Highest posterior mass prediction intervals for binomial and poisson distributions

Abstract

The problems of constructing prediction intervals (PIs) for the binomial and Poisson distributions are considered. New highest posterior mass (HPM) PIs based on fiducial approach are proposed. Other fiducial PIs, an exact PI and approximate PIs are reviewed and compared with the HPM-PIs. Exact coverage studies and expected widths of prediction intervals show that the new prediction intervals are less conservative than other fiducial PIs and comparable with the approximate one based on the joint sampling approach for the binomial case. For the Poisson case, the HPM-PIs are better than the other PIs in terms of coverage probabilities and precision. The methods are illustrated using some practical examples.

Appendices

Appendix A

We shall now provide an exact expression for the probability mass function of $Y^*$ defined in (8). Note that

$$\begin{aligned} P(Y^* = y)= & {} E_UP(Y^*=y|W)\\= & {} E\left\{ {m\atopwithdelims ()y}W^y(1-W)^{m-y} \right\} \\= & {} {m\atopwithdelims ()y}\frac{1}{B(x+.5,n-x+.5)}\int _0^1 w^{y+x-.5}(1-w)^{n+m-y-x-.5}dw\\= & {} {m\atopwithdelims ()y}\frac{B(y+x+.5,n+m-y-x+.5)}{B(x+.5,n-x+.5)}\\= & {} \frac{\Gamma (x+y+.5)}{\Gamma (y+1)\Gamma (x+.5)} \frac{\Gamma (m+n-x-y+.5)}{\Gamma (n-x+.5)\Gamma (m-y+1)}\bigg /{m+n \atopwithdelims ()m}. \end{aligned}$$

Appendix B

We shall now provide an exact expression for the probability mass function of $Y^*$ defined in (14) for predicting a future observation from a Poisson distribution. We first note that $\chi ^2_a/b$ is distributed as a gamma random variable with shape parameter a / 2 and the scale parameter 2 / b. Letting $a=2x+1$ and $b=2n$, we see that

$$\begin{aligned} P(Y^* = y)= & {} E_WP(Y^*=y|W)\\= & {} \frac{(b/2)^{\frac{a}{2}}}{\Gamma \left( a/2\right) }\int _0^\infty \frac{e^{-mw}(mw)^y}{y!} e^{-bw/2}w^{a/2-1}dw \\= & {} \frac{m^y(b/2)^{\frac{a}{2}}}{y!\Gamma \left( \frac{a}{2}\right) }\int _0^\infty e^{-w(m+b/2)}w^{a/2+y-1}dw \\= & {} \frac{m^y(b/2)^{\frac{a}{2}}\Gamma (a/2+y)}{y!\Gamma (a/2)(m+b/2)^{a/2+y}}\\= & {} \frac{\Gamma (x+y+.5)}{\Gamma (x+.5)\Gamma (y+1)}\left( \frac{m}{m+n}\right) ^y \left( \frac{n}{m+n}\right) ^{x+.5}. \end{aligned}$$