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This question is an extension of this question, but more specific.

This paper E. Bradlow et al is a Weibull counting model which I am using to estimate how many failures will happen between preventive maintenance schedules (my data is the system's mean time to failure that is Weibull-distributed with shape parameter less than one).

In page 12, it shows the derivation of the final results:

$$\begin{align} \Pr(N(t)=n) &= \sum_{j=n}^\infty{\frac{(-1)^{j+n}(\lambda t^c)^j \alpha_j^n}{\Gamma(cj+1)}} \\ E(N(t)) &= \sum_{n=1}^\infty{\sum_{j=n}^\infty{\frac{n(-1)^{j+n}(\lambda t^c)^j \alpha_j^n}{\Gamma(cj+1)}}} \end{align}$$

Where $Pr(N(t) = n)$ is the probability that the count will be $n$, and $E(N(t))$ is the expected value. What I'm having trouble understanding is the vector $\alpha^n_j$, which has the following definitions:

$$\begin{align} \alpha_m^l &= \sum_{m=0}^{l-1}{\frac{\Gamma(cm+1)\Gamma(cl-cm+1)}{\Gamma(m+1)\Gamma(l-m+1)}} \\ \alpha_j^0 &= \frac{\Gamma(cj+1)}{\Gamma(j+1)}, \quad j=0,1,2,\ldots \\ \alpha_j^{n+1} &= \sum_{m=n}^{j-1}{\alpha_m^n \frac{\Gamma(cj-cm+1)}{\Gamma(j-m+1)}} \end{align}$$

I understand $\alpha_j^0$ and $\alpha_j^{n+1}$, but $\alpha_j^{n+1}$ uses $\alpha_m^n$ which is my problem. I don't understand that at all. The external subscript is $m$, but internally, $m$ is the one that increments. So what does that mean? Is the $m$ subscript just a marker if which formula to use, without need for a value?

But if so, then in the formula $\alpha_j^{n+1} = \sum_{m=n}^{j-1}{\alpha_m^n \frac{\Gamma(cj-cm+1)}{\Gamma(j-m+1)}}$, since the value of $m$ doesn't matter for $\alpha_m^n$ and since $n$ is already fixed (it doesn't increment), then it's a constant. Does this mean I can take $\alpha_m^n$ out the summation to make it

$$\alpha_j^{n+1} = \alpha_m^n\sum_{m=n}^{j-1}{ \frac{\Gamma(cj-cm+1)}{\Gamma(j-m+1)}}$$

This looks a little wrong to me due to the subscripts and I must be misunderstanding something. Any hints will be greatly appreciated!

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I read that paper in 2012. It was difficult to me to understand their work, I tried to made a counting model before reading the paper but with the no-solution for the convolution of Weibull PDF was so difficult to me. Using the model I could not get the same results comparing to simulation methods. I totally recommend you to use simulation instead of using the model, its easier, safer and allows to consider risk in your solution.

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  • $\begingroup$ Actually I ended up not using it at all, it was indeed very difficult. But, I do think it's very useful and worth poring over because there's not really another formula I know of that can do what this one does -- ie estimate probabilities for Weibulls like how we do it for normal curves. $\endgroup$ – markovchain Jul 31 '14 at 12:51

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