Dear List-members, I have a problem where I have to estimate a mean, or a sum of a population but for some reason it contains a huge amount of zeros. I cannot give real data but I constructed a toy example as follows N1 <- 100000 N2 <- 3000 x1 <- rep(0,N1) x2 <- rnorm(N2,300,100) x <- c(x1,x2) n <- 1000 x_sample <- sample(x,n,replace=FALSE) I want to estimate the sum of x based on x_sample (not knowing N1 and N2 but their sum (N) only). The sample mean has a huge standard deviation I am looking for a better estimator. I was thinking about trimmed (or "left trimmed" as my numbers are all positive) means or something similar, but if I calculate trimmed mean I do not know N2 to multiply with. Do you have any idea or could you give me some insight? Thanks a lot: Daniel
Although you have provided R code to illustrate your problem, it is
fundamentally a statistics theory question, and belongs somewhere else like
stats.stackexchange.net.
When you post there, I recommend that you spend more effort to identify why the
zeros are present. If they are indicators of unknown values, that will be very
different than if zeros are valid members of the population.
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Sent from my phone. Please excuse my brevity.
"Kehl D?niel" <kehld at ktk.pte.hu> wrote:
>Dear List-members,
>
>I have a problem where I have to estimate a mean, or a sum of a
>population but for some reason it contains a huge amount of zeros.
>I cannot give real data but I constructed a toy example as follows
>
>N1 <- 100000
>N2 <- 3000
>x1 <- rep(0,N1)
>x2 <- rnorm(N2,300,100)
>x <- c(x1,x2)
>
>n <- 1000
>
>x_sample <- sample(x,n,replace=FALSE)
>
>I want to estimate the sum of x based on x_sample (not knowing N1 and
>N2
>but their sum (N) only).
>The sample mean has a huge standard deviation I am looking for a better
>
>estimator.
>I was thinking about trimmed (or "left trimmed" as my numbers are
all
>positive) means or something similar,
>but if I calculate trimmed mean I do not know N2 to multiply with.
>
>Do you have any idea or could you give me some insight?
>
>Thanks a lot:
>Daniel
>
>______________________________________________
>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.
Dear Jeff, thank you for the response. Of course I know this is a theory question still I hope to get some comments on it (if somebody already dealt with alike problems might suggest a package and it would not take longer than saying this is a theoretical question) The values are counts, so 0 means those cases do not have this item, they have 0, as such it means a "real zero", they are valid members. thanks, daniel 2012.05.03. 16:42 keltez?ssel, Jeff Newmiller ?rta:> Although you have provided R code to illustrate your problem, it is fundamentally a statistics theory question, and belongs somewhere else like stats.stackexchange.net. > > When you post there, I recommend that you spend more effort to identify why the zeros are present. If they are indicators of unknown values, that will be very different than if zeros are valid members of the population. > --------------------------------------------------------------------------- > Jeff Newmiller The ..... ..... Go Live... > DCN:<jdnewmil at dcn.davis.ca.us> Basics: ##.#. ##.#. Live Go... > Live: OO#.. Dead: OO#.. Playing > Research Engineer (Solar/Batteries O.O#. #.O#. with > /Software/Embedded Controllers) .OO#. .OO#. rocks...1k > --------------------------------------------------------------------------- > Sent from my phone. Please excuse my brevity. > > > > "Kehl D?niel"<kehld at ktk.pte.hu> wrote: > >> Dear List-members, >> >> I have a problem where I have to estimate a mean, or a sum of a >> population but for some reason it contains a huge amount of zeros. >> I cannot give real data but I constructed a toy example as follows >> >> N1<- 100000 >> N2<- 3000 >> x1<- rep(0,N1) >> x2<- rnorm(N2,300,100) >> x<- c(x1,x2) >> >> n<- 1000 >> >> x_sample<- sample(x,n,replace=FALSE) >> >> I want to estimate the sum of x based on x_sample (not knowing N1 and >> N2 >> but their sum (N) only). >> The sample mean has a huge standard deviation I am looking for a better >> >> estimator. >> I was thinking about trimmed (or "left trimmed" as my numbers are all >> positive) means or something similar, >> but if I calculate trimmed mean I do not know N2 to multiply with. >> >> Do you have any idea or could you give me some insight? >> >> Thanks a lot: >> Daniel >> >> ______________________________________________ >> 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. > >
On Thu, May 03, 2012 at 03:08:00PM +0200, Kehl D?niel wrote:> Dear List-members, > > I have a problem where I have to estimate a mean, or a sum of a > population but for some reason it contains a huge amount of zeros. > I cannot give real data but I constructed a toy example as follows > > N1 <- 100000 > N2 <- 3000 > x1 <- rep(0,N1) > x2 <- rnorm(N2,300,100) > x <- c(x1,x2) > > n <- 1000 > > x_sample <- sample(x,n,replace=FALSE) > > I want to estimate the sum of x based on x_sample (not knowing N1 and N2 > but their sum (N) only). > The sample mean has a huge standard deviation I am looking for a better > estimator.Hi. I do not know the exact answer, but let me formulate the following observation. If the question is redefined to estimate the mean of nonzero numbers, then an estimate is mean(x_sample[x_sample != 0]). Its standard deviation in your situation may be estimated as res <- rep(NA, times=1000) for (i in seq.int(along=res)) { x_sample <- sample(x,n,replace=FALSE) res[i] <- mean(x_sample[x_sample != 0]) } sd(res) [1] 18.72677 # this varies with the seed a bit The observation is that this cannot be improved much, since the estimate is based on a very small sample. The average size of the sample of nonzero values is N2/(N1+N2)*n = 29.1. So, the standard deviation should be something close to 100/sqrt(29.1) = 18.5376. Petr Savicky.
Dear Petr,
thank you for your input.
I tried to experiment with (probably somewhat biased) truncated means
like in the following code.
How I got the 225 as a truncation limit is a good question. :)
REPS1 <- REPS2 <- 1000
N1 <- 100000
N2 <- 30000
N <- N1+N2
x1 <- rep(0,N1)
x2 <- rnorm(N2,300,100)
x <- c(x1,x2)
n <- 1000
for (i in 1:REPS1){
x_sample <- sort(sample(x,n,replace=FALSE),TRUE)
x_trunc <- x_sample[1:225]
REPS1[i] <- mean(x_sample)*N
REPS2[i] <- sum(x_trunc)/n*N
}
sum(x2)
mean(REPS1)
mean(REPS2)
sd(REPS1)
sd(REPS2)
sd(REPS2)/sd(REPS1)
Best,
daniel
2012.05.03. 17:45 keltez?ssel, Petr Savicky ?rta:> On Thu, May 03, 2012 at 03:08:00PM +0200, Kehl D?niel wrote:
>> Dear List-members,
>>
>> I have a problem where I have to estimate a mean, or a sum of a
>> population but for some reason it contains a huge amount of zeros.
>> I cannot give real data but I constructed a toy example as follows
>>
>> N1<- 100000
>> N2<- 3000
>> x1<- rep(0,N1)
>> x2<- rnorm(N2,300,100)
>> x<- c(x1,x2)
>>
>> n<- 1000
>>
>> x_sample<- sample(x,n,replace=FALSE)
>>
>> I want to estimate the sum of x based on x_sample (not knowing N1 and
N2
>> but their sum (N) only).
>> The sample mean has a huge standard deviation I am looking for a better
>> estimator.
> Hi.
>
> I do not know the exact answer, but let me formulate the following
observation.
> If the question is redefined to estimate the mean of nonzero numbers, then
> an estimate is mean(x_sample[x_sample != 0]). Its standard deviation in
your
> situation may be estimated as
>
> res<- rep(NA, times=1000)
> for (i in seq.int(along=res)) {
> x_sample<- sample(x,n,replace=FALSE)
> res[i]<- mean(x_sample[x_sample != 0])
> }
> sd(res)
>
> [1] 18.72677 # this varies with the seed a bit
>
> The observation is that this cannot be improved much, since the estimate
> is based on a very small sample. The average size of the sample of nonzero
> values is N2/(N1+N2)*n = 29.1. So, the standard deviation should be
> something close to 100/sqrt(29.1) = 18.5376.
>
> Petr Savicky.
>
> ______________________________________________
> 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.
On Fri, May 04, 2012 at 07:43:32PM +0200, Kehl D?niel wrote:> Dear Petr, > > thank you for your input. > I tried to experiment with (probably somewhat biased) truncated means > like in the following code. > How I got the 225 as a truncation limit is a good question. :) > > REPS1 <- REPS2 <- 1000 > N1 <- 100000 > N2 <- 30000 > N <- N1+N2 > x1 <- rep(0,N1) > x2 <- rnorm(N2,300,100) > x <- c(x1,x2) > > n <- 1000 > > for (i in 1:REPS1){ > x_sample <- sort(sample(x,n,replace=FALSE),TRUE) > x_trunc <- x_sample[1:225] > REPS1[i] <- mean(x_sample)*N > REPS2[i] <- sum(x_trunc)/n*N > } > > sum(x2) > mean(REPS1) > mean(REPS2) > sd(REPS1) > sd(REPS2) > sd(REPS2)/sd(REPS1)Dear Daniel. Thank you for your reply. In the original question, you used the parameters N1 <- 100000 N2 <- 3000 and now the parameters N1 <- 100000 N2 <- 30000 My remark was that with the original parameters, there are only 29.1 nonzero elements on average. Now, there are 230.8 nonzero elements on average, which is significantly better. Discussion of the use of the truncated mean is probably a question to other members of the list. I do not feel to be an expert on this. Best, Petr.