4
$\begingroup$

You are throwing a party and want to know how many guests you can expect.

You are inviting $N$ persons, each person has a probability $x_i$, $ 1 \leq i \leq N$ of showing up and $x_i$ is between 0 and 1. The probabilities $x_i$ are given. I want to write a small script (preferably in Python), to calculate a $1-\alpha$ - confidence interval for the number of guests to be expected ($\alpha = 0.1 $ or $ 0.05$).

Could somebody point me to some resources for this problem? I feel like this is a very simple, introductory problem, yet, I could not find anything. Alternatively, could somebody explain (conceptually) a solution?

$\endgroup$
4
  • $\begingroup$ In my opinion, you are looking for a prediction interval, not a confidence interval $\endgroup$
    – John M
    Sep 6 '20 at 20:55
  • $\begingroup$ See stats.stackexchange.com/search?q=poisson+binomial. $\endgroup$
    – whuber
    Sep 6 '20 at 21:38
  • $\begingroup$ Why do you have the "mcmc" tag? How does your question relate specifically to Markov chain Monte Carlo? $\endgroup$
    – Glen_b
    Sep 7 '20 at 0:22
  • $\begingroup$ @Glen_b I gave it the mcmc tag, because I thought this kind of problem would be a prime candidate for bein solved by mcmc. Evidently, I was wrong, because the posted solutions work out nicely. $\endgroup$
    – r0f1
    Sep 7 '20 at 12:53
3
$\begingroup$

My computations will be in R (not Python), I hope you can translate. In R, functions dbinom,pbinom, and qbinom designate a binomial PDF, CDF, or quantile (inverse CDF) function. Also, with appropriate parameters, rbinom simulates binomial and Bernoulli samples.

Begin by assuming $n=20$ guests are invited, of whom each has probability $p = .7$ of attending the party. Then the binomial probability distribution of the number who attend is $X \sim \mathsf{Binom}(n, p).$ It's distribution (to 5 decimal places) is tabled below. (Ignore row labels in brackets [ ].)

k = 0:20; pdf = round(dbinom(k, 20, .7), 5)
cbind(k, pdf)
       k     pdf
 [1,]  0 0.00000
 [2,]  1 0.00000
 [3,]  2 0.00000
 [4,]  3 0.00000
 [5,]  4 0.00001
 [6,]  5 0.00004
 [7,]  6 0.00022
 [8,]  7 0.00102
 [9,]  8 0.00386
[10,]  9 0.01201
[11,] 10 0.03082
[12,] 11 0.06537
[13,] 12 0.11440
[14,] 13 0.16426
[15,] 14 0.19164
[16,] 15 0.17886
[17,] 16 0.13042
[18,] 17 0.07160
[19,] 18 0.02785
[20,] 19 0.00684
[21,] 20 0.00080

In order to get a 95% confidence interval (CI), you need to 'trim' $2.5\%$ of the probability from each tail of this (discrete) distribution, to leave $95\%$ on both sides of $X = 14,$ where the highest probability lies.

Roughly, the boundaries of the CI will be 10 and 18.

qbinom(c(.025,.975), 20, .7)
[1] 10 18

It seems you have a choice between intervals $[11,18]$ or $[10,17].$

diff(pbinom(c(10,18), 20, .7))
[1] 0.9444008
diff(pbinom(c(9,17), 20, .7))
[1] 0.9473721

If the $n$ people invited may each have a different attendance probability $p_i, i = 1, 2, \dots n,$ then a simulation may be helpful (as suggested by tags on your Question). Suppose the 20-vector p is defined as below:

 p = seq(.5,1, len=20); p
 [1] 0.5000000 0.5263158 0.5526316 0.5789474 0.6052632
 [6] 0.6315789 0.6578947 0.6842105 0.7105263 0.7368421
[11] 0.7631579 0.7894737 0.8157895 0.8421053 0.8684211
[16] 0.8947368 0.9210526 0.9473684 0.9736842 1.0000000

Then the R function rbinom(20, 1, p) simulates individual attendees at one particular party and sum(rbinom(20, 1, p)) simulates the total number attending.

set.seed(906)
a = rbinom(20, 1, p);  a
[1] 1 0 1 1 1 0 1 1 1 1 1 1 1 1 1 1 1 1 1 1
x = sum(a);  x
[1] 18

Replicating this kind of count for $100\,000$ gives us the simulated distribution of the random number $X$ attending such parties. (Because of the set.seed statement, we have already seen that $X = 18$ attend the first of these many parties. An approximate 95% CI for the number attending is $[12, 18].$

set.seed(906)
p = seq(.5,1, len=20)
x = replicate(10^5,  sum(rbinom(20, 1, p)))
summary(x)
   Min. 1st Qu.  Median    Mean 3rd Qu.    Max. 
   7.00   14.00   15.00   15.01   16.00   20.00 
quantile(x, c(.025, .975))
 2.5% 97.5% 
   11    18 
mean(x >= 12 & x <= 18)
[1] 0.95119

hist(x, prob=T, br = (6:20)+.5, col="skyblue2", 
     main="Simulated Attendees")

enter image description here

Note: If you're only interested in getting a 95% (or 99%) CI for the number $X$ attending, you might get $E(X)$ and $V(X)$ by considering $X$ as a sum $X = \sum_{i=i}^n B_i$ of independent Bernoulli random variables each with its own $p_i.$ Then assume approximate normality (note the symmetry in the summary above). Finally, make a confidence interval based on the normal distribution.

$\endgroup$
2
  • $\begingroup$ (+1) great effort $\endgroup$
    – gunes
    Sep 6 '20 at 21:23
  • 1
    $\begingroup$ Simulation is the tedious approach, because exact calculation is fast and easy. This is explored in various other threads. See, inter alia, stats.stackexchange.com/questions/41247 . $\endgroup$
    – whuber
    Sep 6 '20 at 21:41
3
$\begingroup$

You'll generate $N$ Bernoulli RVs, and count how many guests are present. Then, you'll repeat this experiment, say, $T$ times, and record every result. This will be your distribution. Then, pick the percentiles $\alpha/2$ and $1-\alpha/2$ to come up with a confidence interval. Theoretically, you can compare to Binomial distribution's CI when probabilities are equal.

$\endgroup$
2
  • $\begingroup$ Ok, so this is a simulation approach. I think that the percentiles give you a prediction interval, not a confidence interval though. $\endgroup$
    – John M
    Sep 6 '20 at 20:58
  • 1
    $\begingroup$ Yes, I did it via only simulation because the tag was mcmc. $\endgroup$
    – gunes
    Sep 6 '20 at 21:22

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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