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What is the formula (approximate or exact) for a prediction interval for a Binomial random variable?

Assume $Y \sim \mathsf{Binom}(n, p)$, and we observe $y$ (drawn from $Y$). The $n$ is known.

Our goal is to obtain a 95% prediction interval for a new draw from $Y$.

The point estimate is $n\hat{p}$, where $\hat{p}=\frac{y}{n}$. A confidence interval for $\hat{p}$ is straightforward, but I cannot find a formula for a prediction interval for $Y$. If we knew $p$ (rather than $\hat{p}$), then a 95% prediction interval just involves finding the quantiles of a binomial. Is there something obvious I'm overlooking?

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    $\begingroup$ See What non-Bayesian methods are there for predictive inference?. In this case the method using pivots isn't available (I don't think) but you could use one of the predictive likelihoods. Or of course, a Bayesian approach. $\endgroup$ – Scortchi - Reinstate Monica Jan 11 '17 at 9:40
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    $\begingroup$ Hi Guys, I'd like to take a moment to address the concerns that were raised. - regarding confidence for p: I am not interested for that. - regarding predictions being 95% of distribution: yes, that is exactly what prediction intervals are regardless of the context (in regression you have to assume normal errors, where as confidence intervals rely on CLT - yes, the example of predicting number of heads in a coin flip is correct. What makes this problem hard is that we don't now "p", jut have an estimate. $\endgroup$ – Statseeker Jan 11 '17 at 19:38
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    $\begingroup$ @Addison Read the book Statistical Intervals by G. Hahn and W. Meeker. They explain the difference between confidence intervals, prediction intervals, tolerance intervals and Bayesian credible intervals. A 95% prediction interval does not contain 95% of the distribution. It does what most frequentist intervals do. If you repeatedly sample from the B(n,p) and use the same method each time to produce a 95% prediction interval for p then 95% of the prediction intervals you will contain the true value of p. If you want to cover 95% of the distribution construct a tolerance interval. $\endgroup$ – Michael Chernick Jan 13 '17 at 19:59
  • $\begingroup$ Tolerance intervals cover a percentage of the distribution. For a 95% tolerance interval for 90% of the distribution you again repeat the process many times and use the same method to generate the interval each time then in approximately 95% of the cases at least 90% of the distribution will fall in the interval and 5% of the time less than 90% of the distribution will be contained in the interval. $\endgroup$ – Michael Chernick Jan 13 '17 at 20:06
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    $\begingroup$ Lawless & Fredette (2005), "Frequentist Prediction Intervals and Predictive Distributions", Biometrika, 92, 3 is another good reference, in addition to those at the link I gave. $\endgroup$ – Scortchi - Reinstate Monica Jan 16 '17 at 14:53
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Ok, let's try this. I'll give two answers - the Bayesian one, which is in my opinion simple and natural, and one of the possible frequentist ones.

Bayesian solution

We assume a Beta prior on $p$, i,e., $p \sim Beta(\alpha,\beta)$, because the Beta-Binomial model is conjugate, which means that the posterior distribution is also a Beta distribution with parameters $\hat{\alpha}=\alpha+k,\hat{\beta}=\beta+n-k$, (I'm using $k$ to denote the number of successes in $n$ trials, instead of $y$). Thus, inference is greatly simplified. Now, if you have some prior knowledge on the likely values of $p$, you could use it to set the values of $\alpha$ and $\beta$, i.e., to define your Beta prior, otherwise you could assume a uniform (noninformative) prior, with $\alpha=\beta=1$, or other noninformative priors (see for example here). In any case, your posterior is

$Pr(p|n,k)=Beta(\alpha+k,\beta+n-k)$

In Bayesian inference, all that matters is the posterior probability, meaning that once you know that, you can make inferences for all other quantities in your model. You want to make inference on the observables $y$: in particular, on a vector of new results $\mathbf{y}=y_1,\dots,y_m$, where $m$ is not necessarily equal to $n$. Specifically, for each $j=0,\dots,m$, we want to compute the probability of having exactly $j$ successes in the next $m$ trials, given that we got $k$ successes in the preceding $n$ trials; the posterior predictive mass function:

$Pr(j|m,y)=Pr(j|m,n,k)=\int_0^1 Pr(j,p|m,n,k)dp = \int_0^1 Pr(j|p,m,n,k)Pr(p|n,k)dp$

However, our Binomial model for $Y$ means that, conditionally on $p$ having a certain value, the probability of having $j$ successes in $m$ trials doesn't depend on past results: it's simply

$f(j|m,p)=\binom{j}{m} p^j(1-p)^j$

Thus the expression becomes

$Pr(j|m,n,k)=\int_0^1 \binom{j}{m} p^j(1-p)^j Pr(p|n,k)dp=\int_0^1 \binom{j}{m} p^j(1-p)^j Beta(\alpha+k,\beta+n-k)dp$

The result of this integral is a well-known distribution called the Beta-Binomial distribution: skipping the passages, we get the horrible expression

$Pr(j|m,n,k)=\frac{m!}{j!(m-j)!}\frac{\Gamma(\alpha+\beta+n)}{\Gamma(\alpha+k)\Gamma(\beta+n-k)}\frac{\Gamma(\alpha+k+j)\Gamma(\beta+n+m-k-j)}{\Gamma(\alpha+\beta+n+m)}$

Our point estimate for $j$, given quadratic loss, is of course the mean of this distribution, i.e.,

$\mu=\frac{m(\alpha+k)}{(\alpha+\beta+n)}$

Now, let's look for a prediction interval. Since this is a discrete distribution, we don't have a closed form expression for $[j_1,j_2]$, such that $Pr(j_1\leq j \leq j_2)= 0.95$. The reason is that, depending on how you define a quantile, for a discrete distribution the quantile function is either not a function or is a discontinuous function. But this is not a big problem: for small $m$, you can just write down the $m$ probabilities $Pr(j=0|m,n,k),Pr(j\leq 1|m,n,k),\dots,Pr(j \leq m-1|m,n,k)$ and from here find $j_1,j_2$ such that

$Pr(j_1\leq j \leq j_2)=Pr(j\leq j_2|m,n,k)-Pr(j < j_1|m,n,k)\geq 0.95$

Of course you would find more than one couple, so you would ideally look for the smallest $[j_1,j_2]$ such that the above is satisfied. Note that

$Pr(j=0|m,n,k)=p_0,Pr(j\leq 1|m,n,k)=p_1,\dots,Pr(j \leq m-1|m,n,k)=p_{m-1}$

are just the values of the CMF (Cumulative Mass Function) of the Beta-Binomial distribution, and as such there is a closed form expression, but this is in terms of the generalized hypergeometric function and thus is quite complicated. I'd rather just install the R package extraDistr and call pbbinom to compute the CMF of the Beta-Binomial distribution. Specifically, if you want to compute all the probabilities $p_0,\dots,p_{m-1}$ in one go, just write:

library(extraDistr)  
jvec <- seq(0, m-1, by = 1) 
probs <- pbbinom(jvec, m, alpha = alpha + k, beta = beta + n - k)

where alpha and beta are the values of the parameters of your Beta prior, i.e., $\alpha$ and $\beta$ (thus 1 if you're using a uniform prior over $p$). Of course it would all be much simpler if R provided a quantile function for the Beta-Binomial distribution, but unfortunately it doesn't.

Practical example with the Bayesian solution

Let $n=100$, $k=70$ (thus we initially observed 70 successes in 100 trials). We want a point estimate and a 95%-prediction interval for the number of successes $j$ in the next $m=20$ trials. Then

n <- 100
k <- 70
m <- 20
alpha <- 1
beta  <- 1

where I assumed a uniform prior on $p$: depending on the prior knowledge for your specific application, this may or may not be a good prior. Thus

bayesian_point_estimate <- m * (alpha + k)/(alpha + beta + n) #13.92157

Clearly a non-integer estimate for $j$ doesn't make sense, so we could just round to the nearest integer (14). Then, for the prediction interval:

jvec <- seq(0, m-1, by = 1)
library(extraDistr)
probabilities <- pbbinom(jvec, m, alpha = alpha + k, beta = beta + n - k)

The probabilities are

> probabilities
 [1] 1.335244e-09 3.925617e-08 5.686014e-07 5.398876e-06
 [5] 3.772061e-05 2.063557e-04 9.183707e-04 3.410423e-03
 [9] 1.075618e-02 2.917888e-02 6.872028e-02 1.415124e-01
[13] 2.563000e-01 4.105894e-01 5.857286e-01 7.511380e-01
[17] 8.781487e-01 9.546188e-01 9.886056e-01 9.985556e-01

For an equal-tail probabilities interval, we want the smallest $j_2$ such that $Pr(j\leq j_2|m,n,k)\ge 0.975$ and the largest $j_1$ such that $Pr(j < j_1|m,n,k)=Pr(j \le j_1-1|m,n,k)\le 0.025$. This way, we will have

$Pr(j_1\leq j \leq j_2|m,n,k)=Pr(j\leq j_2|m,n,k)-Pr(j < j_1|m,n,k)\ge 0.975-0.025=0.95$

Thus, by looking at the above probabilities, we see that $j_2=18$ and $j_1=9$. The probability of this Bayesian prediction interval is 0.9778494, which is larger than 0.95. We could find shorter intervals such that $Pr(j_1\leq j \leq j_2|m,n,k)\ge 0.95$, but in that case at least one of the two inequalities for the tail probabilities wouldn't be satisfied.

Frequentist solution

I'll follow the treatment of Krishnamoorthy and Peng, 2011. Let $Y\sim Binom(m,p)$ and $X\sim Binom(n,p)$ be independently Binominally distributed. We want a $1-2\alpha-$prediction interval for $Y$, based on a observation of $X$. In other words we look for $I=[L(X;n,m,\alpha),U(X;n,m,\alpha)]$ such that:

$Pr_{X,Y}(Y\in I)=Pr_{X,Y}(L(X;n,m,\alpha)\leq Y\leq U(X;n,m,\alpha)]\geq 1-2\alpha$

The "$\geq 1-2\alpha$" is due to the fact that we are dealing with a discrete random variable, and thus we cannot expect to get exact coverage...but we can look for an interval which has always at least the nominal coverage, thus a conservative interval. Now, it can be proved that the conditional distribution of $X$ given $X+Y=k+j=s$ is hypergeometric with sample size $s$, number of successes in the population $n$ and population size $n+m$. Thus the conditional pmf is

$Pr(X=k|X+Y=s,n,n+m)=\frac{\binom{n}{k}\binom{m}{s-k}}{\binom{m+n}{s}}$

The conditional CDF of $X$ given $X+Y=s$ is thus

$Pr(X\leq k|s,n,n+m)=H(k;s,n,n+m)=\sum_{i=0}^k\frac{\binom{n}{i}\binom{m}{s-i}}{\binom{m+n}{s}}$

The first great thing about this CDF is that it doesn't depend on $p$, which we don't know. The second great thing is that it allows to easily find our PI: as a matter of fact, if we observed a value $k$ of X, then the $1-\alpha$ lower prediction limit is the smallest integer $L$ such that

$Pr(X\geq k|k+L,n,n+m)=1-H(k-1;k+L,n,n+m)>\alpha$

correspondingly, the the $1-\alpha$ upper prediction limit is the largest integer such that

$Pr(X\leq k|k+U,n,n+m)=H(k;k+U,n,n+m)>\alpha$

Thus, $[L,U]$ is a prediction interval for $Y$ of coverage at least $1-2\alpha$. Note that when $p$ is close to 0 or 1, this interval is conservative even for large $n$, $m$, i.e., its coverage is quite larger than $1-2\alpha$.

Practical example with the Frequentist solution

Same setting as before, but we don't need to specify $\alpha$ and $\beta$ (there are no priors in the Frequentist framework):

n <- 100
k <- 70
m <- 20

The point estimate is now obtained using the MLE estimate for the probability of successes, $\hat{p}=\frac{k}{n}$, which in turns leads to the following estimate for the number of successes in $m$ trials:

frequentist_point_estimate <- m * k/n #14

For the prediction interval, the procedure is a bit different. We look for the largest $U$ such that $Pr(X\leq k|k+U,n,n+m)=H(k;k+U,n,n+m)>\alpha$, thus let's compute the above expression for all $U$ in $[0,m]$:

jvec <- seq(0, m, by = 1)
probabilities <- phyper(k,n,m,k+jvec)

We can see that the largest $U$ such that the probability is still larger than 0.025 is

jvec[which.min(probabilities > 0.025) - 1] # 18

Same as for the Bayesian approach. The lower prediction bound $L$ is the smallest integer such that $Pr(X\geq k|k+L,n,n+m)=1-H(k-1;k+L,n,n+m)>\alpha$, thus

probabilities <- 1-phyper(k-1,n,m,k+jvec)
jvec[which.max(probabilities > 0.025) - 1] # 8

Thus our frequentist "exact" prediction interval is $[L,U]=[8,18]$.

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