---
title: "Take-all units"
vignette: >
  %\VignetteIndexEntry{Take-all units}
  %\VignetteEngine{quarto::html}
  %\VignetteEncoding{UTF-8}
knitr:
  opts_chunk: 
    collapse: true
    comment: '#>'
---

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.^[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.]

Sequential poisson sampling is a bit more complicated because 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 $\alpha$).

```{r}
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

# 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.

```{r}
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, 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 $1 - \alpha$.

```{r, fig.width=8, fig.height=5.33}
#| 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 $i$, we can also identify the unique value for $n$ when a unit enters the take-all stratum.

```{r, fig.width=8, fig.height=5.33}
#| 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"))
```
