2.1. Stein’s method for distributional comparisons

To obtain explicit bounds on the distance between a probability distribution $\mathcal{L}(X)$ of interest and a usually well-understood approximating distribution $\mathcal{L}_0(Z)$ , often called the target distribution, Stein’s method connects the test function $h \in \mathcal{H}$ in (2.1) to the distribution of interest through a Stein equation

(2.3) \begin{equation} h(x) - \mathbb{E} h(Z) = \mathcal{T}g(x).\end{equation}

In (2.3), $\mathcal{T}$ is a Stein operator for the distribution $\mathcal{L}_0 (Z)$ , with an associated Stein class $ \mathcal{F}(\mathcal{T})$ of functions such that $\mathbb{E}[\mathcal{T}g(Z)] = 0 \mbox{ for all } g \in \mathcal{F}(\mathcal{T}) $ if and only if $ Z \sim \mathcal{L}_0 (Z);$ thus, a Stein operator characterises the distribution. The distance (2.1) can then be bounded by $ d_{\mathcal{H}} (\mathcal{L}(X),\mathcal{L}_0(Z)) \le \sup_{g \in \mathcal{F}(\mathcal{H})}|\mathbb{E}\mathcal{T}g(X)| $ where $ \mathcal{F}(\mathcal{H}) = \{g_h \,:\, h \in \mathcal{H}\}$ is the set of solutions of the Stein equation (2.3) for the set of test functions $h \in \mathcal{H}$ .

The Stein operator $\mathcal{T}$ for a probability distribution is not unique; see e.g. [Reference Ley, Reinert and Swan18]. In this paper, we employ the so-called density method, which uses the score function of a probability distribution to construct a Stein operator, called a score Stein operator. Following [Reference Ley and Swan19, Reference Stein, Diaconis, Holmes and Reinert33], a score Stein operator for a continuous distribution with probability density function (PDF) p and support $[a,b] \in \mathbb{R}$ acts on functions g for which the derivative exists by

(2.4) \begin{equation} {\mathcal T}_p g (x) = \frac{ (gp)' (x) }{p (x)},\end{equation}

where we take $0/0 =0$ . For differentiable g and p, (2.4) simplifies to ${\mathcal T}_p g (x) = g'(x) + g(x) \rho (x)$ , where $\rho = p'/p$ is the score function of p. The Stein class ${\mathcal F}(\mathcal{T}_p)$ is the collection of functions $ g\, :\, \mathbb{R} \rightarrow \mathbb{R}$ such that g(x) p(x) is differentiable with integrable derivative and $ \lim_{x \to {a,b} } g(x)p(x)=0 $ . It is straightforward to see that for a PDF p on ${[a,b] \subset \mathbb{R}}$ ,

(2.5) \begin{align} g(x) = g_h(x) = \frac{ 1 }{p (x)} \int_0^x [h(t)-\mathbb{E} h(X)] p(t) \,\mathrm{d}t \end{align}

solves (2.3) for h and that if h is bounded then $g \in \mathcal{F}(\mathcal{T}_p)$ .

To use the Stein equation for a distributional comparison, let X and Y be two random variables with PDFs $p_X$ and $p_Y$ , defined on the same probability space and with nested supports ${\textrm{supp}}(p_Y) \subset {\textrm{supp}}(p_X) =[a,b] \subset \mathbb{R}$ , score functions $\rho_X$ and $\rho_Y$ , and corresponding score Stein operators $\mathcal T_X$ and $\mathcal T_Y$ . Then

(2.6) \begin{equation} \mathbb{E} h(Y)-\mathbb{E} h(X) = \mathbb{E} g_{X, h}(Y)(\rho_Y(Y)-\rho_X(Y)),\end{equation}

where $g_{X, h}(x)$ is the solution of the Stein equation for h and $\mathcal T_X$ . Equation (2.6) is a special case of Equation 23 in Section 2.5 of [Reference Ley, Reinert and Swan18].

For a discrete distribution with probability mass function (PMF) q having support $\mathcal{I}=[a,b] \subset {\mathbb{N}}$ , the discrete backward score function is $\frac{ \Delta^- q(y)}{q(y)}$ , and, as in Remark 3.2 and Example 3.13 of [Reference Ley, Reinert and Swan18], a discrete backward score Stein operator is

(2.7) \begin{equation}{\mathcal T}_q g (y) = \Delta^- g(y) + \frac{ \Delta^- q(y)}{q(y)} g(y-1), \quad y \in \mathcal{I}.\end{equation}

With an abuse of notation, we often refer to a Stein operator for the distribution of X as the Stein operator for X, and similarly refer to the score function of the distribution of X as the score function of X. Further, if $X \sim p$ , we also write $\mathcal{T}_X$ for $\mathcal{T}_p$ .

2.2. CCR distributions

Let $N \in \mathbb{N}$ be a random variable with finite second moment and let ${\textbf{Y}}= (Y_1, Y_2, \ldots)$ be a sequence of positive i.i.d. random variables, with cumulative distribution function (CDF) $F_Y$ independent of N. Then

(2.8) \begin{equation} W_{\alpha} =W_{\alpha}(N, {\textbf{Y}})= \begin{cases} \min\{Y_1, Y_2, \ldots, Y_N\} & \mbox{ if } \alpha = -1, \\ \max\{Y_1, Y_2, \ldots, Y_N\} & \mbox{ if }\alpha = 1 \end{cases}\end{equation}

is called a CCR random variable with type indicator $\alpha \in \{-1, 1\}.$ Setting

(2.9) \begin{equation} U_Y^{\alpha}(\cdot) = \begin{cases} 1- F_Y(\cdot) & \mbox{ if }\alpha = -1, \\ F_Y(\cdot) & \mbox{ if }\alpha = 1\end{cases}\end{equation}

and writing $G_N(x) = \mathbb{E}\, x^N$ , the probability generating function of N, $W_{\alpha} >0$ has CDF

(2.10) \begin{align} F_{W_{\alpha}}(w) &= \mathbb{P} (W_\alpha \le w) \end{align}

(2.11) \begin{align}&= \begin{cases} 1-\sum_{n}\mathbb{P} (N =n)U_Y^{\alpha}(w)^{n} = 1-(G_N\circ U_Y^{\alpha})(w) & \mbox{ if } \alpha = -1, \\ \sum_{n}\mathbb{P} (N =n)U_Y^{\alpha}(w)^{n} = (G_N\circ U_Y^{\alpha})(w) & \mbox{ if } \alpha = 1. \end{cases} \\[8pt]\nonumber\end{align}

If the $Y_i$ have a continuous distribution with PDF $f_Y$ , then $W_\alpha$ has PDF

(2.12) \begin{equation} {f}_{W_{\alpha}}(w) = \sum_{n} \mathbb{P} (N =n)\, n ({U_Y^{\alpha}}(w))^{n-1} {f}_Y(w) = f_Y(w)(G_N'\circ U_Y^{\alpha})(w);\end{equation}

see also [Reference Tahir and Cordeiro34]. Here we have used that $(U_Y^\alpha)'(w) = \alpha f_Y(w)$ .

If the $Y_i$ are discrete with PMF $p_Y$ on $\mathbb{N},$ the resulting random variable $W_{\alpha}$ has PMF $p_{W_{\alpha}}$ , which can be expressed in terms of $G_N(\cdot)$ as

(2.13) \begin{equation} p_{W_{\alpha}}(x) = \alpha \bigl(G_N (U_Y^{\alpha}(x)) - G_N (U_Y^{\alpha}(x-1)) \bigr) = \alpha\Delta^- (G_N \circ U_Y^{\alpha}) (x).\end{equation}

2.3. Stein’s method for the CCR class of distributions

To obtain a Stein operator for the CCR random variable $W_{\alpha} = W_{\alpha}(N, \textbf{Y})$ in (2.8), we use the density method. First, we assume that the $Y_i$ are continuous with differentiable PDF $f_Y$ . From (2.12), the score function for the distribution of $W_{{\alpha}}$ is

(2.14) \begin{equation} \rho_{W_{{\alpha}}}({w}) = {\alpha} {f}_Y({w}) \frac{(G_N'' \circ U_Y^{\alpha})(w)}{(G_N'\circ U_Y^{\alpha})({w})} + \rho_Y(w),\end{equation}

where $\rho_Y = {f_Y'}/{f_Y}$ is the score function of Y. Hence $\mathcal{T}_{W_{\alpha}}$ given by

(2.15) \begin{equation} \mathcal{T}_{W_{\alpha}}g(w) = g'(w) + \rho_{W_{\alpha}}(w) g(w)\end{equation}

for differentiable g is a Stein operator acting on the functions $g \in {\mathcal F}(\mathcal{T}_{W_{\alpha}})$ . For a test function $h \in \mathcal{H}$ , the corresponding Stein equation is

(2.16) \begin{equation} g'({w}) + \rho_{{W_{{\alpha}}}}({w}) g({w}) = h({w}) - \mathbb{E} h ( {W_{\alpha}}).\end{equation}

Thus, for any random variable X, the distance $d_{\mathcal{H}}$ from (2.1) between the distributions of X and $W_{\alpha}$ can be bounded by bounding the expectation of the left-hand side of (2.16). For $Y_i$ taking values in $\mathbb{N}$ with PMF $p_Y$ , the backward score function is

(2.17) \begin{equation} \rho_{W_{{\alpha}}}(w) = \frac{\Delta^- p_{W_{\alpha}}(w)}{p_{W_{\alpha}}(w)} = \frac{\Delta^-( \Delta^- (G_N \circ U_Y^{\alpha}) )(w) }{\Delta^- (G_N \circ U_Y^{\alpha})(w)},\end{equation}

with corresponding discrete backward score Stein operator $\mathcal{T}_{W_{\alpha}}$ operating as

(2.18) \begin{align} \mathcal{T}_{W_{\alpha}} g (w) = \Delta^- g(w) + \rho_{W_{\alpha}}(w) g(w-1).\end{align}

4.1. Approximating the distribution of the maximum of a random number of i.i.d. random variables by a PE distribution

Let $M\in \mathbb{N}$ be independent of ${\textbf{X}} = (X_1, X_2, \ldots)$ , a sequence of i.i.d. random variables, and let $W=W_1(M, {\textbf{X}})$ have a PDF of the form (2.12) for $\alpha = 1$ . Our first comparison result employs Stein’s method.

Corollary 1. Assume that the $X_i$ have differentiable PDF $p_X$ , CDF $F_X$ , and score function $\rho_X$ . Let $W = W_1(M, {\textbf{X}}) = \max \{X_1, \ldots, X_M\}$ . Then

(4.11) \begin{equation} \begin{split} & d_{\rm TV}\big(\mathrm{PE}(\theta, \lambda), \mathcal{L}(W_1(M, {\textbf{X}}))\big) \\ & \quad\le \frac{4}{\lambda} \left(\mathbb{E}\left|\theta \lambda {\mathrm{e}}^{-\lambda W}- \frac{(G_M'' \circ F_X)(W)}{(G_M' \circ F_X)(W)} p_X(W) \right| + \mathbb{E} \bigl| - \lambda - \rho_X (W) \bigr|\right) . \end{split}\end{equation}

If M is a zero-truncated Poisson $(\theta_m)$ random variable, then (4.11) reduces to

\begin{align*} d_{\rm TV}\big(\mathrm{PE}(\theta, \lambda), \mathcal{L}(W_1(M, {\textbf{X}}))\big) \le \frac{4}{\lambda} \Big(\mathbb{E} \bigl|\theta\lambda {\mathrm{e}}^{-\lambda W} - \theta_mp_X(W) \bigr| + \mathbb{E} \bigl| - \lambda - \rho_X (W) \bigr| \Big) . \end{align*}

The next result is based on the Lindeberg argument.

Corollary 2. Let $W_1(N, {\textbf{E}}) \sim \mathrm{PE}(\theta, \lambda)$ and let $W_{1}(M, {\textbf{X}}) = \max(X_1, \ldots, X_M)$ have a CCR distribution. Then

\begin{align*} d_{\rm BW}\bigl(\mathrm{PE}(\theta, \lambda), \mathcal{L}(W_{1}(M, {\textbf{X}})\bigr) & \le \mathbb{E} M \, \mathbb{E}| E_1 - X_1| \nonumber \\ & \quad + \sum_{m,n=1}^\infty \mathbb{P} (M=m, N=n) \left( H_{\max\{m,n\}} - H_{\min\{m,n\}}\right), \end{align*}

where $H_n$ is the nth harmonic number.

If M is also a zero-truncated $\mathrm{Poisson}(\theta)$ random variable, then

\begin{align*} d_{\rm BW}(\mathrm{PE}\bigl(\theta, \lambda), \mathcal{L}(W_{1}(M, {\textbf{X}})\bigr) \le \frac{\theta}{1 - {\mathrm{e}}^{-\theta} } \,\mathbb{E} | E_1 - X_1|. \end{align*}

Proof. Proposition 3 gives

\begin{align*} & \big|\mathbb{E} h(W_{1}(N, {\textbf{E}})) - \mathbb{E} h (W_{1}(M, {\textbf{X}})) \big| \\ & \quad \le \sum_{i=1}^\infty \mathbb{E} | E_i - X_i| \,\mathbb{P} (M \ge i)\, \| h'\| \\ &\qquad + \sum_{m,n=1}^\infty \mathbb{P} (M=m, N=n) \,\big|\mathbb{E} h(W_{1}(n, {\textbf{E}})) - \mathbb{E} h(W_{1}(m, {\textbf{E}}))\big|\, \| h'\|. \end{align*}

Since the random variables in ${\textbf{X}}$ are i.i.d., those in ${\textbf{E}}$ are i.i.d. exponential with parameter $\lambda$ , and the expectation of the maximum of n exponential random variables is $\sum_{i=1}^n \frac1i = H_n$ , the assertion follows.

4.2. Approximating the generalised Poisson–exponential distribution

The PE distribution has an increasing or constant failure rate. To model decreasing failure rate as well, Fatima & Roohi [Reference Fatima and Roohi12] introduced the family of generalised Poisson–exponential (GPE) distributions. The differentiable PDF of a GPE distribution with parameters $\beta, \theta,\lambda > 0 $ , denoted by $\mathrm{GPE}(\theta,\lambda,\beta)$ , is

(4.12) \begin{equation} p(x| \theta, \lambda, \beta)= \frac{\beta \theta \lambda {\mathrm{e}}^{-\lambda x - \theta \beta {\mathrm{e}}^{-\lambda x}}}{(1 - {\mathrm{e}}^{-\theta})^{\beta}}\, \Big(1-{\mathrm{e}}^{-\theta + \theta {\mathrm{e}}^{-\lambda x}} \Big)^{\beta -1} , \quad x > 0.\end{equation}

For $\beta = 1$ the density of the GPE distribution simplifies to that of the PE distribution given in (4.1). For ${0 < \beta < 1}$ the PDF of the GPE distribution is monotonically decreasing, while for $\beta \ge 1$ it is unimodal positively skewed with skewness depending on the shape parameters $\beta$ and $\theta$ . The shape of the hazard function also depends on these two shape parameters. For example, for $\theta = 1$ and $\lambda =2$ , the failure rate is decreasing for $ 0 < \beta <1$ and increasing for $\beta \ge 1$ , as in Figure 2 of [Reference Fatima and Roohi12].

For a data set from [Reference Aarset1] which consists of 50 observations on the time to first failure of devices, in [Reference Fatima and Roohi12] it was shown that the GPE distribution provides a better fit than the PE and some other candidate distributions. However, a GPE distribution is not as easy to manipulate and interpret as a PE distribution. Therefore, a natural question is how to quantify the sacrifice when approximating a GPE distribution with a PE distribution. Here we note that GPE distributions are not from the CCR family, and hence Proposition 3 cannot be applied. Instead we use Stein’s method to bound the approximation error.

For such an approximation to be intuitive, the failure rate of the approximating distribution should be qualitatively similar. Hence we restrict attention to the case of $\beta \ge 1$ , for which both the GPE and the PE distributions have an increasing failure rate. As an aside, we note that for $\beta \ge 1$ , the limit of the PDF at 0 is 0, while for $0 < \beta < 1$ it is undefined; the latter condition leads to a Stein class for the GPE Stein score operator which differs from the Stein class for the PE Stein score operator.

We note here that if $X \sim \mathrm{GPE}(\theta, \lambda, \beta)$ , then for $\beta \ge 1$ we have

(4.13) \begin{equation} \mathbb{E} X \le \frac{\beta \theta}{\lambda(1 - {\mathrm{e}}^{-\theta})^{\beta}}.\end{equation}

The proof of (4.13) is given in the appendix.

The GPE random variable is not of the CCR form (2.8), but its score function, given in (4.14) below, can be derived from its PDF (4.12) with parameters $\beta, \theta, \lambda > 0$ as

(4.14) \begin{equation} \rho(x) = \frac{p'(x)}{p(x)} = \lambda \theta {\mathrm{e}}^{-\lambda x} \Bigg[\beta +\frac{(\beta-1) \,{\mathrm{e}}^{-\theta + \theta {\mathrm{e}}^{-\lambda x}}}{1-{\mathrm{e}}^{-\theta + \theta {\mathrm{e}}^{-\lambda x}}}\Bigg] - \lambda, \quad x > 0.\end{equation}

In the following theorem we bound the distance between a GPE with $\beta \ge 1$ and a PE distribution, using their corresponding score Stein operators in (2.6).

Theorem 1. Let $W \sim \mathrm{PE}(\theta_1, \lambda_1)$ and $X \sim \mathrm{GPE}(\theta_2, \lambda_2, \beta)$ , let $h\,:\, {\mathbb{R}^{+}_{>}0} \rightarrow \mathbb{R}$ be bounded, and let $\tilde h(x) = h(x) - \mathbb{E} h(W)$ . Then for $\lambda_1 \le \lambda_2$ and $\beta \ge 1$ ,

(4.15) \begin{equation} \big|\mathbb{E} h(X)-\mathbb{E} h(W)\big| \le \|\tilde{h}\| \left\{ \frac{\lambda_2}{\lambda_1}|\beta-1| + \left|\frac{\lambda_2 \theta_2}{\lambda_1 \theta_1}\beta - 1 \right| + \left( \frac{\lambda_2}{\lambda_1} - 1\right) \left( \lambda_1 \mathbb{E} X + 2\right) \right\}.\end{equation}

Proof. Let $p_W$ and $p_X$ denote the PDFs of W and X, and let $\mathcal{T}_{X}$ denote a score Stein operator for a GPE distribution. To employ (2.6), we first check that $f_W$ as in (2.5) for the PE distribution, for h bounded, is in the Stein class of $\mathcal{T}_X$ . Invoking Lemma 1, $f_W$ is bounded. Now $ {\mathbb{E}} [{\mathcal T}_X\ f_W (X) ]= \int_0^\infty \frac{(f_W p_X)' (x)}{p_X(x)} p_X (x) \,\mathrm{d} x= \lim_{x \to \infty } (f_W p_X) (x) - \lim_{x \rightarrow 0} (f_W p_X) (x),$ and for $\beta \ge 1$ we have that $ {(f_Wp_{X})(x)} \rightarrow 0$ as $x \rightarrow 0$ and as $x \rightarrow \infty$ , showing that $f_W$ is in the Stein class for $\mathcal{T}_{X}$ . Applying (2.6) with the score functions from (4.2) and (4.14), we have

\begin{align*} & \mathbb{E} h(X) -\mathbb{E} h(W) \nonumber \\ &= \mathbb{E} \bigg[\lambda_2 \theta_2 {\mathrm{e}}^{-\lambda_2 X} \Bigg(\beta +\frac{(\beta-1) {\mathrm{e}}^{-\theta_2 + \theta_2 {\mathrm{e}}^{-\lambda_2 X}}}{1-{\mathrm{e}}^{-\theta_2 + \theta_2 {\mathrm{e}}^{-\lambda_2 X}}}\Bigg) - \lambda_2 - \lambda_1 (\theta_1 {\mathrm{e}}^{-\lambda_1 X}-1) \bigg]f_{W}(X) \\ &\le \mathbb{E} \left[\lambda_2 \theta_2 (\beta-1) {\mathrm{e}}^{(\lambda_1-\lambda_2) X} \frac{ {\mathrm{e}}^{-\theta_2 + \theta_2 {\mathrm{e}}^{-\lambda_2 X}}}{1-{\mathrm{e}}^{-\theta_2 + \theta_2 {\mathrm{e}}^{-\lambda_2 X}}} \,{\mathrm{e}}^{-\lambda_1 X} |f_W(X)|\right] \nonumber \\ &\quad+ \mathbb{E}\left|\lambda_2\theta_2\beta {\mathrm{e}}^{-\lambda_2 X} - \lambda_1 \theta_1 {\mathrm{e}}^{-\lambda_1 X}\right| |f_W(X)| + \mathbb{E}|\lambda_1 - \lambda_2|\,|f_W(X)|.\end{align*}

Next,

(4.16) \begin{equation}{ |\theta_2 \lambda_2 {\mathrm{e}}^{-\lambda_2 x} - \theta_1 \lambda_1 {\mathrm{e}}^{-\lambda_1 x}|} \le |\theta_2 \lambda_2 \beta -\theta_1 \lambda_1 |\,{\mathrm{e}}^{(\lambda_1 - \lambda_2) x} {\mathrm{e}}^{-\lambda_1 x} + (\lambda_2-\lambda_1) x \theta_1 \lambda_1 {\mathrm{e}}^{-\lambda_1 x}, \end{equation}

since $|{\mathrm{e}}^{-x} - 1 | \le x$ for all $x \ge 0$ . With ${\mathrm{e}}^{(\lambda_1 - \lambda_2) x} \le 1$ for $\lambda_1 \le \lambda_2$ , we write

(4.17) \begin{align} & |\mathbb{E} h(X) -\mathbb{E} h(W)| \nonumber \\ &\le \lambda_2 \theta_2 |\beta-1| \,\mathbb{E} \left|\frac{ {\mathrm{e}}^{-\theta_2 + \theta_2 {\mathrm{e}}^{-\lambda_2 X}}}{1-{\mathrm{e}}^{-\theta_2 + \theta_2 {\mathrm{e}}^{-\lambda_2 X}}} {\mathrm{e}}^{-\lambda_1 X} f_W(X)\right| + ( \lambda_2 - \lambda_1) \,\mathbb{E}|f_W(X)| \nonumber \\ &\quad + \lambda_1 \theta_1 \left|\frac{\lambda_2 \theta_2}{\lambda_1 \theta_1}\beta - 1 \right|\mathbb{E}|{\mathrm{e}}^{-\lambda_1 X}f_W(X)| + \lambda_1 \theta_1 ( \lambda_2 - \lambda_1) \,\mathbb{E} X |{\mathrm{e}}^{-\lambda_1X}f_W(X)| . \end{align}

Now, using (4.5) as well as ${\mathrm{e}}^{x} - 1 \ge x$ and $1-{\mathrm{e}}^{-x} \le x$ for $x \ge 0$ , we have

(4.18) \begin{align} \left|\frac{{\mathrm{e}}^{-\theta_2 + \theta_2 {\mathrm{e}}^{-\lambda_2 x}}}{1-{\mathrm{e}}^{-\theta_2 + \theta_2 {\mathrm{e}}^{-\lambda_2 x}}}{\mathrm{e}}^{-\lambda_1 x} f_W(x)\right| &\le \frac{\|\tilde{h}\|}{\lambda_1 \theta_1} \frac{{\mathrm{e}}^{-\theta_2 + \theta_2 {\mathrm{e}}^{-\lambda_2 x}}}{1-{\mathrm{e}}^{-\theta_2 + \theta_2 {\mathrm{e}}^{-\lambda_2 x}}}\, (1-{\mathrm{e}}^{-\theta_1(1 - {\mathrm{e}}^{-\lambda_1 x})}) \nonumber \\ & \le \frac{\|\tilde{h}\|}{\lambda_1 \theta_2} .\end{align}

Using (4.6), (4.8), (4.13) and (4.18) in (4.17) gives the bound (4.15).

Remark 3. For $\lambda_1=\lambda_2$ and $\theta_1=\theta_2$ in Theorem 1, the bound (4.15) can be improved by a factor of 2, using that with $f_{W}$ being the solution of the PE score Stein equation,

\begin{align*} \big|\mathbb{E} h(X) -\mathbb{E} h(W)\big| = \mathbb{E} \! \left(\lambda \theta |\beta - 1|\frac{{\mathrm{e}}^{-\lambda X}}{1-{\mathrm{e}}^{-\theta + \theta {\mathrm{e}}^{-\lambda X}}}f_{W}(X) \right).\end{align*}

With (4.5) and $\tilde h(x) = h(x) - \mathbb{E} h(W)$ we obtain

(4.19) \begin{equation} \big|\mathbb{E} h(X)-\mathbb{E} h(W)\big|\le \|\tilde{h}\|\, |\beta -1 |.\end{equation}

This bound depends solely on the parameter $\beta$ and tends to 0 as $\beta\rightarrow 1$ , which is in line with the fact that for $\beta \rightarrow 1$ , $\mathrm{GPE}(\theta, \lambda, \beta)$ converges to $\mathrm{PE}(\theta, \lambda)$ ; see [Reference Fatima and Roohi12].

The next result follows immediately from Theorem 1 and (4.13).

Corollary 3. Let $W_1 \sim \mathrm{PE}(\theta_1, \lambda_1)$ and $W_2 \sim \mathrm{PE}(\theta_2, \lambda_2)$ with $\lambda_1 \le \lambda_2$ . Let $\mathcal{H} = \{h\,:\, \mathbb{R}^+_{>0} \rightarrow \mathbb{R}, \,\|h\| \le 1\}$ . Then for all $h \in \mathcal{H}$ , letting $\tilde h(w) = h(w) - \mathbb{E} h(W)$ , we have from (4.15) with $\beta = 1$ that

(4.20) \begin{equation} \big|\mathbb{E} h(W_2)-\mathbb{E} h(W_1)\big| \le \|\tilde{h}\| \left\{ \left|\frac{\lambda_2 \theta_2}{\lambda_1 \theta_1} - 1 \right| + \left( \frac{\lambda_2}{\lambda_1} - 1\right) \left( \frac{\lambda_1 \theta_2}{\lambda_2 (1 - {\mathrm{e}}^{-\theta_2})} + 2\right) \right\}.\end{equation}

4.3. Approximating the Poisson–geometric distribution

Next we consider the distribution of $W_G = \max\{T_1, \ldots, T_M\}$ , where $T_1, T_2, \ldots \in \{1, 2, \ldots\}$ are i.i.d. Geometric(p) random variables and M is an independent zero-truncated Poisson( $\theta$ ) random variable; the distribution of $W_G$ is the Poisson–geometric (PG) distribution; it has PMF

(4.21) \begin{align} \mathbb{P}(W_G=w) = \frac{{\mathrm{e}}^{-(1-p)^w\theta}- {\mathrm{e}}^{-(1-p)^{w-1}\theta}}{(1-{\mathrm{e}}^{-\theta})}, \quad {w} \in \mathbb{N} \end{align}

and the discrete backward score function

\begin{align*}\rho_{W_{G}}(w) = 1- \frac{{\mathrm{e}}^{-(1-p)^{w-1}\theta}- {\mathrm{e}}^{-(1-p)^{w-2}\theta}}{{\mathrm{e}}^{-(1-p)^{w}\theta}- {\mathrm{e}}^{-(1-p)^{w-1}\theta}}, \quad w \in \mathbb{N}.\end{align*}

For $T \sim \mathrm{Geometric}(\lambda/n)$ , the distribution of $n^{-1} {T}$ converges to $\mathrm{Exp}(\lambda)$ in probability, and hence it is plausible to approximate the distribution of $Z_n={W_{G,n}}/n$ , for ${W_{G,n}} \sim \mathrm{PG}(\theta, \lambda/n)$ , by a corresponding PE distribution. With $q_n = 1-\lambda/n $ and $\tilde{\rho}_n (z) = \rho_{W_G} (nz)$ ,

\begin{align*} \tilde{\rho}_n (z) = 1- \frac{{\mathrm{e}}^{-q_n^{nz-1}\theta}- {\mathrm{e}}^{-q_n^{nz-2}\theta}}{{\mathrm{e}}^{-q_n^{nz}\theta}- {\mathrm{e}}^{-q_n^{nz-1}\theta}}, \quad nz \in \mathbb{N}.\end{align*}

This function is the ratio of two exponential functions, complicating the comparison using (2.6). To simplify the comparison, we use Proposition 2. From (3.4), a standardised PG Stein operator for $Z_n$ is

(4.22) \begin{equation} {\mathcal{T}_{Z_n} ^{(d)}} (g(z)) = g(z) d(z) - g \!\left( z - \frac1n \right) d\! \left( z - \frac1n \right) +\tilde{\rho}_n(z) g \!\left( z - \frac1n \right)d \!\left( z - \frac1n \right);\end{equation}

here we choose

(4.23) \begin{equation} d(z) = {\mathrm{e}}^{-q_n^{nz + 1}\theta}- {\mathrm{e}}^{-q_n^{nz}\theta}= {\mathrm{e}}^{ -\theta {q_n^{nz}}} \bigl( {\mathrm{e}}^{ -\theta (q_n-1) q_n^{nz}} -1 \bigr)\end{equation}

so that $ \tilde{\rho}_n(z) d\left( z - \frac1n \right) = {\mathrm{e}}^{-q_n^{nz }\theta}- 2 {\mathrm{e}}^{-q_n^{nz-1}\theta} + {\mathrm{e}}^{-q_n^{nz -2}\theta}.$ As $nd(z)\rightarrow \lambda \theta {\mathrm{e}}^{-\lambda z - \theta {\mathrm{e}}^{-\lambda z}}$ as $n \rightarrow \infty$ , for the approximating PE distribution we choose

\begin{align*} c(z) = \frac{\lambda \theta}{n}{\mathrm{e}}^{-\lambda z - \theta {\mathrm{e}}^{-\lambda z}}, \quad z > 0\end{align*}

as a standardisation function in (3.3), giving rise to the standardised PE Stein equation

(4.24) \begin{equation} c(w) g'(w) + \big[c(w) \lambda(\theta {\mathrm{e}}^{-\lambda w}-1)+ c'(w)\big] g(w) = h(w) - \mathbb{E} h(W)\end{equation}

for $W \sim \mathrm{PE}(\theta, \lambda)$ . Again we can bound the solution of this Stein equation as follows.

Lemma 2. For $W \sim \mathrm{PE}(\theta, \lambda)$ , the solution g(w) of the Stein equation (4.24), bounded differentiable h such that $\|h\| \le 1$ and $\|h'\| \le 1$ with $\tilde{h} (x) = h(x) - \mathbb{E} h(W)$ , and $ c(w) = \frac{\lambda \theta}{n}\,{\mathrm{e}}^{-\lambda {w} - \theta {\mathrm{e}}^{-\lambda {w}}}$ , we have

(4.25) \begin{align} \big|{\mathrm{e}}^{-2\lambda w - \theta {\mathrm{e}}^{-\lambda w}} g(w)\big| &\le \frac{n}{\lambda^2\theta^2} \|\tilde{h}\|, \end{align}

(4.26) \begin{align} \big|{\mathrm{e}}^{-\lambda w - \theta {\mathrm{e}}^{-\lambda w}} (\theta {\mathrm{e}}^{-\lambda w} - 1)g(w) \big| &\le \frac{n}{\lambda^2 \theta}\|\tilde{h}\|, \end{align}

(4.27) \begin{align} \big|{\mathrm{e}}^{-\lambda w - \theta {\mathrm{e}}^{-\lambda w}}g(w)\big| &\le 2\frac{n}{\lambda^2\theta} \|\tilde{h}\|,\end{align}

(4.28) \begin{align} \big|{\mathrm{e}}^{-\lambda w - \theta {\mathrm{e}}^{-\lambda w}}g'(w) \big| &\le 3 \frac{n}{\lambda \theta} \|\tilde{h}\|, \end{align}

(4.29) \begin{align} \bigg|\frac{\lambda \theta}{n}{\mathrm{e}}^{-\lambda w - \theta {\mathrm{e}}^{-\lambda w}}g''(w)\bigg| &\le \|h'\| + 9 \lambda\theta \|\tilde{h}\| + 11 \lambda \|\tilde{h}\|, \\[8pt]\nonumber \end{align}

and

(4.30) \begin{equation} \frac{c(x)}{ c\left( x + \frac{\rho}{n}\right)} \le \begin{cases}\! {\mathrm{e}}^{\theta ({\mathrm{e}}^{ \frac{\lambda}{n}} -1)}, & \rho = -1, \\ {\mathrm{e}}^{\frac{\lambda}{n}}, & 0 < \rho <1. \end{cases}\end{equation}

Proof. In (4.24), $cg = f$ is the solution of the Stein equation (4.4), so we use the bounds for f to bound g. The bound (4.6) in Lemma 1 immediately gives (4.25). Also $c'(w) = c(w) \lambda (\theta {\mathrm{e}}^{-\lambda w} - 1) $ so that $c'(w)g(w) = \lambda (\theta {\mathrm{e}}^{-\lambda w} - 1) f(w)$ ; using (4.7) we get (4.26). Combining (4.25), (4.26), and the triangle inequality gives (4.27).

Since $(cg)' = cg' + c'g$ , rearranging and using (4.9) and (4.7) with the triangle inequality gives (4.28). For $(cg)'' - c''g - 2c'g' = cg''$ , using (4.28) we have

\begin{align*} |c'(w)g'(w)| \le \lambda|\theta {\mathrm{e}}^{-\lambda w} - 1|c(w)g'(w) \le 3\lambda(\theta+1) \|\tilde{h}\|.\end{align*}

For $c''(w)g(w) = \lambda^2(\theta {\mathrm{e}}^{-\lambda w} -1)^2 c(w)g(w) - \lambda^2\theta {\mathrm{e}}^{-\lambda w}c(w)g(w)$ , using the triangle inequality, (4.26), and (4.25) yields

\begin{align*} |c''(w)g(w)| \le \lambda|\theta {\mathrm{e}}^{-\lambda w} - 1|\|\tilde{h}\| + \lambda \|\tilde{h}\| \le \lambda\theta \|\tilde{h}\| + 2\lambda \|\tilde{h}\|.\end{align*}

These two results along with (4.10) give (4.29). Now, for any $0 < \rho < 1$ ,

\begin{align*} \frac{c(x)}{ c\left( x + \frac{\rho}{n}\right)} = {\mathrm{e}}^{\lambda \frac{\rho}{n} - \theta {\mathrm{e}}^{-\lambda x} (1- {\mathrm{e}}^{-\lambda \frac{\rho}{n}}) } \le {\mathrm{e}}^{\lambda \frac{\rho}{n}};\end{align*}

for $\rho = -1$ we have $\frac{c(x)}{ c\left( x - \frac{1}{n}\right)} = {\mathrm{e}}^{- \frac{\lambda}{n} + \theta {\mathrm{e}}^{-\lambda x} ({\mathrm{e}}^{ \frac{\lambda}{n}} -1) } \le {\mathrm{e}}^{ \theta ({\mathrm{e}}^{\frac{\lambda }{n}} -1)}$ , yielding (4.30).

Theorem 2. Let $W_{G,n} \sim \mathrm{PG}(\theta, p_n) $ with $p_n = \lambda/n$ where $ 0< \lambda < n$ , and let $W \sim \mathrm{PE}(\theta,\lambda)$ . Then for the scaled PG random variable $Z_n = W_{G,n}/n$ and any bounded function h with bounded first derivative, we have

(4.31) \begin{align} &\big|\mathbb{E} h(Z_n) - \mathbb{E} h(W)\big| \nonumber\\ & \le \frac{{\mathrm{e}}^{\frac{\lambda}{n}}}{n} \biggl(1+\frac{{\mathrm{e}}^{ \frac{\theta\lambda}{n {\mathrm{e}}}}}{n}B_1(\theta, \lambda)\biggr)\|h'\| \nonumber \\ & \quad + \frac{1}{n} \left(\left(1-\frac{\lambda}{n}\right)^{-6} {\mathrm{e}}^{\theta {\mathrm{e}}^{ \frac{\lambda}{n}} + \frac{\theta\lambda}{n} + 2\frac{\theta\lambda}{n {\mathrm{e}}} + \frac{\lambda}{n}}B_2(\theta, \lambda) + \frac{1}{2n} {\mathrm{e}}^{\frac{\lambda}{n} + \frac{\theta\lambda}{n {\mathrm{e}}}}B_3(\theta, \lambda)\right)\|\tilde{h}\| , \end{align}

where

\begin{align*} B_1(\theta, \lambda) &= \lambda \max(1,\theta) + \frac{{\mathrm{e}}^{\lambda \theta + {\frac{\theta}{{\mathrm{e}}}}}}{2\theta\lambda } + \frac{\theta\lambda}{2{\mathrm{e}}}\frac{(1+{\mathrm{e}}-{\mathrm{e}}^{-\theta})}{(1-{\mathrm{e}}^{-\theta})} ,\\B_2(\theta, \lambda) &= 11 \lambda + 9 \lambda \theta + 6\lambda \max(1,\theta) + \frac{3{\mathrm{e}}^{\lambda \theta + {\frac{\theta}{{\mathrm{e}}}}}}{\theta \lambda} + 3\theta\lambda \frac{(1+{\mathrm{e}}-{\mathrm{e}}^{-\theta})}{(1-{\mathrm{e}}^{-\theta}){\mathrm{e}}} \nonumber \\&\quad + 4\lambda {\mathrm{e}}^{-\theta}\left(1+ 2 \max(1,\theta) + \frac{(3{\mathrm{e}} + 1) \theta}{{\mathrm{e}}}+ \frac{(5{\mathrm{e}} + 3)\theta^2}{3{\mathrm{e}}} + \frac{\theta(2\theta +1)}{(1-{\mathrm{e}}^{-\theta})} \right),\\B_3(\theta, \lambda) & = (9 \lambda \theta + 11 \lambda ) \left(2\lambda \max(1,\theta) + \frac{{\mathrm{e}}^{\lambda \theta + {\frac{\theta}{{\mathrm{e}}}}}}{\theta \lambda} + \theta\lambda \frac{(1+{\mathrm{e}}-{\mathrm{e}}^{-\theta})}{(1-{\mathrm{e}}^{-\theta}){\mathrm{e}}} \right).\end{align*}

Proof. We employ Proposition 2. First we note that $ c(w) \rho(w) = c'(w)$ and that $ {\tilde \rho}_n (w) d\bigl( w - \frac1n \bigr)= d \bigl( w - \frac1n \bigr) - d \bigl( w - \frac2n \bigr)$ . Thus, for (3.6), with $h \in \textrm{Lip}_b(1)$ we have

(4.32) \begin{align} & {\big|\mathbb{E} h(Z_n) - \mathbb{E} h(W)\big|} \nonumber \\ & \le \bigg| \mathbb{E} [ n \Delta^{-n} (dg)(Z_n) - (cg)'(Z_n) ] + \mathbb{E}\Big[ n {\tilde \rho}_n (Z_n) (dg)\Bigl( Z_n - \frac1n \Bigr) - (cg)\left( Z_n\right) \rho(Z_n)\Big] \bigg| \nonumber \\ & \le \mathbb{E}\left| n \left\{ g(Z_n) - g \left( Z_n - \frac1n \right)\right\} c(Z_n) - g'(Z_n) c(Z_n) \right| \end{align}

(4.33) \begin{align} & \quad + \mathbb{E} \left| n \left\{ g(Z_n) - g \left( Z_n - \frac1n \right)\right\} \left(d(Z_n) - c(Z_n) -\frac{2}{n} c'(Z_n)\right) \right|\end{align}

(4.34) \begin{align} & \quad + \mathbb{E} \left| d(Z_n) - d \left( Z_n - \frac2n \right) - \frac{2}{n} c'(Z_n)\right| \left|ng\left( Z_n - \frac1n \right)\right| .\\[8pt]\nonumber\end{align}

To bound the term (4.32), for some $0 < \rho < 1$ we write

\begin{align*} \mathbb{E}\left| n \left\{ g(Z_n) - g \left( Z_n - \frac1n \right)\right\} c(Z_n) - g'(Z_n) c(Z_n) \right| & \le \frac1n \,\mathbb{E} \left|c(Z_n) g''\left( Z_n + \frac{\rho}{n}\right) \right| \\ &\le \frac1n \| c g''\| \,\mathbb{E} \left| \frac{c(Z_n)}{ c\left( Z_n + \frac{\rho}{n}\right)} \right|.\end{align*}

Then (4.29) and (4.30) give the bound

(4.35) \begin{align} \mathbb{E}\left| n \{ g(Z_n) - g ( Z_n - n^{-1})\} c(Z_n) - g'(Z_n) c(Z_n) \right| \le \frac{{\mathrm{e}}^{\frac{\lambda}{n}}}{n} \big\{ \| h'\| + 9 \lambda \theta \| {\tilde h}\| + 11 \lambda \| {\tilde h}\|\big\}. \nonumber \\\end{align}

To bound (4.33), we let $\tau(z) = {\mathrm{e}}^{\lambda z + \theta {\mathrm{e}}^{-\lambda z}}$ , so that $\tau^{-1} g = \frac{n}{\theta\lambda}cg$ can be bounded as in Lemma 2, and write

\begin{align*} &\mathbb{E} \left| n \left\{ g(Z_n) - g \left( Z_n - \frac1n \right)\right\} \left(d(Z_n) - c(Z_n) -\frac{2}{n} c'(Z_n)\right) \right| \\ & \quad = \mathbb{E} \left| l(Z_n) n \left\{ g(Z_n) - g \left( Z_n - \frac1n \right)\right\}\tau^{-1}(Z_n) \right|,\end{align*}

where $l(Z_n) = \tau(Z_n)\bigl(d(Z_n) - c(Z_n) -\frac{2}{n} c'(Z_n)\bigr)$ . We show in Appendix that

(4.36) \begin{align} |l(Z_n)| \le \frac{2\lambda^2 \theta}{n^2} \max(1,\theta) + \frac{1}{n^2} {\mathrm{e}}^{\lambda \theta + {\frac{\theta}{{\mathrm{e}}}}} + \frac{\theta^2\lambda^2}{n^2 } {\mathrm{e}}^{ \frac{\theta\lambda}{n {\mathrm{e}}}-1} + \frac{\theta\lambda^3Z_n}{n^2} {\mathrm{e}}^{ \frac{\theta\lambda}{n {\mathrm{e}}}}.\end{align}

By Taylor expansion, for some $0 < \epsilon < 1$ we have

(4.37) \begin{align} n \left\{ g(Z_n) - g \left( Z_n - \frac1n \right)\right\}\tau^{-1}(Z_n) &= g'(Z_n)\tau^{-1}(Z_n) + \frac{1}{2n}g''\left(Z_n + \frac{\epsilon}{n}\right)\tau^{-1}(Z_n) \nonumber \\ &= \frac{n}{\theta\lambda}\|cg'\| + \frac{1}{2\theta\lambda} \|cg''\| \frac{c(Z_n)}{ c\left(Z_n + \frac{\epsilon}{n}\right)}. \end{align}

Using (4.36) and (4.37) along with (4.28), (4.29) and (4.30) yields

(4.38) \begin{align} &{\mathbb{E} \left| n \left\{ g(Z_n) - g \left( Z_n - \frac1n \right)\right\} \left(d(Z_n) - c(Z_n) -\frac{2}{n}c'(Z_n)\right) \right|} \nonumber \\ &\le \left(\frac{2\lambda^2 \theta}{n^2} \max(1,\theta) + \frac{1}{n^2} {\mathrm{e}}^{\lambda \theta + {\frac{\theta}{{\mathrm{e}}}}}+ \frac{\theta^2\lambda^2}{n^2 } {\mathrm{e}}^{ \frac{\theta\lambda}{n {\mathrm{e}}}-1} + \frac{\theta\lambda^3 \mathbb{E} Z_n}{n^2} {\mathrm{e}}^{ \frac{\theta\lambda}{n {\mathrm{e}}}}\right) \nonumber \\ &\quad\times \left\{ \frac{n}{\theta\lambda}3\|\tilde{h}\| + \frac{1}{2\theta\lambda} \{ \| h'\| + 9 \lambda \theta \| {\tilde h}\| + 11 \lambda \| {\tilde h}\|\} {\mathrm{e}}^{\frac{\lambda}{n}} \right\} . \end{align}

Finally, to bound (4.34), we show in Appendix that

(4.39) \begin{align} & \left|\frac{2}{n} c'(z) - d(z) + d \left( z - \frac2n \right)\right|\left|ng\left( z - \frac1n\right)\right| \\ &\le 2\lambda \left\{ \left(1-\frac{\lambda}{n}\right)^{-2} -1 \right\} \max(1,\theta) c(z) \left| g\left( z - \frac1n\right)\right| \nonumber \\ &\quad + \frac{\theta\lambda^2}{n} \left(1-\frac{\lambda}{n}\right)^{-2} {\mathrm{e}}^{{\frac{\theta\lambda}{n {\mathrm{e}}}}} \Biggl\{ \frac{1}{3} \left(1-\frac{\lambda}{n}\right)^{-1} \nonumber \biggl[\theta + \theta {\mathrm{e}}^{\frac{\theta\lambda}{n}} + 6 {\mathrm{e}}^{\frac{\theta\lambda}{n}} + 12 \left(1-\frac{\lambda}{n}\right)^{-1} \\ & \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad\; + 8\theta \left(1-\frac{\lambda}{n}\right)^{-3}\biggr] + 2\left({\frac{\theta}{{\mathrm{e}}}} + 2\lambda z \right)\Biggr\}\nonumber c(z) \left| g\left( z - \frac1n\right)\right| \\ &\quad + \frac{2\lambda^2}{n}\left( 1- \frac{\lambda}{n}\right)^{-2}{\mathrm{e}}^{{\frac{\theta\lambda}{n {\mathrm{e}}}}}\left[ \left(1-\frac{\lambda}{n}\right)^{-1}{\mathrm{e}}^\frac{\theta \lambda}{n} + \lambda z + {\frac{\theta}{{\mathrm{e}}}} \right] c(z) \left|g\left( z - \frac1n\right)\right|. \nonumber\end{align}

Note that

(4.40) \begin{equation} 0 \le \left(1-\frac{\lambda}{n}\right)^{-2} -1 = \frac{\lambda}{n} \left(1-\frac{\lambda}{n}\right)^{-2}\left( 2 - \frac{\lambda}{n} \right)\end{equation}

and that, using (4.27) and (4.30),

\begin{align*}c(z) \left|g\left( z - \frac1n\right)\right|= \left|c(z) g\left( z - \frac1n\right)\right| \le \|cg\|\left| \frac{c(z)}{ c\left( z - \frac{1}{n}\right)} \right| \le \frac{2}{\lambda} {\mathrm{e}}^{\theta ({\mathrm{e}}^{ \frac{\lambda}{n}} -1)}\|\tilde{h}\|.\end{align*}

Combining this with (4.39) and (4.40), we bound (4.34) as

(4.41) \begin{align} & \mathbb{E} \left| d(Z_n) - d \left( Z_n - \frac2n \right) - \frac{2}{n} c'(Z_n)\right| \left|ng\left( Z_n - \frac1n \right)\right| \\ & \le \frac{4 \lambda}{n}\max(1,\theta) \left(1-\frac{\lambda}{n}\right)^{-2}\left|\frac{\lambda}{n} - 2 \right| {\mathrm{e}}^{\theta ({\mathrm{e}}^{ \frac{\lambda}{n}} -1)}\|\tilde{h}\| + \frac{2\theta\lambda}{n} \left(1-\frac{\lambda}{n}\right)^{-2} {\mathrm{e}}^{{\frac{\theta\lambda}{n {\mathrm{e}}}}} {\mathrm{e}}^{\theta ({\mathrm{e}}^{ \frac{\lambda}{n}} -1)}\|\tilde{h}\| \nonumber \\ &\quad \times \bigg\{ \frac{1}{3} \left(1-\frac{\lambda}{n}\right)^{-1} \left[\theta + \theta {\mathrm{e}}^{\frac{\theta\lambda}{n}} + 6 {\mathrm{e}}^{\frac{\theta\lambda}{n}} + 12 \left(1-\frac{\lambda}{n}\right)^{-1} + 8\theta \left(1-\frac{\lambda}{n}\right)^{-3}\right] \nonumber \\ &\qquad + 2\left({\frac{\theta}{{\mathrm{e}}}} + 2\lambda \mathbb{E} Z_n \right) + \frac{2}{\theta} \bigg[ \left(1-\frac{\lambda}{n}\right)^{-1}{\mathrm{e}}^{\frac{\theta \lambda}{n}} + \lambda \mathbb{E} Z_n + {\frac{\theta}{{\mathrm{e}}}} \bigg] \bigg\}. \nonumber\end{align}

To bound $\mathbb{E} Z_n $ , we argue as for (4.13); $W_{G,n} = \max \{ T_1, \ldots, T_M\} \le \sum_{i=1}^{M} T_i$ and so $ \mathbb{E} {W_{G,n}} \le \mathbb{E} \sum_{i=1}^{M} T_i = \frac{n\theta}{\lambda(1-{\mathrm{e}}^{-\theta})}$ , giving that $ \mathbb{E} Z_n = \frac{1}{n} \mathbb{E} W_{G,n}\le \frac{\theta}{\lambda(1-{\mathrm{e}}^{-\theta})}.$ Adding (4.35), (4.38), and (4.41) and simplifying gives (4.31).

The next result instead uses Proposition 3 to bound the distance between a PG and a PE distribution.

Corollary 4. Let $W = W_1(M,{\boldsymbol{E}}) \sim \mathrm{PE}(\theta, \lambda)$ and ${W_{G,n}} = W_1(M,{\boldsymbol{G}}) \sim \mathrm{PG}(\theta, \frac{\lambda}{n})$ with $G_i \sim \mathrm{Geometric}(\lambda/n)$ be two CCR random variables, and let $Z_n = \frac{{W_{G,n}}}{n}$ . Then, for all bounded Lipschitz functions $h\,:\, \mathbb{R}_{>0}^{+} \rightarrow \mathbb{R}$ ,

(4.42) \begin{equation} \big| \mathbb{E} h(W) - \mathbb{E} h (Z_n ) \big| \le \frac{\theta}{1 - {\mathrm{e}}^{-\theta}} \frac{2}{n}\left(\frac{n - {\mathrm{e}}^{\frac{\lambda}{n}}(n-\lambda)}{\lambda({\mathrm{e}}^{\frac{\lambda}{n}}-1)}\right)\|h'\| .\end{equation}

Proof. Using (3.9) in Proposition 3 gives

\begin{align*} \big| \mathbb{E} h(W) - \mathbb{E} h (Z_n ) \big| = \left| \mathbb{E} h(W) - \mathbb{E} h \left(\frac{W_{G,n}}{n} \right) \right| &\le \frac{\theta}{1 - {\mathrm{e}}^{-\theta} } \,\mathbb{E} \left|E - \frac{G}{n}\right| \| h'\|.\end{align*}

With the coupling $\tilde{G} = \lceil n{E}\rceil \sim \mathrm{Geometric}(1-{\mathrm{e}}^{-\lambda/n})$ , $\tilde{G}$ is stochastically greater than or equal to G and $\mathbb{E} |n{E} - G| \le \mathbb{E} |n{E}-\tilde{G}| + \mathbb{E} (\tilde{G} - G)$ . Moreover,

\begin{align*} \mathbb{E} |nE - \tilde{G}| = \sum_{k=0}^{\infty} \int^{\frac{k}{n}}_{\frac{(k-1)}{n}} \lambda {\mathrm{e}}^{-\lambda x} \left(k - nx \right) {\mathrm{d}} x = \frac{n - {\mathrm{e}}^{\frac{\lambda}{n}}(n-\lambda)}{\lambda({\mathrm{e}}^{\frac{\lambda}{n}}-1)}\end{align*}

and $\mathbb{E}(\tilde{G} - G) = ({1-{\mathrm{e}}^{-\frac{\lambda}{n}}})^{-1} - {n}/{\lambda}.$ Hence the bound (4.42) follows.

Table 1. Values of n above which the coefficient of $\|h'\|$ is smaller in (4.31) than in (4.42)

5.1. The minimum of a geometric number of i.i.d. random variables

Let $N\in \mathbb{N}$ be a random variable which is independent of ${\textbf{X}} = (X_1, X_2, \ldots)$ , a sequence of i.i.d. random variables, and let $W=W_{-1}(N, {\textbf{X}})$ have PDF of the form (2.12) for $\alpha = -1$ . In this subsection we approximate its distribution by an EG distribution. First we apply Proposition 1.

Corollary 5. Assume that the $X_i$ have CDF $F_X$ , differentiable PDF $p_X$ , and score function $\rho_X$ , and let $N \sim \mathrm{Geometric}(p)$ . Then

\begin{align*} d_{\rm TV}\bigl(\text{EG}(\lambda,p),\mathcal{L}(W)\bigr)\le \frac{2}{\lambda}\,\biggl|\, & \mathbb{E} (\! -\lambda- \rho_X(W)) \\ & - \mathbb{E} \left( \frac{2(1-p)\lambda {\mathrm{e}}^{-\lambda W}}{1-(1-p){\mathrm{e}}^{-\lambda W}} - \frac{2(1-p)f_X(W)}{1-(1-p)(1-F_X(W))} \right) \biggr|.\end{align*}

Next, we instead use Proposition 3; the corollary follows immediately from (3.9).

Corollary 6. For $W = W_{-1}(N, {\boldsymbol{X}})$ with $N \sim \mathrm{Geometric}(p)$ ,

\begin{align*} d_{\rm BW}\bigl(\mathrm{EG}(\lambda,p), \mathcal{L}(W)\bigr) \le {\mathbb{E} | E - X|}/{p} . \end{align*}

5.2. Approximating the extended exponential–geometric distribution

Motivated by population heterogeneity, Adamidis et al. [Reference Adamidis, Dimitrakopoulou and Loukas2] developed the extended exponential–geometric (EEG) distribution by assuming that individual units in a population have increasing failure rates that depend on a random scale parameter A. Their lifetimes $X/A$ are modelled by a modified extreme value distribution; if $A = \alpha$ , then the PDF is $f(x|\alpha;\beta) = \alpha \beta {\mathrm{e}}^{\beta x + \alpha(1-{\mathrm{e}}^{\beta x})}$ , where $x, \alpha, \beta \in \mathbb{R}^+_{>0}$ ; it is assumed that A has an $\mathrm{Exp}(\gamma)$ distribution. Then the unconditional lifetime distribution X has PDF

(5.7) \begin{equation} f(x;\,\beta, \gamma) = \frac{\beta \gamma {\mathrm{e}}^{-\beta x}} {(1- (1-\gamma) {\mathrm{e}}^{-\beta x})^2}, \quad x>0,\end{equation}

with $\beta, \gamma \in \mathbb{R}^+_{>0}$ ; we use the notation $X \sim \mathrm{EEG}(\beta, \gamma)$ . Its score function is

(5.8) \begin{equation} \rho(x) = \frac{f'(x)}{f(x)} = - \beta \left(\frac{1+(1-\gamma){\mathrm{e}}^{-\beta x}}{1-(1-\gamma){\mathrm{e}}^{-\beta x}}\right).\end{equation}

This distribution is not in the CCR family. However, the EG distribution $\mathrm{EG}(\beta, \gamma)$ is a special case when $\gamma \in (0,1)$ . To assess the total variation distance between the distributions, we use the general approach developed in Section 2.

Theorem 3. For $X \sim \mathrm{EEG}(\beta, \gamma)$ and $W \sim \mathrm{EG}(\lambda, p)$ , with $\beta, \gamma, \lambda \in \mathbb{R}^+_{>0}$ and $p \in (0,1)$ , and for a bounded test function h, we have

(5.9) \begin{align} \big|\mathbb{E} h(X)-\mathbb{E} h(W)\big| \le \begin{cases} \frac{\|\tilde{h}\|}{\lambda}\left[| \lambda - \beta| \left( \frac{2-p}{p} + \frac{2(1-\gamma)}{{\mathrm{e}} \gamma^2} \right) + \frac{2\beta} {\min(\gamma, p)^2} | p - \gamma| \right], & 0 < \gamma <1 ,\\ \frac{\|\tilde{h}\|}{\lambda p } \left[\left|\lambda - \beta\right|\left(1+(\gamma-1)(1-p)\right) + (\lambda+\beta)\left(\gamma -p \right)\right], & \gamma \ge 1. \end{cases}\end{align}

Proof. Using the score functions (5.2) and (5.8) in (2.6) yields

(5.10) \begin{align} & |\mathbb{E} h(X)-\mathbb{E} h(W)| = |\mathbb{E}(\rho_X(X)-\rho_Y(X))g(X)| \nonumber \\ &\quad= \left|\mathbb{E} \left(\!- \beta \left(\frac{1+(1-\gamma){\mathrm{e}}^{-\beta X}}{1-(1-\gamma){\mathrm{e}}^{-\beta X}}\right)+ \lambda \left(\frac{1+ (1-p){\mathrm{e}}^{-\lambda X}}{1- (1-p){\mathrm{e}}^{-\lambda X}}\right)\right)g(X)\right| \nonumber \\ &\quad\le \frac{\|\tilde{h}\|}{\lambda} \mathbb{E} \left| - \beta \left(\frac{1+(1-\gamma){\mathrm{e}}^{-\beta X}}{1-(1-\gamma){\mathrm{e}}^{-\beta X}}\right)+ \lambda \left(\frac{1+ (1-p){\mathrm{e}}^{-\lambda X}}{1- (1-p){\mathrm{e}}^{-\lambda X}}\right)\right|. \end{align}

To bound the expectation in the above equation, we let

\begin{align*}R(x) = - \beta \left(\frac{1+(1-\gamma){\mathrm{e}}^{-\beta x}}{1-(1-\gamma){\mathrm{e}}^{-\beta x}}\right)+ \lambda \left(\frac{1+ (1-p){\mathrm{e}}^{-\lambda x}}{1- (1-p){\mathrm{e}}^{-\lambda x}}\right).\end{align*}

Case 1: $0 < \gamma <1$ . In this case we decompose $|R(x)|$ as

(5.11) \begin{align} |R(x)| & \le \left|(\lambda - \beta) \frac{1+ (1-p){\mathrm{e}}^{-\lambda x}}{1- (1-p){\mathrm{e}}^{-\lambda x}}\right| \end{align}

(5.12) \begin{align} &\quad + \left|\beta \left( \frac{1+ (1-p){\mathrm{e}}^{-\lambda x}}{1- (1-p){\mathrm{e}}^{-\lambda x}} - \frac{1+ (1-\gamma){\mathrm{e}}^{-\lambda x}}{1- (1-\gamma){\mathrm{e}}^{-\lambda x}}\right) \right| \end{align}

(5.13) \begin{align} &\quad + \left|\beta \left( \frac{1+ (1-\gamma){\mathrm{e}}^{-\lambda x}}{1- (1-\gamma){\mathrm{e}}^{-\lambda x}} - \frac{1+ (1-\gamma){\mathrm{e}}^{-\beta x}}{1- (1-\gamma){\mathrm{e}}^{-\beta x}}\right) \right| \\[8pt]\nonumber\end{align}

and bound these terms separately. For (5.11) we use that $ \frac{1 + \alpha q}{1 - \alpha q} \le \frac{1+q}{1-q}$ when $\alpha \in [0,1]$ and $q \in (0,1).$ Hence

\begin{eqnarray*} \left\vert (\lambda - \beta) \frac{1+ (1-p){\mathrm{e}}^{-\lambda x}}{1- (1-p){\mathrm{e}}^{-\lambda x}}\right\vert &\le& | \lambda - \beta| \frac{2-p}{p}.\end{eqnarray*}

For (5.12), Taylor expansion about $1-\gamma$ of the function $f(q) = \frac{1+aq }{1-aq} $ with $a = {\mathrm{e}}^{-\lambda x} \in (0,1)$ and $q=1-p$ gives $f'(q) = \frac{a}{1 - aq} + \frac{a(1+aq)}{(1-aq)^2} = \frac{2a}{(1-aq)^2}>0$ . Moreover, for $0< a< 1$ we have $ \frac{2a}{(1-aq)^2} \le \frac{2}{(1-q)^2}$ , and hence for $\theta \in (0,1)$ we have $0< f'(\theta (1-p) + (1- \theta) (1-\gamma)) \le \frac{2}{(1- \max(1-\gamma, 1-p))^2} = \frac{2}{(\min (\gamma, p))^2}$ . Therefore,

\begin{eqnarray*} \left\vert \beta \left( \frac{1+ (1-p){\mathrm{e}}^{-\lambda x}}{1- (1-p){\mathrm{e}}^{-\lambda x}} - \frac{1+ (1-\gamma){\mathrm{e}}^{-\lambda x}}{1- (1-\gamma){\mathrm{e}}^{-\lambda x}}\right)\right\vert &\le& \beta \frac{2| p - \gamma|}{(\min (\gamma, p))^2}.\end{eqnarray*}

For (5.13), first-order Taylor expansion of the function $f(\beta) = \frac{1+(1-\gamma) {\mathrm{e}}^{-\beta x}}{1-(1-\gamma) {\mathrm{e}}^{-\beta x}} $ gives

\begin{eqnarray*}\left\vert \beta \left( \frac{1+ (1-\gamma){\mathrm{e}}^{-\lambda x}}{1- (1-\gamma){\mathrm{e}}^{-\lambda x}} - \frac{1+ (1-\gamma){\mathrm{e}}^{-\beta x}}{1- (1-\gamma){\mathrm{e}}^{-\beta x}}\right) \right\vert&\le& \beta | \lambda - \beta|\, | f'(\theta \lambda + (1-\theta) \beta)|\end{eqnarray*}

for some $\theta \in (0,1).$ Now the function $f'(\beta) = \frac{{2 x (1-\gamma)} {\mathrm{e}}^{-\beta x}}{(1 - (1-\gamma) {\mathrm{e}}^{-\beta x})^2}$ is positive; moreover, $x {\mathrm{e}}^{-\beta x} \le ({\mathrm{e}} \beta)^{-1}$ for $x \ge 0$ , so that $f'(\beta) \le \frac{1}{\beta {\mathrm{e}}} \frac{2(1-\gamma)}{ (1 - (1-\gamma) {\mathrm{e}}^{-\beta x})^2} \le \frac{2(1-\gamma)}{{\mathrm{e}} \beta \gamma^2}.$ Hence we can bound

\begin{eqnarray*}\left\vert \beta \left( \frac{1+ (1-\gamma){\mathrm{e}}^{-\lambda x}}{1- (1-\gamma){\mathrm{e}}^{-\lambda x}} - \frac{1+ (1-\gamma){\mathrm{e}}^{-\beta x}}{1- (1-\gamma){\mathrm{e}}^{-\beta x}}\right) \right\vert&\le& \frac{2(1-\gamma)}{{\mathrm{e}} \gamma^2} | \lambda - \beta|.\end{eqnarray*}

This gives as overall bound

\begin{eqnarray*} | R(x) | &\le& | \lambda - \beta| \left( \frac{2-p}{p} + \frac{2(1-\gamma)}{{\mathrm{e}} \gamma^2} \right) + \frac{2\beta} {\min(\gamma, p)^2} | p - \gamma| .\end{eqnarray*}

Substituting this bound into (5.10) gives the final bound for the case where $0 < \gamma <1$ .

Case 2: For $\gamma \ge 1$ , we have

\begin{align*} |R(x)| & \le \left|\frac{(\lambda - \beta)(1-(1-\gamma)(1-p){\mathrm{e}}^{-(\lambda+\beta)x})}{\left(1-(1-\gamma){\mathrm{e}}^{-\beta x}\right) \left(1- (1-p){\mathrm{e}}^{-\lambda x}\right)}\right| \\ &\quad + \left|\frac{(\lambda+\beta)\left((1-p){\mathrm{e}}^{-\lambda x} - (1-\gamma) {\mathrm{e}}^{-\beta x}\right)}{\left(1-(1-\gamma){\mathrm{e}}^{-\beta x}\right) \left(1- (1-p){\mathrm{e}}^{-\lambda x}\right)}\right| \\ &\le \frac{\left|\lambda - \beta\right|\left(1+(\gamma-1)(1-p)\right)}{p} + \frac{(\lambda+\beta)\left(\gamma -p \right)}{p}.\end{align*}

Substituting this result into (5.10) gives the bound for $\gamma \ge 1$ in (5.9).