Chapter 42 — Joint Densities Challenge Solutions

Chapter 42 · Solutions

Joint Densities, Worked Answers ✅

Full solutions to the five challenges. Try them yourself first, then compare.

Statistics, Data Science and AI: A Visual Handbook · John Fisher · 2026

⚙️ Setup¶

In [1]:

import numpy as np
rng = np.random.default_rng(421)
joint = np.array([[0.10, 0.05, 0.05],
                  [0.05, 0.20, 0.10],
                  [0.05, 0.10, 0.30]])   # rows = X in {0,1,2}, cols = Y in {0,1,2}
print("joint sums to", joint.sum())

joint sums to 1.0

CHALLENGE 1 · READ THE JOINT

A specific pair

Using the 3x3 joint above (X, Y each in {0,1,2}), find P(X=2, Y=2) and P(X=Y).

In [2]:

print(f"P(X=2, Y=2) = {joint[2,2]:.2f}")
print(f"P(X=Y) = {np.trace(joint):.2f}")

P(X=2, Y=2) = 0.30
P(X=Y) = 0.60

Answer: P(X=2, Y=2) is the bottom-right cell, 0.30. P(X=Y) sums the diagonal: 0.10 + 0.20 + 0.30 = 0.60.

CHALLENGE 2 · MARGINALS

Each variable alone

Find the marginal distributions of X and Y.

In [3]:

pX = joint.sum(axis=1); pY = joint.sum(axis=0)
print(f"P(X): {pX.round(2)}")
print(f"P(Y): {pY.round(2)}")

P(X): [0.2  0.35 0.45]
P(Y): [0.2  0.35 0.45]

Answer: summing each row gives P(X) = [0.20, 0.35, 0.45]; summing each column gives P(Y) = [0.20, 0.35, 0.45]. Both marginals are valid distributions summing to 1.

CHALLENGE 3 · CONDITIONAL

Y given X

Find the conditional distribution P(Y | X=2).

In [4]:

pX = joint.sum(axis=1)
cond = joint[2]/pX[2]
print(f"P(Y | X=2) = {cond.round(3)}")

P(Y | X=2) = [0.111 0.222 0.667]

Answer: take the X=2 row [0.05, 0.10, 0.30] and divide by P(X=2)=0.45, giving P(Y | X=2) ≈ [0.111, 0.222, 0.667]. Given X=2, Y is most likely 2 as well, the variables are positively associated.

CHALLENGE 4 · INDEPENDENT?

Test the factorization

Are X and Y independent? Compare the joint to the product of the marginals.

In [5]:

pX=joint.sum(axis=1); pY=joint.sum(axis=0)
indep = np.outer(pX,pY)
print(f"product of marginals[0,0] = {indep[0,0]:.3f}, actual joint[0,0] = {joint[0,0]:.3f}")
print(f"independent? {np.allclose(joint, indep)}")

product of marginals[0,0] = 0.040, actual joint[0,0] = 0.100
independent? False

Answer: the product of marginals gives 0.04 in the top-left cell, but the actual joint has 0.10, so the joint does not factor into the marginals. X and Y are dependent (they tend to move together, as the conditional in Challenge 3 showed).

CHALLENGE 5 · CORRELATION

From a simulated joint

Simulate 5000 draws from a bivariate normal with correlation 0.6 and recover the correlation from the sample.

In [6]:

cov=[[1,0.6],[0.6,1]]
xy=rng.multivariate_normal([0,0], cov, size=5000)
print(f"sample correlation = {np.corrcoef(xy.T)[0,1]:.3f}  (target 0.6)")
print(f"sample covariance matrix:\n{np.cov(xy.T).round(3)}")

sample correlation = 0.587  (target 0.6)
sample covariance matrix:
[[0.995 0.579]
 [0.579 0.98 ]]

Answer: the recovered correlation is close to the target 0.6, and the off-diagonal of the sample covariance matrix is near 0.6 as well. Correlation is the standardized covariance, a unitless measure of how two variables move together, and it is read straight off the joint distribution.

🎉 Nicely done

You read joint probabilities, recovered marginals and conditionals, tested independence by factorization, and measured correlation from a simulated joint. The joint distribution holds all the information; marginals and conditionals are just two ways of looking at it.