Joint pdf a(x+y): marginals, probability

Find the constant a

For a joint pdf, ∫∫ f(x, y) dy dx = 1.

∫₀¹ ∫₁² a(x + y) dy dx = a ∫₀¹ [xy + y²/2]₁² dx = a ∫₀¹ (x + 3/2) dx = a [x²/2 + 3x/2]₀¹ = a(1/2 + 3/2) = 2a

2a = 1, so a = 1/2. Thus f(x, y) = (x + y)/2, 0 < x < 1, 1 < y < 2.

i) Marginal functions

Marginal of X:

g(x) = ∫₁² (x + y)/2 dy = (1/2) [xy + y²/2]₁² = (1/2)(x + 3/2) = (2x + 3)/4, 0 < x < 1

Marginal of Y:

h(y) = ∫₀¹ (x + y)/2 dx = (1/2) [x²/2 + xy]₀¹ = (1/2)(1/2 + y) = (2y + 1)/4, 1 < y < 2

Check: ∫₀¹ (2x + 3)/4 dx = (1 + 3)/4 = 1 and ∫₁² (2y + 1)/4 dy = (3 + 1)/4 = 1.

ii) P(0.5 < X < 0.8 and Y > 1.5)

P = ∫₀.₅^₀.₈ ∫₁.₅² (x + y)/2 dy dx

Inner integral: ∫₁.₅² (x + y) dy = [xy + y²/2]₁.₅² = 0.5x + (2 − 1.125) = 0.5x + 0.875

P = (1/2) ∫₀.₅^₀.₈ (0.5x + 0.875) dx = (1/2) [0.25x² + 0.875x]₀.₅^₀.₈

= (1/2) [(0.16 + 0.7) − (0.0625 + 0.4375)] = (1/2)(0.86 − 0.50) = 0.18

Answer: a = 1/2; g(x) = (2x + 3)/4; h(y) = (2y + 1)/4; P = 0.18.