Working on a reliability project and ran across this problem of parametric degradation over time relative to a failure threshold.
Need to calculate the distribution of failure times $t_{f,i}$ for a set of devices $i=1...N $ where failure is defined parametrically as $|S_{i}(t)|\ge S_{f}$ and failure times per device as $|S_{i}(t_{f,i})|=S_f$ where $S_{i}(t) = A_{i} \ln (1+bt)$ and $\{S_{i}(t_1)\}$ , where at a time $t=t_1<t_{f,i}$ for all $i$, is Gaussian distributed with mean $\mu $ and standard deviation $\sigma $, i.e. $N(\mu,\sigma).$
In contrast, it is relatively easy to show that if instead the degradation path were a simple power law, $S_{t}(t)=At^{n}$ that the resultant failure time distribution is lognormal with log-mean $\ln t_{1} + \frac{1}{n}\ln(\frac{S_{f}}{\mu})$ and log-standard deviation $\frac{1}{n}\ln(1+\frac{\sigma}{\mu})$
The failure theshold of a normal distribution is related to the divergence of the generating function. The generating function is an expression which arrives at the maximal changes between data points as the sample space expands.
In this case the data is the parametic representation of some function
$ S(t_{n}) = Aln(1+bt_{n}) $
The $ k^{th} $ moment is the $ k^{th} $ derivative of the generating function for the sample space, usually in the form of an exponential expansion of the function being used. (This way it is infinitely continuous and has uniform and proportional continuity throughout)
$$ e^{S(t_{n})} = \Sigma_{n=1}^{k}\frac{(S(t_{n}))^{n}}{n!} $$
Noting that $ S(t_{n})^{2} = S_{n}^{2} = A^{2}ln(1+bt)^{2} = 2A^{2}ln(1+bt) $
and that $ ln(1 + x) = x $ for $ x < N $ where $ N $ is a large number.
we can say that $ (S(t_{n}))^{n} = S_{n}^{n} = nbtA^{n} $
Thus $ e^{S(t_{n})} = \Sigma_{n = 1}^{k} \frac{nbtA^{n}}{n!} $
To get the generating function for the $ k^{th} $ values of the sample space, we take the $k^{th} $ derivative of this exponential expansion.
$$ \frac{d^{k}}{dx^{k}} \Sigma_{n = 1}^{k} \frac{nbtA^{n}}{n!} = S_{k}^{k}$$
Then $ \mu_{k} = \Sigma_{n = 1}^{N}\frac{S_{k_{n}}}{n} $
The $ k^{th} $ sigma, goes as followed. $ k = 1 $ is the variance, $ k = 2 $ is the standard deviation, and $ k = 3 $ is:
$ \sigma^{3} = \sqrt[3]{\Sigma_{n=1}^{\infty}\frac{\left(t_{n} - t\right)^{3}}{n}} $
The normal distribution
$$ N(\mu_{k},\sigma_{k}) = \frac{1}{\sqrt{2\pi\sigma^{2}}}e^{-\frac{(y-\mu_{k})^{k}}{2\sigma^{k}}} $$
Will then represent the failure threshold when $ k $ exceeds the inequalities for the initial distribution:
$$ \mu_{k} - k\sigma < S_{1} $$ $$ \mu_{k} + k\sigma > S_{1} $$