Parton distribution function - dependence on $Q$? In one of my previous questions I define the parton distribution function, following that of D.Stump as follows:

$f_i(x,Q^2)dx$ is the mean number of the $i$th type of patron with longitudinal momentum fraction from $x$ to $x+dx$ appropriate to a scattering experiment with momentum transfer $Q$.

I am slightly confused by the last term 'appropriate to a scattering experiment with momentum transfer $Q$'. How can the 'mean number of the $i$th type of patron with longitudinal momentum fraction from $x$ to $x+dx$'  (which seems to be an property of the proton) change depending on how we probe it? And why can we take:
$$\sum_i\int f_i(x,Q^2)dx=1~\forall~Q$$
 A: Consider the diagram below: 

With a small momentum transfer $Q$ the resolution is low and you in essence interpret everything in $A$ as a single entity (a valance quark say) and as such associate with it a high Bjorken parameter $x$. As $Q$ increases your resolution increases and you see the quarks, antiquarks and gluons that once you thought made up this single entity. Naturally with each of these you associate a smaller value of $x$ and thus in this high $Q$ case you see lots of particles at low $x$ values and fewer at high $x$ values compared with the low $Q$ case. This should explain why you still require the total momentum fraction of all the partons to equal 1 (as explained by @Lê Dũng).
A: The PDF $f_i(x,Q^2)$ is defined as the probability density (number densities as they are normalised to the number of partons - this can be the meaning of "mean number" in the definition of D.Stump) to find a parton of type i carrying a momentum fraction x at a the energy scale $Q^2$ (momentum transfer). 
In QCD, there is a phenomenon called "asymptotic freedom", namely the interaction between partons inside a nucleon changes according to the distance (higher distance, stronger interaction and vice versa), or, equivalently, according to the energy scale. That is to say, the PDF should depend on the energy scale (as confirmed by DIS experiments, which you can search easilly).
And, I think the sum rule is not as you presented, but as follows:
$$\sum_i\int xf_i(x,Q^2)dx=1$$
This rule is from the fact that, the total momentum of all partons should be equal to the momentum of the nucleon.
