Hi Tal,
basically, by summing over the (pointwise) density, you are
"approximating" the integral over the density (which should be around 1)
- but to really do a rectangular approximation, you will of course need
to multiply each function value by the width of the corresponding
rectangle. I'd recommend reading through Wikipedia's article on the
Riemann integral, you may be enlightened.
http://en.wikipedia.org/wiki/Riemann_integral
HTH,
Stephan
Tal Galili schrieb:
Hi all,
A friend send me a question on why does this:
x<-rpois(100,1)
sum( hist(x)$density )
Gives out "2"
I tried this:
sum( hist(x, freq =T)$density )
It didn't help.
Then he came back with the following insight:
# with breaks
b<-c(0,0.9,1:8)
sum(hist(x,breaks=b)$density) # Much more then 2
# but if we add weights according to the interval length
sum(hist(x,breaks=b)$density * diff(b))
# it works
What do you think ?
Tal
----------------Contact
Details:-------------------------------------------------------
Contact me: tal.gal...@gmail.com | 972-52-7275845
Read me: www.talgalili.com (Hebrew) | www.biostatistics.co.il (Hebrew) |
www.r-statistics.com (English)
----------------------------------------------------------------------------------------------
[[alternative HTML version deleted]]
______________________________________________
R-help@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.
______________________________________________
R-help@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.
