Using R, I want to estimate two curves using points from two vectors, and then find the x and y coordinates where those estimated curves intersect.
In a strategic setting with players "t" and "p", I am simulating best responses for both players in response to what the other would pick in a strategic setting (game theory). The problem is that I don't have functions or lines, I have two sets of points originating from simulation, with one set of points corresponding to the player's best response to given actions by the other player. The actual math was too difficult for me (or matlab) to solve, which is why I'm using this simulated visual approach. I want to estimate best response functions (i.e. create non-linear curves) using the points, and then take the two estimated curves and find where they intersect in order to identify nash equilibrium (where the best response curves intersect).
As an example, here are two such vectors I am working with:
t=c(10.0,10.0,10.0,10.0,10.0,10.0,10.0,10.0,10.1,10.1,10.1,10.1,10.1,10.1,10.1,10.1,10.1,10.1,10.1,10.1,10.0,10.0,10.0,10.0,10.0,10.0,10.0,10.0,10.0,10.0,10.0,10.0,10.0,10.0,10.0,10.0,10.0,10.0,10.0,10.0,10.0)
p=c(12.3,12.3,12.3,12.3,12.3,12.3,12.4,12.4,12.4,12.5,12.5,12.5,12.6,12.6,12.7,12.7,12.8,12.8,12.9,12.9,13.0,13.1,13.1,13.2,13.3,13.4,13.5,13.4,13.5,13.6,13.6,13.7,13.8,13.8,13.9,13.9,13.9,14.0,14.0,14.0,14.0)
For the first line, the sample is made up of (t,a), and for the second line, the sample is made up of (a,p) where a is a third vector given by
a = seq(10, 14, by = 0.1)
For example, the first point for the sample corresponding to the first vector would be (10.0,10.0) and the second point would be (10.0,10.1). The first point for the sample corresponding to the second vector would be (10.0,12.3) and the second point would be (10.1,12.3).
What I originally tried to do is estimate the lines using polynomials produced by lm models, but those don't seem to always work:
plot(a,t, xlim=c(10,14), ylim=c(10,14), col="purple")
points(p,a, col="red")
fit4p <- lm(a~poly(p,3,raw=TRUE))
fit4t <- lm(t~poly(a,3,raw=TRUE))
lines(a, predict(fit4t, data.frame(x=a)), col="purple", xlim=c(10,14), ylim=c(10,14),type="l",xlab="p",ylab="t")
lines(p, predict(fit4p, data.frame(x=a)), col="green")
fit4pCurve <- function(x) coef(fit4p)[1] +x*coef(fit4p)[2]+x^2*coef(fit4p)[3]+x^3*coef(fit4p)[4]
fit4tCurve <- function(x) coef(fit4t)[1] +x*coef(fit4t)[2]+x^2*coef(fit4t)[3]+x^3*coef(fit4t)[4]
a_opt1 = optimise(f=function(x) abs(fit4pCurve(x)-fit4tCurve(x)), c(10,14))$minimum
b_opt1 = as.numeric(fit4pCurve(a_opt1))
EDIT: After fixing the type, I get the correct answer, but it doesn't always work if the samples don't come back as cleanly.
So my question can be broken down a few ways. First, is there a better way to accomplish what I'm trying to do. I know what I'm doing isn't perfectly accurate by any means, but it seems like a decent approximation for my purposes. Second, if there isn't a better way, is there a way I could improve on the methodology I have listed above.
Restart your R session, make sure all variables are cleared and copy/paste this code. I found a few mistakes in referenced variables. Also note that R is case sensitive. My suspicion is that you've been overwriting variables.
plot(a,t, xlim=c(10,14), ylim=c(10,14), col="purple")
points(p,a, col="red")
fit4p <- lm(a~poly(p,3,raw=TRUE))
fit4t <- lm(t~poly(a,3,raw=TRUE))
lines(a, predict(fit4t, data.frame(x=a)), col="purple", xlim=c(T,P), ylim=c(10,14),type="l",xlab="p",ylab="t")
lines(p, predict(fit4p, data.frame(x=a)), col="green")
fit4pCurve <- function(x) coef(fit4p)[1] +x*coef(fit4p)[2]+x^2*coef(fit4p)[3]+x^3*coef(fit4p)[4]
fit4tCurve <- function(x) coef(fit4t)[1] +x*coef(fit4t)[2]+x^2*coef(fit4t)[3]+x^3*coef(fit4t)[4]
a_opt = optimise(f=function(x) abs(fit4pCurve(x)-fit4tCurve(x)), c(T,P))$minimum
b_opt = as.numeric(fit4pCurve(a_opt))
As you will see:
> a_opt
[1] 12.24213
> b_opt
[1] 10.03581