X and Y are independent if:

p(x,y) = p_X(x) + p_Y(y) for one pair

Not independent, since p(1,1) = 0.5 but 0.6 x 0.6 = 0.36

Independent, since marginals are equal

Not independent, since p(1,1) = 0.36

Joint pmf: marginals and independence

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.

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?

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.

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

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.)

✓ Answer checked by Claude Sonnet 5.5