根據(jù)概率密度函數(shù)的積分=1,可以算出A的值.即：∫ ∫ f(x,y) dx dy = 1 （∫ 均從-∞積分到+∞）.
則從題中已知條件可得,
+∞ π/2
∫ ∫ f(x,y) dx dy = ∫ ∫ A sin(x+y) dx dy (x,y 均從0到π/2積分)
-∞ 0
π/2
= A ∫ -[cos(π/2+y) - cos(0+y)] dy
0
π/2
= A ∫ [sin(y) + cos(y)] dy
0
= A [cos0 - cos(π/2) + sin(π/2) - sin0]
= 2A = 1
所以 A = 1/2.
f(x,y) = 1/2 sin(x+y),...
+∞
x的邊緣分布密度 fX(x) = ∫ f(x,y) dy
-∞
π/2
= 1/2 ∫ sin(x+y) dy
0
= 1/2 [-cos(x+π/2) + cos(x+0)]
= 1/2 [sin(x) + cos(x)]
+∞
同理可以算得y的邊緣分布密度 fY(y) = ∫ f(x,y) dx
-∞
π/2
= 1/2 ∫ sin(x+y) dx
0
= 1/2 [-cos(y+π/2) + cos(y+0)]
= 1/2 [sin(y) + cos(y)]
對(duì)于x不在 [0,π/2],y不在 [0,π/2]的,都等于0.