Single change-point detection methods for small lifetime samples

Abstract

In this paper, we address the problem of deciding if either n consecutive independent failure times have the same failure rate or if there exists some $k\in \{1,\ldots ,n\}$ such that the common failure rate of the first k failure times is different from the common failure rate of the last $n-k$ failure times, based on an exponential lifetime distribution. The statistical test we propose is based on the empirical average ratio under the assumption of exponentiality. The proposed test is compared to the one based on the Mann–Whitney statistic for which no parametric assumption on the underlying distribution is necessary. The proposed statistics are free of the unknown underlying distribution under the null hypothesis of homogeneity of the n failure times which enables the determination of critical values of the proposed tests by Monte Carlo methods for small sample sizes.

Appendix: Proofs

1.1 Proof of Proposition 1

In the case of exponential distribution, the statistic $S_{n,k}^{(1)}$ involves ratio of independent Erlang distributed random variables. Recall that random variable X follows the Erlang distribution with parameters $r \in \mathbb {N}^*$ and $\alpha >0$ if its probability density function f is given by

$$\begin{aligned} f(x;r,\alpha )=\frac{x^{r-1}}{\alpha ^r\varGamma (r)}\exp (-x/\alpha )1_{\{x>0\}} , \end{aligned}$$

where $\varGamma (r)=(r-1)!$. We denote it by $X\sim Erl(r,\alpha )$. In particular, we use the simplified notation $Erl(r)\equiv Erl(r,1)$.

Calculation of the covariance matrix of $\mathbf{S}_n$ requires the calculation of $\mathbb {E}[1/(X(X+Y))]$, where X and Y are two independent Erlang distributed random variables.

Lemma 2

Let $X\sim Erl(r)$ and $Y\sim Erl(s)$ be independent. If $r\geqslant 2$ and $s\geqslant 1$, then

$$\begin{aligned} \mathbb {E}\left[ \frac{1}{X(X+Y)}\right] = \frac{1}{(r-1)(r+s-2)}. \end{aligned}$$

Proof

Let $U=\tfrac{X}{X+Y}$ and $V=X+Y$. It is well-known that U and V are independent. Moreover, U is Beta distributed with parameters (r, s) and V is Erlang distributed with parameter $r+s$. Then, we have

$$\begin{aligned}&\mathbb {E}\left[ \frac{1}{X(X+Y)}\right] = \mathbb {E}\left[ \frac{1}{U}\frac{1}{V^2}\right] = \mathbb {E}\left[ \frac{1}{U}\right] \mathbb {E}\left[ \frac{1}{V^2}\right] \\&\quad = \frac{(r-2)!(r+s-1)!}{(r+s-2)!(r-1)!} \frac{1}{(r+s-1)(r+s-2)} = \frac{1}{(r-1)(r+s-2)} \end{aligned}$$

using expressions for the moments of the Beta distribution in Johnson et al. (1995b) and of the gamma distribution in Johnson et al. (1995a). $\square$

Now, we shall present the proof of Proposition 1. We have

$$\begin{aligned} S_{n,k}^{(1)}=\frac{n-k-1}{k}\frac{T_k}{T_n-T_k}=\frac{n-k-1}{k}\frac{T_k}{R_k}, \end{aligned}$$

where $R_k=T_n-T_k=\sum _{j=k+1}^n X_j$. Let $(k,k') \in \{3,\dots ,n-3\}^2$, then we find

$$\begin{aligned} {\text{ cov }}\left( S_{n,k}^{(1)},S_{n,k'}^{(1)}\right) =\frac{(n-k-1)(n-k'-1)}{kk'}\mathbb {E}\left[ \frac{T_kT_{k'}}{R_kR_{k'}}\right] -1. \end{aligned}$$

Assuming that $3\leqslant k < k'\leqslant n-3$, we write

$$\begin{aligned} R_k= & {} X_{k+1}+\cdots +X_{k'}+X_{k'+1}+\cdots +X_n\\= & {} X_{k+1}+\cdots +X_{k'}+R_{k'}\\\equiv & {} U_{k,k'}+R_{k'},\\ \end{aligned}$$

and $T_{k'} = U_{k,k'}+T_{k}$. Then, we need to find the expectation

$$\begin{aligned} \mathbb {E}\left[ \frac{T_kT_{k'}}{R_kR_{k'}}\right] =\mathbb {E}\left[ \frac{T_k(T_k+U_{k,k'})}{R_{k'}(R_{k'}+U_{k,k'})}\right] = \mathbb {E}\left[ \frac{X(X+Z)}{Y(Y+Z)}\right] , \end{aligned}$$

where

$$\begin{aligned} \left\{ \begin{array}{lclll} X=T_k&{}\sim &{} E(k) &{} \text{ with } &{} k\geqslant 3, \\ Y=R_{k'}&{}\sim &{} E(n-k') &{} \text{ with } &{} l=n-k'\geqslant 3, \\ Z=U_{k,k'}&{}\sim &{} E(k'-k) &{} \text{ with } &{} q=k'-k\geqslant 1. \end{array}\right. \end{aligned}$$

These three random variables are clearly independent, and since

$$\begin{aligned} \frac{X(X+Z)}{Y(Y+Z)}=\frac{X^2}{Y(Y+Z)}+\frac{X}{Y}-\frac{X}{Y+Z}, \end{aligned}$$

we readily find

$$\begin{aligned} \mathbb {E}\left[ \frac{X(X+Z)}{Y(Y+Z)}\right]= & {} \mathbb {E}[X^2]\mathbb {E}\left[ \frac{1}{Y(Y+Z)}\right] +\mathbb {E}[X]\mathbb {E}\left[ \frac{1}{Y}\right] -\mathbb {E}[X]\mathbb {E}\left[ \frac{1}{Y+Z}\right] \\= & {} \frac{k(k+1)}{(n-k'-1)(n-k-2)} +\frac{k}{n-k'-1} -\frac{k}{n-k-1} \\= & {} \frac{k(n-1)}{(n-k'-1)(n-k-2)} -\frac{k}{n-k-1} . \end{aligned}$$

Then, we obtain

$$\begin{aligned} {\text{ cov }}\left( S_{n,k}^{(1)},S_{n,k'}^{(1)}\right)= & {} \frac{(n-k-1)(n-k'-1)}{kk'} \left( \frac{k(n-1)}{(n-k'-1)(n-k-2)} -\frac{k}{n-k-1}\right) -1 \\= & {} \frac{n-1}{k'(n-k-2)}. \end{aligned}$$

Similarly, we obtain the variance of $S_{n,k}^{(1)}$ as

$$\begin{aligned} {\text{ var }}\left( S_{n,k}^{(1)}\right)= & {} \mathbb {E}\left( \left( S_{n,k}^{(1)}-1\right) ^2\right) \\= & {} \left( \frac{n-k-1}{k}\right) ^2\mathbb {E}(T_k^2)\mathbb {E}\left( \frac{1}{R_k^2}\right) -2\left( \frac{n-k-1}{k}\right) \mathbb {E}(T_k)\mathbb {E}\left( \frac{1}{R_k}\right) +1\\= & {} \frac{n-1}{k(n-k-2)}. \end{aligned}$$