# 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.

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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.

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I really appreciate you taking the time to answer as thoroughly as you did. Thanks for the help! I think that should be enough to get me started on my own for now. – mathjacks May 26 '12 at 16:28
Your conditional density questions raise new issues. The expectation question is unclear to me, don't know what you mean by $X''Y'''$. The independence question has a formal answer and a somewhat less formal one. The less formal one is that, for example, if you know that $X \le 1/3$, then for sure $Y\le 1/3$, so sometimes knowing about $X$ tells you stuff about $Y$. – André Nicolas May 26 '12 at 16:44