Average Sample Size For Curtailed Double Sampling Plan I am currently implementing double sampling. The goal is to accept or reject a lot by only sampling a few instances from it. In a two-stage testing procedure (double sampling) one first  draws a sample of size $n_1$ and compares the number $K_1$ of
items “of interest” (e.g., “non compliant”) with two integers: $c_1$,
$c_2$. If $K_1 \le c_1$, we accept the null hypotheses ($H_0$) that the batch
is of acceptable quality; and if $X > c_2$, we accept the alternative
hypotheses ($H_1$) that the batch quality is unacceptable.
However, if $c_1 < X \le c_2$, we draw a second sample of size $n_2$. We stop in case we find more than $c_2$. You can play with some plans in here.

My problem is now finding the average sample size (ASN) with partial curtailment, that is we look at the first sample as a whole and then stop looking at the first sample once it is clear that we reject it.
I found this paper that computes it as
$ASN = n_1 + \sum_{i=c_1 + 1}^{c2} P(n_1, i) [ n_2 [ P_1(n_2, c_2 - i) + \frac{c_2 - i + 1}{p} P_2(n + 1, c_2 - i + 2) ]]$
where $P(n_1, i)$ is the probability of observing exactly $j$ defectives in a sample of size $n_1$,
$P_1(n_2, c_2 − i)$ is the probability of observing $c_2 − i$ or fewer defectives in a sample of size $n_2$,
and $P_2(n_2 + 1, c_2 − i + 2)$ is the probability of observing $c_2 − i + 2$ defectives in a sample of size $n_2 + 1$.
From when I read it, then I would think $P$, $P_1$ and $P_2$ are all PMFs, e.g. of Bernoulli $p$, but when I implement it, then I get very different curves for most of the plans compared to here.
I also had a look at

The Average Sample Number for Truncated Single and Double Attributes
Acceptance Sampling Plans Author(s): C. C. Craig Source:
Technometrics, Vol. 10, No. 4 (Nov., 1968), pp. 685-692

and at the end it looks like $P_1$ and $P_2$ are CDFs, but I do not manage to get the same results as their tables. Can you help me understand what I need to put in as $P$, $P_1$ and $P_2$?
 A: This comes down to an analysis of "curtailment" (which is a simplified version of sequential sampling).  Let's focus on that first.  Because I think your reference is incorrect, I will be careful and show the details.
Curtailment occurs during the second stage of double sampling of a process that has an (assumed) probability $q \gt 0$ at each step of yielding a "defective" result.  In this stage we plan to sample for as many as $n$ steps, but we will stop sampling at any step where the cumulative number of defectives reaches a threshold $h \gt 1.$ What is the expected sample size for this stage?
It helps to be formal because we need to track two random variables:

*

*Let $X(i)$ be the number of defectives observed on or before step $i.$

*Let $N$ be the number of steps taken to first observe $h$ defectives with unlimited sampling.  It is related to $X(i)$ by $N = \min\{i\mid X(i) \ge h\}.$

*Write $N\wedge n = \min(N, n)$ for the number of samples limited by $n.$
Even if we were to terminate sampling early ($N \lt n$), it would not change the results of our calculations if we went ahead and obtained a full sample of size $n.$  Breaking up the possible outcomes according to the values of $X(n),$ we may express the expected sample size as a weighted average of the expectations for each of those values:
$$E[N\wedge n] = \sum_{k = 0}^n E[N \wedge n\mid X(n) = k] \Pr(X(n) = k).\tag{*}$$
The probabilities are Binomial: $\Pr(X(n)=k)$ is the chance of seeing exactly $k$ defectives in any sample of size $n.$  A formula is
$$\Pr(X(n) = k) = \binom{n}{k} q^k(1-q)^{n-k} = \frac{n!}{k!(n-k)!} q^k(1-q)^{n-k}.$$
We must figure out the conditional expectations in $(*),$ which for brevity I will write as
$$g(n,h,k) = E[N \wedge n\mid X(n) = k].$$
That is, knowing there are $k$ defectives among the next $n$ observations, when will we stop sampling?  Clearly, when $k \lt h$ we will obtain all $n$ samples.  Otherwise, when $k\ge h,$ we will stop sampling by step $n$ (and likely sooner than that).
Consider the first observation among those $n.$  There are two possibilities: it is defective or not.  The key idea is that because there are $k$ defectives among the next $n$ observations, the chance that the next observation is defective is $k/n.$  (The information given by the numbers $(n,k)$ does not determine where in the sequence $1,2,\ldots, n$ the defectives might occur: all subsets of $k$ of those positions are equally likely.)  Consequently
$$g(n,h,k) = 1 + \frac{i}{n} g(n-1,h-1,k-1) + \frac{n-i}{n} g(n-1, h, k).$$
The initial $1$ counts the first observation and the next two terms use the law of conditional expectations to find the expected number of additional observations: $g(n-1,h-1,k-1)$ is the expected sample size with $n-1$ observations left to go; the threshold to terminate sampling has been reduced to $h-1$ because we just saw a defective; and there remain $k-1$ defectives among these observations.  $g(n-1, h, k)$ is the expected sample size with $n-1$ observations left to go and we haven't yet seen a defective.
The (unique) solution to this recursion (consistent with obvious starting conditions) is
$$g(n, h, k) = \frac{h(n+1)}{k+1}.\tag{**}$$
To see why, note that the formula is correct in all edge cases; e.g., when $k=n,$ every one of the next $n$ observations is a defective and so we stop after seeing the first $h = h(n+1)/(n+1)$ of them.  Then all we need to check is the recursion, finding
$$\begin{aligned}
 g(n, h, k) &= \frac{h(n+1)}{k+1}\\
&= 1 + \frac{k}{n}\frac{(h-1)n}{k} + \frac{n-k}{n} \frac{hn}{k+1}\\
&= 1 + \frac{k}{n} g(n-1,h-1,k-1) + \frac{n-k}{n} g(n-1, h, k).
\end{aligned}$$
with straightforward algebra.  Writing $F(\ ;n,q)$ for the Binomial$(n,q)$ distribution function we may now express $(*)$ in the closed form
$$\begin{aligned}
E[N\wedge n] &= n\Pr(X(n)\lt h) + \sum_{k=h}^n \frac{h(n+1)}{k+1} \binom{n}{k}q^k(1-q)^{n-k}\\
&= nF(h-1;n,q) + h\sum_{k=h}^n \binom{n+1}{k+1}q^k(1-q)^{n-k}\\
&= nF(h-1;n,q) + h\sum_{k=h+1}^{n+1} \binom{n+1}{k}q^{k-1}(1-q)^{n+1-k}\\
&= nF(h-1;n,q) + \frac{h}{q}\sum_{k=h+1}^{n+1} \binom{n+1}{k}q^{k}(1-q)^{n+1-k}\\
&= nF(h-1;n,q) + \frac{h}{q}\left(1 - F(h;n+1,q)\right).
\end{aligned}.$$
The first step breaks the expectation into the cases $X(n)\lt h,$ where the full sample will be obtained, and the remaining cases $X(n)\ge h,$ where early termination is possible.  A little algebra and a change in the index from $k$ to $k+1$ yields the result.

We are ready to compute the expected sample size for double sampling with curtailment.  The initial sample of $n_1$ observations will be completed, contributing $n_1$ to the expectation.  If the number of defectives in that sample is $c_1$ or less or if it exceeds $c_2,$ sampling stops with a size of $n_1.$  The lower (red) and upper (purple) paths in the figure illustrate these scenarios.

Otherwise, additional sampling is needed.  Sampling terminates either when the cumulative number of defectives exceeds $c_2$ (as in the teal path) or the total sample size reaches $n_1 + n_2$ (green path).  For paths like the teal, $N$ (marked with a solid vertical line) can have values between $n_1$ and $n_1 + n_2.$
The expected extra effort is (as before) computed by partitioning the possibilities according to the value of $X(n_1),$ which ranges from $c_1+1$ through $c_2.$  Let this value be $j.$  The threshold to terminate this second stage of sampling with a maximum sample size of $n=n_2$ is $h = c_2+1 - j$ because $j$ defectives have already been observed.  Applying the formula $(**)$ gives
$$\begin{aligned}
&E[N] \\
&= n_1 + \sum_{j = c_1 + 1}^{c_2} E[N-n_1 \mid X(n_1) = j]\Pr(X(n_1) = j)\\
&= n_1 + \sum_{j = c_1 + 1}^{c_2} \left[nF(c_2-j;n_1,q) + \frac{h}{q}\left(1 - F(c_2+1 - j;n+1,q)\right)\right]\binom{n_1}{j}q^j(1-q)^{n_1-j}.
\end{aligned}\tag{***}$$
Although this looks a lot like formula $(13)$ in your reference, it is not the same, because the $F(\cdots)$ values are not the Binomial probabilities described there: they are cumulative probabilities and cannot be equated with or converted into individual probabilities.  (This makes me suspect the Technometrics article, which I haven't seen, is likely correct.)
Here is a plot of $E[N]$ as a function of $q$ for the sampling plan in the previous illustration ($c_1=4,$ $c_2 =14,$ $n_1=30,$ and $n_2=40$).

This is a general pattern: the sampling effort will be greatest when the rate of defectives is in a "sweet spot" enabling the sample path to stay out of the gray areas.  Otherwise it's highly likely sampling will end after the first round of size $n_1.$  Even so, the expectation is substantially less than the maximum planned sample size of $n_1+n_2.$

To settle the question of whose formula (if either) is correct, we need a third approach. Let's simulate some samples in a way that does not depend on much if any analysis.  Small errors will be most apparent in small problems, so here is an example with $n_1=10,$ $n_2=10,$ $c_1=3,$ $c_2=7,$ and $q=0.5.$
#
# The formula.
#
f <- function(n, h, q) n * pbinom(h-1, n, q) + h * (1 - pbinom(h, n+1, q)) / q
a <- Vectorize(function(c, n, q) {
  j <- seq(c[1]+1, c[2])
  n[1] + sum(dbinom(j, n[1], q) * f(n[2], c[2]+1-j, q))
}, "q")
#
# Simulation.
#
cc <- c(3, 7)
nn <- c(10,10)
q <- 0.5
set.seed(17)
N <- replicate(5e3, {
  k1 <- rbinom(1, nn[1], q)         # Take the first sample
  if(k1 <= cc[1] || k1 > cc[2]) nn[1] else {
    pop <- rbinom(nn[2], 1, q)      # Take the second sample, one at a time
    x <- cumsum(pop) + k1           # Compute the sample path
    x[nn[2]] <- cc[2] + 1           # Force termination at the end
    nn[1] + match(TRUE, x > cc[2])  # Find where sampling can first terminate
  }
})
signif(c(Simulation = mean(N), Formula = a(cc, nn, q)), 4)

The output prints the average sample length in 5,000 simulated samples along with what this formula says:

Simulation    Formula 
     14.00      14.04 


They are statistically indistinguishable.  Repeated experiments of this nature bear out the correctness of formula $(***).$
