Problem
Let X and Y be two random variables with the following joint pdf:
$f_{X,Y}(x,y) = 2$ for $x+y<1, x>0, y>0.$
$f_{X,Y}(x,y) = 0, otherwise.$
a) Find the marginal densities of X and Y .
b) Calculate Cov(X, Y) and the correlation coefficient ρ(X, Y).
Attempt
a)
$f_{x}(x) = \int_0^{1-x}2 dy = 2(1-x) = -2x+2$
$f_{y}(y) = \int_0^{1-y}2 dx = 2(1-y) = -2y+2$
b) i) covariance
$Cov(X, Y) = E(XY) - E(X)E(Y) = $
$\int\int xy ⋅ f(x,y) dy dx - \int x ⋅ f(x) dx ⋅ \int y ⋅ f(y) dy = $
$\int_0^1\int_0^{1-x} (2xy) dy dx - \int_0^1 (-2x^2 + 2x) dx ⋅ \int_0^1 (-2y^2 + 2y) dy $
$\int_0^1\int_0^{1-x} (2xy) dy dx = \int_0^1 \frac{2x(1-x)}{2} dx = \int_0^1 x - x^2 dx = \frac{2}{3} $
$\int_0^1 (-2x^2 + 2x) dx = \int_0^1 (-2y^2 + 2y) dy = \frac{-2}{3} + 1 = \frac{1}{3} $
$\frac{2}{3} - \frac{1}{3} ⋅ \frac{1}{3} = \frac{5}{9} = Cov(X, Y)$
ii) correlation coefficient
$ρ(X, Y) = \frac{Cov(X, Y)}{SD_x⋅SD_y}$
$SD_x = \sqrt{Var(X)}$
$SD_y = \sqrt{Var(Y)}$
$E(X^2) = E(Y^2) = \frac{1}{6}$
$E(X) = E(Y) = \frac{1}{3}$ (from part b.i.)
$Var(X) = Var(Y) = E(X^2) - (E(X))^2 = \frac{1}{6} - \frac{1}{9} = \frac{1}{18} $
$ρ(X, Y) = \frac{Cov(X, Y)}{SD_x⋅SD_y} = \frac{5/9}{1/18} = 10$
I know the correlation coefficient cannot be 10, but where did I go wrong?
Only here (it's only a minor calculation error):
The correct result is obviously
$$\int_{0}^{1}[x-x^2]dx=\Bigg(\frac{x^2}{2}-\frac{x^3}{3}\Bigg)\Bigg]_{x=0}^{x=1}=\frac{1}{6}$$
Thus the correlation coefficient $\rho=1$