What the difference between pin and pn diodes i-v curve?
There is indeed no big difference in IV-curve between a PIN-diode and a regular abutted PN-diode when forward biased. For wider I-regions, the current may differ slightly.

I believe the statement


  the bigger resistance of the intrinsic layer


is incorrect though.

The resistivity \$\rho\$ tells us how charges move under the influence of an electric field, or

$$\vec{J} = \sigma \vec{E} = \frac{1}{\rho} \vec{E}$$

This method of carrier transport is called drift. However, diode operation also heavily depends on diffusion. The higher resistivity basically causes the depletion region to be wider as the balance between drift and diffusion tilts more towards diffusion. 

PN-diode current increases as more charges have enough energy to cross the built-in potential barrier. Applying a voltage will lower that potential barrier, causing more electrons to to cross it increasing the current.

A PIN-diode has a barrier height that is the same as for a PN-diode with the same doping concentrations for P and N-regions. A PIN-diode, however, will have a barrier spread out over a longer distance. This may effect the IV-curve, but from what I have found in simulations that effect is rather small.

Other more notable differences:


A PIN-diode has a wider depletion region, so it also has a lower capacitance.
The electric field is also weaker in a PIN diode due the wider depletion region. This means that the breakdown voltage will be larger too.