Dave,
  I am a little confused. You ask about a triangle distribution, but 
describe the situation as a relation between y and x. Do you really want a 
piecewise linear fit of y to x? That could be gotten using nonlinear least 
squares.
Dave



From:
Jinsong Zhao <jszhao@yeah.net>
To:
r-help@r-project.org
Date:
12/23/2010 07:53 AM
Subject:
Re: [R] Fitting a Triangular Distribution to Bivariate Data
Sent by:
r-help-bounces@r-project.org



On 2010-12-23 2:19, David Bapst wrote:
> Hello,
> I have some xy data which clearly shows a non-monotonic, peaked
> triangular trend. You can get an idea of what it looks like with:
>
> x<-1:20
> y<-c(2*x[1:10]+1,-2*x[11:20]+42)
>
> I've tried fitting a quadratic, but it just doesn't the data-structure
> with the break point adequately. Is there anyway to fit a triangular
> or 'tent' function to my data in R?
>
> Some sample code would be appreciated; I'm not new to R, but I
> sometimes have difficulty understanding the model-fitting functions
> (finally figured out how to extrapolate with predict() today!)
>
> Thanks!
> -Dave Bapst, UChicago
>


Hi,

you may try the following code:

 > library(triangle)
 > library(fitdistrplus)
 > summary(fitdist(y, "triangle", start = list(a = 1.9, b= 21.1, c = 
11.5)))
Fitting of the distribution ' triangle ' by maximum likelihood
Parameters :
    estimate Std. Error
a -1.400007  2.3523724
b 23.627448  1.9804026
c 13.000000  0.1107073
Loglikelihood:  -62.41994   AIC:  130.8399   BIC:  133.8271
Correlation matrix:
             a           b           c
a  1.00000000 -0.14537297 -0.01203898
b -0.14537297  1.00000000 -0.01439500
c -0.01203898 -0.01439500  1.00000000

HTH ...

Jinsong Zhao

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