Interpretation of Simple Uniform Marginal Density Example from Mathematical Statistics by Rice

Question

I am stuck trying to visualize Example B from Mathematical Statistics and Data Analysis 3rd ed by Rice. The examples revolve around the concept of independence of Random Variables which is defined in the following way:

I included Example A from which Example B is derived:

A couple of things going on here:

1. He says "rotate the square by $90^{\circ}$" so first off that has to be a typo because that will just give me another square. So we shall assume $45^{\circ}$ to get the desired diamond.

2. He first says "you can see that the marginal density of $X$ is nonegative for $\frac{-1}{2} \leq x \leq \frac{1}{2}$, but is not uniform. Why is it nonnegative? Is it because when we integrate over the bounds of integration our marginal density will be positive? More importantly, why is this marginal not considered uniform anymore? The bounds of integration may have changed, but the joint density is still defined as $f_{XY}(x,y) = 1$.

3. In trying to illustrate the concept of independence he took the result $f_{X}(0.9) > 0$ and $f_{Y}(0.9)>0$. Now using my scattered trig knowledge the bounds of integration would've changed to $\frac{-1}{\sqrt{2}} \leq x \leq \frac{1}{\sqrt{2}}$ and same for $y$. Which would mean $f_{X}(0.9) = 0$ and same for $Y$. I get what he may have been trying to insinuate, if he had chosen $f_{X}(0.65)$ and $f_{Y}(0.65)$ he would've gotten the desired result I believe.

Maybe because of my inability to sketch this out properly or put it into some graphing software I'm not seeing what is occurring, but I don't think that would help too much here. I asked a lot of questions here so if one was of most importance it would be #2 and why is the marginal density not a uniform density anymore?

Answer 1

The answer is very simple....the error you found is not the only one in the text.

Rotating the square of 45° you get a diamond with the same area $A=1$

Where, obviously,

a) $y=-\frac{\sqrt{2}}{2}-x$

b) $y=\frac{\sqrt{2}}{2}+x$

c) $y=-\frac{\sqrt{2}}{2}+x$

d) $y=\frac{\sqrt{2}}{2}-x$

Then the joint density is the following

$$f_{XY}(x,y)=\mathbb{1}_{(-\frac{\sqrt{2}}{2};0)}(x)\mathbb{1}_{(-\frac{\sqrt{2}}{2}-x;\frac{\sqrt{2}}{2}+x)}(y)+\mathbb{1}_{[0;\frac{\sqrt{2}}{2})}(x)\mathbb{1}_{(x-\frac{\sqrt{2}}{2};\frac{\sqrt{2}}{2}-x)}(y)$$

to derive the marginal X it is enough to integrate in Y (with the joint density expressed as I did the integral extremes are written in the joint density)

Thus

$$ f_X(x) = \begin{cases} 2x+\sqrt{2}, & \text{if $ -\frac{\sqrt{2}}{2}<x<0 $ } \\ \sqrt{2}-2x, & \text{if $0 \leq x<\frac{\sqrt{2}}{2}$ } \\ 0, & \text{elsewhere } \end{cases}$$

so X density is non negative for $ -\frac{\sqrt{2}}{2}<x<\frac{\sqrt{2}}{2}$ but it is not uniform....it's a triangle!