---
title: "Joint pmf: marginals and independence"
subject: "Probability and Statistics"
chapter: 2
key: "semester_5/probability_and_statistics:joint-pmf-marginal-independence"
source: https://www.bctengineer.com/q/semester_5/probability_and_statistics/joint-pmf-marginal-independence
site: BCT Engineering
---

# Joint pmf: marginals and independence

Probability and Statistics · Chapter 2: Discrete Probability Distribution

- Asked in: 2068 Jestha (old course), 5 marks
- Video: [38-Joint Probability Mass Function (PMF)](https://www.youtube.com/watch?v=aOVGeF976kY) — Probability Course
- Answer checked by Claude Sonnet 5.5 on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_5/probability_and_statistics/joint-pmf-marginal-independence
- Whole chapter: https://www.bctengineer.com/md/semester_5/probability_and_statistics/chapter-2

## Question

The probability mass function is as given below:
| Y \ X | 1 | 2 |
|---|---|---|
| 1 | 0.5 | 0.1 |
| 2 | 0.1 | 0.3 |
i) Determine the marginal mass distribution for X and Y.
ii) Whether the random variables are independent?

## Short answer

Marginals are p_X(1)=0.6, p_X(2)=0.4, p_Y(1)=0.6, p_Y(2)=0.4. X and Y are not independent because p(1,1)=0.5, while p_X(1)p_Y(1)=0.36.

## Answer

Joint pmf p(x, y) from the table (rows Y, columns X):

| Y \ X | 1 | 2 | Row total p_Y(y) |
|---|---|---|---|
| 1 | 0.5 | 0.1 | 0.6 |
| 2 | 0.1 | 0.3 | 0.4 |
| Column total p_X(x) | 0.6 | 0.4 | 1.0 |

The total is 0.5 + 0.1 + 0.1 + 0.3 = 1, so it is a valid joint pmf.

#### i) Marginal distributions
Marginal of X: p_X(x) = Σ_y p(x, y)

| x | 1 | 2 |
|---|---|---|
| p_X(x) | 0.5 + 0.1 = 0.6 | 0.1 + 0.3 = 0.4 |

Marginal of Y: p_Y(y) = Σ_x p(x, y)

| y | 1 | 2 |
|---|---|---|
| p_Y(y) | 0.5 + 0.1 = 0.6 | 0.1 + 0.3 = 0.4 |

#### ii) Independence
X and Y are independent only if p(x, y) = p_X(x) · p_Y(y) for **every** pair.

Check (1, 1): p_X(1) · p_Y(1) = 0.6 × 0.6 = 0.36, but p(1, 1) = 0.5 ≠ 0.36.

Since the condition fails, **X and Y are not independent**.

(Also, Cov(X, Y) = E(XY) − E(X)E(Y) = (0.5 + 0.2 + 0.2 + 1.2) − 1.4 × 1.4 = 2.1 − 1.96 = 0.14 ≠ 0.)

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. A marginal of X is obtained by:
   A. Summing the joint pmf over y
   B. Multiplying by y
   C. Taking the largest entry
   D. Dividing by P(Y)
   Answer: A. p_X(x) = sum over y of p(x, y). (Key point: Marginals)

2. p_X(1) is:
   A. 0.5
   B. 0.4
   C. 0.6
   D. 0.1
   Answer: C. 0.5 + 0.1 = 0.6. (Key point: Marginal of X)

3. X and Y are independent if:
   A. p(x,y) = p_X(x) + p_Y(y) for one pair
   B. The marginals are equal
   C. p(x,y) = p_X(x) p_Y(y) for every pair
   D. The total is 1
   Answer: C. The product rule must hold for all pairs. (Key point: Independence test)

4. Here X and Y are:
   A. Not independent, since p(1,1) = 0.5 but 0.6 x 0.6 = 0.36
   B. Independent, since marginals are equal
   C. Independent, since total is 1
   D. Not independent, since p(1,1) = 0.36
   Answer: A. One failing pair is enough. (Key point: Conclusion)

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