Solution
= Solution
Under a risk-neutral measure,
$$
\mathbb E_Q\xi_n=1+r.
$$
Writing $q=Q(\xi_n=1+b)$ gives
$$
q(1+b)+(1-q)(1+a)=1+r,
\qquad q=\frac{r-a}{b-a}.
$$
Equivalence requires $0<q<1$, hence $a<r<b$. At every node the risk-neutral condition determines this same conditional probability $q$, so the multipliers remain independent and identically distributed under $Q$.