Appendix: Derivations
1.1 Derivation of formula (21)
To obtain the asymptotic variance of the approximate quantity \(\widetilde{C}_{1;\,r}(\hat{\alpha }_0,\hat{\alpha }_1)\) from (19), we start by defining the vectors
$$\begin{aligned} \varvec{Y}_t^{(r)} \!:= \!\Big ( \frac{(X_t)_{(r)}}{(\alpha _0\!+\!\alpha _1\, X_{t-1})^r} \!-\! 1, X_t \!-\! f_1,\ X_t^2 - f_2-f_1^2,\ X_t X_{t-1} - \alpha _1\, f_2-f_1^2 \Big )^{\top }\nonumber \\ \end{aligned}$$
(28)
with mean \(\varvec{0}\), and by deriving a central limit theorem for \((\varvec{Y}_t^{(r)})_{\mathbb {Z}}\).
Lemma 2
Let \((X_t)_{\mathbb {Z}}\) be a stationary INARCH(1) process, define \(\varvec{Y}_t^{(r)}\) as in formula (28). Denote \(f_k := \alpha _0 / \prod _{i=1}^k (1-\alpha _1^i)\) such that \(\mu =f_1\) and \(\sigma ^2=f_2\): Then
$$\begin{aligned} \begin{array}{l} \frac{1}{\sqrt{T}} \sum _{t=1}^T \varvec{Y}_t^{(r)} \mathop {\longrightarrow }\limits ^{\mathcal {D}} {\text {N}}\big ( \mathbf {0}, \varvec{\varSigma }^{(r)} \big ) \qquad \text {with } \varvec{\varSigma }^{(r)} = \big ( \sigma _{ij}^{(r)} \big ) \text { given by}\\ \sigma _{ij}^{(r)}\ =\ E\big [ Y_{0,i}^{(r)} Y_{0,j}^{(r)}\big ] + \sum _{k=1}^\infty \Big ( E\big [Y_{0,i}^{(r)} Y_{k,j}^{(r)}\big ] + E\big [ Y_{k,i}^{(r)} Y_{0,j}^{(r)}\big ] \Big ), \end{array} \end{aligned}$$
(29)
where \(Y_{k,i}^{(r)}\) denotes the i-th entry of \(\varvec{Y}_k^{(r)}\), and where the entries \(\sigma _{ij}^{(r)}\) of the symmetric matrix \(\varvec{\varSigma }^{(r)}\) are given as follows:
$$\begin{aligned} \begin{array}{l} \sigma _{11}^{(r)}\ =\ \sum _{k=1}^r\ \left( {\begin{array}{c}r\\ k\end{array}}\right) ^2\,k!\, q_{0,k}\qquad \text {(remember (15))}, \quad \sigma _{12}^{(r)}\ =\ \frac{r}{1-\alpha _1}, \\ \sigma _{13}^{(r)}\ =\ \frac{2r\,f_1}{1-\alpha _1} + \frac{r^2}{1-\alpha _1^2} + \frac{r\,\alpha _1}{(1-\alpha _1)(1-\alpha _1^2)}, \quad \sigma _{14}^{(r)}\ =\ \frac{2\,r\,f_1}{1-\alpha _1}\ +\ \frac{r^2\,\alpha _1}{1-\alpha _1^2}\ +\ \frac{r\,\alpha _1^2}{(1-\alpha _1)(1-\alpha _1^2)}, \end{array} \end{aligned}$$
and \(\displaystyle \sigma _{22}^{(r)}\ =\ \frac{f_1}{(1-\alpha _1)^2},\)
$$\begin{aligned} \sigma _{23}^{(r)}= & {} \frac{1+\alpha _1+2 \alpha _1^2}{(1-\alpha _1)(1-\alpha _1^2)}\, f_2 + \frac{2\, f_1^2}{(1-\alpha _1)^2},\\ \sigma _{24}^{(r)}= & {} \frac{\alpha _1(2+\alpha _1+\alpha _1^2)}{(1-\alpha _1)(1-\alpha _1^2)}\, f_2 + \frac{2\, f_1^2}{(1-\alpha _1)^2},\\ \sigma _{33}^{(r)}= & {} \frac{1+2 \alpha _1 +8 \alpha _1^2+9 \alpha _1^3+4 \alpha _1^4+6 \alpha _1^5}{(1-\alpha _1 ^2)^2}\, f_3\\&+ \frac{2(3+4 \alpha _1 +7 \alpha _1 ^2+4 \alpha _1 ^3)}{1-\alpha _1 ^2}\, f_2^2 + \frac{4\, f_1^3}{(1-\alpha _1)^2},\\ \sigma _{34}^{(r)}= & {} \frac{\alpha _1 (2+5 \alpha _1 +8 \alpha _1^2+10 \alpha _1^3+3 \alpha _1^4+2 \alpha _1^5)}{(1-\alpha _1^2)^2}\, f_3\\&+ \frac{2 (1+6 \alpha _1 +6 \alpha _1^2+4 \alpha _1^3+\alpha _1^4)}{1-\alpha _1^2}\, f_2^2 + \frac{4\, f_1^3}{(1-\alpha _1)^2},\\ \sigma _{44}^{(r)}= & {} \frac{\alpha _1(1+3 \alpha _1 +8 \alpha _1^2+8 \alpha _1^3+8 \alpha _1^4+2 \alpha _1^5)}{(1-\alpha _1^2)^2}\, f_3\\&+ \frac{1+8 \alpha _1 +16 \alpha _1^2+8 \alpha _1^3+3 \alpha _1^4}{1-\alpha _1^2}\, f_2^2 + \frac{4\, f_1^3}{(1-\alpha _1)^2}. \end{aligned}$$
Proof
With the same arguments as in Section 2 of Weiß and Schweer (2016), Theorem 1.7 of Ibragimov (1962) is applicable. Furthermore, the expressions for \(\sigma _{kl}^{(r)}\) with \(k,l\ge 2\) are already known from Theorem 2.2 in Weiß and Schweer (2016), and \(\sigma _{11}^{(r)}\) was derived before in the context of formula (11). Hence, to prove Lemma 2, it remains to compute the entries \(\sigma _{12}^{(r)}\), \(\sigma _{13}^{(r)}\) and \(\sigma _{14}^{(r)}\) of the asymptotic covariance matrix \(\varvec{\varSigma }^{(r)}\).
We start with some auxiliary expressions. We have
$$\begin{aligned} \begin{array}{rl} Q_1^{(r)} :=&E\Big [\frac{(X_t)_{(r)}\,X_t}{M_t^r}\Big ] \! =\! E\Big [\frac{E[(X_t)_{(r+1)}+r\,(X_t)_{(r)}\ |\ X_{t-1},\ldots ]}{M_t^r}\Big ] \! =\! E[M_t+r] \! =\! f_1+r. \end{array}\nonumber \\ \end{aligned}$$
(30)
Similarly, using that
$$\begin{aligned} E[M_t^2] = \alpha _0^2+2\alpha _0\alpha _1\,f_1+\alpha _1^2\,(f_2+f_1^2) = (\alpha _0+\alpha _1\,f_1)^2+\alpha _1^2\,f_2\ =\ f_1^2+\alpha _1^2\,f_2, \end{aligned}$$
it follows that
$$\begin{aligned} \begin{array}{rl} Q_2^{(r)} :=&{} E\Big [\frac{(X_t)_{(r)}\,X_t^2}{M_t^r}\Big ] \ =\ E\Big [\frac{E[(X_t)_{(r+2)}+(2r+1)\,(X_t)_{(r+1)}+r^2\,(X_t)_{(r)}\ |\ X_{t-1},\ldots ]}{M_t^r}\Big ] \\[1ex] =&{} E[M_t^2+(2r+1)\,M_t+r^2] \\[1ex] =&{} r^2+ f_1^2+\alpha _1^2\,f_2 + (2r+1)f_1 \ =\ r^2 + 2r\,f_1 + f_2 + f_1^2. \end{array}\nonumber \\ \end{aligned}$$
(31)
Finally,
$$\begin{aligned} \begin{array}{rl} Q_{1,1}^{(r)}\ :=&{} E\Big [\frac{(X_t)_{(r)}\,X_t X_{t-1}}{M_t^r}\Big ] \ =\ E\Big [\frac{X_{t-1}\,E[(X_t)_{(r+1)}+r\,(X_t)_{(r)}\ |\ X_{t-1},\ldots ]}{M_t^r}\Big ] \\[1ex] =&{} E\big [X_{t-1}\,(M_t+r)\big ] \ =\ (r+\alpha _0)\, f_1+\alpha _1\, (f_2+f_1^2) \\[1ex] =&{} r\,f_1+\alpha _1\,f_2+f_1\,(\alpha _0+\alpha _1\,f_1) \ =\ r\,f_1+\alpha _1\,f_2+f_1^2. \end{array}\nonumber \\ \end{aligned}$$
(32)
Now we can start with computing \(\sigma _{1j}^{(r)}\) for \(j=2,3,4\). For \(k\ge 1\), we always have
$$\begin{aligned} E[ Y_{k,1}^{(r)} Y_{0,j}^{(r)}]\ =\ E\big [E[ Y_{k,1}^{(r)} Y_{0,j}^{(r)}\ |\ X_{k-1}, \ldots ]\big ]\ =\ E\big [Y_{0,j}^{(r)}\,\underbrace{E[ Y_{k,1}^{(r)} \ |\ X_{k-1}, \ldots ]}_{=0}\big ]\ =\ 0.\nonumber \\ \end{aligned}$$
(33)
Let us compute \(\sigma _{12}^{(r)}\) first. For \(k\ge 1\), by conditioning and using that \(M_k=\alpha _0+\alpha _1\,X_{k-1}\), we have
$$\begin{aligned} \begin{array}{rl} E[Y_{0,1}^{(r)} Y_{k,2}^{(r)}]\ =&{} E\big [\frac{(X_0)_{(r)}}{M_0^r}\, X_k\big ]\ -\ f_1 \ =\ \alpha _1\,E\big [\frac{(X_0)_{(r)}}{M_0^r}\, X_{k-1}\big ] + \alpha _0\ -\ f_1 \\[1ex] =&{} \ldots \ =\ \alpha _1^k\,E\big [\frac{(X_0)_{(r)}}{M_0^r}\, X_{0}\big ] + \alpha _0\,(1+\alpha _1+\ldots +\alpha _1^{k-1})\ -\ f_1 \\[1ex] =&{} \alpha _1^k\,Q_1^{(r)} + \alpha _0\,\frac{1-\alpha _1^{k}}{1-\alpha _1}\ -\ f_1 \ =\ \alpha _1^k\,(Q_1^{(r)} - f_1) \ \overset{(30)}{=}\ \alpha _1^k\,r, \end{array} \end{aligned}$$
which also holds for \(k=0\). Together with (33), it follows that
$$\begin{aligned} \sigma _{12}^{(r)} \ =\ \sum _{k=0}^\infty E[Y_{0,1}^{(r)} Y_{k,2}^{(r)}] \ =\ \sum _{k=0}^\infty r\,\alpha _1^k \ =\ \frac{r}{1-\alpha _1}. \end{aligned}$$
Concerning \(\sigma _{13}^{(r)}\), first note that the 2nd non-central moment of the Poisson distribution implies
$$\begin{aligned} E[X_t^2\ |\ X_{t-1},\ldots ]\ =\ M_t^2+M_t \ =\ \alpha _1^2\,X_{t-1}^2+\alpha _1(2\alpha _0+1)\,X_{t-1}+\alpha _0(\alpha _0+1). \end{aligned}$$
Then we compute by successive conditioning that
$$\begin{aligned} E[Y_{0,1}^{(r)} Y_{k,3}^{(r)}]= & {} \alpha _1^2\,E\big [\frac{(X_0)_{(r)}}{M_0^r}\, X_{k-1}^2\big ] + \alpha _1(2\alpha _0+1)\,(r\,\alpha _1^{k-1}+f_1)\\&+ \alpha _0(\alpha _0+1)\ -\ f_2-f_1^2 \\= & {} \alpha _1^2\,E\big [\frac{(X_0)_{(r)}}{M_0^r}\, X_{k-1}^2\big ] + (2\alpha _0+1)\,r\,\alpha _1^{k}\\&+\,f_1\,\big (1+f_1(1-\alpha _1^2)\big )\ -\ f_2-f_1^2\\= & {} \ldots \ =\ \alpha _1^{2k}\,E\big [\frac{(X_0)_{(r)}}{M_0^r}\, X_{0}^2\big ] \\&+\, (2\alpha _0+1)\,r\,\alpha _1^{k}(1+\alpha _1+\ldots +\alpha _1^{k-1})\\&+\, f_1\,\big (1+f_1(1-\alpha _1^2)\big )(1+\alpha _1^2+\ldots +\alpha _1^{2(k-1)})\ -\ f_2-f_1^2 \\= & {} \alpha _1^{2k}\,Q_2^{(r)} + (2\alpha _0+1)\,r\,\alpha _1^{k}\,\frac{1-\alpha _1^{k}}{1-\alpha _1}\\&+\, (f_2+f_1^2)(1-\alpha _1^2)\,\frac{1-\alpha _1^{2k}}{1-\alpha _1^2}\ -\ f_2-f_1^2\\= & {} \alpha _1^{2k}\,\big (Q_2^{(r)}-r\,\frac{2\alpha _0+1}{1-\alpha _1}- f_2-f_1^2\big ) \\&+\, r\,\alpha _1^{k} \,\frac{2\alpha _0+1}{1-\alpha _1}\\&\overset{(31)}{=} r\,\alpha _1^{2k}\, (r-\frac{1}{1-\alpha _1}) + r\,\alpha _1^{k} \,(2f_1+\frac{1}{1-\alpha _1}). \end{aligned}$$
So it follows that
$$\begin{aligned} \sigma _{13}^{(r)}= & {} r\,(2f_1+\tfrac{1}{1-\alpha _1})\,\sum _{k=0}^\infty \alpha _1^k\ +\ r\,(r-\tfrac{1}{1-\alpha _1})\,\\ \sum _{k=0}^\infty \alpha _1^{2k}= & {} \frac{2r\,f_1}{1-\alpha _1} + \frac{r^2}{1-\alpha _1^2} + \frac{r\,\alpha _1}{(1-\alpha _1)(1-\alpha _1^2)}. \end{aligned}$$
Finally, combining the previous derivations, we compute \(\sigma _{14}^{(r)}\) as
$$\begin{aligned} E[Y_{0,1}^{(r)} Y_{k,4}^{(r)}]= & {} \alpha _1\,E\big [\frac{(X_0)_{(r)}}{M_0^r}\, X_{k-1}^2\big ] + \alpha _0\,E\big [\frac{(X_0)_{(r)}}{M_0^r}\, X_{k-1}\big ]\ -\ \alpha _1\, f_2-f_1^2\\= & {} \alpha _1\,\Big (r\,\alpha _1^{2(k-1)}\, \left( r-\frac{1}{1-\alpha _1}\right) + r\,\alpha _1^{k-1} \,\left( 2f_1+\tfrac{1}{1-\alpha _1}\right) \ +\ f_2+f_1^2\Big )\\&+\, \alpha _0\,(r\,\alpha _1^{k-1}+f_1)\ -\ \alpha _1\, f_2-f_1^2\\= & {} \frac{r}{\alpha _1}\, \left( r-\frac{1}{1-\alpha _1}\right) \,\alpha _1^{2k} + r\,\alpha _1^{k}\,\big (\frac{1}{1-\alpha _1}+f_1+\frac{f_1}{\alpha _1}\big ) \end{aligned}$$
for \(k\ge 1\), while
$$\begin{aligned} \textstyle E[Y_{0,1}^{(r)} Y_{0,4}^{(r)}]\ =\ Q_{1,1}^{(r)} - \alpha _1\, f_2-f_1^2\ \overset{(32)}{=}\ r\,f_1. \end{aligned}$$
Therefore,
$$\begin{aligned} \begin{array}{rl} \sigma _{14}^{(r)} \ =&{} r\,\big (\frac{1}{1-\alpha _1} + f_1 + \frac{f_1}{\alpha _1}\big )\,\sum _{k=0}^\infty \alpha _1^k\ +\ \frac{r}{\alpha _1}\, \big (r-\frac{1}{1-\alpha _1}\big )\,\sum _{k=0}^\infty \alpha _1^{2k}\\ &{}\qquad -\ \frac{r}{\alpha _1}\, \big (r-\frac{1}{1-\alpha _1}\big ) - r\,\big (\frac{1}{1-\alpha _1}+\frac{f_1}{\alpha _1}\big ) \\[1ex] =&{} r\,\big (\frac{1}{(1-\alpha _1)^2} + \frac{f_1(1+\alpha _1)}{\alpha _1(1-\alpha _1)}\big )\ +\ \frac{r}{\alpha _1(1-\alpha _1^2)}\, \big (r-\frac{1}{1-\alpha _1}\big ) \ -\ \frac{r^2}{\alpha _1} + \frac{r}{\alpha _1}-\frac{r\,f_1}{\alpha _1} \\[1ex] =&{} \frac{2\,r\,f_1}{1-\alpha _1}\ +\ \frac{r^2\,\alpha _1}{1-\alpha _1^2}\ +\ \frac{r\,\alpha _1^2}{(1-\alpha _1)(1-\alpha _1^2)}. \end{array} \end{aligned}$$
This completes the proof. \(\square \)
In the next step, we apply the Delta method to derive the joint distribution of \((\widehat{C}_{1;\,r}, \hat{\alpha }_0, \hat{\alpha }_1)^{\top }\).
Corollary 1
Let \((X_t)_{\mathbb {Z}}\) be a stationary INARCH(1) process. Then the distribution of \((\widehat{C}_{1;\,r}, \hat{\alpha }_0, \hat{\alpha }_1)^{\top }\) is asymptotically approximated by a normal distribution with mean vector \((1, \alpha _0, \alpha _1)^{\top }\) and covariance matrix \(\frac{1}{T-1}\,\tilde{\varvec{\varSigma }}^{(r)}\), where
$$\begin{aligned} \tilde{\varvec{\varSigma }}^{(r)}\ =\ \left( \begin{array}{ccc} \sum _{k=1}^r\ \left( {\begin{array}{c}r\\ k\end{array}}\right) ^2\,k!\, q_{0,k} \quad &{} r \quad &{} 0 \\ r \quad &{} \frac{\alpha _0}{1-\alpha _1}\big ( \alpha _0(1+\alpha _1)+ \frac{1+2 \alpha _1^4}{1+\alpha _1+\alpha _1^2} \big ) \quad &{} - \alpha _0(1+\alpha _1)-\frac{(1+2 \alpha _1) \alpha _1^3}{1+\alpha _1 +\alpha _1^2} \\ 0 \quad &{} - \alpha _0(1+\alpha _1) -\frac{(1+2 \alpha _1) \alpha _1^3}{1+\alpha _1 +\alpha _1^2} \quad &{} (1-\alpha _1^2)\big (1+\frac{\alpha _1 (1+2\alpha _1^2)}{\alpha _0 (1+\alpha _1 +\alpha _1^2) }\big ) \end{array} \right) . \end{aligned}$$
Proof
Define the function \(\varvec{g}: \mathbb {R}^4\rightarrow \mathbb {R}^3\) by
$$\begin{aligned} g_1(\varvec{y})\ :=\ y_1,\quad g_2(\varvec{y})\ :=\ y_2\,\frac{y_3-y_4}{y_3-y_2^2},\quad g_3(\varvec{y}) \ :=\ \frac{y_4-y_2^2}{y_3-y_2^2}. \end{aligned}$$
(34)
Note that \(g_2\big (\cdot ,f_1,f_2+f_1^2,\alpha _1\, f_2+f_1^2\big )\, =\, \alpha _0\) and \(g_3\big (\cdot ,f_1,f_2+f_1^2,\alpha _1\, f_2+f_1^2\big )\, =\, \alpha _1\).
From the proof of Theorem 4.2 in Weiß and Schweer (2016) (see p. 13 in Appendix B.4), we know that the Jacobian of \(\varvec{g}\) equals
$$\begin{aligned} \mathbf{J }_{\varvec{g}}(\varvec{y})\ =\ \left( \begin{array}{cccc} 1 \quad &{} 0 \quad &{} 0 \quad &{} 0 \\ 0 \quad &{} \displaystyle \frac{(y_3-y_4)\big (y_3+y_2^2\big )}{\big (y_3-y_2^2\big )^2} \quad &{} \displaystyle \frac{y_2\big (y_4-y_2^2\big )}{\big (y_3-y_2^2\big )^2} \quad &{} \displaystyle \frac{-y_2}{y_3-y_2^2} \\[3ex] 0 \quad &{} \displaystyle \frac{2y_2(y_4-y_3)}{\big (y_3-y_2^2\big )^2} \quad &{} \displaystyle \frac{y_2^2-y_4}{\big (y_3-y_2^2\big )^2} \quad &{} \displaystyle \frac{1}{y_3-y_2^2} \end{array} \right) , \end{aligned}$$
such that \(\mathbf{D }:= \mathbf{J }_{\varvec{g}}\big (1,f_1,f_2+f_1^2,\alpha _1\, f_2+f_1^2\big )\) is given by
$$\begin{aligned} \begin{array}{rl} \mathbf{D }\ =&{} \left( \begin{array}{cccc} 1 \quad &{} 0 \quad &{} 0\quad &{} 0 \\ 0 \quad &{} \displaystyle \frac{(1-\alpha _1)(f_2+2f_1^2)}{f_2} \quad &{} \displaystyle \frac{\alpha _1\,f_1}{f_2} \quad &{} \displaystyle -\frac{f_1}{f_2} \\[3ex] 0 \quad &{} \displaystyle -\frac{2(1-\alpha _1)\,f_1}{f_2} \quad &{} \displaystyle -\frac{\alpha _1}{f_2} \quad &{} \displaystyle \frac{1}{f_2} \end{array} \right) \\ \\ [-1ex] =&{} \left( \begin{array}{cccc} 1 \quad &{} 0 \quad &{} 0 \quad &{} 0 \\ 0 \quad &{} \displaystyle (1-\alpha _1)\big (1+2(1-\alpha _1^2)\,f_1\big ) \quad &{} \displaystyle \alpha _1(1-\alpha _1^2) \quad &{} \displaystyle -(1-\alpha _1^2) \\ 0 \quad &{} \displaystyle -2(1-\alpha _1)(1-\alpha _1^2) \quad &{} \displaystyle -\frac{\alpha _1}{f_2} \quad &{} \displaystyle \frac{1}{f_2} \end{array} \right) . \end{array} \end{aligned}$$
Now, let us look at
$$\begin{aligned} \tilde{\varvec{\varSigma }}^{(r)} = \big (\tilde{\sigma }_{ij}^{(r)}\big )\ :=\ \mathbf{D }\varvec{\varSigma }^{(r)} \mathbf{D }^{\top }, \end{aligned}$$
where \(\varvec{\varSigma }^{(r)}\) is the covariance matrix from Lemma 2 above. The components \(\tilde{\sigma }_{22}^{(r)},\tilde{\sigma }_{23}^{(r)},\tilde{\sigma }_{33}^{(r)}\) are already known from formula (11) in Weiß (2010) (or from Theorem 4.2 in Weiß and Schweer (2016)), and \(\tilde{\sigma }_{11}^{(r)}=\sigma _{11}^{(r)}\) obviously holds.
So it remains to compute \(\tilde{\sigma }_{12}^{(r)}\!=\!\sum _{j=2}^4\, d_{11}d_{2j}\,\sigma _{1j}^{(r)}\) and \(\tilde{\sigma }_{13}^{(r)}\!=\!\sum _{j=2}^4\, d_{11}d_{3j}\,\sigma _{1j}^{(r)}\). We get
$$\begin{aligned} \tilde{\sigma }_{12}^{(r)}= & {} (1-\alpha _1)\big (1+2(1-\alpha _1^2)\,f_1\big )\,\sigma _{12}^{(r)} +\alpha _1(1-\alpha _1^2)\,\sigma _{13}^{(r)} -(1-\alpha _1^2)\,\sigma _{14}^{(r)} \\= & {} r+2r\,(1-\alpha _1^2)\,f_1 +2r\,f_1\,\alpha _1(1+\alpha _1) +r^2\,\alpha _1+\frac{r\,\alpha _1^2}{1-\alpha _1} -2\,r\,f_1\,(1+\alpha _1) \\&-\,r^2\,\alpha _1 -\frac{r\,\alpha _1^2}{1-\alpha _1} = r, \end{aligned}$$
as well as
$$\begin{aligned} \tilde{\sigma }_{13}^{(r)}= & {} -2(1-\alpha _1)(1-\alpha _1^2)\,\sigma _{12}^{(r)} -\frac{\alpha _1}{f_2}\,\sigma _{13}^{(r)} +\frac{1}{f_2}\,\sigma _{14}^{(r)} \\= & {} -\,2r\,(1-\alpha _1^2) -2r\,\alpha _1(1+\alpha _1) -\frac{r^2\,\alpha _1}{f_1} -\frac{r\,\alpha _1^2}{f_1\,(1-\alpha _1)} +2r\,(1+\alpha _1) \\&+\,\frac{r^2\,\alpha _1}{f_1} +\frac{r\,\alpha _1^2}{f_1\,(1-\alpha _1)} = 0. \end{aligned}$$
This completes the proof. \(\square \)
Using Corollary 1, we are able to approximate the variance of \(\widehat{C}_{1;\,r}(\hat{\alpha }_0,\hat{\alpha }_1)\) by the asymptotic variance \(\tfrac{1}{T-1}\,\sigma _{1;\,r}^2\) of \(\widetilde{C}_{1;\,r}(\hat{\alpha }_0,\hat{\alpha }_1)\) according to (19):
$$\begin{aligned} \begin{array}{rl} \sigma _{1;\,r}^2\ =&{} \tilde{\sigma }_{11}^{(r)} + r^2\, q_{0,1}^2\,\tilde{\sigma }_{22}^{(r)} + r^2\, q_{1,1}^2\,\tilde{\sigma }_{33}^{(r)} -2r\, q_{0,1}\,\tilde{\sigma }_{12}^{(r)} + 2r^2\, q_{0,1} q_{1,1}\tilde{\sigma }_{23}^{(r)}\\ =&{} \sum _{k=1}^r\ \left( {\begin{array}{c}r\\ k\end{array}}\right) ^2\,k!\, q_{0,k}\ -\ 2r^2\, q_{0,1} \ +\ r^2\, q_{0,1}^2\,\frac{\alpha _0}{1-\alpha _1}\big ( \alpha _0(1+\alpha _1)+ \frac{1+2 \alpha _1^4}{1+\alpha _1+\alpha _1^2} \big )\\ &{} \ +\ r^2\, q_{1,1}^2\,(1-\alpha _1^2)\big (1+\frac{\alpha _1 (1+2\alpha _1^2)}{\alpha _0 (1+\alpha _1 +\alpha _1^2) }\big ) \ -\ 2r^2\, q_{0,1}q_{1,1}\,\big (\alpha _0(1+\alpha _1) +\frac{(1+2 \alpha _1) \alpha _1^3}{1+\alpha _1 +\alpha _1^2}\big ). \end{array} \end{aligned}$$
So the proof of formula (21) is complete.
1.2 Derivation of equality (26)
First, we note that if the random variable Z follows a Poisson distribution with mean \(\lambda \), and if \(a>0\), we have for \(k=1,2,\ldots \)
$$\begin{aligned} E\left[ \left( \frac{a}{a+Z}\right) ^{k}\right]= & {} \int _{0}^{1}\exp \big ( -\lambda \left( 1-s\right) \big )\, \frac{a^{k}}{\left( k-1\right) !}\,s^{a-1}\,\log ^{k-1}\left( \frac{1}{s}\right) \, ds \\= & {} \frac{a^{k}}{\left( k-1\right) !}\,\sum _{n=0}^{+\infty }\frac{\left( -1\right) ^{n}\,\lambda ^{n}}{n!}\,\int _{0}^{1}\left( 1-s\right) ^{n}\,s^{a-1}\,\log ^{k-1}\left( \frac{1}{s}\right) \, ds \\= & {} a^{k}\,\sum _{n=0}^{+\infty }\frac{\left( -1\right) ^{n}\,\lambda ^{n}}{n!}\, \sum _{j=0}^{n}\left( {\begin{array}{c}n\\ j\end{array}}\right) \,\frac{\left( -1\right) ^{j}}{\left( a+j\right) ^{k}}, \end{aligned}$$
using the Dominated Convergence Theorem and the following result (formula 16 on page 552 of Gradshteyn and Ryzhic (2007))
$$\begin{aligned} \int _{0}^{1}\left( \log \frac{1}{x}\right) ^{n}\,\left( 1-x^{q}\right) ^{m}\,x^{p-1}\,dx\ =\ n!\,\sum _{k=0}^{m}\left( {\begin{array}{c}m\\ k\end{array}}\right) \,\frac{\left( -1\right) ^{k}}{\left( p+kq\right) ^{n+1}}\qquad \text {with } p,q>0. \end{aligned}$$
We note that for \(k=1\), the expression may be replaced by the equivalent one
$$\begin{aligned} E\left[ \frac{a}{a+Z}\right] \ =\ \varGamma \left( a+1\right) \, \sum _{n=0}^{+\infty } \frac{\left( -1\right) ^{n}}{\varGamma \left( a+n+1\right) }\,\lambda ^{n}, \end{aligned}$$
since
$$\begin{aligned} \frac{\varGamma \left( a+1\right) }{\varGamma \left( a+n+1\right) }\ =\ \frac{a}{n!}\, \sum _{j=0}^{n}\left( {\begin{array}{c}n\\ j\end{array}}\right) \,\frac{\left( -1\right) ^{j}}{a+j} \end{aligned}$$
as may be proved by recurrence.
Let us now consider that the moment generating function of \(M_{1},\)
\({\text {mgf}}_{M_{1}}(u)=E\left[ \exp (uM_{1})\right] \), is defined for every \(u\in (u_{1};u_{2})\), where \(u_{1}<0<u_{2}\) such that \(\min {\{-u_{1},u_{2}\}}=b>2\). With these conditions, we will prove that
$$\begin{aligned} E\left[ \frac{1}{M_{t}^{l}}\right] \ =\ \frac{1}{\alpha _{1}^{l}}\, \sum _{n=0}^{+\infty }\frac{\left( -1\right) ^{n}}{n!}\,E[M_{t-1}^{n}]\,\sum _{j=0}^{n}\left( {\begin{array}{c}n\\ j\end{array}}\right) \,\frac{\left( -1\right) ^{j}}{\left( \frac{\alpha _{0}}{\alpha _{1}}+j\right) ^{l}}, \end{aligned}$$
that is, the change between the expectation and the infinite sum is allowed. For this purpose, let us consider s such that \(0<s<\frac{1}{2}\min {\{-u_{1},u_{2}\}}\) and the function
$$\begin{aligned} H\left( t\right) \ =\ \int \sum _{n=0}^{+\infty }\frac{\left( -1\right) ^{n}\left( tx\right) ^{n}}{n!}\,\sum _{j=0}^{n}\left( {\begin{array}{c}n\\ j\end{array}}\right) \,\frac{\left( -1\right) ^{j}}{\left( \frac{\alpha _{0}}{\alpha _{1}}+j\right) ^{l}}\, dP_{M_{1}}(x),\qquad t\in (-s;s). \end{aligned}$$
Considering the functions
$$\begin{aligned} h_{k}\left( x\right)= & {} \sum _{n=0}^{k}\frac{\left( -1\right) ^{n}\left( tx\right) ^{n}}{n!}\,\sum _{j=0}^{n}\left( {\begin{array}{c}n\\ j\end{array}}\right) \,\frac{\left( -1\right) ^{j}}{ \left( \frac{\alpha _{0}}{\alpha _{1}}+j\right) ^{l}}\qquad \text {with } k\in \mathbb {N}_{0}, \end{aligned}$$
and \(h(x):=h_{\infty }(x)\), we have for every x and for \(k=1,2,\ldots \)
$$\begin{aligned} \left| h_{k}\left( x\right) \right|\le & {} \sum _{n=0}^{k}\frac{ \left| tx\right| ^{n}}{n!}\,\sum _{j=0}^{n}\left( {\begin{array}{c}n\\ j\end{array}}\right) \,\frac{1}{\left( \frac{\alpha _{0}}{\alpha _{1}}+j\right) ^{l}} \le \ \left( \frac{\alpha _{1}}{ \alpha _{0}}\right) ^{l}\,\sum _{n=0}^{k}\frac{\left( 2\left| tx\right| \right) ^{n}}{n!}\\\le & {} \ \left( \frac{\alpha _{1}}{\alpha _{0}} \right) ^{l}\,\exp \left( 2s\left| x\right| \right) , \end{aligned}$$
since \(\left| t\right| <s\), and also \(\lim _{k\rightarrow \infty } h_{k}\left( x\right) =h(x)\). Moreover,
$$\begin{aligned} \int \exp \left( 2s\left| x\right| \right) dP_{M_{1}}(x)\le & {} \int _{-\infty }^{+\infty }\exp \left( 2sx\right) \, dP_{M_{1}}(x)\ +\ \int _{-\infty }^{+\infty }\exp \left( -2sx\right) \, dP_{M_{1}}(x) \\= & {} {\text {mgf}}_{M_{1}}(2s)\ +\ {\text {mgf}}_{M_{1}}(-2s)\ <+\infty . \end{aligned}$$
So, we may apply the Dominated Convergence Theorem and we obtain
$$\begin{aligned} H\left( t\right) \!= \!\int h(x)\,dP_{M_{1}}\left( x\right) \! =\! \lim _{k\rightarrow \infty }\sum _{n=0}^{k}\frac{\left( -1\right) ^{n}\,t^{n}}{n!}\,\sum _{j=0}^{n}\left( {\begin{array}{c}n\\ j\end{array}}\right) \, \frac{\left( -1\right) ^{j}}{\left( \frac{\alpha _{0}}{\alpha _{1}}\!+\!j\right) ^{l}}\,\int x^{n}dP_{M_{1}}(x) , \end{aligned}$$
that is,
$$\begin{aligned}&E\left[ \sum _{n=0}^{+\infty }\frac{\left( -1\right) ^{n}\,\left( tM_{t-1}\right) ^{n}}{n!}\,\sum _{j=0}^{n}\left( {\begin{array}{c}n\\ j\end{array}}\right) \,\frac{\left( -1\right) ^{j}}{\left( \frac{\alpha _{0}}{\alpha _{1}}+j\right) ^{l}}\right] \\&\qquad = \sum _{n=0}^{+\infty }\frac{\left( -1\right) ^{n}}{n!}\,E[t^{n}M_{t-1}^{n}]\,\sum _{j=0}^{n}\left( {\begin{array}{c}n \\ j\end{array}}\right) \,\frac{\left( -1\right) ^{j}}{\left( \frac{\alpha _{0}}{\alpha _{1}} +j\right) ^{l}}, \end{aligned}$$
for \(t\in [-s;s]\). The result is valid for \(t=1\) if and only \(s>1\), which is possible as \(\min {\{-u_{1},u_{2}\}} >2\), and so (26) follows.
