2
$\begingroup$

A simple random sample of size $n=n_1 + n_2$ is drawn without replacement from a finite population of size $N$. Further a simple random sample of size $n_1$ is drawn without replacement from the first sample. Let $\bar y$ and $\bar {y_1}$ be the respective sample means.

Find V($\bar {y_1}$) and V ($\bar {y_2}$), where $\bar {y_2}$ is the mean of the remaining $n_2$ units in the first sample. Also find Cov($\bar {y_1}$,$\bar {y_2}$). Assume that $S^2$ is the population variance.


This is what I've been able to do thus far:

Note that $\bar {y_2}$ = $\frac{(n\bar y) - n_1\bar {y_1}}{n_2}$.

Also, V($\bar {y_1}$) = $E_1[V_2(\bar {y_1})] + V_1[E_2(\bar {y_1})]$. After simplification, this has turned out to be $(\frac1{n_1} - \frac1N)S^2$. Now am I on the right track? Also, some help in finding $V(\bar y_2)$ and the covariance would be appreciated.

$\endgroup$
3
  • $\begingroup$ Hint: you already know $\text{V}(\bar{y}_2)$: the formula is in your question. $\endgroup$
    – whuber
    May 7, 2014 at 15:14
  • $\begingroup$ @whuber, kindly point it out meticulously. It is turning out to be quite complicated. $\endgroup$ May 7, 2014 at 17:40
  • $\begingroup$ Interchanging $1$ and $2$ in your question produces the answer for $\text{V}(\bar{y}_2)$, because the complement of a random sample without replacement is itself a random sample without replacement. This leaves the hard part, which is finding the covariance. $\endgroup$
    – whuber
    May 7, 2014 at 17:54

1 Answer 1

1
$\begingroup$

Following the analysis of a single sample without replacement at https://stats.stackexchange.com/a/622287/919, the variance of the mean of any sample of size $K$ from the population $(x_1, x_2, \ldots, x_N)$ with variance $\sigma^2$ (called $S^2$ in the present question) is

$$\operatorname{Var}(\bar X) = \frac{N-K}{K(N-1)}\,\sigma^2.$$

In the present double sampling scenario, for each subject index $i$ let $J_i$ indicate whether $x_i$ is in the second sample of size $n_1$ and $I_i$ indicate whether $x_i$ is in the remainder of the sample. Translating the description in the question into properties of these random variables gives

$$n_1 = \sum_{i=1}^N J_i,\tag{1a}$$ $$n_2 = \sum_{i=1}^N I_i,\tag{1b}$$ and $$I_iJ_i = 0,\ i = 1, 2, \ldots, N.\tag{1c}$$

We need to compute the second moments (variances and covariances) of these $2N$ variables. As noted in the previous thread, the $(I_i,J_i)$ are exchangeable--all pairs of nested subsets of the population are equally likely to result from this double sample--and we can exploit this in the calculation.

We already know (as shown in the initially referenced post) that $(1a)$ and $(1b)$ imply that for $i\ne j,$

$$\operatorname{Cov}(J_i,J_j) = -\frac{n_1(N-n_1)}{N^2(N-1)}$$

with a similar expression for $\operatorname{Cov}(I_i,I_j).$ The only new calculations needed are for the cross-covariances of $I_i$ and $J_j.$ Consider the expectation of the product of $(1a)$ and $(1b)$ and use $(1c)$ to simplify the diagonal terms:

$$\begin{aligned} n_1n_2 &= E\left[\sum_{j=1}^{N} J_j \sum_{i=1}^N I_i\right]\\ &= E\left[\sum_{i=1}^{N} I_iJ_i\right] + E\left[\sum_{i\ne j}^N I_iJ_j\right]\\ &= \sum_{i=1}^N E[0] + \sum_{i\ne j}^N E[I_iJ_j]\\ &= N(N-1)E[I_1J_2] \end{aligned}$$

The last equality is consequence of the exchangeability.

Solving, we find that for all $i\ne j,$

$$E[I_iJ_j] = \frac{n_1n_2}{N(N-1)}.$$

A standard formula for covariance shows (still for $i\ne j$)

$$\operatorname{Cov}(I_i, J_j) = E[I_iJ_j] - E[I_i]E[J_j] = \frac{n_1n_2}{N(N-1)} - \frac{n_1}{N}\frac{n_2}{N} = \frac{n_1n_2}{N^2(N-1)}.$$

The remaining calculation that is needed is simple:

$$\operatorname{Cov}(I_i,J_i) = E[I_iJ_i] - E[I_i]E[J_i] = -\frac{n_1n_2}{N^2}.$$

Now we can proceed without further thought--entirely mechanically--to find the covariance of the subsample means. Let $$\bar X = \frac{1}{n_2}\sum_{i=1}^N I_i x_i,\quad \bar Y = \frac{1}{n_1}\sum_{j=1}^N J_j x_j$$

be the means of the remainder and of the subsample. Let's clear the fractions and work directly with the sums, writing $\mu$ for the population mean (which, as a check of the preceding work, had better not enter into the result!):

$$\begin{aligned} n_1n_2\operatorname{Cov}(\bar X, \bar Y) &= \operatorname{Cov}(n_2\bar X, n_1\bar Y) = \operatorname{Cov}\left(\sum_{i=1}^N I_i x_i \sum_{j=1}^N J_j x_j\right)\\ &= \sum_{i=1}^N x_i^2 \operatorname{Cov}(I_i,J_i) + \sum_{i\ne j} x_ix_j \operatorname{Cov}(I_i,J_j)\\ &= -\frac{n_1n_2}{N^2}\sum_{i=1}^N x_i^2 + \frac{n_1n_2}{N^2(N-1)} \sum_{i\ne j}x_ix_j\\ &= \frac{n_1n_2}{N}\left[-\frac{1}{N}\sum_{i=1}^N x_i^2 + \frac{1}{N(N-1)}\left(\sum_{i,j=1}^N x_ix_j - \sum_{i=1}^N x_i^2\right)\right]\\ &= \frac{n_1n_2}{N}\left[-(\sigma^2 + \mu^2) + \frac{1}{N(N-1)}\left((N\mu)^2 - N(\sigma^2+\mu^2)\right)\right]\\ &= \frac{n_1n_2}{N}\left[\frac{-N}{N-1}\,\sigma^2\right]\\ &= -\frac{n_1n_2}{N-1}\,\sigma^2. \end{aligned}$$

Thus

$$\operatorname{Cov}(\bar X, \bar Y) = -\frac{\sigma^2}{N-1}.$$

Incidentally, we didn't need much of the preliminary algebra. We could have solved for $\operatorname{Cov}(I_i,J_j),$ where $i\ne j,$ as the unique value that results in the coefficient of $\mu^2$ equalling zero.


I was surprised that this result does not depend on the subsample sizes $n_1$ and $n_2$ or even on their sum, so I tested it with large amounts of simulated data. It looks correct: in the following R code, the covariance matrix estimated from 10,000 independent iterations of this experiment is very close to the variances found here. It outputs the term-by-term ratio of those two matrices, so at a glace (by comparing it to the matrix of $1$s) we can see how accurate the formulas are. Here's the output with the seed set to 17:

     [,1] [,2]
[1,] 1.02 1.02
[2,] 1.02 1.00

The chi-squared distribution tells us these ratios are likely to lie between 0.97 and 1.03, and so they are. Similar results ensue when the $n_i$ are varied (in the vector n), demonstrating the lack of dependence on the subsample sizes.

#
# Create a population.
#
N <- 20      # Population size
n <- c(5, 8) # Subsample sizes: sum cannot exceed `N`
# set.seed(17)

P <- sample.int(2*N, N, replace = TRUE) 
sigma2 <- mean((P -  mean(P))^2)         # The population variance
#
# Sample and subsample.
#
n.iter <- 1e4
xy.bar <- replicate(n.iter, {
  i <- sample.int(N, sum(n))       # Sample n[1] + n[2] indexes without replacement
  j <- sample.int(length(i), n[1]) # Subsample n[1] indexes from the sample
  y <- P[i[j]]                     # The subsample
  x <- P[i[-j]]                    # The remainder
  c(mean(y), mean(x))              # Their means
})
#
# Estimate the variance-covariance matrix relative to the population variance.
#
S <- var(t(xy.bar)) / sigma2
#
# Compare to the theoretical formula.
#
Sigma <- matrix(c((N - n[1])/(n[1] * (N-1)),
         rep(-1 / (N-1), 2),                 # Doesn't depend on `n`!
         (N - n[2])/(n[2] * (N-1))), 2, 2)

ndigits <- -log10((qchisq(1 - 0.05/2, n.iter - 1) / n.iter) - 1) # 95% confidence
(round(S / Sigma, ndigits))
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.