Assessing the robustness of estimators when fitting Poisson inverse Gaussian models

Abstract

The generalized linear mixed model (GLMM) extends classical regression analysis to non-normal, correlated response data. Because inference for GLMMs can be computationally difficult, simplifying distributional assumptions are often made. We focus on the robustness of estimators when a main component of the model, the random effects distribution, is misspecified. Results for the maximum likelihood estimators of the Poisson inverse Gaussian model are presented.

Appendix: Upper and lower bounds for Poisson inverse Gaussian probability ratios

Below we provide proofs to Lemmas 2 and 3, in which we find an upper bound and a lower bound for Poisson inverse Gaussian probability ratios.

Lemma 2

Let $\nu =+\sqrt{\tau ^{-1}(\tau ^{-1} + 2\mu )}$. Then for $y>0,$

$$\begin{aligned} (y+1)\frac{P_{y+1}}{P_y} \le (1+2\tau )^{-1/2}(1+\nu ^{-1})(2y-1). \end{aligned}$$

Proof

Equation (7) gives us the following relationship:

$$\begin{aligned} (y+1)\frac{P_{y+1}}{P_y}= & {} (y+1) \left\{ \frac{(1+2\tau )^{-1/2}}{y+1}\frac{ K_{y+\frac{1}{2}}(\nu )}{K_{y-\frac{1}{2}}(\nu )}\right\} \nonumber \\= & {} (1+2\tau )^{-1/2} \frac{ K_{y+\frac{1}{2}}(\nu )}{K_{y-\frac{1}{2}}(\nu )}. \end{aligned}$$

(13)

From Abramowitz and Stegun (1972), we have that for $k=0,\pm \,1,\pm \,2,\ldots$

$$\begin{aligned} K_{k+\frac{1}{2}}(\nu ) = {C(\nu )}\sum _{r=0}^{k }\frac{(k+r)!}{r!(k -r)!(2\nu )^r}, \end{aligned}$$

(14)

where ${C(\nu )}={\sqrt{\pi /(2\nu )} \exp (-\nu )}$. Therefore, we may write

$$\begin{aligned} \frac{K_{y+\frac{1}{2}}(\nu )}{{C(\nu )}}= & {} \sum _{r=0}^{y}\frac{(y+r)!}{r!(y-r)!(2\nu )^r}\\= & {} \sum _{r=0}^{y}\frac{(y-1+r)!}{r!(y-1-r)!(2\nu )^r} \frac{(y+r)}{(y-r)} \\= & {} \frac{(2y)!}{y! (2\nu )^y} + \sum _{r=0}^{y-1}\frac{(y-1+r)!}{r!(y-1-r)!(2\nu )^r}\frac{(y+r)}{(y-r)}\\\le & {} \frac{(2y)!}{y! (2z)^y} + (2y-1)\sum _{r=0}^{y-1}\frac{(y-1+r)!}{r!(y-1-r)!(2\nu )^r}\\\le & {} \frac{(2y)!}{y! (2\nu )^y} + \frac{(2y-1)K_{y-\frac{1}{2}}(\nu )}{{C(\nu )}}. \end{aligned}$$

Substituting in the numerator of (13) we find that

$$\begin{aligned} (y+1)\frac{P_{y+1}}{P_y}\le & {} (1+2\tau )^{-1/2}\left[ \frac{{C(\nu )}(2y)!/(y!(2\nu )^y) + (2y-1) K_{y-\frac{1}{2}}(\nu )}{K_{y-\frac{1}{2}}(\nu )}\right] \nonumber \\= & {} (1+2\tau )^{-1/2}\left[ \frac{{C(\nu )}(2y)!}{y!(2\nu )^y K_{y-\frac{1}{2}}(\nu ) } + 2y-1\right] \nonumber \\\le & {} (1+2\tau )^{-1/2}(1+\nu ^{-1})(2y-1), \end{aligned}$$

(15)

where the last line uses the following inequality:

$$\begin{aligned} \frac{y!(2\nu )^y K_{y-\frac{1}{2}}(\nu )}{{C(\nu )}(2y)!}= & {} \frac{y!(2\nu )^y}{(2y)!}\sum _{r=0}^{y-1}\frac{(y-1+r)!}{r!(y-1-r)!(2\nu )^r}\\\ge & {} \frac{y!(2\nu )^y}{(2y)!} \frac{(2y-2)!}{(y-1)!(2\nu )^{y-1}} \\= & {} \frac{\nu }{2y-1}. \end{aligned}$$

$\square$

Lemma 3

Let $\nu =+\,\sqrt{\tau ^{-1}(\tau ^{-1} + 2\mu )}$. Then for $y>0,$

$$\begin{aligned} (y+1)\frac{P_{y+1}}{P_y} \ge (1+2\tau )^{-1/2} \left[ 1+\left( \frac{1}{1+2\nu }\right) ^y\right] . \end{aligned}$$

Proof

Recall from Eq. (13) that

$$\begin{aligned} (y+1)\frac{P_{y+1}}{P_y} =(1+2\tau )^{-1/2} \frac{ K_{y+\frac{1}{2}}(\nu )}{K_{y-\frac{1}{2}}(\nu )}. \end{aligned}$$

(16)

Using (14) with ${C(\nu )}={\sqrt{\pi /(2\nu )} \exp (-\nu )}$, we may write

$$\begin{aligned} \frac{K_{y-\frac{1}{2}}(z)}{{C(\nu )}}= & {} \sum _{r=0}^{y-1}\frac{(y-1+r)!}{r!(y-1-r)!(2\nu )^r} \\= & {} \sum _{r=0}^{y-1}\frac{(y+r)!}{r!(y-r)!(2\nu )^r} \frac{(y-r)}{(y+r)}\\\le & {} \sum _{r=0}^{y-1}\frac{(y+r)!}{r!(y-r)!(2\nu )^r}\\= & {} \sum _{r=0}^{y}\frac{(y+r)!}{r!(y-r)!(2z)^r} - \frac{(2y)!}{y! (2\nu )^y} \\= & {} \frac{K_{y+\frac{1}{2}}(\nu )}{{C(\nu )}} - \frac{(2y)!}{y! (2\nu )^y}. \end{aligned}$$

Therefore, substituting into (16), we have the following lower bound:

$$\begin{aligned} (y+1)\frac{P_{y+1}}{P_y}\ge & {} (1+2\tau )^{-1/2} \frac{ K_{y+\frac{1}{2}}(\nu )}{K_{y+\frac{1}{2}}(\nu ) -{C(\nu )}(2y)! / (y!(2\nu )^y)} \nonumber \\= & {} (1+2\tau )^{-1/2} \frac{1}{1 - {C(\nu )}(2y)!/(y!(2\nu )^y K_{y+\frac{1}{2}}(\nu ) )}\nonumber \\\ge & {} (1+2\tau )^{-1/2}\left[ 1+\frac{{C(\nu )}(2y)!}{y!(2\nu )^y K_{y+\frac{1}{2}}(\nu )}\right] , \end{aligned}$$

(17)

where we use the elementary inequality $1/(1-x) \ge 1+x$. Notice that

$$\begin{aligned} \frac{y! (2\nu )^y K_{y+\frac{1}{2}}(\nu ) }{{C(\nu )}(2y)!}= & {} \sum _{r=0}^{y}\frac{(y+r)!}{r!(y-r)!(2\nu )^r} \frac{y! (2\nu )^y }{(2y)!}\\= & {} \sum _{r=0}^{y}\frac{(y+r)!}{(2y)!} \frac{y!(2\nu )^{y-r}}{r!(y-r)!}\\= & {} \sum _{r=0}^{y}\frac{(y+r)!}{(2y)!} \left( {\begin{array}{c}y\\ r\end{array}}\right) (2\nu )^{y-r} \\\le & {} \sum _{r=0}^{y} \left( {\begin{array}{c}y\\ r\end{array}}\right) (2\nu )^{y-r} \\= & {} (1+2\nu )^y. \end{aligned}$$

Therefore, (17) becomes

$$\begin{aligned} (y+1)\frac{P_{y+1}}{P_y} \ge (1+2\tau )^{-1/2} \left[ 1+\left( \frac{1}{1+2\nu }\right) ^y\right] \end{aligned}$$

by substitution. $\square$