Simple conditional continuous distribution Q

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February 20th 2011, 01:58 PM
padawan

Simple conditional continuous distribution Q

Hi all,

Been going through a question I have the answer to but no working from an exercise book... can't see the logic. Wonder if anyone can lend an eye?

Q:Let $(X,Y)$ be a random point chosen uniformly on the region $R = {(x,y): |x| + |y| \le 1}$

a) Sketch R
b) Find the marginal densities of X & Y
c) Find the conditional density of Y given X

My answers:

a) simply a kite shape with vertices at (-1,-1),(-1,1),(1,1) & (1,-1)
b) Evaluate joint density function:

By observation, the area is a tilted square with sides 1 x1 => area = 1 => $f_{XY}(xy) dy = 1$

$f_{X}(x) = \int_{-\infty}^{\infty} f_{XY}(xy) dy$

$= [y]_{0}^{1-|y|}$

$= 1-|y|$ over $-1\le x \le 1$ , now this makes sense to me, as we integrate of the 0 and (1-mod(y)) as limits given by initial conditions. By symetary, marginal of Y is the same except Y instead of X in the expression.

c)

conditional density $f_{Y/X}(y/x) = {f_{YX}(y/x) \over f_{X}(x)}$

surely this gives the reciprocol of the marginals already found ie:

$f_{Y/X}(y/x) = {1 \over 1-|y|}$

The answer in the back of the book is:

$f_{Y/X}(y/x) = {1 \over 2-2|y|}$

over $1-|x| \le y \le 1-|x|$

Now I can understand the range of y, but cant seem to see where the conditional comes from..... would make my day if someone says its a typo lol

Thanks for reading.
February 20th 2011, 05:40 PM
matheagle

What do you mean by a kite shape density?
I don't follow how (-1,-1) satisfies that inequality.
Do you mean the points (1,0), (-1,0), (0,1), (0,-1)?
AND how can the marginal density of X, a function of only x, have y's in it?
A limit of dy cannot have y's in it.
February 20th 2011, 06:22 PM
matheagle

I get the square on my 4 points, which has 4 triangles each with area equal to .5.
So the total area is 2, hence the joint density is .5.

Now on 0<x<1, we have $f_X(x)=\int^{1-x}_{-1+x} .5dy=1-x$

You also need to compute this density on -1<x<0.
February 21st 2011, 01:05 PM
padawan

Apoligies for the typos- even though there were some fundamental maths errors!

Shape of polynomial is square with vertices (1,0)(0,1)(-1,0)(0,-1) and area root 2 X root 2 = 2 => density function is half as explained already.

To finish off what you started matheagle:

Between $-1 \le x \le 0$ the boundries of $y$ are the line $y = x + 1$ for upper bound and $y = -x - 1$ for lower bound; therefore

$f_{X}(x) = \int_{-x-1}^{x+1} 0.5 dy = x+1$

Now representing this over the entire region, is the answer in the book:

$f_{X}(x) = 1 - |x|$ over $-1 \le x \le 1$ .... now that the density function is correctly stated as 1/2, I undertsnad the answers.

It wasnt a typo in the text book afterall!

Thanks a lot Matheagle.