Consider the following list of 5 real numbers
$$
\begin{align}
y_{1}\equiv &q_{1}[ \alpha^2 p_{1}+(1-\alpha)^2(1-p_{1})] \\
y_{2}\equiv &q_{2}[ \alpha^2 p_{2}+(1-\alpha)^2(1-p_{2})] \\
y_{3}\equiv &q_{3}[ \alpha^2 p_{3}+(1-\alpha)^2(1-p_{3})] \\
y_{4}\equiv &q_{4}[ \alpha^2 p_{4}+(1-\alpha)^2(1-p_{4})] \\
y_{5}\equiv &q_{4}[ \alpha(1-\alpha)] \\
\end{align}
$$
where
- $\alpha\in (\frac{1}{2},1)$
- $q_i\in (0,1)$ $\forall i\in \{1,2,3,4\}$
- $\sum_{i=1}^4 q_i=1$
- $p_i\in (0,1)$ $\forall i\in \{1,2,3,4\}$.
I would like to find some sufficient conditions on $\alpha$, $\{p_i\}_{i=1,2,3,4}$, and $\{q_i\}_{i=1,2,3,4}$ such that the 5 numbers above can be strictly ordered. Below, I report my results and what I don't like of them.
My attempt:
There are several ways to strictly order the 5 numbers. Ideally, I would like to find an ordering that imposes "weak" conditions on $\alpha$, $\{p_i\}_{i=1,2,3,4}$, and $\{q_i\}_{i=1,2,3,4}$.
After several attempts, this is an ordering that I am almost satisfied with except for one assumption:
by doing some algebra it can be shown that if
-
$p_1<p_2<p_3<p_4$
-
$q_1<q_2<q_3<q_4$
-
$p_4<1-\alpha$
then, $y_5>y_4>y_3>y_2>y_1$.
What I don't like of this result is Assumption 3 because it imposes an upper bound on the maximum element among $\{p_1,p_2,p_3,p_4\}$, which seems quite strong. Rather, it would be preferable to find an ordering which imposes an upper bound on the minimum element among $\{p_1,p_2,p_3,p_4\}$. Or, alternatively, an ordering which imposes a lower bound on the maximum element among $\{p_1,p_2,p_3,p_4\}$.
I have not been successful in finding an ordering that imposes weaker conditions than above, though. Could you help?
Best Answer
This answer proves that if $p_4\lt 1-\alpha\lt p_3\lt p_2\lt p_1$ and $q_4\lt q_3\lt q_2\lt q_1$, then $y_4\lt y_5\lt y_3\lt y_2\lt y_1$.
Proof :
We have $$\begin{align} y_4\lt y_5&\iff q_{4}[ \alpha^2 p_{4}+(1-\alpha)^2(1-p_{4})]\lt q_{4}[ \alpha(1-\alpha)] \\\\&\iff \alpha^2 p_{4}+(1-\alpha)^2(1-p_{4})\lt \alpha(1-\alpha) \\\\&\iff p_4(2\alpha-1)\lt (1-\alpha)(2\alpha-1) \\\\&\iff p_4\lt 1-\alpha\end{align}$$
Since we have $$\begin{align}&\alpha^2 P+(1-\alpha)^2(1-P)\lt \alpha^2 P'+(1-\alpha)^2(1-P') \\\\&\iff \alpha^2(P-P')+(1-\alpha)^2(1-P-1+P')\lt 0 \\\\&\iff (P-P')(\alpha^2-1+2\alpha-\alpha^2)\lt 0 \\\\&\iff P\lt P'\end{align}$$ we see that if $p_{i+1}\lt p_i$ and $q_{i+1}\lt q_i$ for $i=1,2$ and $3$, then $$\alpha^2 p_{i+1}+(1-\alpha)^2(1-p_{i+1})\lt \alpha^2 p_i+(1-\alpha)^2(1-p_i)\implies y_{i+1}\lt y_i$$
Also, since we get $$\begin{align}&\alpha(1-\alpha)\lt \alpha^2 p_{3}+(1-\alpha)^2(1-p_{3}) \\\\&\iff (2\alpha-1)(1-\alpha)\lt (2\alpha-1)p_3\ \\\\&\iff 1-\alpha\lt p_3 \end{align}$$ we see that if $1-\alpha\lt p_3$ and $q_4\lt q_3$, then $$\alpha(1-\alpha)\lt \alpha^2 p_{3}+(1-\alpha)^2(1-p_{3})\implies y_5\lt y_3\quad\blacksquare$$