Uniform distribution with random support I have a uniform distribution $ X \sim U(A,B) $ where the limits themselves are random: $A \sim N(\mu_A,\sigma_A^2)$ and $B \sim N(\mu_B,\sigma_B^2)$. Hence the support of $X$ is random.

$A$, $B$ are independent. Also consider $\mu_B-\mu_A \gg \sigma_A+\sigma_B$

Any clues on the probability density function of this one?
The pdfs of A and B are known since they are normal distributions

The pdf of X given A and B is
$$
f_{X|A,B}(x|a,b)=1_{x \ge a} 1_{x \le b} \frac{1}{|a-b|}
$$

The joint pdf of X, A and B is 
$$
f_{X,A,B}(x,a,b)=f_{X|A,B}(x|a,b) f_A(a) f_B(b)
$$

And the pdf of X is given as a marginal distribution from the latter
$$
f_X(x)= \int_{D_B} \int_{D_A} \! f_{X,A,B}(x,a,b) \, \mathrm{d}a \mathrm{d}b. 
$$
where $D_A$ and $D_B$ the domains of A and B respectively.
So, let's be a bit more specific. The idea is that you first draw the samples for A and B. Then, you draw the sample for X.

I am assuming that if it turns out to be A>B, then X is set to an undefined value. In that case $\int_{\mathbb{R}} \mathbb{P}(X=x) dx &lt; 1 $, ie. you are losing some probability mass, but since this event is rare (under your assumption), I would say this is ok. Also, A and B are independent.

However, X can't be independent of A,B or I am missing something (here comes my comment on that you should start stating your question by first defining what the underlying $\sigma$-algebras are).

So, using total probability:
$ \mathbb{P}(X=x) = \int_{\mathbb{R}} \int_{\mathbb{R}} \mathbb{P}(X=x|A=a,B=b) \cdot \mathbb{P}(A=a) \cdot\mathbb{P}(B=b) \cdot da \cdot db $

where:
$\mathbb{P}(X=x|A=a,B=b) = 1_{x\geq a} 1_{x\leq b} \frac{1}{|a-b|} $

$\mathbb{P}(A=a) = \frac{1}{\sigma_{A}\sqrt{2\pi}} exp\left(-\frac{(a-\mu_{A})^{2}}{\sigma_{A}^{2}}\right) $

$\mathbb{P}(B=b) = \frac{1}{\sigma_{B}\sqrt{2\pi}} exp\left(-\frac{(b-\mu_{B})^{2}}{\sigma_{B}^{2}}\right) $

I realize that computing the integral might be non-trivial.

I hope this sheds some light, or, again, I might miss something on the statement.