Joint distribution question?

i) To find k, remember that the total area under the joint distribution is 1:
∫(x = 0 to 1) ∫(y = 0 to 1-x) k dy dx = 1
==> ∫(x = 0 to 1) k(1 - x) dx = 1
==> k/2 = 1
==> k = 2.
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ii) To find the marginal distribution for X, we integrate through y:
fx(x) = ∫(y = 0 to 1-x) 2 dy
........= 2(1 - x) if x is in (0, 1), and 0 otherwise.

Similarly,
fy(y) = ∫(x = 0 to 1-y) 2 dx
........= 2(1 - y) if y is in (0, 1), and 0 otherwise.

X and Y are not independent, because f(x,y) ≠ fx(x) fy(y) for all (x, y).
For instance, f(1/2, 1/2) = 2, but fx(1/2) fy(1/2) = 1 * 1 = 1.
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iii) E[X] = ∫ x fx(x) dx
..............= ∫(x = 0 to 1) x * 2(1 - x) dx
..............= ∫(x = 0 to 1) (2x - 2x^2) dx
..............= (x^2 - (2/3)x^3) {for x = 0 to 1}
..............= 1/3.

Similarly,
E[Y] = ∫ y fy(y) dy
........= ∫(y = 0 to 1) y * 2(1 - y) dy
........= 1/3.

E[XY] = ∫∫ xy f(x,y) dA
...........= ∫(x = 0 to 1) ∫(y = 0 to 1-x) xy * 2 dy dx
...........= ∫(x = 0 to 1) xy^2 {for y = 0 to 1-x} dx
...........= ∫(x = 0 to 1) x(1 - x)^2 dx
...........= ∫(x = 0 to 1) (x - 2x^2 + x^3) dx
...........= ((1/2)x^2 - (2/3)x^3 + (1/4)x^4) {for x = 0 to 1}
...........= 1/12.

So, Cov(X, Y) = E[XY] - E[X] E[Y] = (1/12) - (1/3)^2 = -1/36.
(Note: The fact that the covariance is nonzero also shows that X and Y are not independent.)
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I hope this helps!

The double integral within the region needs to equal 1. Since the region is just a triangle:
∫ ∫ ƒ(x, y) dydx = k / 2
Since that needs to equal 1:
k / 2 = 1
k = 2