What happens if at equilibrium pressure is increased by decreasing the volume? I think, according to Le Châtelier's principle both the conditions i.e. increasing the pressure and decreasing the volume must shift reaction towards the side on which there are less amounts of gaseous molecules. And if there are the same stoichiometric coefficients for the gaseous substances (i.e. $∆n_g = 0$) then it must not have any effect on the reaction. 
But increasing the pressure and decreasing the volume increases the number of collisions between the molecules of reacting substances and hence the concentration of products must increase. 

Both these thoughts confuse me.

Questions I encountered:
$$ \ce{H2O(g) + CO(g) &lt;=&gt; H2(g) + CO2(g) + Heat }$$
In this question $∆n_g = 0$ so there must be no effect of increasing pressure and decreasing the volume. But the answer mentions that the concentration of both $\ce{CO}$ and $\ce{CO2}$ will increase.

There was also another question of same type: 
$$ \ce{SO2Cl2(g) &lt;=&gt; SO2(g) + Cl2(g) }$$
Here the reaction must shift backward as pressure is increased and volume is decreased.
But again the answer mentions that concentration of all the gasses increases.
When a reaction is at equilibrium, the forward and reverse rates are equal. Decreasing the volume increases the concentration of all species (both reactants and products). This will result in a higher forward rate (because the concentration of reactants increased) and in a higher reverse rate (because the concentration of products increased). If forward and reverse rate increase by the same factor, the reaction remains at equilibrium.