Find cumulative, marginal densities, moments for $f(x)=2,\le y \le x \le 1$ (multivariate) Im learning multivariate statistics on my own and I have come across some problems I don't understand. Unfortunately there's no solution manual in the back so I thought I might ask here. Its not a homework problem and I feel should be easy enough to get explained.
$f(x)=2,\le y \le x \le 1$
$0$ otherwise
Find
a) $F(x,y)$
b) $F(x)$
c) $f(x)$
d) $G(y)$
e) $g(y)$
f) $f(x|y)$
g) $f(y|x)$
h) moments $X''Y'''$
i) Are X and Y independent?
Im not asking for answers to them all, just some intuition so I can answer it on my own. Im not sure how the answer should be stated even thou I know the definition of each of these.
 A: A lot of questions! Perhaps you could split your question, to increase the likelihood that everything gets fully answered.
Here is a small start.  We will mostly argue geometrically, though for generality an integral may be mentioned.  Drawing a picture is essential to understanding what is going on.
Note that the joint density function "lives" on the region $R$ which is a triangle with corners $(0,0)$, $(1,0)$, and $(1,1)$.
(a) For $F(x,y)$ we want the probability that $X\le x$ and $Y \le y$. 
This is the probability that $(X,Y)$ lies below and to the left of $(x,y)$.
This probability depends very much on what $x$ and $y$ are. For example, if $x\le 0$ or $y\le 0$ (or both), then $F(x,y)=0$. If $x\ge 1$ and $y\ge 1$, then $F(x,y)=1$.  That still leaves several cases. 
The most interesting case is when $0\le y\le x\le 1$.  Look at the part of our triangle that lies below and to the left of $(x,y)$. This is all points $(s,t)$ in the triangle such that $s\le x$ and $t\le y$. Call this region $K$. We want fo find 
$$\int_{-\infty}^y\left(\int_{-\infty}^x f(s,t)\,ds\right) dt.$$
Because our density function is the constant $2$ on the triangle, and $0$ outide, all we need to do is to find the area of $K$ and multiply by $2$. The region $K$ is a trapezoid. The bottom base has length $x$. The top base has length $x-y$. And the height is $y$. So the area is $\frac{2x-y}{2}y$. Multiply by $2$. We get $F(x,y)=(2x-y)y$. Or else we could note that $K$ is an $x\times y$ rectangle minus a small right-angled triangle with legs $y$ and $y$, so the area of $K$ is $xy-\frac{y^2}{2}$. 
There are still two cases to deal with. Suppose that $0\le x \lt 1$ and $y \gt x$. Then the part of the triangle which is below and to the left of $(x,y)$ is a right-angled triangle with legs $x$ and $x$. So it has area $\frac{x^2}{2}$. Thus, for such $(x,y)$, $F(x,y)=x^2$. 
I leave it to you to deal with the case $x\gt 1$, $0 \lt y \lt 1$.
(b) Let's call it $F_X(x)$, like the text maybe should have. Recall that $F_X(x)=P(X\le x)$. For $x\le 0$, $F_X(x)=0$. For $x\ge 1$,
$F_X(x)=1$. For $0 \lt x\lt 1$, we want to integrate the density function over all $(s,t)$ such that $s\le x$. This is twice the area of a right-angled triangle with legs $x$ and $x$. So for $0\lt x\lt 1$, $F_X(x)=x^2$.
(c) I will call it $f_X(x)$. Differentiate the $F_X(x)$ computed above. There are other ways to find $f_X(x)$ which I should mention, but won't.
(d) The calculations are very similar to those of (b). We have $G_Y(y)=P(Y \le y)$. For $y \le 0$, $G_Y(y)=0$. For $y\ge 1$, $G_Y(y)=1$. For $0\lt y\lt 1$, we want to integrate our density function over all $(s,t)$ such that $t\le y$. 
(Well, in all cases that's what we need to do.)
This integral is twice the area of a certain trapezoid, which has bases $1$ and $1-y$, and height $y$. So on $0\lt y\lt 1$, we have $G_Y(y)=(2-y)y$. 
(e) Differentiate. 
Enough for now, need to eat. 
