It is believed that $BPP$ has no complete problems. Even for $BPP^O$ for a suitable oracle $O$ it is believed not to have complete problems, unless P=BPP. I wonder if the class MA (the randomized version of NP) has complete problems. For example, IP has complete problems (given that it is equal to PSPACE). There is a similar post asking about complexity classes with no complete problems. Here, I'm interested specifically on complete problems for MA. If the answer is positive, can you give some examples? I've tried google and the complexity zoo, but with no success.
[Math] Complete problems for randomized complexity classes
computational complexitycomputer science
Related Solutions
First, let's be slightly pedantic and not make statements like P = #P, which cannot possibly be true just because P is a set of decision problems and #P is not. To get a decision version of #P, one can use PP, or something like P#P.
About your question, BPPNP is contained in PPP and P#P by Toda's theorem. On the other hand, if P#P were contained in BPPNP, it would imply that PH is contained in BPPNP, which would collapse the polynomial hierarchy to the third (or second?) level, which is considered unlikely.
In conclusion, P#P is considered to be more powerful than NP, BPP, BPPNP and even NPNPNP.
The zoo of complexity classes extends naturally into the realm of computability theory and beyond, to descriptive set theory, and in these higher realms there are numerous natural classes which have no members that are complete, even with respect to far more generous notions of reduction than the one you mention.
For example, consider the class Dec of all decidable sets of natural numbers. There can be no decidable set $U$ such that every other decidable set $A$ reduces to it in time uniformly bounded by any computable function $f$ (even iterated exponentials, or the Ackerman function, etc.). In particular, it has no member that is complete in your sense. If there were such a member, then we would be able to consruct a computable enumeration $A_0$, $A_1$, $\ldots$ containing all decidable sets, which is impossible since then we could diagonalize against it: the set of $n$ such that $n\notin A_n$ would be computable, but it can't be on the list.
The class of all arithmetically definable sets is obtained by closing the decidable sets (or much less) under projection from $\mathbb{N}^{n+1}\to \mathbb{N}^n$ and under Boolean combinations. The members of this class are exactly the sets that are defined by a first order formula over the structure $\langle\mathbb{N},+,\cdot,\lt\rangle$, and the hierarchy is stratified by the complexity of these definitions. This class has no member that is complete with respect to any computable reduction, and even with respect to any arithmetically definable reduction of bounded complexity, for in this case the hierarchy would collapse to some level $\Sigma_n$, which is known not to occur.
A similar argument works for the hyperarithmetic hiearchy, which can have no universal member hyperarithmetic reductions of any fixed complexity.
And similarly for the projective hiearchy on sets of reals.
The general phenomenon is that there are numerous hierarchies growing from computability theory into descriptive set theory which are all known to exhibit strictly proper growth in such a way that prevents them from having universal members.
Best Answer
In general, for randomized classes complete problems tend to be either promise problems or approximation problems (which means they don't technically satisfy the conditions for being complete problems).
If you allow approximation problems, you can get complete problems in BPP. For example, for BPP you can ask: given a Turing machine $M$, approximate the probability that it accepts within $t$ steps on input $x$, where $t$ is polynomial in the size of the input. It's clear that you can approximate this probability with a BPP machine (using simulation), and it's also obvious that any BPP language can be reduced to this problem. Technically, this problem isn't in BPP since it's not a language (i.e., its output is not $\{0,1\}$), so it's not BPP-complete.
You can turn this into a promise problem by imposing the "promise" that the acceptance probability either be greater than $\frac{2}{3}$ or less than $\frac{1}{3}$.
These complete promise problems or approximation problems play the same role that complete problems play for non-randomized languages, and they really deserve more respect from computer scientists. For quantum computing, the natural complete problems also tend to be approximation (or promise) problems, and they have been quite useful in the theory of quantum computing.
There is a natural promise problem from quantum computing (stoquastic Hamiltonian) which is MA-complete (and not trivially so). However, I don't believe there are any languages (non-promise problems with $\{0,1\}$ answers) known to be MA-complete.