (continuous, like density)

  3. Expectation = Weighted Average:

    • E(X) = Σ x × P(x) - weights are probabilities

  4. Variance Formula:

    • “E of Square minus Square of E”
    • Var(X) = E(X²) - [E(X)]²

  5. CDF = Cumulative = Up to:

    • F(x) = P(X ≤ x) - up to x

  6. PDF and CDF Relationship:

    • f is slope of F (derivative)
    • F is area under f (integral)

Common Mistakes to Avoid

Common Mistakes

  1. P(X = x) for continuous RV: Always 0! Use intervals instead

  2. Confusing PMF and PDF: PMF gives probability directly, PDF needs integration

  3. Forgetting to check Σp = 1: PMF must sum to 1

  4. Wrong variance formula: Var(X) = E(X²) - [E(X)]², NOT E[(X - μ)²] (though they’re equal)

  5. E(XY) ≠ E(X)E(Y): Only true if X and Y are independent

  6. Var(X + Y) ≠ Var(X) + Var(Y): Only true if independent

  7. CDF discontinuity: For discrete RV, CDF has jumps (step function)

  8. Units: Var(X) has squared units, SD(X) has original units

Solved Examples

Example 1: PMF and Expectation (Dice Game)

Problem: A die is rolled. If it shows an even number, you win that number of rupees. If odd, you lose ₹1. Find your expected earnings.

Solution:

Random Variable X = earnings

Outcome123456
X-12-14-16
P(X)1/61/61/61/61/61/6

Expectation:

$$E(X) = (-1) \cdot \frac{3}{6} + 2 \cdot \frac{1}{6} + 4 \cdot \frac{1}{6} + 6 \cdot \frac{1}{6}$$ $$= \frac{-3 + 2 + 4 + 6}{6} = \frac{9}{6} = \frac{3}{2} = ₹1.50$$

Answer: Expected earning = ₹1.50 per game

Example 2: Variance Calculation

Problem: X has PMF: P(X = 0) = 0.3, P(X = 1) = 0.5, P(X = 2) = 0.2. Find Var(X).

Solution:

Step 1: Find E(X)

$$E(X) = 0 \cdot 0.3 + 1 \cdot 0.5 + 2 \cdot 0.2 = 0 + 0.5 + 0.4 = 0.9$$

Step 2: Find E(X²)

$$E(X^2) = 0^2 \cdot 0.3 + 1^2 \cdot 0.5 + 2^2 \cdot 0.2 = 0 + 0.5 + 0.8 = 1.3$$

Step 3: Calculate Var(X)

$$\text{Var}(X) = E(X^2) - [E(X)]^2 = 1.3 - (0.9)^2 = 1.3 - 0.81 = 0.49$$

Example 3: Linear Transformation

Problem: X has E(X) = 10, Var(X) = 4. Find E(Y) and Var(Y) where Y = 3X - 5.

Solution:

E(Y):

$$E(Y) = E(3X - 5) = 3E(X) - 5 = 3(10) - 5 = 25$$

Var(Y):

$$\text{Var}(Y) = \text{Var}(3X - 5) = 3^2 \text{Var}(X) = 9 \cdot 4 = 36$$

(Constant -5 doesn’t affect variance)

Example 4: Continuous RV (Triangular Distribution)

Problem: X has PDF $f(x) = cx$ for $0 \leq x \leq 2$, and 0 otherwise. Find: (a) Value of c (b) E(X) (c) P(X > 1)

Solution:

(a) Find c:

$$\int_0^2 cx \, dx = 1$$ $$c \left[\frac{x^2}{2}\right]_0^2 = 1$$ $$c \cdot 2 = 1$$ $$c = \frac{1}{2}$$

(b) E(X):

$$E(X) = \int_0^2 x \cdot \frac{x}{2} \, dx = \frac{1}{2} \int_0^2 x^2 \, dx = \frac{1}{2} \left[\frac{x^3}{3}\right]_0^2 = \frac{1}{2} \cdot \frac{8}{3} = \frac{4}{3}$$

(c) P(X > 1):

$$P(X > 1) = \int_1^2 \frac{x}{2} \, dx = \frac{1}{2} \left[\frac{x^2}{2}\right]_1^2 = \frac{1}{2} \left(2 - \frac{1}{2}\right) = \frac{1}{2} \cdot \frac{3}{2} = \frac{3}{4}$$

Example 5: CDF and Probability

Problem: X has CDF:

$$F(x) = \begin{cases} 0 & x < 0 \\ x^2 & 0 \leq x < 1 \\ 1 & x \geq 1 \end{cases}$$

Find: (a) PDF, (b) P(0.5 < X < 0.8)

Solution:

(a) PDF: $f(x) = \frac{d}{dx} F(x)$

$$f(x) = \begin{cases} 0 & x < 0 \\ 2x & 0 \leq x < 1 \\ 0 & x \geq 1 \end{cases}$$

(b) Probability:

Method 1 (using CDF):

$$P(0.5 < X < 0.8) = F(0.8) - F(0.5) = (0.8)^2 - (0.5)^2 = 0.64 - 0.25 = 0.39$$

Method 2 (using PDF):

$$P(0.5 < X < 0.8) = \int_{0.5}^{0.8} 2x \, dx = [x^2]_{0.5}^{0.8} = 0.64 - 0.25 = 0.39$$

Example 6: Lottery Problem (Opening Example)

Problem: Lottery ticket costs ₹50. Prizes: ₹10,000 (p=0.001), ₹100 (p=0.01), ₹0 (p=0.989). Is it worth buying?

Solution:

Random Variable X = net profit (prize - cost)

PrizeProbabilityNet Profit X
₹10,0000.001₹9,950
₹1000.01₹50
₹00.989-₹50

Expected Profit:

$$E(X) = 9950 \cdot 0.001 + 50 \cdot 0.01 + (-50) \cdot 0.989$$ $$= 9.95 + 0.50 - 49.45 = -39$$

Answer: Expected loss = ₹39. Not worth buying! (On average, you lose money)

Practice Problems

Level 1: Foundation (JEE Main)

  1. Two coins are tossed. X = number of tails. Find PMF, E(X), and Var(X).

  2. X has PMF: P(X = 1) = 0.4, P(X = 2) = 0.6. Find E(X).

  3. If E(X) = 5, find E(2X + 3).

Solutions

  1. PMF: P(X=0)=1/4, P(X=1)=1/2, P(X=2)=1/4 E(X) = 0(1/4) + 1(1/2) + 2(1/4) = 1 E(X²) = 0(1/4) + 1(1/2) + 4(1/4) = 1.5 Var(X) = 1.5 - 1² = 0.5

  2. E(X) = 1(0.4) + 2(0.6) = 0.4 + 1.2 = 1.6

  3. E(2X + 3) = 2E(X) + 3 = 2(5) + 3 = 13

Level 2: Intermediate (JEE Main/Advanced)

  1. X has PDF $f(x) = kx^2$ for $0 \leq x \leq 1$. Find k and E(X).

  2. X and Y are independent with E(X) = 3, E(Y) = 5, Var(X) = 2, Var(Y) = 4. Find E(XY) and Var(X + Y).

  3. CDF of X is $F(x) = 1 - e^{-2x}$ for $x \geq 0$. Find PDF and P(X > 1).

Solutions

  1. ∫₀¹ kx² dx = 1 → k[x³/3]₀¹ = 1 → k/3 = 1 → k = 3 E(X) = ∫₀¹ x·3x² dx = 3∫₀¹ x³ dx = 3[x⁴/4]₀¹ = 3/4

  2. E(XY) = E(X)E(Y) = 3·5 = 15 (independent) Var(X+Y) = Var(X) + Var(Y) = 2 + 4 = 6 (independent)

  3. f(x) = F’(x) = 2e⁻²ˣ for x ≥ 0 P(X > 1) = 1 - F(1) = 1 - (1 - e⁻²) = e⁻² ≈ 0.135

Level 3: Advanced (JEE Advanced)

  1. Show that Var(X) = E(X²) - [E(X)]² starting from Var(X) = E[(X - μ)²].

  2. X has PDF $f(x) = \frac{3}{8}(4x - 2x^2)$ for $0 \leq x \leq 2$. Find median of X (value m where P(X ≤ m) = 0.5).

  3. If X and Y are independent, prove that E(XY) = E(X)E(Y).

Solutions

  1. Var(X) = E[(X-μ)²] = E[X² - 2μX + μ²] = E(X²) - 2μE(X) + μ² = E(X²) - 2μ² + μ² = E(X²) - μ² = E(X²) - [E(X)]² ✓

  2. CDF: F(x) = ∫₀ˣ (3/8)(4t - 2t²) dt = (3/8)[2t² - 2t³/3]₀ˣ = (3/8)(2x² - 2x³/3) Set F(m) = 0.5: (3/8)(2m² - 2m³/3) = 0.5 2m² - 2m³/3 = 4/3 → 6m² - 2m³ = 4 → m³ - 3m² + 2 = 0 Solve numerically: m ≈ 0.71

  3. For independent X, Y: E(XY) = ΣΣ xy P(X=x,Y=y) = ΣΣ xy P(X=x)P(Y=y) [independence] = Σx P(X=x) · Σy P(Y=y) = E(X)·E(Y) ✓

Real-Life Applications

1. Finance

  • Stock Returns: Continuous RV (price can be any value)
  • Expected Return: E(X) guides investment decisions
  • Risk: Measured by variance/standard deviation

2. Insurance

  • Claim Amount: Random variable
  • Premium Calculation: Based on E(X) + buffer
  • Risk Assessment: Var(X) quantifies uncertainty

3. Quality Control

  • Number of Defects: Discrete RV
  • Process Control: Monitor E(X) and σ
  • Six Sigma: ±6σ from mean

4. Games and Gambling

  • Casino Games: Calculate expected winnings
  • Fair Game: E(X) = 0
  • House Edge: Negative E(X) for player

5. Physics

  • Quantum Mechanics: Wave function = probability density
  • Measurement: Random variable
  • Expectation Value: Observable quantity

Connection to Other Topics

Cross-Topic Links

Related JEE Topics:

Advanced Connections:

  • Integration → Computing E(X) for continuous RV
  • Summation → Computing E(X) for discrete RV
  • Differentiation → PDF from CDF

Formula Quick Reference

Formula Summary

Discrete Random Variable

PropertyFormula
PMF$P(X = x_i) = p_i$, $\sum p_i = 1$
CDF$F(x) = P(X \leq x) = \sum_{x_i \leq x} p_i$
Expectation$E(X) = \sum x_i p_i$
Variance$\text{Var}(X) = E(X^2) - [E(X)]^2$

Continuous Random Variable

PropertyFormula
PDF$f(x) \geq 0$, $\int_{-\infty}^{\infty} f(x) dx = 1$
CDF$F(x) = \int_{-\infty}^{x} f(t) dt$
PDF from CDF$f(x) = F'(x)$
Expectation$E(X) = \int_{-\infty}^{\infty} x f(x) dx$
Variance$\text{Var}(X) = \int x^2 f(x) dx - \mu^2$

Properties (Both Types)

  • $E(aX + b) = aE(X) + b$
  • $\text{Var}(aX + b) = a^2 \text{Var}(X)$
  • $E(X + Y) = E(X) + E(Y)$
  • $\text{Var}(X + Y) = \text{Var}(X) + \text{Var}(Y)$ (if independent)

Quick Revision Checklist

Quick Revision
✓ Random Variable: Function from sample space to real numbers ✓ Discrete: Countable values (PMF, summation) ✓ Continuous: Interval values (PDF, integration) ✓ PMF: $\sum p_i = 1$, gives probability directly ✓ PDF: $\int f(x) dx = 1$, need integral for probability ✓ CDF: F(x) = P(X ≤ x), always between 0 and 1 ✓ Expectation: E(X) = weighted average ✓ Variance: Var(X) = E(X²) - [E(X)]² ✓ Linear Transform: E(aX+b) = aE(X)+b, Var(aX+b) = a²Var(X)

Exam Tips

  1. Identify type: Discrete or continuous? Determines which formulas to use
  2. Check normalization: PMF must sum to 1, PDF must integrate to 1
  3. Use alternative variance formula: Var(X) = E(X²) - [E(X)]² is faster
  4. For continuous: P(X = a) = 0, always use intervals
  5. CDF properties: Non-decreasing, limits are 0 and 1
  6. Independence: Simplifies E(XY) and Var(X+Y)
  7. Units: Var has squared units, SD has original units

Next Steps

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Last updated: December 20, 2025 Master random variables at JEENotes Practice Portal