Are distributions of position and momentum assumed to be independent in quantum mechanics? Given a wave-function of a single particle we can calculate probability density for positions. We can also calculate probability density for momenta. Are these probability densities assumed to be always independent?
Or, in other words, if we measure position and momentum of a particle (for example electron in hydrogen being in the ground state), should we expect that these two random quantities independent?
 A: To say that position and momentum are not independent random variables in QM really undersells the point. It is not that we cannot measure them independently, the Heisenberg uncertainty principle states that we cannot measure both quantities simultaneously at all.
What this means is that the distribution of momentum depends not only on the outcome of a measurement of position, it depends on whether we measured position at all. We cannot measure momentum and then ask what we would have measured if we had also tried to measure position, because that would change the outcome of the momentum measurement.
A: 
if we measure position and momentum of a particle (for example electron in hydrogen being in the ground state), should we expect that these two random quantities independent?

No. The momentum probability distribution and the position probability distribution have an inverse relationship with each other, which is a direct consequence of the Heisenberg's uncertainty relation.
Suppose you have a certain wave packet. What is the special thing about a wave packet? Well, it's confined to a certain region of space. Let's say that it is confined within a length $a$. So, you know that the particle can not exist outside of that region of space, and so the uncertainty principle puts you into a position where you can not measure the momentum arbitrarily accurately. This is why, its probability distribution is also dependent upon that of the position. Mathematically speaking, the momentum and position wavefunctions are related by a fourier transform-
$$\psi(x)=\dfrac{1}{\sqrt{2\pi \hbar}}\displaystyle \int_{-\infty}^{+\infty}\Psi(p)\text{e}^{ipx/\hbar}\ dp$$
From this formula, we can see that the position probability density, which is $|\Psi(x)|^2$ is obviously dependent on the functional form of $\Psi(p)$. I hope this answers your question.
A: Any real world position measurement result also implicitly includes a momentum measurement. Why? Any measurement of x results in some psi(x), its Fourier transform is psi(p), both are measurement results and quantum states, they are simply represented in different bases and have inversely related widths, neither width can be zero. If the position measurement result is accurate (ie, psi(x) is narrow), the momentum measurement is commensurately not accurate (ie, psi(p) is wide). This is the stipulation of the Heisenberg uncertainty principle, it distinctly does not say x and p cannot be measured simultaneously, it simply says the accuracies of the simultaneous x and p results are inversely related, and neither width can be zero, as the product of delta x and delta p must be greater than or equal to hbar/2. This product cannot be zero, so neither delta can be zero. Thus, ideal quantum pure states, and single real numbered eigenvalues, are not physically realizable by any means whether manmade or natural. So every physical realization of a position measurement is also a momentum measurement.
Re independence of x and p, a free particle at position x’ may have any value of p independent of x’, and a free particle with momentum value p’ may have any value of x independent of p’. So x and p may be considered independent, while their wave function widths (in their respective eigenvalue domain spaces) are inversely related, as any Fourier transform pairs.
