[R] Johnson Distribution Fit

[David Winsemius]

On Oct 6, 2010, at 7:38 PM, Abey George wrote:

> I am trying to only fit this distribution to the data to see how  
> well it fits it. Getting the parameters is secondary. Did correct  
> the square brackets in the exponential equation but still get the  
> same result. There is only 1 variable to data2 and that is points.

I was asking for the data. The process of trying to guess what other  
errors you are making is tedious.

> I am trying to fit this distribution to points data scored by teams.
>
> The result for any(data2 < 0.5 | data2 > 0.5+50) is TRUE.

It should NOT be TRUE if you are using those limits as your bounding  
parameters. So there is a further source of problem ....  very  
possibly why you are getting NaNs.

-- 
David.

>
> Thanks
> Abey
>
>
> On Mon, Oct 4, 2010 at 11:33 AM, David Winsemius <dwinsemius at comcast.net 
> > wrote:
>
> On Oct 3, 2010, at 3:47 PM, Abey George wrote:
>
> Hi,
>     I am trying to fit a Johnson SB distribution using fitdist  
> function in
> fitdistrplus Library. I have defined the Johnson SB distribution  
> from (
> http://www.ntrand.com/johnson-sb-distribution/) . But it gives me the
> follwing errors. Any help would be appreciated
>
> Are you really trying to estimate the bounding values as well as the  
> gamma and delta parameters. Those would seem to be more likely  
> determined by the nature of the problem, e.g., policy limits on the  
> insured sums if this were a financial problem.
>
>
>
> #xi = xi
>
> #lambda =l
> #delta =d
> #gamma = g
>
> djohn = function(x,xi,l,d,g)
> (d/(l*sqrt(2*pi)*((x-xi)/l)*(1-((x-xi)/l))))*exp[-0.5*(g +
> d*log(((x-xi)/l)/(1-((x-xi)/l))))^2]
>
> You used exp[ ] where you probably wanted exp().
>
>
>
> pjohn = function(x,xi,l,d,g)   pnorm(g + d*log(((x-xi)/l)/(1-((x-xi)/ 
> l))))
>
> qjohn = function(p,xi,l,d,g)   xi + (l*exp((qnorm(p) - g)/d))/(1 +
> exp((qnorm(p) - g)/d))
>
> f1c <- fitdist(data2,"john",start=list(xi = 0.5 ,l = 50, d = 1, g =  
> 1))
>
> You have not given us the "data2" variables, so we have no way of  
> checking whether any of them appear outside the range [epsilon,  
> lambda+epsilon]. Using your data2 vector, what are the results of :
>
> any(data2 < 0.5 | data2 > 0.5+50)   #?
>
>
> Error in fitdist(data2, "john", start = list(xi = 0.5, l = 50, d =  
> 1,  :
>  the function mle failed to estimate the parameters,
>               with the error code 100
> In addition: Warning message:
> In log(((x)/l)/(1 - ((x)/l))) : NaNs produced
>
>
> Cheers
> AG

David Winsemius, MD
West Hartford, CT