Prediction interval for binomial random variable

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

Assume YBinom(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 nˆp, where ˆp=yn. A confidence interval for ˆp is straightforward, but I cannot find a formula for a prediction interval for Y. If we knew p (rather than ˆp), then a 95% prediction interval just involves finding the quantiles of a binomial. Is there something obvious I’m overlooking?

Answer

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., pBeta(α,β), because the Beta-Binomial model is conjugate, which means that the posterior distribution is also a Beta distribution with parameters ˆα=α+k,ˆβ=β+nk, (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 α and β, i.e., to define your Beta prior, otherwise you could assume a uniform (noninformative) prior, with α=β=1, or other noninformative priors (see for example here). In any case, your posterior is

Pr(p|n,k)=Beta(α+k,β+nk)

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 y=y1,,ym, where m is not necessarily equal to n. Specifically, for each j=0,,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)=10Pr(j,p|m,n,k)dp=10Pr(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].

Attribution
Source : Link , Question Author : Statseeker , Answer Author : Community

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