Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

share|improve this question
add comment

1 Answer 1

up vote 1 down vote accepted

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.

share|improve this answer
    
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. –  flapjackery 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
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.