Archive for survival analysis

survivalists [a Riddler’s riddle]

Posted in Books, Kids, R, Statistics with tags , , , , , , on April 22, 2019 by xi'an

A neat question from The Riddler on a multi-probability survival rate:

Nine processes are running in a loop with fixed survivals rates .99,….,.91. What is the probability that the first process is the last one to die? Same question with probabilities .91,…,.99 and the probability that the last process is the last one to die.

The first question means that the realisation of a Geometric G(.99) has to be strictly larger than the largest of eight Geometric G(.98),…,G(.91). Given that the cdf of a Geometric G(a) is [when counting the number of attempts till failure, included, i.e. the Geometric with support the positive integers]

F(x)=\Bbb P(X\le x)=1-a^{x}

the probability that this happens has the nice (?!) representation

\sum_{x=2}^\infty a_1^{x-1}(1-a_1)\prod_{j\ge 2}(1-a_j^{x-1})=(1-a_1)G(a_1,\ldots,a_9)

which leads to an easy resolution by recursion since

G(a_1,\ldots,a_9)=G(a_1,\ldots,a_8)-G(a_1a_9,\ldots,a_8)

and G(a)=a/(1-a)

and a value of 0.5207 returned by R (Monte Carlo evaluation of 0.5207 based on 10⁷ replications). The second question is quite similar, with solution

\sum_{x=2}^\infty a_1^{x-1}(1-a_1)\prod_{j\ge 1}(1-a_j^{x})=a^{-1}(1-a_1)G(a_1,\ldots,a_9)

and value 0.52596 (Monte Carlo evaluation of 0.52581 based on 10⁷ replications).

David Cox gets the first International Prize in Statistics

Posted in pictures, Statistics, University life with tags , , , , , , , , on October 20, 2016 by xi'an

Just received an email from the IMS that Sir David Cox (Nuffield College, Oxford) has been awarded the International Prize in Statistics. As discussed earlier on the ‘Og, this prize is intended to act as the equivalent of a Nobel prize for statistics. While I still have reservations about the concept. I have none whatsoever about the nomination as David would have been my suggestion from the start. Congratulations to him for the Prize and more significantly for his massive contributions to statistics, with foundational, methodological and societal impacts! [As Peter Diggle, President of the Royal Statistical Society just pointed out, it is quite fitting that it happens on European Statistics day!]

a ghastly ghost

Posted in Books, Kids, Mountains with tags , , , , , , , , , , , , , , , , , , , , on March 13, 2016 by xi'an

My daughter sort of dragged me to watch The Revenant as it just came out in French cinemas and I reluctantly agreed as I had read about magnificent winter and mountain sceneries, shot in an unusually wide format with real light. And indeed the landscape and background of the entire movie are magnificent, mostly shot in the Canadian Rockies, around Kananaskis and Canmore, which is on the way to Banff. (Plus a bit in Squamish rain forest.) The story is however quite a disappointment as it piles up one suspension of disbelief after another. This is a tale of survival (as I presume everyone knows!) but so implausible as to cancel any appreciation of the film. It may be the director Iñárritu is more interested in a sort of new age symbolism than realism, since there are many oniric passages with floating characters and falling meteors, desecrated churches and pyramids of bones, while the soundtrack often brings in surreal sounds, but the impossible survival of Hugh Glass made me focus more and more on the scenery… While the true Hugh Glass did manage to survive on his own, fixing his broken leg, scrawling to a river, and making a raft that brought him to a fort downstream, [warning, potential spoilers ahead!] the central character in the movie takes it to a fantasy level as he escapes hypothermia while swimming in freezing rapids, drowning while wearing a brand new bearskin, toxocariasis while eating raw liver,  bullets when fleeing from both Araka Indians and French (from France, Louisiana, or Québec???) trappers, a 30 meter fall from a cliff with not enough snow at the bottom to make a dent on, subzero temperatures while sleeping inside a horse carcass [and getting out of it next morning when it should be frozen solid], massive festering bone-deep wounds, and the deadly Midwestern winter… Not to mention the ability of make fire out of nothing in the worst possible weather conditions or to fire arrows killing men on the spot or to keep a never ending reserve of bullets. And while I am at it, the ability to understand others: I had trouble even with the French speaking characters, despite their rather modern French accent!

generating from a failure rate function [X’ed]

Posted in Books, Kids, Statistics, University life with tags , , , , , , , , on July 4, 2015 by xi'an

While I now try to abstain from participating to the Cross Validated forum, as it proves too much of a time-consuming activity with little added value (in the sense that answers are much too often treated as disposable napkins by users who cannot be bothered to open a textbook and who usually do not exhibit any long-term impact of the provided answer, while clogging the forum with so many questions that the individual entries seem to get so little traffic, when compared say with the stackoverflow forum, to the point of making the analogy with disposable wipes more appropriate!), I came across a truly interesting question the other night. Truly interesting for me in that I had never considered the issue before.

The question is essentially wondering at how to simulate from a distribution defined by its failure rate function, which is connected with the density f of the distribution by

\eta(t)=\frac{f(t)}{\int_t^\infty f(x)\,\text{d}x}=-\frac{\text{d}}{\text{d}t}\,\log \int_t^\infty f(x)\,\text{d}x

From a purely probabilistic perspective, defining the distribution through f or through η is equivalent, as shown by the relation

F(t)=1-\exp\left\{-\int_0^t\eta(x)\,\text{d}x\right\}

but, from a simulation point of view, it may provide a different entry. Indeed, all that is needed is the ability to solve (in X) the equation

\int_0^X\eta(x)\,\text{d}x=-\log(U)

when U is a Uniform (0,1) variable. Which may help in that it does not require a derivation of f. Obviously, this also begs the question as to why would a distribution be defined by its failure rate function.