calculate the Probability density function of the absolute difference of two random variable If $X$ and $Y$ are two independent random variables with probability density functions $f$ and $g$, respectively, then the probability density of the difference $Y − X$ is given by the cross-correlation . In contrast, the convolution f * g gives the probability density function of the sum $X + Y$.
what i would like to have is the probability density functions of $|y-x|$ or $(y-x)^2$. Is that possible ? 
is that the equivalent as doing the following with two histogram ?

for (bin b1 in Histogram1)
  for (bin b2: Histogram2)
        prob = b1.probability * b2.probability
        distance += prob * abs(b1.value - b2.value)

 A: Expanding on Robert's answer (adapted to $Y-X$, in view of the question). 
Let $h(t)=\int_{ - \infty }^\infty  {f(x)g(t + x)dx}$, $t \in \mathbb{R}$, be the cross-correlation of the densities $f$ and $g$ (of $X$ and $Y$, respectively), that is the probability density function of $Y-X$.
We want to show that, for any $t > 0$,
$$
{\rm P}(|Y-X| \leq t) = \int_0^t {[h(u) + h( - u)]du} 
$$
(implying that the density of $|Y-X|$ is $h(t)+h(-t)$ for $t > 0$; for $t < 0$ the density is trivially $0$). 
Indeed,
$$
\int_0^t {[h(u) + h( - u)]du} = \int_0^t {h(u)du}  + \int_0^t {h( - u)du}  = \int_0^t {h(u)du}  + \int_{ - t}^0 {h(u)du} ,
$$
and so from
$$
\int_0^t {h(u)du} = {\rm P}(0 \leq Y-X \leq t) \;\; {\rm and} \;\; \int_{ - t}^0 {h(u)du} = {\rm P}(-t \leq Y-X \leq 0),
$$
it follows that
$$
\int_0^t {[h(u) + h( - u)]du} = {\rm P}(-t \leq Y-X \leq t) = {\rm P}(|Y-X| \leq t).
$$
Having shown that $h(t)+h(-t)$ is the density of $|Y-X|$ (for $t > 0$), the density of $(Y-X)^2$ can be found easily as follows. For any $t > 0$,
$$
{\rm P}((Y-X)^2 \leq t) = {\rm P}(|Y-X| \leq \sqrt t) = \int_0^{\sqrt t } {[h(u) + h( - u)]du} .
$$
A change of variable $u \mapsto u^2$ then gives
$$
{\rm P}((Y-X)^2 \leq t) = \int_0^t {[h(\sqrt u ) + h( - \sqrt u )]\frac{{du}}{{2\sqrt u }}} ,
$$
implying that the density of $(Y-X)^2$ is $(h(\sqrt t)+h(-\sqrt t))/(2\sqrt t)$ for $t > 0$; for $t < 0$ the density is trivially $0$.
EDIT: A key point here is that (a nonnegative measurable function) $f_Z$ is a probability density function of $Z$ if and only if $F_Z (z) = \int_{ - \infty }^z {f_Z (} u)du$ $\forall z \in \mathbb{R}$, where $F_Z$ is the distribution function of $Z$. (In this case, $Z$ is said to be absolutely continuous with density function $f_Z$.) 
EDIT: In the case of continuously differentiable distribution functions (rather than the general case absolutely continuous distribution functions), you can obtain the above results simply as follows. Let $H$ denote the distribution function of $Y-X$ and, as above, $h$ its probability density function.
Then, for any $t > 0$,
$$
{\rm P}(|Y-X| \leq t) = {\rm P}(-t \leq Y-X \leq t) = {\rm P}(Y-X \leq t) - {\rm P}(Y-X \leq -t) = H(t) - H(-t).
$$
Hence 
$$
\frac{d}{{dt}}{\rm P}(|Y - X| \le t) = h(t) + h( - t).
$$
As for the density function of $(Y-X)^2$, first note that, for any $t > 0$,
$$
{\rm P}((Y-X)^2 \leq t) = {\rm P}(|Y-X| \leq \sqrt{t}) = H(\sqrt t ) - H( - \sqrt t ).
$$
Hence
$$
\frac{d}{{dt}}{\rm P}((Y - X)^2  \le t) = \frac{{h(\sqrt t )}}{{2\sqrt t }} + \frac{{h( - \sqrt t )}}{{2\sqrt t }} = \frac{{h(\sqrt t ) + h( - \sqrt t )}}{{2\sqrt t }}.
$$
A: As you said, the density of $X-Y$ is the cross-correlation $h(t) = \int_{-\infty}^\infty f(t+y) g(y) \, dy$.  The density of $|X-Y|$ is $f_{|X-Y|}(t) = h(t) + h(-t)$ for $t > 0$, $0$ for $t < 0$.  The density of $(X-Y)^2$ is $f_{(X-Y)^2}(t) = (h(\sqrt{t}) + h(-\sqrt{t}))/(2 \sqrt{t})$ for $t > 0$, $0$ for $t < 0$.
A: It is a very big question.
Yes you can find the pdf of any combination of variables, there are many ways. 
The one which works most of the time use the characteristic function:
http://en.wikipedia.org/wiki/Characteristic_function_(probability_theory)#Basic_manipulations_of_distributions
You have many complex and exotic tools to exploit the properties of CF.
Anyhow this process greatly depends from the type of variable. E.g.: It is much easier with exponential variables than with non exponential variables.
