Given are the following two functions: $ g(x,\theta)=1-\frac{\left( 1-\theta\right) }{\theta}\left( \frac{(x-1)R} {(1-(1-\pi)i)R +\left( 1-\pi\right) i x}\right) \tag{NAG} $ and $f(x)=\frac{\pi xR}{\left[ \left( 1-i\right) R+ix\right] } \cdot\frac{(1-i)}{\left( 1-(1-\pi)(1-i)\right) } \tag{CMP}$
What I want to show is that the intersection point of $g(x,\theta)$ and $f(x)$ increases with $\theta$. I mean that an increase in $\theta$ leads to a new intersection which is characterized by both higher x and higher function value.
This can be shown graphically but what I need is a formal proof. The problem is that solving for the equilibrium is a solution of a complex (but just) quadratic equation.
Note that the CMP is no function of $\theta$ but an increasing function of x $\forall x \in \mathbb{R}_{\leq0}$, thus a change in $\theta$ moves the intersection along CMP. Additionally, NAC is a decreasing function of $x$ $(\frac{\partial g(x,\theta)}{\partial p_{n}}<0$) but the slope of the tangent increases with $\theta$ $(\frac{\partial^2 g(x,\theta)}{\partial p_{n} \partial \theta}>0$). Since the NAC passes independently of the parameter setting through $(1,1)$, an increase in $\theta$, from $\theta^{**}$ to $\theta^{*} $leads to a raise of the angel $\alpha$ moving a low intersection point $(x^{**},y^{**})$ to a higher $(x^{*},y^{*})$. To summarize, what I need is a formal proof that shows that the intersection point increases with an increase in $\theta$. I hope someone can help me to show this with a without the need of solving for the intersection point. (I actually do not know how to start). Thx for your help!!!
Okay, you can do this in three steps (BTW, you need to assume also that $\theta\in (0,1)$; I take it that assumption is okay since you call $\theta$ a probability).
Step 1: As you noted, $f(x)$ is an increasing function of $x$ when $x > 0$. In fact, $f(x)$ is strictly increasing: its slope is always positive (using that $\pi,i\in (0,1)$ and $R \geq 1$ by assumption). This implies
Step 2: As you also noted $g(x,\theta)$, for fixed value of $\theta$, is a decreasing function of $x$. In particular, it is monotonic. Which leads us to
and hence
Proof: by monotonicity, for every $0 \leq y \leq x_1$ we have $f(y) \leq f(x_1)$. Similarly we also have $g(y,\theta_2) \geq g(x_1,\theta_2)$. So if $g(x_1,\theta_2) > f(x_1)$, it is impossible that $f(y) = g(y,\theta_2)$. Since the two curves must intersect, they have to intersect somewhere $y > x_1$. q.e.d.
Note that similarly if $g(x_1,\theta_2) < g(x_1,\theta_1)$, the intersection point $x_2$ must be less than $x_1$.
Step 3: So all you need to show now is that if $(x,\theta)$ is such that $f(x) = g(x,\theta)$, for any $\theta' > \theta$, you have that $g(x,\theta') > g(x,\theta)$. Looking at the equation for $g(x,\theta)$, you see that $g(x,\theta)$, for fixed $x > 1$, is an increasing function of $\theta$ (just compute $\partial g / \partial \theta$). So you'd be done if you can show that the intersection point of $f$ and $g$ cannot be a point where $x \leq 1$.
Using the monotonicity again, it suffices to compute $g(1,\theta)$ and $f(1)$. If $g(1,\theta)> f(1)$, then from the same argument as in step 2, we know that the intersection point must be $x > 1.
You see immediately that $g(1,\theta) = 1$. On the other hand it is a simple algebraic computation to see that $f(1) < 1$: you just need to show that $\pi (1-i) R < [ (1-i)R + i ][ 1- (1-\pi)(1-i)]$. If you multiply out the right hand side, and subtract from it the left hand side, you get that you just need the inequality $0 < (1-\pi)(1-i)i R + (1- (1-\pi)(1-i))i$ to hold, which follows from the assumed range of $\pi,i$ and that $R$ is positive.