This question is probably easy but I couldn't find the answer, nor remember my lectures in statistic.

I have an (infinite) bag of red (A) and blue (B) chips, i.e.

$P(A) = p = 1 - P(B)$

I want to estimate the minimum red chips $k$ that I should expect in a random sample of size $m$, for a given confidence level $\alpha$, i.e. $P(k<=x<=m) = \alpha$, where $x$ is the number of red chips out of the total sample $m$

I know the answer to the problem for a known $p$ value (used this lecture), except in this case I don't know it. I estimated it from a large sample $n$, where i took $j$ red chips, i.e. $\hat{p}=\frac{j}{n}$

What is the formula for linking $k = f(m, \alpha, j, n)$ or $\alpha = f(m, k, j, n)$, considering that the underlying probability is an estimate with uncertainty?

  • 2
    $\begingroup$ This sounds like a binomial prediction interval (do you want it to be one-sided) and there are several possible approaches. stats.stackexchange.com/questions/255570/… is a similar two-sided version of the same question. The easiest approach would be Bayesian with a Beta conjugate prior leading to use of a Beta-binomial cumulative distribution function for your interval $\endgroup$
    – Henry
    Jul 15, 2021 at 22:18
  • 1
    $\begingroup$ Is $m$ largish or smallish? Do you want an approximate or exact solution? One idea is to go bayes, with a prior on $p$. $\endgroup$ Jul 16, 2021 at 17:18

1 Answer 1


Your situation is that $X \sim \mathcal{Binomial}(n,p)$ is observed, and based on that you want somehow to predict the future $Y \sim \mathcal{Binomial}(m,p)$. $X, Y$ are stochastically independent, but linked by sharing the same value of $p$. You have proposed to find a predictive density for $Y$ by substituting the MLE of $p$ based on $X$, $x/n$, but notes that that forgets about the estimation uncertainty of $p$. So what to do?

First, a Bayesian solution. Assume a conjugate prior (for simplicity, the principle will be the same for any prior), that is, a beta distribution. We do it in two stages, first we find the posterior distribution of $p$ given that $X=x$ is observed, and then we use that as a prior which we combine with the likelihood based on $Y$. $$ \pi(p \mid x)=\mathcal{beta}(\alpha+x,\beta+n-x) $$ Then we find the posterior predictive density (well, in this case probability mass) of $Y=y$ by integrating out $p$: $$ f_\text{pred}(y \mid x)=\frac{\binom{m}{y}p^y(1-p)^{m-y}\mathcal{beta}(p,\alpha+x-1,n+\beta-x-1)}{\int_ 0^1 \binom{m}{y}p^y(1-p)^{m-y}\mathcal{beta}(p,\alpha+x-1,n+\beta-x-1)\; dp} \\ =\frac{\binom{m}{y}B(\alpha+x+y,n+m+\beta-x-y)}{B(\alpha+x,n+\beta-x)} $$ where $B$ is the beta function. I left out the intermediate algebra as this is standard manipulations.

But what to do if we do not want to go Bayes? There is a concept of predictive likelihood, a review is in Predictive Likelihood: A Review by Jan F. Bjørnstad, in Statist. Sci. 5(2): 242-254 (May, 1990). DOI: 10.1214/ss/1177012175 . This is somewhat more complicated and controversial as there are multiple versions, not essential uniqueness as with parametric likelihood. We start with the joint likelihood of $y,p$ based on $X=x$, but then we must find a way of eliminating $p$. We will look at three ways (there are more):

  1. substituting for $p$ its MLE, maximum likelihood estimator. This is what you have done.

  2. Profiling, that is, eliminating $p$ by maxing it out.

  3. Conditioning on a minimal sufficient statistic

Then the solutions.

  1. $$f_0(y\mid x)=\binom{m}{y}\hat{p}_x^y (1-\hat{p}_x)^{m-y}; \quad \hat{p}_x=x/n $$

  2. The mle of $p$ based on $X, Y$ is $\hat{p}_y=\frac{x+y}{n+m}$. Substituting this gives the profile predictive likelihood $$ f_\text{prof}(y \mid x) = \binom{m}{y} \hat{p}_y^y (1-\hat{p}_y)^{m-y} $$ Note that this profile likelihood does not necessarily sum to 1, so to use it as a predictive density it is usual to renormalize.

  3. The minimal sufficient statistic for $p$ based on the joint sample $X,Y$ is $X+Y$. By calculating the conditional probability $\DeclareMathOperator{\P}{\mathbb{P}} \P(Y=y \mid X+Y=x+y)$ by sufficiency the unknown parameter $p$ will be eliminated, and we find a hypergeometric distribution $$ f_\text{cond}(y \mid x) = \frac{\binom{m}{y}\binom{n}{x}}{\binom{n+m}{x+y}}$$ Note that this will also need renormalization to be used directly as a density. Note that this coincides with the bayesian conjugate solution for the case $\alpha=1, \beta=1$.

Let us look at some numerical examples (in the plots the densities are renormalized, the bayes solution shown is for the Jeffrey's prior $\alpha=1/2, \beta=1/2$):

First with $n=m=10; x=3$:

enter image description here

Then an example with larger $n$, so more precisely estimated $p$:

enter image description here

The code for the last plot is below:

p_0 <- function(n, x, m) function(y, log=FALSE) {
    phat <- x/n
    dbinom(y, m, phat, log=log)
p_prof <- function(n, x, m) function(y, log=FALSE) {
    phat <- (x + y)/(n + m)
    dbinom(y, m, phat, log=log)
p_cond <- function(n, x, m) function(y, log=FALSE) {
    dhyper(y, m, n, x + y, log=log)

n <- 100; m <- 20; x <- 30

plot(0:m,  p_0(n, x, m)(0:m), col="red", main="Predictive densities",
     ylab="density", xlab="y", type="b",
     sub="n=100, m=20;  x=30")
points(0:m,  p_prof(n, x, m)(0:m)/ sum( p_prof(n, x, m)(0:m)), col="blue", type="b")
points(0:m,  p_cond(n, x, m)(0:m)/ sum( p_cond(n, x, m)(0:m)), col="orange", type="b")
legend("topright", c("naive", "profile", "cond"),
       col=c("red", "blue", "orange"),
       text.col=c("red", "blue", "orange"), lwd=2)

Prediction intervals can then be constructed based on this predictive likelihoods.

  • $\begingroup$ The profile likelihood formula is incorrect. I'm reading "In All Likelihood" and the binomial predictive likelihood is derived on pp. 432. I won't type the formula as it's a bit scary but $\operatorname{L}(y)$ is not (proportional to) the binomial pmf. Before normalization, the formula is dbinom(y, m, p = (x + y) / (n + m)) * dbinom(x, n, p = (x + y) / (n + m)). This effectively weights each term $pˆ{y}(1-p)ˆ{m-y}$ differently than the usual binomial coefficient. $\endgroup$
    – dipetkov
    Mar 24, 2023 at 22:15
  • 1
    $\begingroup$ @dipetkov: thanks. I'm i a chess tournament now, will have a look later $\endgroup$ Mar 24, 2023 at 22:42

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