> The distribution of the randomly truncated variable has thus four
> parameters: a, b, mu and sigma. I was able to write down the likelihood
> and attempted to maximise it
I read the Wikipedia article more carefully.
The formula is relatively simple, and is based on the application of Bayes
On 13/07/19 10:54 AM, Spencer Graves wrote:
Hello:
What do you suggest I do about modeling random truncation?
Good question! Probably the best answer is "Give up and go to the pub!" :-)
But seriously, there is a package DTDA on CRAN which purports to analyse
randomly truncated
Did you search on e.g. "model truncation" at rseek.org? Several packages
came up that appear to deal with truncated data, though I have no clue
whether in the way you specify.
-- Bert
Bert Gunter
"The trouble with having an open mind is that people keep coming along and
sticking things into
> It would be nice if I had an R
> package that would make it relatively easy to model the truncation as a
> function of "d"
I suspect that R has everything you need, already.
However, I suspect you may need to reformulate your question to find what
you need.
> I assume that the probability of
Hello:
What do you suggest I do about modeling random truncation?
I have data on a variable Y in strata S[0], S[1], ..., S[n],
where Y is always observed in S[0] but is less often observed in the
other strata. I assume that the probability of observing Y is a
monotonically
Oh, sorry, I think I see what you have tried to do. You want yearly
ticks but month-day labels. These won't mean much unless you also have
the year. If you ask for a date with just the year, as.Date will give
you a date in the middle of that year:
as.Date("2002","%Y")
[1] "2002-07-13"
So making
Hi Steven,
year1 is a number (e.g. 1993), monthday (e.g. 05-01) is not.
Jim
On Fri, Jul 12, 2019 at 10:56 PM wrote:
>
> Thanks Jim.
>
> I am trying to apply this to my version with plot_ly, and couldn't make it to
> work.
> The sydf$year1 field is numeric, so the min() and max() works, but
"ifelse is very slow"? Benchmark time.
> x <- runif(100)
> y <- runif(100)
> system.time(ifelse(x < y, x, y))
user system elapsed
0.403 0.044 0.448
> system.time(y + (x < y)*(x - y))
user system elapsed
0.026 0.012 0.038
This appears to be a quality-of-implementation
Thanks Jim.
I am trying to apply this to my version with plot_ly, and couldn't make it to
work.
The sydf$year1 field is numeric, so the min() and max() works, but when I tried
to use your formula for the sydf$monthday field I get an error:
yrticks <-
Por si es de vuestro inter�s...
https://www.r-users.gal/
Pd.- Lanzado el "call for papers", el programa todav�a est� por determinar.
Un saludo.
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