Notes on stochastic count processes with independent non stationary increments

Problem Statement

Let $\mathbb{P}(X_{ij}>t)=e^{-\lambda_i t}$ thus $X_{ij} \sim Exponential(\lambda_i)$ , we construct a stochastic process as $T_{jn} = \sum_{i=0}^n X_{ij}$ , this kind of random variable are called hypoexponential random variables. We define a counting process such that $N_j(t) = max\{n| T_{jn} < t\}$ . This allows to model a certain relation between events, for example to have a shorter expected waiting time for a second event if the first event was observed.

Statement 1: $N(t)$ is no Poisson process.
To see this let $0<T_1<t_1<T_2<T_3<t_2$ and $t_2-t_1<T_3$ then $N(t_2)-N(t_1)=X_2 + X_3$ while $N(t_2 - t_1)=X_1 + X_2$ . Since the density of $X_i + X_{i+1}$ is given by $\frac{\lambda_i}{\lambda_i- \lambda_{i+1}} \lambda_{i+1} e^{- \lambda_{i+1} t } + \frac{\lambda_{i+1}}{\lambda_{i+1} - \lambda_i} \lambda_i e^{- \lambda_i t }$ with $\lambda_1 \neq \lambda_3$ the process is non stationary and can thus not be a Poisson process.

Statement 2: The (hypoexponential) distribution of $T_{jn}$ is given by $f(t)= \sum_{i=1}^n C_{i,n} \lambda_i e ^{-\lambda_it}$ for $t>0$ and $0$ else where $C_{i,n} = \prod_{i \neq j} \frac{\lambda_j}{\lambda_j - \lambda_i}$ and thus $\mathbb{E}(T_{jn})= \int_{0}^\infty \sum_{i=1}^n t C_{i,n} \lambda_i e ^{-\lambda_it} dt = \sum_{i=1}^n \frac{C_{i,n}}{\lambda_i}$ .

(1) $\begin{eqnarray*}\mathbb{E}(N_j(t))=\sum_m m \mathbb{P}(N_j(t)=m)= \sum_m m \mathbb{P}\left(T_{jm}<t \leq T_{jm+1}\right) \\= \sum_m m \left(1-\sum_{k=0}^{m} C_{k,m} e^{-\lambda_k t} \right) \left(\sum_{k=0}^{m+1} C_{k,m+1} e^{-\lambda_k t} \right)\end{eqnarray*}$

Estimation

To estimate $\lambda_i$ we construct a different process. For fixed $t$ , let $N_{(j)}(t)$ be $N_j(t)$ sorted descending by size. $S_i(t)=max\{j| N_{(j)}(t) \leq i\}$ this is the amount of all processes where the $i$ -th event was reached at time $t$ . Let $E_i(t) = \sum_{j=0}^{S_i(t)} X_{i(j)}$ , here $(j)$ is determined by the ordering of $N_{(j)}(t)$ , this is also often referred to as exposure. $\lambda_i$ is now estimated with $\hat{\lambda}_i=\frac{S_i(t)}{E_i(t)}$ .

The sum of order statistics exponential random variables is given by $E_i(t) \sim \Gamma(\alpha= S_i(t), \beta= \lambda_i)$ [1] thus the joined distribution is given by

(2) $\begin{eqnarray*}f_{E_i(t)}(x) = \sum_m \frac{\lambda_i^m x^{m-1}}{\Gamma(m)} e^{-\lambda_i x} \mathbb{P}\left(S_i(t)=m\right).\end{eqnarray*}$

Note that $X_{ij}$ and $S_i(t)$ are not independent, in particular

(3) $\begin{eqnarray*}\mathbb{E}(S_i(t)) = \sum_{j=0}^n \mathbb{P}(\sum_{k=0}^i X_{kj} \leq t) = n \int_0^t \sum_{k=0}^i C_{k,i} \lambda_k e ^{-\lambda_kx} dx = n -n \sum_{k=0}^i C_{k,i} e^{-\lambda_k t} \\\mathbb{P}\left(S_i(t)=m\right) = \binom{n}{m} \left(1 - \sum_{k=0}^i C_{k,i} e^{-\lambda_k t}\right)^m \left(\sum_{k=0}^i C_{k,i} e^{-\lambda_k t} \right)^{n-m}\end{eqnarray*}$

(see Statement 2), if the realisation of $X_{ij}$ is relatively small than for given $t$ the probability that the event was observed is higher. For simplicity, in the following, we will consider a special case where $t^*_i = max_j(T_{ji})$ , then $S_i(t^*_i)=n$ and the sum of exponential random variables is given by $y = E_i(t^*_i) \sim \Gamma(\alpha= n, \beta= \lambda_i)$ , thus

(4) $\begin{eqnarray*}\mathbb{E}\left(\hat{\lambda}_i\right)= &\int_0^\infty \frac{n}{y}\frac{\lambda_i^n}{\Gamma(n)}y^{n-1}e^{-\lambda_i y}dy = \frac{n \lambda_i^{n}_i \Gamma(n-1)}{\Gamma(n) \lambda_i^{n-1}} = \frac{n}{n-1} \lambda_i. \\Var\left(\hat{\lambda}_i\right)= &\int_0^\infty \left(\frac{n}{y}\right)^2 \frac{\lambda_i^n}{\Gamma(n)}y^{n-1}e^{-\lambda_i y}dy = \frac{n^2}{(n-1)^2(n-2)} \lambda_i^2.\end{eqnarray*}$

Therefore the estimator $\mathbb{E}(\frac{n-1}{n} \hat{\lambda}_i ) = \lambda_i$ is unbiased while the MSE is given by $\mathbb{E}(\hat{\lambda}_i - \lambda_i)^2 = \frac{\lambda_i^2 (n+2)}{(n-1)(n-2)}$ . Which reflects the well-known dichotomy between unbiasedness and minimal MSE.

For the general case

(5) $\begin{eqnarray*}\mathbb{E}\left(\frac{S_i(t)-1}{S_i(t)} \hat{\lambda}_i |S_i(t)>0 \right) = \sum_{m=1}^n \int_0^\infty \frac{m-1}{m} \frac{m}{y} \frac{\lambda_i^m y^{m-1}}{\Gamma(m)} e^{-\lambda_i y} \mathbb{P}\left(S_i(t)=m\right) dy \\= \lambda_i \left(1- \mathbb{P}\left(S_i(t)=0\right) \left) = \lambda_i \left(1 - \left(\sum_{k=0}^i C_{k,i} e^{-\lambda_k t} \right)^{n} \right)\end{eqnarray*}$

thus we face a slight bias, the cause of which is when there are too few observations available.

[1]

H. N. Nagaraja, “Order statistics from independent exponential random variables and the sum of the top order statistics,” in Advances in distribution theory, order statistics, and inference, N. Balakrishnan, J. M. Sarabia, and E. Castillo, Eds., Boston, MA: Birkhäuser boston, 2006, p. 173–185.
[Bibtex]

@Inbook{Nagaraja2006,
author="Nagaraja, H. N.",
editor="Balakrishnan, N.
and Sarabia, Jos{\'e} Mar{\'i}a
and Castillo, Enrique",
title="Order Statistics from Independent Exponential Random Variables and the Sum of the Top Order Statistics",
bookTitle="Advances in Distribution Theory, Order Statistics, and Inference",
year="2006",
publisher="Birkh{\"a}user Boston",
address="Boston, MA",
pages="173--185",
abstract="Let X(1)<...

Problem Statement

Estimation

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