If a protocol has a statistical security parameter, does this mean it is only computationally secure?

Question

if a cryptographic protocol has a computational security parameter and a statistical security parameter, does this mean it is only computationally secure instead of information-theoretically secure?

I am wondering because this answer says that statistical indistinguishability is when the adversary is computationally unbounded: https://crypto.stackexchange.com/a/11790 That would imply that the presence of a statistical security parameter means that a protocol is information-theoretically secure. Is that true?

Thank you!

Answer 1

Yes, security parameters imply we are dealing with computational security not with information theoretical security. But there could be other usages for security parameters.

When we use one time pad or secret share we have information theoretical security. It can't be broken with any computational power and there are no security parameters.

When we do computational security, we often have a security parameter, to give e.g key size or iteration count, and we hope as we increase it so do we increase the computational cost for attacker.

However there could be other cases, we could have proofs or signature which are secure even against an unbounded adversary but still have a probability of failing (e.g probability of a forgery not being detected) and these may be controlled with a security parameter as well.

With some quantum systems, there seems to be a guarantee of perfect security regardless of adversary, e.g with Quantum Key Distribution yet the actual system may fail to full meet the requirements for the proof to actually hold. Again we may have a security parameter (though probably not one we control and pick) which could describe e.g the probability of getting more than a single photon. This as well can be a quantifier of system security.