A uniformisation-driven algorithm for inference-related estimation of a phase-type ageing model

Abstract

We develop an efficient algorithm to compute the likelihood of the phase-type ageing model. The proposed algorithm uses the uniformisation method to stabilise the numerical calculation. It also utilises a vectorised formula to only calculate the necessary elements of the probability distribution. Our algorithm, with an error’s upper bound, could be adjusted easily to tackle the likelihood calculation of the Coxian models. Furthermore, we compare the speed and the accuracy of the proposed algorithm with those of the traditional method using the matrix exponential. Our algorithm is faster and more accurate than the traditional method in calculating the likelihood. Based on our experiments, we recommend using 20 sets of randomly-generated initial values for the optimisation to get a reliable estimate for which the evaluated likelihood is close to the maximum likelihood.

Appendix

Proof of Theorem 2.1

Suppose $h_1=h_m=h$. It follows that $h_i=h$ for $i=1,\ldots ,m$.

Suppose $h_1<h_m$. When $s>0$, $h_1^s< h_m^s$. For any $1 \le i<j \le m$,

$$\begin{aligned} h_i^s&=h_1^sp_1+h_m^s(1-p_1) < h_1^sp_2+h_m^s(1-p_2) =h_j^s, \end{aligned}$$

where $\displaystyle p_1=\frac{m-i}{m-1}>\frac{m-j}{m-1}=p_2$ and the inequality holds using the lemma below. Therefore, $h_i<h_j$ using the fact that $x^s$ is an increasing function of x when $x,s>0$.

Similarly, when $s<0$, $h_1^s>h_m^s$, we have, for any $1 \le i<j \le m$,

$$\begin{aligned} h_i^s&=h_1^sp_1+h_m^s(1-p_1)> h_1^sp_2+h_m^s(1-p_2) =h_j^s. \end{aligned}$$

Hence, $h_i<h_j$ since $x^s$ is a decreasing function of x when $x>0$ and $s<0$.

When $s=0$,

$$\begin{aligned} \log (h_i)=\frac{\log (h_m)-\log (h_1)}{m-1}i+\frac{m\log (h_1)-\log (h_m)}{m-1}, \end{aligned}$$

which is an increasing function of i if and only if $h_1<h_m$.

In summary, for any value of s, $h_i$ is an increasing function of i if and only if $h_1<h_m$. Likewise, it may be shown that $h_i$ is a decreasing function with respect to i if and only if $h_1>h_m$. $\square$

Lemma

For any $0<a<b$ and $0\le p_2<p_1 \le 1$, the inequality $ap_1+b(1-p_1)<ap_2+b(1-p_2)$ holds.

Proof

$$\begin{aligned} ap_2+b(1-p_2)-ap_1-b(1-p_1)=(p_2-p_1)(a-b)>0. \end{aligned}$$

$\square$