It is a well-known theorem by Kuhn in game theory that if a player in a game has perfect recall, then each of his behavioral strategies has an equivalent mixed strategy, and vice-versa: each of his mixed strategy has an equivalent behavioral strategy. Examples are known of games where some players lack perfect recall and each of the statements above doesn't hold: There are some games where some behavioral strategies of some players don't have any equivalent mixed strategies, and there are some games where some mixed strategies of some players don't have any equivalent behavioral strategies. (See "Game Theory" by Maschler, Solan, & Zamir: Example 6.8 for the second statement, Example 6.9 for the first statement.) Nevertheless, even to such players in such games, there are some behavioral strategies that have equivalent mixed strategies, and some mixed strategies that have equivalent behavioral strategies.
So I'm interested on having a way to check whether a particular behavioral/mixed strategy of a player has an equivalent mixed/behavioral strategy, and if it exists, finding that strategy, even in cases where the player lacks perfect recall.
Converting a behavioral strategy b to a mixed strategy m is easy: For every pure strategy p, multiply the probabilities b(d) of each decision d in p to obtain its probability m(p).
For (a very simple) example, suppose that the edges of a perfect-information one-player game are A->B, A->C, B->D, and B->E. Suppose that b assigns the probabilities: 0.4 to A->B, 0.6 to A->C, 0.5 to B->D, and 0.5 B->E. (Notice that b is a valid behavioral strategy, because for every vertex, the probabilities that b assigns to the edges emanating from it rightly add up to 1: 0.4 + 0.6 = 0.5 + 0.5 = 1.) Then m assigns the probabilities to each pure strategy: 0.4 x 0.5 = 0.2 to A->B, B->D; 0.4 x 0.5 = 0.2 to A->B, B->E; 0.6 x 0.5 = 0.3 to A->C, B->D; 0.6 x 0.5 = 0.3 to A->C, B->E (that is, 0.6 to A->C; if you regard the strategies A->C, B->D and A->C, B->E as the same strategy because the irrelevant decision at B is ignored). Notice that these probabilities rightly add up to 1 (0.2 + 0.2 + 0.3 + 0.3 = 1).
But what if upon this construction, the probabilities don't add up to 1--that is, the m constructed is not a mixed strategy? My conjecture is: there is a mixed strategy equivalent to b if and only if the probabilities in m constructed as above rightly add up to 1. This is interesting, because if m is a mixed strategy, there can also be other mixed strategies m' that are equivalent to b other than m (not in my particular over-simplistic example, sadly). So my idea is: test a (specifically constructed) candidate; if it fails then none succeeds. Is this true?
Converting a mixed strategy to a behavioral strategy is more complicated. It is found in Kuhn's "Lectures on the Theory of Games" (interestingly, he puts it as a definition, as if every behavioral strategy has an equivalent mixed strategy). Here's an excerpt of the book where he defines (that is, constructs) a behavioral strategy from a mixed strategy.
I'd like to show that if a behavioral strategy is equivalent to a mixed strategy, then it must satisfy the equation given in this "definition" by Kuhn (for each decision of each information set). (Again, Kuhn didn't bother to prove this, because he gave it as a definition...) So, if this construction by Kuhn from a mixed strategy doesn't result in a valid behavioral strategy (that is, for some information set, the decisions' probabilities don't add up to 1), then no behavioral strategy is equivalent to that mixed strategy. Again, one candidate suffices.
Any suggestion?