PMF Joint and Marginal PMFs 1 — Questions and Answers
Question 1: For two discrete random variables X and Y, the joint PMF p(x, y) must satisfy which condition?
- ∑_x ∑_y p(x, y) = 1 and p(x, y) ≥ 0 for all x, y (Correct answer)
- ∑_x p(x, y) = 1 for each fixed y
- p(x, y) = p(x) · p(y) for all x, y
- ∑_x ∑_y p(x, y) = 0
Correct answer: ∑_x ∑_y p(x, y) = 1 and p(x, y) ≥ 0 for all x, y
A valid joint PMF must be non-negative everywhere and sum to 1 over all possible pairs (x, y).
Question 2: The marginal PMF of X is obtained from the joint PMF p(x, y) by:
- Multiplying p(x, y) by p(y) for each y
- Summing p(x, y) over all values of y (Correct answer)
- Dividing p(x, y) by p(y) for each y
- Summing p(x, y) over all values of x
Correct answer: Summing p(x, y) over all values of y
The marginal PMF p_X(x) = ∑_y p(x, y) is obtained by summing the joint PMF over all values of the other variable Y.
Question 3: X and Y are discrete random variables with joint PMF p(x, y). They are independent if and only if:
- p(x, y) = p_X(x) + p_Y(y) for all x, y
- p(x | y) = p(y | x) for all x, y
- p(x, y) = p_X(x) · p_Y(y) for all x, y (Correct answer)
- E[XY] = 0
Correct answer: p(x, y) = p_X(x) · p_Y(y) for all x, y
X and Y are independent if and only if their joint PMF factors as the product of their marginal PMFs for every pair (x, y).
Question 4: Suppose p(x, y) = c · x · y for x ∈ {1, 2} and y ∈ {1, 2}, and 0 otherwise. What is the value of c?
- 1/9 (Correct answer)
- 1/6
- 1/4
- 1/3
Correct answer: 1/9
Summing c·x·y over all pairs: c(1·1 + 1·2 + 2·1 + 2·2) = c(1+2+2+4) = 9c = 1, so c = 1/9.
Question 5: Given joint PMF p(1,1)=0.1, p(1,2)=0.2, p(2,1)=0.3, p(2,2)=0.4, what is the marginal PMF p_X(1)?
- 0.1
- 0.4
- 0.3 (Correct answer)
- 0.7
Correct answer: 0.3
p_X(1) = p(1,1) + p(1,2) = 0.1 + 0.2 = 0.3, summing over all values of Y.
Question 6: For the joint PMF p(x, y) = 1/9 · x · y with x, y ∈ {1, 2}, are X and Y independent?
- Yes, because p(x, y) = p_X(x) · p_Y(y) for all x, y (Correct answer)
- No, because E[XY] ≠ E[X]E[Y]
- No, because the marginals are not uniform
- Yes, because the joint PMF is symmetric
Correct answer: Yes, because p(x, y) = p_X(x) · p_Y(y) for all x, y
With p_X(1)=1/3, p_X(2)=2/3, p_Y(1)=1/3, p_Y(2)=2/3, we verify p(x,y) = p_X(x)·p_Y(y) holds for all pairs, confirming independence.
Question 7: The conditional PMF of X given Y = y is defined as:
- p(x | y) = p(x, y) · p_Y(y)
- p(x | y) = p(x, y) / p_Y(y), provided p_Y(y) > 0 (Correct answer)
- p(x | y) = p_X(x) / p(x, y)
- p(x | y) = p_X(x) · p(x, y)
Correct answer: p(x | y) = p(x, y) / p_Y(y), provided p_Y(y) > 0
The conditional PMF is defined as p(x | y) = p(x, y) / p_Y(y), analogous to conditional probability P(A|B) = P(A∩B)/P(B).
PMF Joint and Marginal PMFs 1
For two discrete random variables X and Y, the joint PMF p(x, y) must satisfy which condition?