Convergence and inference for mixed Poisson random sums

Abstract

We study the limit distribution of partial sums with a random number of terms following a class of mixed Poisson distributions. The resulting weak limit is a mixture between a normal distribution and an exponential family, which we call by normal exponential family (NEF) laws. A new stability concept is introduced and a relationship between $\alpha$-stable distributions and NEF laws is established. We propose the estimation of the NEF model parameters through the method of moments and also by the maximum likelihood method via an Expectation–Maximization algorithm. Monte Carlo simulation studies are addressed to check the performance of the proposed estimators, and an empirical illustration of the financial market is presented.

Appendix

Proof of Theorem 2

Consider the change of variable $W_\phi ^*=\phi W_\phi$. With this, we obtain that the density function of $W_\phi ^*$ assumes the form (3) with $\theta =\xi _0$, $B(\theta )=-\phi b(\theta )$, $R(y)=-\log \phi +c(y;\phi )$, and support ${\mathbb {S}}={\mathbb {R}}^+$. Then, we use Theorem 1 above to obtain the ch.f. of $W_\phi ^*$. This immediately gives us the ch.f. of $W_\phi$ as follows:

$$\begin{aligned} \varPsi _{W_\phi }(t) = \exp \left\{ -\phi \left[ b(\xi _0) - b\left( \xi _0 + \frac{it}{\phi }\right) \right] \right\} ,\quad t\in {\mathbb {R}}. \end{aligned}$$

(13)

From (13), we obtain that

$$\begin{aligned} \varPsi _{N_{\lambda }}(t) = \varPsi _{W_{\phi }}\left( \lambda (e^{it} - 1)\right) = \exp \left\{ -\phi \left[ b(\xi _0) - b\left( \xi _0 + \frac{i}{\phi }\lambda \left( e^{it} - 1\right) \right) \right] \right\} ,\quad t\in {\mathbb {R}}. \end{aligned}$$

(14)

The Tower Property of conditional expectations giwsves

$$\begin{aligned} \varPsi _{{\widetilde{S}}_{\lambda }}(t) = E\left[ E\left( \exp \left\{ it\left( a_{\lambda }\sum \limits _{i = 1}^{N_{\lambda }}(X_i + d_{\lambda })\right) \right\} \bigg | N_{\lambda }\right) \right] . \end{aligned}$$

Let $G_{N_{\lambda }}(\cdot )$ denote the probability generating function of $N_{\lambda }$. Then, we use (14) to obtain

$$\begin{aligned} \varPsi _{{\widetilde{S}}_{\lambda }}(t)= & {} G_{N_{\lambda }}\left( \varPsi _{X_1 - \mu }\left( \frac{t}{\sqrt{\lambda }}\right) e^{i\frac{\mu }{\lambda }t}\right) = \varPsi _{N_{\lambda }}\left( \frac{1}{i}\log \left\{ \varPsi _{X_1 - \mu }\left( \frac{t}{\sqrt{\lambda }}\right) e^{i\frac{\mu }{\lambda }t}\right\} \right) \nonumber \\= & {} \exp \left\{ - \phi \left[ b(\xi _0) - b\left( \xi _0 + \frac{i\lambda }{\phi }\left( \varPsi _{X_1 - \mu }\left( \frac{t}{\sqrt{\lambda }} \right) e^{i\frac{\mu }{\lambda }t} - 1\right) \right) \right] \right\} . \end{aligned}$$

(15)

Taking $\lambda \rightarrow \infty$ and applying L’Hôpital’s rule twice (in the second-order derivative we are using the assumption of finite variance of $X_1$) we obtain

$$\begin{aligned} \lim \limits _{\lambda \rightarrow \infty } \varPsi _{{\widetilde{S}}_{\lambda }}(t) = \exp \left\{ -\phi \left[ b(\xi _0) - b\left( \xi _0 + \phi ^{-1}\left( it\mu - \frac{t^2\sigma ^2}{2}\right) \right) \right] \right\} \equiv \varPsi _Y(t), \quad \forall \; t \in {\mathbb {R}}. \end{aligned}$$

This completes the proof. $\square$

Proof of Theorem 3

If Y is MP-stable, then ${\mathcal {D}}(G)\ne \emptyset$. By Lévy’s Continuity Theorem and (15), the convergence in Expression (5) holds if and only if

$$\begin{aligned} \exp \left\{ -\phi \left[ b(\xi _0) - b\left( \xi _0 + \frac{\lambda }{\phi }(\varphi _{\lambda }(t) - 1)\right) \right] \right\} \xrightarrow [\lambda \rightarrow \infty ]{} \varPsi _Y(t), \end{aligned}$$

(16)

where $\varphi _{\lambda }(t) = \varPsi _{X_1}(a_{\lambda }t)\exp (ita_{\lambda }d_{\lambda })$.

The first derivative of the function $b(\cdot )$ is positive since it is equal to the mean of the exponential family (its support here is assumed to be ${\mathbb {R}}^+$, so that the mean is positive). This implies that $b(\cdot )$ is strictly monotone increasing and therefore it is invertible. Using this invertibility, it follows that (16) is equivalent to

$$\begin{aligned} \lambda (\varphi _{\lambda }(t)-1) \xrightarrow [\lambda \rightarrow \infty ]{}\phi \left\{ b^{-1}\left( b(\xi _0) + \frac{1}{\phi }\log {\varPsi _Y(t)}\right) - \xi _0\right\} , \quad \forall t\in {\mathbb {R}}. \end{aligned}$$

We can take the limit above with $\lambda = \lambda _n=n\in {\mathbb {N}}$ instead of $\lambda \in {\mathbb {R}}$; $\{\lambda _n\}$ can be seen as a subsequence. In this case, letting $a_{\lambda }=a_n$ and $d_{\lambda }=d_n$, it follows that

$$\begin{aligned} n(\varphi _{n}(t)-1) \xrightarrow [n\rightarrow \infty ]{}\phi \left\{ b^{-1}\left( b(\xi _0) + \frac{1}{\phi }\log {\varPsi _Y(t)}\right) - \xi _0\right\} ,\quad \forall t\in {\mathbb {R}}. \end{aligned}$$

From Theorem 1 in Chapter XVII in Feller (1971), the above limit implies that

$$\begin{aligned} (\varphi _{n}(t))^n \xrightarrow [n\rightarrow \infty ]{}\exp \left\{ \phi \left\{ b^{-1}\left[ b(\xi _0) + \frac{1}{\phi }\log {\varPsi _Y(t)}\right] - \xi _0\right\} \right\} ,\quad \forall t\in {\mathbb {R}}. \end{aligned}$$

(17)

Since the left side of (17) is the ch.f. of

$$\begin{aligned} a_{n}\sum \limits _{i = 1}^n(X_i + d_n), \end{aligned}$$

it follows that the right side of that equation is the ch.f. of some $\alpha$-stable random variable A [see Theorem 3.1, Chapter 9 in Gut (2013)], say $\varPsi _A(t)$. Hence, we obtain

$$\begin{aligned} \varPsi _Y(t) = \exp \left\{ -\phi \left[ b(\xi _0) - b\left( \xi _0 + \frac{1}{\phi }\log {\varPsi _A(t)}\right) \right] \right\} , \end{aligned}$$

which gives us the sufficiency part of the theorem.

Conversely, if (6) holds and $\varPsi _A(t)$ is the ch.f. of some $\alpha$-stable random variable A, then there exist an i.i.d. sequence $\{X_n\}_{n\in {\mathbb {N}}}$ of random variables and real sequences $\{a_n\}_{n\in {\mathbb {N}}}$ and $\{d_n\}_{n\in {\mathbb {N}}}$ such that

$$\begin{aligned} a_{n}\sum \limits _{i = 1}^n(X_i + d_n)\xrightarrow [n\rightarrow \infty ]{d}U. \end{aligned}$$

(18)

Denote by $\varPsi _{X_1}(t)$ the ch.f. of $X_1$ and let $\gamma _{n}(t) = \varPsi _{X_1}(a_n t)\exp (ita_n d_n)$, $n\in {\mathbb {N}}$. Then, the weak limit in (18) is equivalent to

$$\begin{aligned} (\gamma _{n}(t))^n \xrightarrow [n\rightarrow \infty ]{}\varPsi _A(t),\quad \forall t\in {\mathbb {R}}. \end{aligned}$$

From Feller (1971), we have that the above limit implies that

$$\begin{aligned} n(\gamma _{n}(t)-1) \xrightarrow [n\rightarrow \infty ]{}\log \varPsi _A(t). \end{aligned}$$

Since by hypothesis

$$\begin{aligned} \varPsi _A(t) = \exp \left\{ \phi \left\{ b^{-1}\left( b(\xi _0) + \frac{1}{\phi }\log {\varPsi _Y(t)}\right) - \xi _0\right\} \right\} , \end{aligned}$$

we obtain that

$$\begin{aligned} n(\gamma _{n}(t)-1) \xrightarrow [n\rightarrow \infty ]{}\phi \left\{ b^{-1}\left( b(\xi _0) + \frac{1}{\phi }\log {\varPsi _Y(t)}\right) - \xi _0\right\} , \end{aligned}$$

which is equivalent to

$$\begin{aligned} \lambda (\gamma _{\lambda }(t)-1) \xrightarrow [\lambda \rightarrow \infty ]{}\phi \left\{ b^{-1}\left( b(\xi _0) + \frac{1}{\phi }\log {\varPsi _Y(t)}\right) - \xi _0\right\} . \end{aligned}$$

The above limit gives Eq. (16) with $\gamma _{\lambda }(t)$ instead of $\varphi _{\lambda }(t)$. This shows that ${\mathcal {D}}(G)\ne \emptyset$ and the proof is complete. $\square$

Proof of Proposition 1

By standard properties of conditional expectation and using the ch.f. of $W_{\phi }$ given in (13), we obtain that

$$\begin{aligned}&\varPsi _{\mu W_{\phi } + \sigma \sqrt{W_{\phi }}Z}(t)\\&\quad =E\left( \exp \left\{ it(\mu W_{\phi } + \sigma \sqrt{W_{\phi }}Z)\right\} \right) \\&\quad = E\left[ E\left( \exp \{it(\mu W_{\phi } + \sigma \sqrt{W_{\phi }}Z)\}| W_{\phi }\right) \right] \\&\quad = \int \limits _{0}^{\infty }e^{it\mu w}E\left( e^{it\sigma \sqrt{w}Z}\right) f_{W_\phi }(w)dw\\&\quad = \int \limits _{0}^{\infty }\exp \left\{ w\left( it\mu -\frac{1}{2}t^2\sigma ^2\right) \right\} f_{W_\phi }(w)dw\\&\quad = \varPsi _{W_\phi }\left( t\mu + \frac{it^2\sigma ^2}{2}\right) \\&\quad = \exp \left\{ -\phi \left[ b(\xi _0) - b\left( \xi _0 + \frac{1}{\phi }\left( it\mu - \frac{1}{2}t^2\sigma ^2 \right) \right) \right] \right\} , \end{aligned}$$

which is the characteristic function given in Theorem 2. $\square$

Proof of Proposition 3

We have that

$$\begin{aligned}&E\left( W_{\phi }^Kg(W_{\phi })^L\Big |Y\right) = \int _{0}^{\infty } w^Kg(w)^L \dfrac{f_{Y|W}(y|w)f_W(w)}{f_Y(y)}dw=\\&\quad \dfrac{1}{f_Y(y)} \int _{0}^{\infty } \frac{w^Kg(w)^L}{\sqrt{2\pi \sigma ^2w}}\exp \left\{ -\frac{(y-\mu w)^2}{2\sigma ^2w} \right. \\&\quad \left. + \phi [-w + \log (\phi ) + \log (w)] - \log \varGamma (\phi )-\log w\right\} dw. \end{aligned}$$

By using the explicit form of the normal-gamma distribution density given in Expression (8), we get

$$\begin{aligned}&E\left( W_{\phi }^Kg^L(W_{\phi })\Big |Y\right) \\&\quad = \dfrac{\left( \frac{\mu ^2 + 2\phi \sigma ^2}{y^2}\right) ^{\frac{\phi }{2} - \frac{1}{4}}}{2{\mathcal {K}}_{\phi - \frac{1}{2}}\left( \sqrt{\left[ \frac{\mu ^2}{\sigma ^2}+2\phi \right] \frac{y^2}{\sigma ^2}}\right) }\\&\qquad \times \int _{0}^{\infty }g(w)^L\underbrace{w^{\left( \phi + K - \frac{1}{2}\right) - 1}\exp \left\{ -\frac{1}{2}\left[ \left( \frac{\mu ^2}{\sigma ^2} + 2\phi \right) w + \left( \frac{y^2}{\sigma ^2}\right) \frac{1}{w}\right] \right\} }_{\text {Kernel GIG}\left( \frac{\mu ^2}{\sigma ^2} + 2\phi , \frac{y^2}{\sigma ^2}, \phi + K - \frac{1}{2}\right) }dw. \end{aligned}$$

Denoting $U \sim \text{ GIG }\left( a, b, p\right)$ with $a = \frac{\mu ^2}{\sigma ^2} + 2\phi$, $b = \frac{y^2}{\sigma ^2}$ and $p = \phi + K - \frac{1}{2}$ and noting that the integrand above is the kernel of a GIG density function, we obtain the desired result. $\square$

Proof of Proposition 4

We have that

$$\begin{aligned}&E\left( W_{\phi }^Kg(W_{\phi })^L\big |Y=y\right) \\&\quad = \int _{0}^{\infty } w^Kg(w)^L \dfrac{f_{Y|W}(y|w)f_W(w)}{f_Y(y)}dw\\&\quad \dfrac{1}{f_Y(y)} \int _{0}^{\infty } w^Kg(w)^L \frac{1}{\sqrt{2\pi \sigma ^2w}}e^{-\frac{(y-\mu w)^2}{2\sigma ^2w}}e^{-\frac{\phi }{2}\left( w+\frac{1}{w}\right) + \frac{1}{2}\left( \log \phi - \log (2\pi w^3)\right) }dw. \end{aligned}$$

Let $f_Y(y)$ be the NIG density function as given in (9) and denote $U \sim \text{ GIG }\left( a, b, p\right)$ with $a = \frac{\mu ^2}{\sigma ^2} + \phi$, $b = \frac{y^2}{\sigma ^2} + \phi$ and $p = K - 1$. It follows that

$$\begin{aligned}&E\left( W_{\phi }^Kg(W_{\phi })^L\big |Y=y\right) \\&\quad = \dfrac{\left( \frac{y^2 + \phi \sigma ^2}{\mu ^2 + \phi \sigma ^2}\right) ^{\frac{1}{2}}}{{\mathcal {K}}_{-1}\left( \sqrt{\left( \frac{\mu ^2}{\sigma ^2} + \phi \right) \left( \frac{y^2}{\sigma ^2} + \phi \right) }\right) }\\&\qquad \times \int _{0}^{\infty }g(w)^L\underbrace{w^{\left( K - 1\right) - 1}\exp \left\{ -\frac{1}{2}\left[ \left( \frac{\mu ^2}{\sigma ^2} + \phi \right) w + \left( \frac{y^2}{\sigma ^2} + \phi \right) \frac{1}{w}\right] \right\} }_{\text {Kernel GIG}\left( \frac{\mu ^2}{\sigma ^2} + \phi , \frac{y^2}{\sigma ^2} + \phi , K - 1\right) }dw\\&\quad = \dfrac{{\mathcal {K}}_{K - 1}(\sqrt{ab})}{{\mathcal {K}}_{-1}(\sqrt{ab})} \left( \dfrac{b}{a}\right) ^{\frac{K}{2}}E\left( g(U)^L\right) . \end{aligned}$$

This completes the proof. $\square$