Can Nondet Rabin Tree automaton be determinized?

Question

In other words, are they equally powerful?

(for word automata the answer is "yes"; this question is about tree automata).

(i am talking about tree automata that work on $in$finite trees)

Answer 1

The answer is no, nondeterministic Rabin automata are more expressive than deterministic ones.

As an example, consider the language of $\{a,b\}$-labeled binary trees that contain at least one $b$.

It's easy to construct a two-state Büchi tree automaton for this language (and hence a Rabin automaton), but there is no deterministic equivalent.

You can look at Lemma 6.1 here, for a start.