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= E2is equivalent to
E1 = (T)((E1) op (E2)), where
Tis the type of
E1, except that
E1is 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.