Let the probability density function of two-dimensional random variable (x, y) be f (x, y) = {asin (x + y), 0

Let the probability density function of two-dimensional random variable (x, y) be f (x, y) = {asin (x + y), 0

According to the integral of probability density function = 1, the value of a can be calculated, that is: ∫ ∫ f (x, y) DX dy = 1 (∫ all from - ∞ integral to + ∞)
From the known conditions in the question,
+∞ π/2
∫∫ f (x, y) DX dy = ∫∫ a sin (x + y) DX Dy (x, y are integral from 0 to π / 2)
-∞ 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
So a = 1 / 2
f(x,y) = 1/2 sin(x+y),...
+∞
The edge distribution density of X 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)]
+∞
Similarly, we can calculate the edge distribution density of Y, 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)]
If x is not in [0, π / 2], y is not in [0, π / 2], it is equal to 0