As always with these questions, the JLS holds the answer. In this case §15.26.2 Compound Assignment Operators. An extract:

A compound assignment expression of the form E1 op= E2 is equivalent to E1 = (T)((E1) op (E2)) , where T is the type of E1 , except that E1 is evaluated only once.

An example cited from §15.26.2

[...] the following code is correct: short x = 3; x += 4.6; and results in x having the value 7 because it is equivalent to: short x = 3; x = (short)(x + 4.6);

In other words, your assumption is correct.