Alternative definition of conditional independence I'm reading a paper named "Conditional Independence in Statistical Theory" by A. P. Dawid. The paper shows some alternative definitions of independence:
X and Y are independent iff:
1. $p(x, y) = p(x)p(y)$
2. there exist two function $a(x)$ and $b(y)$ such that: $p(x, y) = a(x)b(y)$.
Dawid said that it is simple to verify that the two definitions are equivalent. However, I have spent several hours to solve this but failed. 
 A: One implication is trivial (hereafter, I assume $X$ and $Y$'s probability distributions are continuous wrt Lebesgue, ie have densities — this is implicit in your question). As for the other, suppose there exist $a$,$b$ satisfying (2). As a first consequence, $a,b\geq 0$.
Fix any $x$. By definition,
$$
 p_X(x) = \int_\mathbb{R} p(x,y)dy = \int_\mathbb{R} a(x)b(y)dy =  a(x)\underbrace{\int_\mathbb{R} b(y)dy}_{K_b}
$$
so in particular $b$ must be in $L_1$, and $a\propto p_X$ (up to a factor $K_b$). Similarly, for any $y$
$$
 p_Y(y) = \int_\mathbb{R} p(x,y)dx = \int_\mathbb{R} a(x)b(y)dx =  b(y)\underbrace{\int_\mathbb{R} a(x)dx}_{K_a}
$$
so that $\forall(x,y)$ 
$$
p(x,y) = \frac{1}{K_a K_b} p_X(x)p_Y(y)
$$
Integrating both sides, we get
$$
 1 = \int_{\mathbb{R}\times\mathbb{R}} p(x,y)dxdy = \int_{\mathbb{R}\times\mathbb{R}} a(x)b(y)dxdy = K_a K_b
$$
and thus 
$$
p(x,y) = p_X(x)p_Y(y) \qquad \forall x,y
$$
A: They are not--the first one is more restrictive than the second, asking that the random variables X and Y are independent and with the same distribution.
One may want to replace 1. by 3. $p(x,y)=p(x)q(y)$, then 3. and 2. are equivalent. To show this, note that 3. obviously implies 2. and that, if 2. holds, then $\alpha=\sum\limits_xa(x)$ and $\beta=\sum\limits_yb(y)$ are such that $\alpha\beta=1$ hence $p(x)=a(x)/\alpha$ and $q(y)=b(y)/\beta$ solve 3. 
(For continuous distributions, use $\alpha=\int a(x)\mathrm dx$ and $\beta=\int b(y)\mathrm dy$ instead.)
A: Suppose
$$
p(x,y) = a(x)b(y)
$$
Then
$$
\int_{\text{$x$-space}} a(x)\,dx\cdot\int_{\text{$y$-space}} b(y)\,dy = \iint_\text{product space} a(x)b(y)\,dy\,dx = \iint_\text{product space} p(x,y)\,dy\,dx = 1.
$$
Let
$$
A = \int_{\text{$x$-space}} a(x)\,dx.
$$
Then
$$
\int_{\text{$y$-space}} b(y)\,dy
$$
must be $1/A$ since the product is $1$.  Therefore $x\mapsto \dfrac{a(x)}{A}$ and $y\mapsto Ab(y)$ are density functions, and when you multiply them you get $p(x,y)$.
Suppose there were other denisty functions $c$, $d$ such that $c(x)d(y)= a(x)b(y)$.  Then we would have $\dfrac{c(x)}{a(x)} = \dfrac{d(y)}{b(y)}$, at least at points where the denominator is not $0$.  Then we would have
$$
\frac{c(x)}{a(x)} = \text{something not depending on $x$}
$$
(since $d(y)/b(y)$ does not depend on $x$).  In other words $c(x)/a(x)$ is a constant.  Thus $c(x)$ and $a(x)$ are essentially the same function.
