# Estimating Sample CDF from 1st Order Statistics

I have a process where I can only measure the 1st order statistics, but would like to know something about the underlying sample CDF. I understand that I can calculate the CDF of my 1st order statistics, $$\Phi_1$$, and then calculate the CDF of the underlying sample process, $$\Phi$$, using $$\Phi=1-(1-\Phi_1)^{1/n}$$ where n is sample size.

However, I've done some Monte Carlo simulations of this and need very large set of data to calculate $$\Phi_1$$ with enough accuracy to get a good estimate of $$\Phi$$ (especially toward the right side of the CDF where $$\Phi_1$$ approaches 1). (I'm using Matlab's Empirical Kaplan-Meier algorithm to calculate $$\Phi_1$$.)

Is there a more efficient way of doing this?

For background, I'm trying to measure the distribution of layer strengths of a FDM 3D-printed part. From a tensile test I can measure the strength of the weakest layer in a test section (with n=100 to 400 layers). Monte Carlo simulations indicate that I'd need 1000-10000 tests to get a reasonable estimate of the shape of the CFD if it was a normal distribution (I don't know exactly what kind of distribution it is).

• This is the problem with extreme values : they have the fatest tails in the sample... Feb 12, 2021 at 19:38

Comment. I don't know what kind of precision you require. Also, if the part becomes useless when its weakest layer fails, isn't looking at minima the most practical measurement? Depending on the actual situation, maybe the results of this comment are good news, bad news, or irrelevant.

If the distribution of breaking strengths of individual layers happens to be exponential with rate $$\lambda,$$ then let's see how this method of measuring minima works. The distribution of the minimum of $$m$$ layers is exponential with rate $$m\lambda.$$ With an exponential model for layers, you are not speculating about the distribution of minima, the exact distribution (again exponential) is known.

The simulation below illustrates what you'd get with as few as 100 layers. My actual rate per layer is $$\lambda = 1/100,$$ so $$\mu = 100.$$ The minimum of 100 has mean $$1.$$ That is, $$V \sim\mathsf{Exp}(1).$$

set.seed(2021)
min = replicate(10^6, min(rexp(100, .01)))
mean(min)
[1] 0.9998841
sd(min)
[1] 0.9996088


If I had $$n = 50$$ minima $$V_i$$ with $$\bar V = 0.97$$ then $$\frac{\bar V}{\mu_v} \sim\mathsf{Gamma}(50,\mathsf{rate}=50)$$ so 95% CI is of the form $$(\bar V/U,\, \bar V/L),$$ where $$L$$ and $$U$$ cut probability 0.025 from lower and upper tails respectively of $$\mathsf{Gamma}(50,50).$$ So $$V = 0.97$$ gives CI $$(75,131)$$ for the mean breaking strength of a single layer.

0.97/qgamma(c(.975,.025), 50,50)
[1] 0.7486809 1.3068914

• I want the underlying single-layer properties so that it can be extrapolated to other geometries. You've done the analysis assuming you know it's an exponential. I want to find out if it's exponential, Weibull, bimodal, etc. Feb 12, 2021 at 21:34
• It sounds more like Weibull - extreme value type I. Why not for competitors to your data and look at BIC /AIC to say which fits best? Feb 26, 2021 at 19:01
• The dist'n theory is easy with exponential failure dist'n for individual parts. Min of $k$ exponentials is again exponential; mean of indep exponentials all with same rate is gamma with shape param $k.$ // Rather than a shopping list of many dist'ns at the start, it's likely best to explore one dist'n at a time. Perhaps some have easy analytic solutions; perhaps others best simulated for specific data at hand. // This site works best with one carefully posed question at a time; you shouldn't expect chapter-length answers here. Feb 26, 2021 at 20:11