Analyzing joint uniform random variables geometrically I am analyzing a simple problem that looks at independence between a pair of jointly distributed continuous uniform random variables $X$ and $Y$.
The example from the text (Mathematical Statistics and Data Analysis Third Ed. by John A. Rice) seems straightforward enough, but I'm a bit baffled by the values that the random variables are allowed to take on in the second part of the example.
Here is the example:
Suppose that a point $(X,Y)$ is uniformly distributed on the square $S = \{(x,y) \mid -1/2 \leq x \leq 1/2, -1/2 \leq y \leq 1/2\}$. Then the joint density function is given by $f_{XY}(x,y)=1$ for $(x,y)\in S$ and $0$ elsewhere.
The example then goes on to state the obvious independence of $X$ and $Y$ in this case. 
However, at this point in the text we are asked to consider a rotated version of the square above (a diamond formed by rotating by $90$ degrees). It is then argued that the distribution of the marginal densities are no longer uniform (this is clear to me). The author then states: Thus, for example, $f_X(.9) > 0$ and $f_Y(.9) > 0$. But from the sketch you can see that $f_{XY}(.9,.9)=0$. (note that the sketch that is referenced is one the reader is supposed to draw).
My question is, how is $f_X(.9)>0$ or $f_Y(.9)>0$? If we rotate a unit square the diagonal will be $\sqrt{2}$. I assume that we are supposed to expand the respective range of $X$ and $Y$. But clearly $-\frac{\sqrt{2}}{2}\leq X \leq \frac{\sqrt{2}}{2}$ and likewise for $Y$. But then the maximum value for $X$ is $X\approx .71$, and $.9 > .71$. Am I supposed to be thinking about this differently or assuming an absolute value here?
EDIT: I guess it would also seem to me that rotating a square $90$ degrees would also produce another square and not a diamond. I feel as if a diamond would require a $45$ degree rotation.
 A: If I understand your description correctly, you are right about the extents
of the values after rotation and about 45-degrees vs. 90 degrees of rotation.
However, let's not lose sight of the main point of the example. Before rotation
you have independence of $X$ and $Y,$ but after 45 degree rotation you do
not have independence of $X^\prime = (X + Y)/\sqrt{2}$ and
$Y^\prime = (X - Y)/\sqrt{2}.$
I use simulation in R as a quick way to illustrate using a figure (denoting primes as 1's).
m = 10^5; a = 1/2                         # 100,000 points
x = runif(m, -a, a); y = runif(m, -a, a)
x1 = (x+y)/sqrt(2);  y1 = (x-y)/sqrt(2)   # rotation
plot(x1, y1, pch=".")
abline(v=-.4, col="red", lwd = 2);  abline(h=.4, col="red", lwd = 2)


The point is that the narrow strips at top and left both have positive
probability, but that their intersection (square at upper-left) has
probability $0.$ Thus $X^\prime$ and $Y^\prime$ cannot be independent.
mean(x1 < -0.4);  mean(y1 > .4);  mean(x1< -.4 & y1 > .4)
## 0.09601    # aprx P(X1 < -.4)
## 0.09487    # aprx P(Y1 > .4)
## 0          # P(Intersection)

However, because of the symmetry, we have both $\rho(X, Y) = 0$ and $\rho(X^\prime, Y^\prime) = 0.$
cor(x,y);  cor(x1,y1)
## 0.003507761        # consistent with 0 correlation before rotation
## -0.00117999        # consistent with 0 correlation after rotation

