I am trying to find if there exists a relation (> or <) between two equations:
$$X = (I\mathbb{P}-I\mathbb{P}x^{*})+q(I-I\mathbb{P})+(cI-qcI)$$
and
$$X' = (I\mathbb{P}'-I\mathbb{P}'x^{**})+q'(I-I\mathbb{P}')+(c'I-q'c'I)$$
Here are some facts about the variables:
$$I \in \mathbb{R}_{++}$$
$$0 < q < q' < 1$$
$$0 < x^{*} < x^{**} < 1$$
$$0 < \mathbb{P} < \mathbb{P}' < 1$$
$$c,c' \in (0,1)$$
My initial attempt was to treat each equation as $A+B+C$ and $A'+B'+C'$ respectively.
However, comparing $A \stackrel{?}{>} A'$ essentially brings me to: $x-xy \stackrel{?}{>} x'-x'y'$ which I know can be either > or < depending on the specific values.
Trying to instead expand them out is leaving me similarly hopeless. It's entirely possible I can not simply put < or > between the two, but I figured I would reach out here before I accept that fate (I have not been able to produce a counterexample, but perhaps someone else can).
If there exists a relation, my intuition suggests $X' > X$
No, I cannot say > or < here.
Consider:
$$I = 1$$
$$q = 0.5$$
$$x^{*} = 0.5$$
$$\mathbb{P} = 0.5$$
$$c = 0.5$$
Then $X = 0.75$
$$I' = 1$$
$$q' = 0.51$$
$$x^{**} = 0.51$$
$$\mathbb{P}' = 0.51$$
$$c' = 0.01$$
Now, $X' \approx 0.50$
So here $X' < X$
Now, suppose $c' = 0.52$,
Now $X' \approx 0.755$
So, $X' > X$
So one is not less than the other for all possible values of the constants.