Take-all units
In sps: Sequential Poisson Sampling

#| echo: false
litedown::reactor(print = NA)

When drawing a sample of size $n$ from a population of size $N$, where each unit is drawn proportional to some measure of size $x_{i}$, $i=1,\ldots,N$, the probability that the $i$th unit is included in the sample is

$$ \pi_{i} = n x_{i} / \sum_{i=1}^{N} x_{i}. $$

In theory these inclusion probabilities should be less than 1; in practice units with a large $x_{i}$ can have an inclusion probability greater than 1, especially when $n$ is large. The usual procedure to deal with these units is to put them in a special take-all stratum so that they are always included in the sample, essentially fixing their inclusion probabilities at 1, with the remaining units (the take-some units) drawn at random. The usual algorithm used by, say, sampling::inclusionprobabilities() repeatedly moves units into the take-all stratum and recalculates the inclusion probabilities for the remaining units until all inclusion probabilities are less than 1.[^1] See, e.g., Ohlsson (1998) or Tillé (2020, Chapter 5).

[^1]: To see why this needs to be done repeatedly, consider a population with three units with sizes 1, 2, and 3 and find which units have an inclusion probability greater than 1 when the sample size is 3.

Generalizing the usual approach

Sequential poisson sampling is a bit more complicated because, as noted by Ohlsson (1998), it can be useful to place units with an inclusion probability greater than $1 - \alpha$, for some small $\alpha$, into the take-all stratum. Unless $\alpha = 0$, the usual algorithm for finding take-all units can put units into the take-all stratum that have an inclusion probability less than $1 - \alpha$. We can see this with the following example (with a larger value for $\alpha$ for illustration).

pi <- function(x, n) {
  n * (x / sum(x))
}

# Population with 13 units.
x <- c(1:8, 9.5, 10, 20, 20, 30)

alpha <- 0.15

# With a sample size of 8, units 11, 12, and 13
# have an inclusion probability greater than 1 - alpha.
which(pi(x, 8) >= 1 - alpha)

# Now units 9 and 10 have an inclusion probability
# greater than 1 - alpha.
which(pi(x[1:10], 5) >= 1 - alpha)

# After two rounds of removal all inclusion probabilities
# are less than 1 - alpha.
any(pi(x[1:8], 3) >= 1 - alpha)

Although all inclusion probabilities are less than $1 - \alpha$ after two rounds of placing units into a take-all stratum, the inclusion probability for unit 9 would also be less than $1 - \alpha$ despite it being in the take-all stratum.

pi(x[1:9], 4)[9] >= 1 - alpha

This means that units have to be considered one at a time, from largest to smallest, to determine if they belong in the take-all stratum. Rather than doing this as a loop and potentially calculating the inclusion probabilities many times for most units (see vignette("performance")), we can calculate a sequence of inclusion probabilities for the $n$ units with the largest sizes

$$ p_{i} = \frac{x_i i}{\bar{x} + \sum_{j=1}^{i} x_j}, $$

where $\bar{x}$ is the total size for all remaining units (those always in the take-some stratum) in the population. If units are sorted in increasing order of size, then the $i$th element of this sequence gives the largest inclusion probability after putting the $n - i$ largest units in the take-all stratum.

Finding the take-all units is now simply a case of finding the smallest integer $i^{}$ (if one exists, for otherwise there are no take-all units) such that $p_{i^{}} \geq 1 - \alpha$. Although the sequence given by $p_{1}, \ldots, p_{n}$ is not increasing, it is monotonic enough to get a unique set of take-all units. To see this, write the $i+1$th element of this sequence as

$$ \begin{align} p_{i+1} &= \frac{x_{i+1} (i + 1)}{\bar{x} + \sum_{j=1}^{i+1} x_j} \ &= \frac{x_{i+1} (i + 1)}{x_{i} i / p_{i} + x_{i+1}}. \end{align} $$

It is straightforward to verify that

$$ p_{i+1} > p_{i} \Leftrightarrow p_{i} < \frac{x_{i+1} - x_{i}}{x_{i+1}} i + 1 $$

and

$$ p_{i+1} \geq 1 \Leftrightarrow p_{i} \geq x_{i} / x_{i+1}. $$

This means that $p_{i+1}>p_{i}$ if $p_{i} < 1$ and $p_{i+1}\geq 1$ if $p_{i}\geq 1$---the sequence is strictly increasing whenever it is less than 1 and does not drop below 1 after reaching this point. Therefore if $p_{i^{}} \geq 1- \alpha$ then all units $i > i^{}$ have $p_{i} \geq 1- \alpha$.

Returning to the previous example, we can correctly find the four take-all units from the elements of $p_{i}$ that are greater than $1 - \alpha$.

#| fig.alt: "Diagram showing how to find units that belong in the take-all
#|   stratum."
p <- function(x, n) {
  ord <- order(x, decreasing = TRUE)
  s <- seq_len(n)
  possible_ta <- rev(ord[s])
  x_ta <- x[possible_ta] # ties are in reverse
  definite_ts <- ord[seq.int(n + 1, length.out = length(x) - n)]

  x_ta * s / (sum(x[definite_ts]) + cumsum(x_ta))
}

plot(
  1:4,
  p(x, 8)[1:4],
  xlab = "",
  ylab = "p",
  xlim = c(1, 8),
  ylim = c(0, 2),
  pch = 20
)
points(5:8, p(x, 8)[5:8], pch = 19)
abline(1 - alpha, 0, lty = 2)
legend(1, 2, c("take-some", "take-all"), pch = c(20, 19))

As each element $p_{i}$ in the sequence is a strictly increasing function of $n$, we can also identify the unique value for $n$ when a unit enters the take-all stratum.

#| fig.alt: "Diagram showing when units first enter the take-all stratum."
plot(
  2:5,
  p(x, 8)[1:4],
  xlab = "",
  ylab = "p",
  xlim = c(1, 9),
  ylim = c(0, 2.5),
  pch = 20
)
points(6:9, p(x, 8)[5:8], pch = 19)
points(1:3, p(x, 9)[1:3], pch = 20, col = "red")
points(4:9, p(x, 9)[4:9], pch = 19, col = "red")
abline(1 - alpha, 0, lty = 2)
legend(1, 2.5, c("take-some", "take-all"), pch = c(20, 19))
legend(1, 2, c("n = 8", "n = 9"), pch = 20, col = c("black", "red"))
symbols(4.5, 0.85, circles = 1, inches = FALSE, add = TRUE, lty = 2)

And this can be calculated for each unit in the population.

sps::becomes_ta(x, alpha = alpha)

Knowing when a unit enters the take-all stratum also makes it possible to draw units one at a time with the sequential Poisson method, as with the sps_iterator() function.