On Fri, May 04, 2012 at 07:43:32PM +0200, Kehl Dániel 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.
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