Proving independence of two variables in a joint distribution, using cumulative distribution functions

I have two variables $X$ and $Y$ with the following joint probablity density function

$$f (x,y) = \begin{cases} \frac14 (1+xy) & \text{if } |x| < 1, |y| < 1\\ &\\ 0 & \text{otherwise} \end{cases}$$

The problem is to prove that $X$ and $Y$ are not independent, but that $X^2$ and $Y^2$ are. I calculated the marginal density functions of both $X$ and $Y$ and since their product doesn't equal the marginal density function, I proved they are not independent.

However, I wasn't sure about the second part and reviewed the solution. In the given solution, independence was not proven by the following.

$$f_{x^2,y^2} (u,v) = f_{x^2} (u) \centerdot f_{y^2} (v)$$

Instead, it was proven that the cumulative distribution functions exhibit this property and that this implies independence.

$$P (X^2 \leq u \cap Y^2 \leq v) = P (X^2 \leq u) \centerdot P (Y^2 \leq v)$$

I couldn't find any reference that said this implied independence. I believed such implication only worked with the density functions. This is also not mentioned on the wikipedia page for cumulative distribution function.

So, I'm wondering, is this also a way to prove independence? Can I use this technique when possible?

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In this particular instance, it is not necessary to calculate the marginal density functions and then check whether $f_{X,Y}(x,y)$ equals $f_X(x)f_Y(y)$ in order to verify non-independence. Since $f(x,y)$ does not factor into $g(x)h(y)$ (where $g(x)$ and $h(y)$ are not required to be valid pdfs), $X$ and $Y$ cannot be independent. –  Dilip Sarwate May 15 '12 at 12:57
@DilipSarwate, Thanks for that tip. Didn't know such a deduction was possible. Can you point me to some reading on that method? –  Shrikant Sharat May 15 '12 at 13:13
See for example Lecture 34 in this collection of lectures –  Dilip Sarwate May 15 '12 at 15:04

For real-valued random variables, the definition in terms of cdf is the usual definition. When moreover the random variables have densities, one can prove the density version that you quoted, but the cdf is more fundamental.

Added: See for example Wikipedia for a somewhat sketchy discussion of independence of random variables.

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The CDF makes more sense in real-valued variables, but the definition of independence is described with PDF's, no? Also, does proving that the product of marginal PDF's equal the joint PDF imply that the product of marginal CDF's equal the joint CDF? This seems slightly non-obvious to me. –  Shrikant Sharat May 15 '12 at 12:41
It is straightforward. We want $P((X\le u)\cap (Y\le v))$. Integrate the product of the marginals, so double integral $-\infty \lt x\le u$, $-\infty \lt y \le v$. The double integral is just the product of the integrals, it is the nice case of double integral, where you can integrate separately. –  André Nicolas May 15 '12 at 12:50

To elaborate on Dilip's comment, if we could write $f(x,y)$ in product form, then $$f(x,y)f(w,z)-f(x,z)f(w,y)=0\tag1$$ for $(x,y), (w,z)\in A\subset \mathbb{R}^2$ where $A$ has full Lebesgue measure. But for your function $f$, equation (1) says $${1\over 16}(x-w)(y-z)=0\tag2$$ which is only true if $x=w$ or $y=z$. Thus for each $(x,y)$, equation (2) only holds for $(w,z)$ in a set of Lebesgue measure zero. This proves that $X$ and $Y$ are not independent.

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As mathematicians we should always be precise. None of the above claims about independenc is correct unless you add that it holds for every u and v. –  Michael Chernick May 17 '12 at 14:48