Talk:Ratio estimator

This article is very poorly written, and no assumptions are stated to justify the following claims:

• "θy is known to be asymptotically normally distributed." - asymptotics requires a parameter going to infinity, which is never stated. Is the resulting normality a result of the Central Limit Theorem? If so, an independence assumption must be made, as well as a finite-variance assumption.
• "E(x*1/y) = E(x)*E(1/y)" - this requires independence of x & y, which is never stated.

--65.209.72.194 (talk) 15:02, 25 July 2014 (UTC)

I bet the description of Lahiri's method in this article is wrong. I don't know Lahiri's method but I'm guessing it's just the Midzuno-Sen method using rejection sampling. If that's correct I would move the description of Midzuno-Sen before the description of Lahiri and replace the description of the Lahiri method with a brief statement that it's Midzuno-Sen using rejection sampling to pick the first item. 2620:0:1003:1019:24E6:C515:AC70:A1BB (talk) 18:30, 3 September 2015 (UTC)

I have corrected the application of Lahiri's method and fixed poor citations of a couple of other references. The Lahiri method is based upon the textbook by Lohr, cited. Incidently, Lahiri's method is not limited to ratio estimators but is a general sampling technique.

empirical_bayesian@ieee.org

This user is a member of WikiProject Statistics.

19:42, 1 July 2016 (UTC)

I indicated that the Lahiri estimator is biased and recommended that the Midzuno-Sen technique be used exclusively. See code below.

# Lahiri algorithm, own implementation. Jan Galkowski.
# empirical_bayesian@ieee.org, 3rd July 2016
# Last changed 3rd July 2016

is.natural<- function(x)
{
  x<- (0 < x) & (x == floor(x))
  return(x)
}

lahiri.sampling<- function(x, n, per=10)
{
  stopifnot(is.natural(per))
  stopifnot(all(is.natural(x)))
  M<- sum(x)
  stopifnot( is.natural(n) )
  N<- length(x)
  y.i<- rep(NA,n)
  y<- rep(NA,n)
  for (k in (1:n))
  {
    j<- sample(N, 1)
    z<- sample(M, 1)
    while( z > x[j] )
    {
      j<- sample(N, 1)
      z<- sample(M,1)
    }
    y.i[k]<- j
    y[k]<- x[j]
    if (0 == k%%per)
    {
      cat(sprintf("Lahiri sampling: Did %.0f\n", k))
    }
  }
  return(list(indices=y.i, sizes=y))
}

lahiri.Midzuno.Sen.sampling<- function(x, n)
{
  # Called this by Sarndahl, Swensson, and Wretman
  stopifnot(all(is.natural(x)))
  stopifnot( is.natural(n) )
  N<- length(x)
  y.i<- rep(NA,n)
  y<- rep(NA,n)
  p<- x/sum(x)
  y.i[1]<- sample.int(N, 1, prob=p)
  y[1]<- x[y.i[1]]
  y.i[2:N]<- sample((1:N)[-y.i[1]], (N-1), replace=FALSE)
  y[2:N]<- x[y.i[2:N]]
  return(list(indices=y.i, sizes=y)) 
}

# Test.

# General sample from a Gamma distribution with shape 2 and scale 10,
# meaning it has a mean of 20, and make sure it consists of positive 
# integers.

X<- ceiling(rgamma(10000, shape=2, scale=10))

# Empirical mean and median:

cat(sprintf("Mean[X]: %.3f, Median[X]: %.3f\n", mean(X), median(X)))

# Lahiri (runs for a while):

L<- lahiri.sampling(X, 100, per=20)


# Lahiri-Midzuno-Sen:

LMS<- lahiri.Midzuno.Sen.sampling(X, 100)

cat(sprintf("Lahiri Mean[X]: %.3f, Lahiri Median[X]: %.3f\n", mean(L$sizes), median(L$sizes)))

cat(sprintf("Lahari-Midzuno-Sen Mean[X]: %.3f, Lahiri-MidzunoSen Median[X]: %.3f\n", mean(LMS$sizes), median(LMS$sizes)))

empirical_bayesian@ieee.org

This user is a member of WikiProject Statistics.

15:39, 3 July 2016 (UTC)