I have the following exercise:
(i) Find $k$ such that $f(x,y) = k, 0<x<y<1$ is a probability joint density function. ($f(x,y) = 0$ otherwise).
$\displaystyle \int_{0}^{1} \int_{0}^{y} f(x,y) dxdy = \int_{0}^{1} kydy = \frac{k}{2}$.
If $f$ is a joint density, then $\frac{k}{2} = 1 \iff k = 2$.
(ii) Find the marginal densities $f_X(x)$ and $f_Y(y)$.
$f_X(x) = \displaystyle \int_{x}^{1} 2 dy = 2(1-x), 0<x<1$
$f_Y(y) = \displaystyle \int_{0}^{y} 2 dx = 2y, 0<y<1$
I believe so far what I did is correct (but please tell if there's any mistake), my problem is in the following:
(iii) Find the probability joint function $F(x,y)$.
$F(x,y) = \displaystyle\int_{0}^{x}\int_{0}^{y} 2 du dv = 2yx, 0<x<y<1$
$ 0$, otherwise.
I can see that $\displaystyle \frac{\partial^2}{\partial{x}\partial{y}} F(x,y) = f(x,y)$, as it should be, but something feels odd to me in this domain of definition "$0<x<y<1$". Is what I'm doing correct? Thanks.
Pick any point in the square. 2 cases: (1) $x<y\ then\ F(x,y)=2xy-x^2$, since x lower limit is 0, while y lower limit is x. (2) $x\ge y \ then\ F(x,y)=y^2$, since x upper limit is y, while y lower limit is 0.
Outside the square you need to work with pieces separately.