[R] Solving two nonlinear equations with two knowns

[Berend Hasselman]

yhsu6 wrote:
> 
> 
> ########## 
> #R code: 
> mu2=0.4 
> sigma2=4 
> tau=0.75 
> mean.diff=2 
> var.ratio=3 
> 
> #Use optim: 
> parameter<-function(c) 
> { 
> a<-c[1] 
> b<-c[2] 
> u<-runif(10000) 
> Y<-qnorm(u,mean=mu2,sd=sigma2) 
> u<-runif(10000) 
> X<-qnorm(u,mean=mu2,sd=sigma2)+a*abs(u-tau)^b*(u>tau) 
> return((abs(mean(X)-mean(Y))-mean.diff)^2+(var(X)/var(Y)-var.ratio)^2) 
> } 
> 
> 
> c0<-c(3,1) 
> cstar<-optim(c0,parameter)$par 
> astar<-cstar[1] #4.1709 
> bstar<-cstar[2] #-0.2578 
> 
> 

Your problem is ill posed.
In your function you have two u <- runif(10000) statements
This means that every time your function is called, u changes.
You can see this happening by repeating parameter(c0) several times

parameter(c0)
[1] 7.346233
parameter(c0)
[1] 7.204457

So you are trying to optimise a function that gives different results for
the same parameter vector every  time it is called. You can't determine a
and b that way.
This will never work.

The two u<-runif(10000) should be removed from the function body and should
be done only once before calling the function.
Your function then becomes

zparameter<-function(c) 
{ 
a<-c[1] 
b<-c[2] 
Y<-qnorm(u,mean=mu2,sd=sigma2) 
X<-qnorm(u,mean=mu2,sd=sigma2)+a*abs(u-tau)^b*(u>tau) 
z<-numeric(2)
z[1] <- abs(mean(X)-mean(Y))-mean.diff
z[2] <- var(X)/var(Y)-var.ratio
z
} 

BTW: you can make this function more efficient by moving the expression for
Y outside the function body and writing X<=Y+.....

If you do this, you should get sensible results. For solving a system of
equation I would not use optim.
Use an equation solver (nleqslv or BB).

Berend

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