A previous version of this question asked about a complexity class I called $MA^*$, which has been recognized by users to be $\exists BPP$.
The difference between $MA$ and $\exists BPP$ is that in the latter class the probabilistic verifier decides a BPP language; while the probabilistic verifier in $MA$, when $x\in L$, is required to accept with bounded error only for one certificate and can answer just randomly for all the others, thus violating the BPP promise.
What is the natural reason to allow this "strange" power to the class? E.g. an $MA$ language which is not (trivially) in $\exists BPP$ would answer the question. More generally, in which contexts the class $MA$ comes up naturally but $\exists BPP$ wouldn't (trivially) work?
P.S. For now, I'm answering me as follows. A set $C$ of certificates $x,c$ for $x \in L$ makes sense only in reference to a verification procedure for $(x,c)$. If the verifier is probabilistic, and we want bounded error, there could be some $x,c$ that aren't decided. For $x \notin L$, it remains defined that $(x,c) \notin C$, or soundness is compromised. But what should define $(x,c) \in C$ for $x \in L$, if not the verification procedure itself? So in this sense $MA$ is more natural than $\exists BPP$, because $MA$ doesn't assume $C$ before the verification procedure.