The resistor noise index provides a lower bound to how precise a resistor can be measured, even using ideal instrumentation. Effectively, it even makes no sense to define the resistance any more precise than given by its noise index. Intuitively, this can be understood as more probe charge (either due to higher current or longer integration time) also influences the resistor more.
However, by increasing the volume of the resistor, one can reduce the noise indexby increasing the volume of the resistor, one can reduce the noise index: Below in Fig. 1, both resistances are equal to $R$, but the right assembly uses four identical resistors each contributing uncorrelated noise. It can be shown, that the 4 resistor assembly has a twice lower total excess noise and thus lower noise index. Therefore, the resistance of the right side assembly can be defined twice more precise than the single resistor on the left side. By quadrupling the volume of the resistor and leaving everything else unchanged, the excess noise drops to half demonstrating the relation $n\propto V^{-1/2}$.
My question is:
How far can this be taken ? One could always add more resistors achieving extremely precise assemblies. But fundamentally, to probe the resistance electrically one relies on electron-electron interaction which are quantum mechanical processes. So I figure that the uncertainty principle might sneak into the backyard somehow.
Figure 1: Using 4 identical resistors, one can make an assembly with the same resistance, but twice lower excess noise
