Show $0<d<1+r_1<u$

Question

Suppose $0<d<1+r<u$ for a fixed $r$, $d$, and $u$. Let $\tilde{p} = \frac{(1+r)-d}{u-d}$ and let $\tilde{q} = \frac{u-(1+r)}{u-d}$. Suppose $0<p_1<1$, $0<q_1<1$, and $p_1 + q_1 = 1$. Further, $p_1 \ne \tilde{p}$. Let $r_1 = p_1u + q_1d -1$

Show that $0<d<1+r_1<u$

Answer 1

Hint: Rewrite the expression $r_1 = p_1u + q_1d -1$ as $1+r_1 = d + p_1 (u-d)$. From there it should be trivial. The expression $p_1u+q_1d$ with $p_1+q_1=1$ and $0<p_1<1$ is also known as a convex combination.