# 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

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.