# formal proof: Intersection of two function

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!!!

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what is $i$ in your equation? Is it $\sqrt{-1}$? If so, what do you mean when you say the function increases when the function takes complex values? – Willie Wong Feb 10 '12 at 23:27
sorry I forgot to say that the functions evolve form a theoretical model where $\pi$ and $i$ are only defined at the open interval $(0,1)$, R>=1, $x$>=0. – Armin Feb 11 '12 at 9:06
I still don't get what are $\pi$ and $i$. Are they functions? What are they functions of? Do they depend on $\theta$ or $x$? For that matter, what is $R$? And please edit clarifications into the question, in addition to responding in the comments. – Willie Wong Feb 11 '12 at 11:23
These are just constants, thus they do not depent on $\theta$ or $x$. To get an idea what the functions are good for. This is a micro economic model where $\pi$ denotes the fraction of naive households in the economy, $i$ is the fraction of early consuming households and R is the return on an asset. Finally, $\theta$ is a crisis probility. All of these constants are given exogeniously. Finally, the intersection of both functions leads to the the equilibrium price. Thus x is the price in the state without crisis and the function value is the price in the crisis. – Armin Feb 11 '12 at 11:36
Moreover, what I want to show is that both prices increase with a higher crisis probilty $\theta$, i.e. an increase of $\theta$ leads to an increase of the intersection point and the value of the functions. I hope this answers your questions. – Armin Feb 11 '12 at 11:39

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

If the intersection point increases in $x$, the intersection value $f(x)$ must also increases.

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

$f(x)$ and $g(x,\theta)$ only intersect once. (Where $x\geq 0$.)

and hence

Consider $\theta_1$ and $\theta_2$. Let $x_1$ be the value such that $f(x_1) = g(x_1,\theta_1)$. Suppose that $g(x_1,\theta_2) > g(x_1,\theta_1)$, then necessarily the intersection point $x_2$ which solve $f(x_2) = g(x_2,\theta_2)$ must satisfy $x_2 > x_1$.

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. - Thx this really helped a lot. One more question: Do you think it is also possible to show the effect of a change in$i$on the intersection point. The problem I face is that both functions change with a varying$i$. Intuitively, what should happen with an increasing$i$is that the function value decreases while$x$increases at the new intersection point. I guess it will not be possible to show this with a similar proof. Nonetheless, thanks for your help and effort. – Armin Feb 12 '12 at 16:39 Initially I would guess that behaviour with regards to$i$would depend on whether$x \geq R$or the other way around. Note that when$x = R$,$g$is independent of$i$and$f$is monotonic in$i$. In each of the regimes$R > x$,$R = x$, and$R < x\$ it may be able to argue similarly, but I am not 100% sure: it requires computing more. – Willie Wong Feb 13 '12 at 8:45