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3 messages · Jim Silverton, Pascal Oettli, William Dunlap
Hello, The value for 'start' is not correct. According to the help for "fitdistr", 'start' should contain the paramaters used by the random function (?rchisq) Hope this helps, Pascal
On 20 November 2013 09:29, Jim Silverton <jim.silverton at gmail.com> wrote:
I generated some random data using R and then trying to see if it came from
the same distribution but the erros keep piling. Can anyone help?
y=rchisq(100000,1)
mean(y)
var(y)
library(MASS)
fitdistr(y, "chi-squared", start = list(4), method = "Brent", lower=0.9,
upper=3)
--
Thanks,
Jim.
[[alternative HTML version deleted]]
______________________________________________ R-help at r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Pascal Oettli Project Scientist JAMSTEC Yokohama, Japan
fitdistr(y, "chi-squared", start = list(4), method = "Brent", lower=0.9, upper=3)
First, it would be nice if your mail had a subject line and you showed
the error message you got from this command.
I got:
> z <- fitdistr(y, "chi-squared", start = list(4), method="Brent", lower=0.9, upper=3)
> z
Error in dn[[2L]] : subscript out of bounds
> str(z)
List of 5
$ estimate: num 1
$ sd : num 0.00286
$ vcov : num [1, 1] 8.16e-06
$ loglik : num -79007
$ n : int 100000
- attr(*, "class")= chr "fitdistr"
showing that the error was in the printing of the result, not in computing
the result.
There are a few problems here. First, the print method assumes that the
optimizer attaches parameter names to the parameter vector. Second,
optim(method="Brent") does not attach a name to its output parameter the
way the other optimization method do (I suppose that is because it only
works on univariate problems.) Third, you didn't supply a parameter name
in 'start'.
You can get a printable answer by using the default method and supplying
a name for the parameter.
> zz <- fitdistr(y, "chi-squared", start = list(df=4), lower=0.9, upper=3)
> zz
df
1.003840114
(0.002856363)
> str(zz)
List of 5
$ estimate: Named num 1
..- attr(*, "names")= chr "df"
$ sd : Named num 0.00286
..- attr(*, "names")= chr "df"
$ vcov : num [1, 1] 8.16e-06
..- attr(*, "dimnames")=List of 2
.. ..$ : chr "df"
.. ..$ : chr "df"
$ loglik : num -79007
$ n : int 100000
- attr(*, "class")= chr "fitdistr"
By the way, giving a starting value that is not between the upper and lower
limits that you also gave is asking for trouble of a different sort.
Bill Dunlap
Spotfire, TIBCO Software
wdunlap tibco.com
-----Original Message----- From: r-help-bounces at r-project.org [mailto:r-help-bounces at r-project.org] On Behalf Of Jim Silverton Sent: Tuesday, November 19, 2013 4:30 PM To: r-help at r-project.org Subject: [R] (no subject) I generated some random data using R and then trying to see if it came from the same distribution but the erros keep piling. Can anyone help? y=rchisq(100000,1) mean(y) var(y) library(MASS) fitdistr(y, "chi-squared", start = list(4), method = "Brent", lower=0.9, upper=3) -- Thanks, Jim. [[alternative HTML version deleted]]
______________________________________________ R-help at r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.