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

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 => $\displaystyle f_{XY}(xy) dy = 1$

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

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

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

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

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

The answer in the back of the book is:

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

over $\displaystyle 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.