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A Naturalist's Guide to the Insects of Britain and Northern Europe
Author Robert Read
Publisher John Beaufoy Publishing
Size 180 x 128mm
Images Colour images
A Naturalist's Guide to the Insects of Britain and Northern Europe is a photographic identification guide to 280 most commonly seen species. High quality photographs from Britain's top nature photographers are accompanied by detailed species descriptions, which include common and scientific name, height, distribution and habitat. The user-friendly introduction covers the different habitats, life-cycles and the study and conservation of insects.
Robert Read is a professional photographer with an emphasis on the natural history of Britain and Northern Europe. He has had a lifelong interest in insects and nature and he also manages the Nature Photographers image library.
Paul Sterry is one of the UK's leading wildlife and natural history photographers, with more than 40 highly successful books to his name. Originally trained as a biologist, he uses his knowledge of natural history and ecology to take images that depict wildlife in a natural context and celebrate its beauty.
field guide / Insecten / insects / insekten / Veldgids
Field Guide to the Micro-moths of Great Britain and Ireland
ANWB Insect Guide
With over 1,500 images including larvae and dolls.
Field Guide to the Ladybirds of Great Britain and Ireland
Field Guide to the Ladybirds of Great Britain and Ireland is a beautiful field guide with all the ladybugs of Great Britain that you can also use in the Netherlands.


Hoodia gordonii (common name: Ghaap) is a leafless spiny succulent plant with medicinal properties. It grows naturally in South Africa and Namibia. In 1977, (CSIR) isolated the ingredient in hoodia—now known as P57—which is responsible for its appetite-suppressant effect.
Pelargonium sidoides (Umckaloabo, South African Geranium) is a medicinal plant native to South Africa. Studies have suggested that extracts from the plant could be used in treating acute bronchitis, acute non-GABHS tonsillopharyngitis (sore throat) in children, and the common cold.
Aloe ferox (common name: Cape Aloe), also known as Bitter Aloe is a species of aloe indigenous to South Africa's Western Cape, Eastern Cape. The bitter yellow juice found just below the skin has been harvested as a renewable resource. The hard, black, resinous product as aloe lump and is used mainly as laxative properties but is also taken for arthritis.
Kigelia africana (common name: Sausage Tree) has historically been used by African people to rejuvenate and heal the skin as well as for stomach ulcers.
Bulbine natalensis (common name: Rooi Wortel) is popular among the traditional healers. The roots are used, taken orally to quell vomiting and diarrhoea, but also for a number of other ailments.
(Common name: Devil's Claw) The most important components are iridoid glycosides (mainly harpagoside, harpagide and procumbide). Some of its properties are listed: analgesic, anti-arrhythmic, antibacterial, anti-inflammatory, antirheumatic, diuretic, hypotensive, laxative, purgative, sedative, uterocontractant and a febrifuge, cholelogue and bitter tonic. In western medicine it is mainly used for arthritis and rheumatism.
Sutherlandia frutescens (common name: Cancer Bush) is a southern African legume. S. frutescens has a reputation as a cure for cancer and for treating the symptoms of HIV/AIDS. However, there is no scientific support for the claims that this plant can cure cancer or AIDS.
K.Paulsen Botanicals is a Company that deals in Herbal/Medicinal plants from Southern Africa. Our C.E.O Kersten Paulsen has been in the Medicinal plant trade for over 14 years and has been sourcing and supplying to clients from all over the world. With our many networks of collectors throughout South Africa we are able to source and supply most of your needs from around the country. We harvest sustainably from the wild all under permit and have benefit sharing agreements for certain products. We are compliant with the new Bio-diversity Act.
Check our Facebook page for regular updates.


Gatley's offers the small holder a great selection of feeds and products.
For keepers of poultry we supply corn and pellets including layers pellets and organic poultry feeds. By way of accessories we sell drinkers, feeders and incubators, hay, bedding, wheelbarrows, health and well-being products.
If your small holding has other types of animals, we stock a range of cattle feeds, sheep and goat feeds, hay and other bedding.
You can either come in to the country store or give us a ring and as long as your order comes to over £50, we will deliver to you for free (within a 15 mile radius of Gatley's Country Store, Storrington).
If you need any advice, we will also do our best to help you.


Government Authority located in Bismarck, North Dakota; USA based organization looking for generator fed portable power network services.
(1) Vendor needs to provide generator fed portable power network services to the government authority located in Bismarck, ND.
(2) A mandatory pre-bid meeting will be held on February 15, 2019.


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2009 Top 20 Singles
December 24, 2009 December 24, 2009 2009 Top 20 Singles 2 Comments 2000s2009adobeanimal collectivefrankie knucklesgrizzly bearmerriweather post pavilionmusicmy girlsnoughtiesorbitalpanda bearprimary coloursthe beach boysthe horrorstop 50 singlestwo dancersveckatimestwild beasts
[1] Animal Collective, 'My Girls'
For all its ecstatic brilliance, it's annoying that I've known what the single of the year is for most of 2009. But it seemed so obvious when I heard it. I hadn't been expecting truly great things from Animal Collective – maybe more of the same quirky pop, bellowed harmonies, abstract lyrics and squelchy textures – so when they came up with this dense, sticky, Beach Boys rave track that actually seemed to be about something (Panda Bear's kids and – erm – Adobe slabs), it was as welcome as a fat cheque on a… well, right about now, please.
It's struck me that I might simply be a sucker for this because I have daughters (and everyone likes to think that song is about them, don't they? Don't they?), but listen to those Frankie Knuckles-nabbing synths and the slow introduction of the bass that makes it sound like Orbital – and then the steady rise and layering of the sonics, the two different hooks that could stand as a chorus. And the "Woo!"s. Your hands are in the air, aren't they?
On this play, Junior pranced like a deer from kitchen to living room, but she's been tuned into 'My Girls' all year, along with its album Merriweather Post Pavilion, a mainstay of the car in 2009 – and probably the album of the year too. The Horrors' one was good too, mind you. And Grizzly Bear's. And Wild Beasts'.
We're not going to do an album chart though. We're going to do the Top 50 Best Singles of the 2000s, and we'll start next week. Merry Christmas, all you cats.
I don't mean to seem like I care about material things…
https://jukeboxjunior.files.wordpress.com/2009/12/02-my-girls.mp3
December 23, 2009 December 22, 2009 2009 Top 20 Singles Leave a comment 2009beroccablondiejulian casablancaskaren omusicnoughtiesyeah yeah yeahszero
[2] Yeah Yeah Yeahs, 'Zero'
And now we come to everyone's second favourite single of the year. 'Zero' is brilliant, a long-, long-awaited blossoming of Karen O and co's always obvious pop chops; a striding, mammoth synth sledgehammer here to deliver us from indie wetness; a brazen bit of late-Noughties electro land-grabbing; a bassline-bouncing hot rock in leathers; a massive sigh of relief in the face of hitherto diminishing returns; and a Blondie-on-Berocca zig-zag through Julian Casablancas' electric dreams.
Junior wasn't in the mood.
Was it the cure?
https://jukeboxjunior.files.wordpress.com/2009/12/01-zero.mp3
December 22, 2009 December 22, 2009 2009 Top 20 Singles Leave a comment bulletproofcasioelly jacksoneuropean unionhot chipla rouxmusicready for the floortony blair
[3] La Roux, 'Bulletproof'
Junior sings, "This time, baby, I'll be blaaaaaiiiiiiirrrrproof". It sounds like an anthem for a resistant European Union.
She loves this song, loves its chorus whatever its words. This is understandable – gratifying, even – because 'Bulletproof''s toytown hook is one of the most delightful I've heard in years. It seems based on minimal effort (Elly Jackson pouting, sullen, around the studio) but the results are perfect, like an even more immediate 'Ready For The Floor'. And like Hot Chip, La Roux's cheap Casio sound is just that bit too tacky to sustain an album. Let's hear it for the singles.
Time is running out:
https://jukeboxjunior.files.wordpress.com/2009/12/04-bulletproof.mp3
December 21, 2009 2009 Top 20 Singles 1 Comment 2009christmas number onejoy divisionkrautrockmusicprimary colourssea within a seasimon cowellthe horrors
[4] The Horrors, 'Sea Within A Sea'
This would have been a better Christmas No.1. It has droney vocals, seemingly endless monotony, a metronomic rhythm like the slow trundle around the Boxing Day M25, that pervading sense of doom – in all, it's a real festive cockle-warmer. But of course it neither has *GASP* swearwords nor the relentless Cowell machine behind it, so there was never a snowball's chance. And no one thought about it. Next year then.
The Horrors can comfort themselves with all the critical garlands they received for an impressive step forward. Only the gloom and occasional tough riff remained from their (pretty funny) goth-garage debut; otherwise, Primary Colours was a fug of glacial synths, Krautrock basslines and happy-go-lucky Joy Division ambience. 'Sea Within A Sea' was the astonishing curtain-raiser, galloping in on a Satanic groove, hanging around for five teasing minutes, then sailing away between banks of tinkling keyboards. It's good.
"I like the singing," came an atypical response, "but I don't like the music." She said that, but she performed a wafty indiegirl dance for the full seven minutes, with some slapstick falling-over thrown in. Slapstick. It's what The Horrors are all about.
Wicked stone (man):
https://jukeboxjunior.files.wordpress.com/2009/12/10-sea-within-a-sea.mp3
December 18, 2009 2009 Top 20 Singles Leave a comment 2009jamie tjamie treaysmike skinnermusicsticks n stonesthe clashwimbledon
[5] Jamie T, 'Sticks 'N' Stones'
Jamie Treays is a national treasure, only no one realises it yet. He's smart, wily and witty, and is all the best bits of The Clash rolled into one rat-faced, half-cut Wimbledon marauder. 'Sticks 'N' Stones' was going to be the record that did it; the first fruits of a difficult second album (so difficult he made it twice) and a boisterous tune blessed with at least two different choruses along with a serious treatise on the ideal circumstances for a "crap". Forgive the unlikely compliment, but it's Mike Skinner gone plastic punk.
And did the record do it? Was it T's big break? Barely. Just another No.15 hit. You should take a long, hard look at yourselves. When she'd finished pogoing (a natural response, even to the plastic variety), Junior took a long, hard look at the sleeve and declared Jamie "green". And that's not all: he even "sounds green, like a tree."
Let's go out and find some trouble:
https://jukeboxjunior.files.wordpress.com/2009/12/02-sticks-n-stones-explicit-version.mp3
December 17, 2009 December 17, 2009 2009 Top 20 Singles 3 Comments 2009apollo 440groove armadakylie minogueliverpool fcmorcheebamusicpeter crouchrobynroyksoppstock aitken watermanthe girl and the robot
[6] Röyksopp featuring Robyn, 'The Girl And The Robot'
Röyksopp get away with being coffee table dance bores because they're Norwegian. Stick them in the Home Counties, and they'll be held in the same esteem as Groove Armada, Morcheeba, Apollo christing 440. "Didn't you make a good record once?" an aging hipster will say. "Yeah, that one on the advert," will be their chipper reply. "Oh, and the one that Robyn rescued."
And she does rescue it. This would be a hi-NRG SAW-era Kylie record (OK at the time, but in 2009?) if it wasn't for the "killingest pop star on the planet" and her ability to make the mundane sound heartbreaking. She could make the Liverpool FC year-end financial figures sound dramatic and devastating. But instead, it's the story of a poor girl ignored by her cold, workaholic boyfriend. He's like a robot. He may even be a robot. After all, as well as sounding like 1989 this sounds like The Future. It's a wormhole.
It's also the key to unlock Junior's dynamic array of disco moves, which now appear to include The Robot. Fortunately, she doesn't look like Peter Crouch. My brother looks like Peter Crouch.
Never know when you'll return:
https://jukeboxjunior.files.wordpress.com/2009/12/02-girl-and-the-robot.mp3
December 16, 2009 December 16, 2009 2009 Top 20 Singles 3 Comments 2009glasvegaskasabianldnlily allenmercury prizemusicthe fear
[7] Lily Allen, 'The Fear'
Too many people give Lily Allen a hard time, because of who her father is, because she's got an unfettered mouth, because she writes songs that tell how it is – at least for a young LDN lady. She's undervalued as an artist. She runs the full gamut of pop – straightforward catchy songs, ska, hip hop, even country – and somehow is taken less seriously for it. Where the hell was her Mercury nomination? Sweet Jesus, Glasvegas got a nomination. Kasabian got one.
'The Fear' is Lil' at her best: sharp, witty, personal, snippy and set to the best pop hook of the year. For Junior, it means earnest school disco dance moves and her parents coughing loudly over the swearwords.
Packing plastic:
https://jukeboxjunior.files.wordpress.com/2009/12/01-the-fear.mp3
December 15, 2009 2009 Top 20 Singles Leave a comment 2009blue skiescharlie finkdfalaura marlingmusicnoah and the whaleremixthe first days of springyacht
[8] Noah And The Whale, 'Blue Skies (YACHT Remix)'
Everyone loves a good volte-face in music, particularly when it's a rather wacky folk-pop outfit deciding to ditch the jarring ditties in favour of widescreen heartbreak-infused indie-country-soul. So imagine the world's delight when Laura Marling dumped chief Noah head Charlie Fink, forcing him to lick his wounds and make an excellent album detailing his "I'm fine, really. Really, I am. OK, I'm not. But I will be" stance. That's what The First Days Of Spring turned out to be, and lead single 'Blue Skies' was its ray of hope.
But we're concerned with YACHT's remix here. The DFA crew took Noah And The Whale's big music, stripped it back and turned it Balearic – yet still retained the beaten-and-bruised hopefulness of the source material. Junior does jazz hands to it, and a lengthy impersonation of one squeaky vocal effect, sweetly undercutting any grand moodiness. I play her the original too, which she describes as "mad". It's what this blog's all about: an unfathomable perspective.
This is a song for anyone with a broken heart:
https://jukeboxjunior.files.wordpress.com/2009/12/blue-skies-yacht-remix.mp3
December 15, 2009 2009 Top 20 Singles 1 Comment 2009alicia keysempire state of mindisaac hayesjay-zmusicnativitynew yorkrihannathe momentswhen harry met sally
[9] Jay-Z featuring Alicia Keys, 'Empire State Of Mind'
New York reminds me of Christmas anyway, possibly through seeing When Harry Met Sally a dozen too many times, so it feels right to eulogise this love letter right now. But is it a no-holds-barred tribute? Alicia Keys' commanding, spine-tingling holler sounds like a throaty homage, but The Hov swings between wallowing in the neighbourhood and caution-poem acknowledgement of those who have fallen by the wayside. It adds bite – "Mommy took a bus trip/Now she got her bust out" – and intrigue – "Good girls gone bad/The city's filled with them" – to a big, ballsy anthem.
As I cue 'Empire…' up, Junior asks, "Will it make my shoulders go funky?" Built around samples from The Moments and Isaac Hayes, I should say so. She bounds about, dangerously overexcited, and seems to know every word of Keys' contribution – until she starts singing her Nativity song over the top, sampling anew. Anyway, this one couldn't fail to move her; it's a tune as massive as Alicia's champion asset.
Don't bite the apple, Eve:
https://jukeboxjunior.files.wordpress.com/2009/12/empire-state-of-mind.mp3
December 14, 2009 2009 Top 20 Singles 1 Comment 2009black waxdananananaykroydfight-popmusicscotland
[10] Dananananaykroyd, 'Black Wax'
Not 'til this year did I realise I liked shouty Scottish "fight-pop". Thanks to Dananananaykroyd for leading me into the light. 'Black Wax' is the closest thing to a proper song on their debut album, and of course they find it a bit wimpy with hindsight. It's a kind of surf-pop with screaming, a real chorus and some lovely guitar meanders at the close – and seen live, they're a blast.
On this play for Junior, the CD skipped like billy-o, but somehow it seemed appropriate, unhinged and jumpy. She clapped along to the choppy riffs and went for a bit of hoarse-throated yelling, just to get into the spirit of the thing. Tags like "fight-pop" conjure playground knockabouts. I doubt the Dananans are long out of primary school themselves.
Highlights in yellow:
https://jukeboxjunior.files.wordpress.com/2009/12/04-black-wax.mp3

redpajama_set_name: RedPajamaCommonCrawl


text:
Journal of Statistical Distributions and Applications
The linearly decreasing stress Weibull (LDSWeibull): a new Weibull-like distribution
Roger W. Barnard1,
Chamila Perera1,
James G. Surles1 &
A. Alexandre Trindade ORCID: orcid.org/0000-0003-3133-56231
Journal of Statistical Distributions and Applications volume 6, Article number: 11 (2019) Cite this article
Motivated by an engineering pullout test applied to a steel strip embedded in earth, we show how the resulting linearly decreasing force leads naturally to a new distribution, if the force under constant stress is modeled via a three-parameter Weibull. We term this the LDSWeibull distribution, and show that inference on the parameters of the underlying Weibull can be made upon collection of data from such pullout tests. Various classical finite-sample and asymptotic properties of the LDSWeibull are studied, including existence of moments, distribution of extremes, and maximum likelihood based inference under different regimes. The LDSWeibull is shown to have many similarities with the Weibull, but does not suffer from the problem of having an unbounded likelihood function under certain parameter configurations. We demonstrate that the quality of its fit can also be very competitive with that of the Weibull in certain applications.
Mechanically stabilized earth is a method of constructing vertical retaining walls which is often seen in overpasses in populated metropolitan areas where space is at a premium. It consists of reinforcements which are buried in soil in layers. These reinforcements are attached to a vertical facing wall. The types of reinforcements vary, but are generally classified as either inextensible (steel) or extensible (polymeric). Of interest here are the steel reinforcements which are generally flat steel strips, flat steel strips with ribs on them, or welded wire mats which look like a ladder or a grid. The strip reinforcements are generally 50 mm (2 inches) wide and 4 mm (0.16 inches) thick.
Consider the case of the smooth steel strip. These types of reinforcements are not generally used in construction, but are often studied in the laboratory. If a smooth steel strip were used, while in service the stress that it would be subject to would be equal along its entire length (nominally, assuming a constant soil pressure). To establish the serviceability of these reinforcements, they are subjected to what is known as a pullout test. That is, they are embedded in backfill and an axial force is applied to the head of the reinforcement. A frictional stress is generated along the reinforcement and soil interface, and this frictional stress is cumulative so that the stress at the head of the reinforcement is equal to the total frictional stress that the entire strip is experiencing, while the stress at the middle is half of that, and so on. This results in a continuous, linearly decreasing force within the reinforcement from the head to the tail. This is much like the linkage between two cars in a train, which must withstand only the stress placed upon it by the cars behind. Thus if a train is being pulled by only locomotives in the front, the linkages between cars at the front of the train are subjected to much more stress than those near the rear of the train. (This train analogy is apt because ribbed strip or the wire mat reinforcements will largely behave in this manner.)
The data gathered from pullout tests can be used to estimate the survival distribution of reinforcements under the conditions of the pullout test, but how can it be used to estimate the survival distribution under actual service conditions?
The Weibull distribution, named for Waloddi Weibull, was popularized by Weibull in his papers from 1939 through 1961, the key paper being (Weibull 1951). It has found wide applicability in engineering practice. In his work, Weibull was studying the strength of materials, but the distribution actually appeared somewhat earlier than that in the late 1920's in the study of extreme values; see (Rinne 2009) for a thorough review. (It should be noted that Weibull was unaware of this earlier work and derived his distribution independently.) In particular, it arises as the minimum (or maximum) of a random sample with support that is bounded below (for the minimum) or above (for the maximum). The old proverb is that the strength of a chain is equal to the strength of its weakest link (the minimum). The proverb may also be applied to the strength of materials in that the strength of the material is equal to the strength of its weakest point. So it is no surprise that the Weibull distribution arises in the study of the strength of materials and has found wide applicability.
Suppose that it is reasonable to assume that a smooth steel strip reinforcement has a Weibull survival distribution were it exposed to a constant stress along its length. It is well known that the minimum of independent and identically distributed (iid) Weibull random variables has a Weibull distribution. That is, suppose that Y1,Y2,…,Yn are iid Weibull with shape β, location/threshold μ, and scale σ under the following pasteurization for the cumulative distribution function (cdf):
$$ F(y;\beta, \mu, \sigma) = 1-\exp\left\{-[(y-\mu)/\sigma]^{\beta}\right\}, \qquad y>\mu,\ \beta>0,\ \sigma>0,\ \mu>0. $$
This 3-parameter Weibull will be referred to as Weibull (μ,β,σ). Then Y(1)=min(Y1,Y2,…,Yn) has cdf given by:
$$F_{1}(y;\beta, \mu, \sigma) = 1-\left\{1-F(y;\beta, \mu, \sigma)\right\}^{n} = 1-\exp\left\{-\left[n^{1/\beta}(y-\mu)/\sigma\right]^{\beta}\right\}. $$
That is, Y(1) is Weibull with shape β, location μ, and scale σ/n1/β.
Consider now a continuous system of length L that is viewed as being composed of n independent "links" of equal length. Assume that the strength of the entire system is Weibull(β,μ,σ), and that the strengths of the individual links are also Weibull. Then each link must have a Weibull (β,μ,σn(1/β)) distribution. Note that this requires that as the number of links, n, increases, the scale increases in a corresponding fashion. That is, shorter links are stronger links (stochastically).
One end is denoted the "head" (location 0) and the other the "tail" (location L). The head is exposed to a stress S0, which decreases linearly along the system to 0 at the tail. The stress at location l is thus \(S_{l}=S_{0}\left (1-\frac {l}{L}\right)\). If we view the system as before; that is, as having a Weibull strength with it being viewed as n independent "links" that are also Weibull, what is the distribution of the system under these conditions as a function of S0?
Suppose that we have Y1,Y2,…,Yn that are iid Weibull (β,μ,σn(1/β)). The system reliability is given by:
$$ \begin{aligned} R(S_{0}) &= P(\text{All segments survive})\\ &= P\left(Y_{1}>S_{0}\left[1-\frac{0}{n}\right], Y_{2}>S_{0}\left[1-\frac{1}{n}\right], \ldots, Y_{n}>S_{0}\left[1-\frac{n-1}{n}\right]\right)\\ &= \prod_{i=0}^{n-1}P\left(Y_{i+1}>S_{0}\left[1-\frac{i}{n}\right]\right).\\ \end{aligned} $$
Note that if \(S_{0}\left [1-\frac {i}{n}\right ]<\mu \), then the probability associated with that i is 1 since the strength Yi must be greater than μ. so the product need only run to the largest k such that
$$ S_{0}\left(1-\frac{k}{n}\right) > \mu \iff k < n\left(1-\frac{\mu}{S_{0}}\right) \iff k=\text{min}\left\{n-1, \left\lfloor n\left(1-\frac{\mu}{S_{0}}\right)\right\rfloor\right\}, $$
where ⌊·⌋ is the floor function.
$$ \begin{aligned} R(S_{0}) &= \prod_{i=0}^{k}\exp\left\{-\left[\left(S_{0}\left(1-\frac{i}{n}\right)-\mu\right)/\left(\sigma n^{1/\beta}\right)\right]^{\beta}\right\}, S_{0} > \mu\\ &= \exp\left\{-\sum\limits_{i=0}^{k}\left[\left(S_{0}\left(1-\frac{i}{n}\right)-\mu\right)/\left(\sigma n^{1/\beta}\right)\right]^{\beta}\right\}.\\ \end{aligned} $$
Taking the natural log and the limit as n tends to infinity:
$$ {\lim}_{n\to\infty}\log R(S_{0}) = -{\lim}_{n\to\infty}\sum\limits_{i=0}^{k}\left[\left(S_{0}\left(1-\frac{i}{n}\right)-\mu\right)/\left(\sigma n^{1/\beta}\right)\right]^{\beta}, $$
where \(k=\text {min}\left \{n-1, \left \lfloor n\left (1-\frac {\mu }{S_{0}}\right)\right \rfloor \right \}\). For sufficiently large n, we can approximate this sum using an integral as:
$$ \begin{aligned} {\lim}_{n\to\infty}\log R(S_{0}) &\approx \frac{1}{\sigma^{\beta}}\int\nolimits_{0}^{1-\mu/S_{0}}\left[S_{0}(1-x)-\mu\right]^{\beta}dx\\ &= -\frac{\left(S_{0}-\mu\right)^{\beta+1}}{\sigma^{\beta}(\beta+1)S_{0}}, S_{0}>\mu.\\ \end{aligned} $$
Thus, \(R(S_{0}) = \exp \left \{-\frac {\left (S_{0}-\mu \right)^{\beta +1}}{\sigma ^{\beta }(\beta +1)S_{0}}\right \}, S_{0}>\mu \). Note that in the case of the standard two-parameter Weibull (μ=0), the resulting reliability is Weibull with shape β and scale σ(β+1)1/β.
Consider a reparametrization with θ=μ, γ=β+1, and δ=[σβ(β+1)]1/(β+1). This yields a cdf (which is one minus the reliability) of the form:
$$ F(x;\theta,\gamma,\delta) = 1-\exp\left\{\frac{-(x-\theta)^{\gamma}}{x\delta^{\gamma}}\right\}, \qquad\text{for \(x\geq\theta\)}. $$
Note that this bridges the gap and allows us to estimate the reliability of in-service steel strip reinforcements (which are exposed to a constant stress along its length and have a Weibull survival distribution) via the results of pullout tests (which expose the strip to a linearly decreasing stress along its length and have the distribution derived above). That is, if we obtain a sample under pullout test conditions and estimate the parameters θ, γ, and δ via maximum likelihood estimation (MLE) and obtain \(\hat {\theta }\), \(\hat {\gamma }\), and \(\hat {\delta }\), then by invariance the MLEs for the parameters of the Weibull are:
$$ \hat{\mu}=\hat{\theta}, \qquad \hat{\beta}=\hat{\gamma}-1, \qquad \hat{\sigma}=\left[\hat{\delta}^{\hat{\gamma}}/\hat{\gamma}\right]^{1/(\hat{\gamma}-1)}. $$
We term the distribution arrived at in the above discussion the linearly decreasing stress Weibull (LDSWeibull); a new Weibull-like distribution. A formal definition along with the derivation of basic and classical properties is presented in "Formal definition, basic properties, and results" section. Maximum likelihood and other types of estimation procedures, along with accompanying asymptotic results, are developed in "Estimation procedures and asymptotic results" section. These procedures are subsequently investigated with simulation studies in "Simulation results" section. We conclude the paper with an application on real data in "Real data application" section.
Formal definition, basic properties, and results
This section formally introduces the LDSWeibull and derives basic properties. It is obvious from (7) that θ is a pseudo-location parameter, γ is a shape parameter, and δ a pseudo-scale parameter. For ease of handling, it will be more convenient to work with the following one-to-one reparametrization, (θ,γ,δ)↦β=(θ,γ,τ), where τ=δγ⇔δ=τ1/γ, whence τ>0 is more obviously seen to be a pseudo-scale parameter.
Definition 1
The LDSWeibull (θ,γ,τ) has parameter space: Ω={(θ,γ,τ):θ≥0, γ>1, τ>0}. Its cdf is given by:
$$ F(x;\theta,\gamma,\tau) = \int\limits_{\theta}^{\infty}f(x;\theta,\gamma,\tau)\,dx =1-\exp\left\{-\frac{(x-\theta)^{\gamma}}{x\tau}\right\}, \qquad\text{for }x\geq\theta. $$
The density function is therefore:
$$ f(x;\theta,\gamma,\tau) = \frac{(x-\theta)^{\gamma}}{x\tau}\left[\frac{\gamma}{x-\theta}-\frac{1}{x}\right]\exp\left\{-\frac{(x-\theta)^{\gamma}}{x\tau}\right\} I_{[\theta,\infty)}(x). $$
Note that the LDSWeibull inherits parameter identifiability from the Weibull, since the transformation (8) is one-to-one.
For θ=0 the LDSWeibull (0,γ,τ) is a two-parameter Weibull with shape γ−1, and scale τ1/(γ−1).
Apart from its intimate connection with the Weibull, there are possibly many related distributions that overlap with the proposed LDSWeibull. Note that written in the form
$$G(x;\alpha,\boldsymbol{\beta}) = 1-\exp\left\{-\alpha H(x;\boldsymbol{\beta})\right\},\qquad\alpha>0, $$
the cdf can more generally be seen to be a member of the very broad class of distributions introduced by (Gurvich et al. 1997), that can be generated from the Weibull by taking H(x;β) to be a non-negative and monotone increasing function, possibly depending on the vector of parameters β. In our case, the LDSWeibull (θ,γ,τ) is obtained by setting α=1/τ and H(x;θ,γ)=(x−θ)γ/x. (Bourguignon et al. 2014) introduce an interesting variant of G(x;α,β) by taking H(x;β) to be a positive power of the ratio of any continuous cdf and its survival function, but the LDSWeibull (θ,γ,τ) does not appear to obey that particular construct.
Existence of moment generating function
We have managed to determine necessary and sufficient conditions for existence of the moment generating function (mgf). These conditions impose a restriction on the shape parameter γ.
Theorem 1
The mgf of the LDSWeibull (θ,γ,τ) satisfies \(M(t) = \mathbb {E} e^{tX}<\infty \), for all \(t\in \mathbb {R}\) in the restricted parameter range:
$$\Omega_{C}=\{(\theta,\gamma,\tau):\theta\geq 0,\ \gamma>2,\ \tau>0\}\subset\Omega. $$
Otherwise, M(t)=∞ for all \(t\in \mathbb {R}\) if γ≤2.
See Appendix: Proof of Theorem 1. □
Attempts at finding a closed form for M(t) (in terms of known special functions) have not, however, yielded positive results. This leads to challenges in devising preliminary parameter estimators such as method of moments. Due to the lack of an analytic expression for the quantile function, and the usual intractability of moments of order statistics, alternative treatments such as probability weighted moments (Greenwood et al. 1979) and L-moments (Hosking 1990), do not appear to be feasible either.
Completeness and minimal sufficiency
There is little hope in being able to determine a complete statistic, but it's not hard to show that the order statistics are minimal.
For a random sample X=(X1,…,Xn) from the LDSWeibull (θ,γ,τ), the order statistics T(X)=(X(1),…,X(n)) are minimal sufficient.
Let x=(x1,…,xn) and y=(y1,…,yn) denote two independent random samples from the LDSWeibull (θ,γ,τ), where the log-density of f(x) is given by:
$$ {\begin{aligned} \log f(\boldsymbol{x}) = \sum\log f(x_{i}) = (\gamma-1)\sum\log(x_{i}-\theta) + \sum\log [(\gamma-1)x_{i}+\theta ]-\sum\frac{(x_{i}-\theta)^{\gamma}}{\tau x_{i}}-n\log\tau-\sum\log{x_{i}}^{2}. \end{aligned}} $$
We note that the family is (trivially) dominated by Lebesgue measure, and hence invoking (Schervish (1995), Theorem 2.29), we need only show that, for any fixed choice of (θ,γ,τ):
$$\log f(\boldsymbol{x})-\log f(\boldsymbol{y}) = h(\boldsymbol{x},\boldsymbol{y})\quad\Longleftrightarrow\quad T(\boldsymbol{x})=T(\boldsymbol{y}). $$
To this end, and ignoring summands that depend on x and/or y only, note that
$$ {\begin{aligned} \log f(\boldsymbol{x})-\log f(\boldsymbol{y}) = (\gamma-1)\sum\log\left(\frac{x_{i}-\theta}{y_{i}-\theta}\right) + \sum\log\left(\frac{(\gamma-1)x_{i}+\theta}{(\gamma-1)y_{i}+\theta}\right) + \frac{1}{\tau}\sum\left[\frac{(y_{i}-\theta)^{\gamma}}{y_{i}}-\frac{(x_{i}-\theta)^{\gamma}}{x_{i}}\right]. \end{aligned}} $$
Now, it is obvious that T(x)=T(y) immediately implies logf(x)− logf(y)=0, which is therefore independent of the parameters. To see that the converse is also true, note that the only way (11) can be independent of (θ,γ,τ), is if each of the three summands is itself free of (θ,γ,τ), whence we must have
$$\begin{array}{@{}rcl@{}} \sum\log(x_{i}-\theta) &=& \sum\log(y_{i}-\theta)\\ \sum\log [(\gamma-1)x_{i}+\theta] &=& \sum\log [(\gamma-1)y_{i}+\theta ]\\ \sum \frac{(x_{i}-\theta)^{\gamma}}{x_{i}} &=& \sum \frac{(y_{i}-\theta)^{\gamma}}{y_{i}} \end{array} $$
Because of the intimate connections with θ and γ, these three requirements can only be met if T(x)=T(y). □
Distribution of the extremes
For a random sample X1,…,Xn from the LDSWeibull (θ,γ,τ), we now consider the distributions of the minimum and maximum order statistics, X(1) and X(n), respectively. Some exact results can be obtained using the usual techniques. Specifically, the survival function of X(1) is given by
$$P(X_{(1)}>x) = [1-F(x)]^{n} = \exp\left\{-\frac{(x-\theta)^{\gamma}}{x(\tau/n)}\right\}, $$
which implies that X(1) is therefore LDSWeibull (θ,γ,τ/n). The cdf of X(n) is of course F(x)n, but this does not appear to have an immediately recognizable form.
It is also possible to obtain the asymptotic distribution of the (appropriately normalized) extremes by invoking the Fisher-Tippett Theorem; see e.g., David and Nagaraja (2003, §10.5). The following theorem reveals that the extremes of the LDSWeibull are in the domain of attraction of the Gumbel.
Let X(1) and X(n) denote the minimum and maximum order statistics, respectively, in a random sample from the LDSWeibull (θ,γ,τ) with parameter space Ω, as in Definition 1. Then we have the following convergence in distribution results, for any \(x\in \mathbb {R}\).
For the maximum,
$${\lim}_{n\rightarrow\infty}P\left(\frac{X_{(n)}-a_{n}}{b_{n}}\leq x\right) = \exp\left\{-e^{-x}\right\}. $$
For the minimum,
$${\lim}_{n\rightarrow\infty}P\left(\frac{X_{(1)}-a_{n}}{b_{n}}\leq x\right) = 1-\exp\left\{-e^{x}\right\}. $$
In each case, the normalizing constants an and bn can be chosen to satisfy the pair of equations
$$(a_{n}-\theta)^{\gamma}-\tau a_{n}\log(n)=0, \qquad\text{and}\qquad b_{n}=\frac{a_{n}(a_{n}-\theta)}{[a_{n}(\gamma-1)+\theta]\log(n)}.$$
Note that the derivative of the inverse of the hazard function is
$$ {\begin{aligned} \frac{\partial}{\partial x}{\Bigg[\frac{1-F(x)}{f(x)}\Bigg]} &= \frac{\partial}{\partial x}{\Bigg[\frac{x^{2}\tau}{(x-\theta)^{\gamma-1}[x(\gamma-1)+\theta]}\Bigg]}\\ &= \frac{2x\tau}{(x-\theta)^{\gamma-1}(x(\gamma-1)+\theta)}-\frac{x^{2}\tau(\gamma-1)}{(x-\theta)^{\gamma}[x(\gamma-1) +\theta]}-\frac{x^{2}\tau(\gamma-1)}{(x-\theta)^{\gamma-1}[x(\gamma-1)+\theta]^{2}}. \end{aligned}} $$
Since γ>1, each of these terms is of order O(1/xε), for some ε>0, and thus they all converge to zero as x→∞. The required result now follows by invoking (David and Nagaraja (2003), Theorem 10.5.2). □
Estimation procedures and asymptotic results
For a random sample x1,…,xn from the LDSWeibull (θ,γ,τ), we have the log-likelihood function:
$$ {\begin{aligned} \ell(\boldsymbol{\beta}) = \left\{\begin{array}{lr} (\gamma-1)\sum\log(x_i-\theta) + \sum\log [(\gamma-1)x_i+\theta ]-\sum\frac{(x_i-\theta)^{\gamma}}{\tau x_i}-n\log\tau-\sum\log{x_i}^{2}, & \theta\leq x_{(1)}, \\ -\infty, & \theta>x_{(1)}. \end{array}\right. \end{aligned}} $$
Denote by β0=(θ0,γ0,τ0)T the true parameter vector.
Note that, unlike the Weibull, this log-likelihood function is bounded and will therefore have a non-degenerate extremum for all (θ,γ,τ)∈Ω. As discussed in (Rinne (2009), §11.3.2), a known issue with the 3-parameter Weibull in (1) is that as μ→x(1), (β−1) log(x(1)−μ)→∞ for β<1, and therefore the MLE of β does not exist when β<1.
To demonstrate the spectrum of possibilities for the various regimes of the MLEs, we will now consider the following subset of just three special cases taken from the exhaustive list of all 7 possible combinations of known and unknown parameter values.
Case 1: (γ,τ) known
It would appear that the maximizer of θ would occur at the boundary value of x(1), however the first two derivatives yield:
$$\begin{array}{@{}rcl@{}} \psi_{1}(\boldsymbol{\beta})\equiv\frac{\partial\ell(\boldsymbol{\beta})}{\partial\theta} &=& -\sum\frac{(\gamma-1)}{x_i-\theta}+\sum\frac{1}{(\gamma-1)x_i+\theta}+\sum\frac{\gamma(x_i-\theta)^{\gamma-1}}{\tau x_i} \\ \frac{\partial^{2}\ell(\boldsymbol{\beta})}{\partial\theta^{2}} &=& -\sum\frac{(\gamma-1)}{(x_i-\theta)^{2}}-\sum\frac{1}{[(\gamma-1)x_i+\theta]^{2}}-\sum\frac{\gamma(\gamma-1)(x_i-\theta)^{\gamma-2}}{\tau x_i}. \end{array} $$
Since each of the summands in the second derivative is positive, it follows that ∂2ℓ/∂θ2<0, whence ℓ(β) is concave in θ, and thus the MLE is the unique maximum of ℓ(β), and occurs at an interior point, albeit close to x(1) (which can therefore be used as an initial estimate).
Let β0 be in the restricted parameter range ΩC, with γ0 and τ0 known. Then the MLE \(\hat {\theta }_{{\gamma }_{0},\tau _{0}}\) of θ0 is consistent.
Take the (continuous) estimating equation (of which \(\hat {\theta }_{{\gamma }_{0},\tau _{0}}\) is the unique root by the above argument) to be
$$\begin{array}{@{}rcl@{}} g_{\gamma,\tau}(\theta)\!\equiv\frac{1}{n}\frac{\partial\ell(\boldsymbol{\beta})}{\partial\theta}&\,=\,&\frac{1}{n} \sum\limits_{i=1}^{n}\frac{1}{(\gamma-1)x_i+\theta} + \frac{\gamma}{\tau}\frac{1}{n}\sum\limits_{i=1}^{n}\frac{(x_i-\theta)^{\gamma-1}}{x_i}\!-(\gamma-1)\frac{1}{n}\sum\limits_{i=1}^{n}\frac{1}{x_i-\theta}\\ &{\xrightarrow{p}} & \mathbb{E}\left[\frac{1}{(\gamma-1)X+\theta}\right]+\frac{\gamma}{\tau}\mathbb{E}\left[\frac{(X-\theta)^{\gamma-1}}{X}\right]- (\gamma-1)\mathbb{E}\left[\frac{1}{X-\theta}\right], \end{array} $$
by the weak law of large numbers applied to each of the sample averages. Since this limiting value is well-defined for each θ, consistency of the unique root \(\hat {\theta }_{{\gamma }_{0},\tau _{0}}\) follows by (Van der Vaart (1998), Lemma 5.10). □
Case 2: θ known
With the argument r as placeholder for γ, first define the terms:
$$ Q_{\theta}=\sum\limits_{i=1}^{n}\log(x_i-\theta), \quad R_{\theta}(r)=\sum\limits_{i=1}^{n}\frac{x_i}{(r-1)x_i+\theta}, \quad S_{\theta}(r)=\sum\limits_{i=1}^{n}\frac{(x_i-\theta)^{r}}{x_i}, $$
and note that
$$S^{\prime}_{\theta}(r)=\frac{\partial S_{\theta}(r)}{\partial r}=\sum\frac{(x_i-\theta)^{r}\log(x_i-\theta)}{x_i}. $$
The corresponding score functions are then:
$$\begin{array}{@{}rcl@{}} \psi_{2}(\boldsymbol{\beta})\equiv\frac{\partial\ell(\boldsymbol{\beta})}{\partial\gamma} &=& Q_{\theta}+R_{\theta}(\gamma)-\frac{1}{\tau}S^{\prime}_{\theta}(\gamma) \\ \psi_{3}(\boldsymbol{\beta})\equiv\frac{\partial\ell(\boldsymbol{\beta})}{\partial\tau} &=& \sum\frac{(x_i-\theta)^{\gamma}}{\tau^2x_i} - \frac{n}{\tau} \end{array} $$
Solving ψ3(β)=0 leads to the profile MLE for τ, \(\hat {\tau }_{\theta }=S_{\theta }(\gamma)/n\), whence substitution into ψ2 leads to the profile score equation for γ
$$ \psi_{2}(\theta,\gamma,\hat{\tau}_{\theta})=Q_{\theta} + R_{\theta}(\gamma) - \frac{nS'_{\theta}(\gamma)}{S_{\theta}(\gamma)}=0, $$
with solution \(\hat {\gamma }_{\theta }\).
It will be more convenient to write the score function (15) in normalized form:
$$ h_{\theta}(r)\equiv\frac{1}{n}\psi_{2}(\theta,\gamma,\hat{\tau}_{\theta})=\frac{1}{n}Q_{\theta} + \frac{1}{n}R_{\theta}(r) - \frac{S'_{\theta}(r)}{S_{\theta}(r)}. $$
Thus the MLEs for γ0 and τ0 satisfy the equations
$$h_{\theta_{0}}(\hat{\gamma}_{\theta_{0}})=0, \qquad\text{and} \qquad \hat{\tau}_{\theta_{0}} = \frac{1}{n}S_{\theta_{0}}(\hat{\gamma}_{\theta_{0}}), $$
where, due to the monotonicity property in Proposition 1, \(\hat {\gamma }_{\theta _{0}}\) is easily determined as either the boundary value \(\hat {\gamma }_{\theta _{0}}=1\), or as the unique root of \(h_{\theta _{0}}(r)\) in (16).
The function hθ(r) in (16) is monotone decreasing over the interval r>1.
First note that Rθ(r) is monotone decreasing, since
$$R^{\prime}_{\theta}(r) = -\sum\left[\frac{x_i}{(r-1)x_i+\theta}\right]^2<0. $$
We now show that the term Tθ(r)=Sθ′(r)/Sθ(r) is monotone increasing, whence the desired result will follow since hθ(r) will then be the sum of a constant term, Qθ, and two monotone decreasing functions. To this end, write \(S_{\theta }(r)=\sum y_{i}^r/x_i=tM(r)\), where \(y_i=x_i-\theta >0, t=\sum x_{i}^{-1}\), and \(M(r)=\sum p_ie^{rz_i}\) corresponds to the moment generating function of a discrete random variable (say Z), with values zi= logyi, and masses 0≤pi=(txi)−1≤1, i=1,…,n. This is sufficient to establish the result, since noting that the cumulant generating function K(r)= logM(r) is convexFootnote 1, we have
$$T^{\prime}_{\theta}(r) = \frac{S_{\theta}(r)S_{\theta}^{\prime\prime}(r)-S_{\theta}^{\prime}(r)^{2}}{S_{\theta}(r)^{2}} = \frac{M_{\theta}(r)M_{\theta}^{\prime\prime}(r)-M_{\theta}^{\prime}(r)^{2}}{M_{\theta}(r)^{2}} = K^{\prime\prime}(r)>0. $$
The question of whether or not \(\hat {\gamma }_{\theta _{0}}\) ever attains the boundary value of 1 is interesting. It is certainly possible to construct a set of real values x1,…,xn such that hθ(1)<0, but whether or not such values correspond to bona fide realizations from an LDSWeibull over some set with positive measure remains an open issue. It is however possible to establish a limiting result as follows.
Let β0 be in the restricted parameter range ΩC, and suppose that plimn→∞hθ(1)>0. Then, with probability 1 in the limit as n→∞, \(\hat {\gamma }_{\theta _{0}}\) is the unique root of \(h_{\theta _{0}}(r)\).
Due to Proposition 1, it suffices to show that plimn→∞hθ(∞)<0. Defining Y=X−θ, note that Y>0 a.s., and \({\text {plim}_{n\rightarrow \infty }} Q_{\theta }/n=\mathbb {E}\log Y\) by the weak law of large numbers. (Note that the finiteness of all moments for logY follows from the finiteness of all moments for X with parameters in ΩC.) Since \(R_{\theta }(\infty)={\lim }_{r\rightarrow \infty }R_{\theta }(r)/n=0\), it follows that \(R_{\theta }(\infty)/n{\xrightarrow {p}} 0\). Now assume (without loss of generality) that 0<y1=y(1)≤⋯≤y(n)<∞ are ordered, and note that in view of the representation
$$T_{\theta}(r)= \sum\limits_{i}\left(\frac{y_{i}^{r} /x_i}{\sum\nolimits_{j} y_{j}^{r} /x_j}\right)\log y_{i} =\sum\limits_{i}\frac{\left(\frac{y_{i}}{y_{1}}\right)^{r} /x_i}{\sum\nolimits_{j}\left(\frac{y_{j}}{y_{1}}\right)^{r} /x_j}\log y_{i}, $$
Lemma 1 in the Appendix is applicable with \(c_{i}(r)=(\frac {y_i}{y_1})/x_{i}^{1/r}\), since for sufficiently large r, we have for i<j:
$$c_{i}(r)^r=\left(\frac{y_i}{y_1}\right)^{r}\frac{1}{y_i+\theta} < \left(\frac{y_j}{y_1}\right)^{r}\frac{1}{y_j+\theta}=c_{j}(r)^{r}, $$
whence it follows that Tθ(∞)= logy(n) and therefore plimn→∞Tθ(∞)=∞. Putting everything together gives:
$$h(\infty){\xrightarrow{p}} \mathbb{E}\log Y + 0 - \infty < 0. $$
Identifiability of the LDSWeibull model combined with third order differentiability of logf(x;β), plus domination of appropriate derivatives of the latter as well as f(x;β) by integrable functions, establishes consistency and asymptotic efficiency of the MLEs directly via classical conditions.
Let β0 be in the restricted parameter range ΩC, with θ0 known. Then, the MLEs \(\hat {\gamma }_{\theta _{0}}\) and \(\hat {\tau }_{\theta _{0}}\) of γ0 and τ0, respectively, satisfy:
$$\sqrt{n}\left(\left(\begin{array}{c} \hat{\gamma}_{\theta_{0}} \\ \hat{\tau}_{\theta_{0}}\end{array}\right) -\left(\begin{array}{c} {\gamma}_{0} \\ \tau_{0} \end{array}\right)\right) \xrightarrow{d}\mathcal{N}\left({0}, J^{-1}(\boldsymbol{\beta}_{0})I(\boldsymbol{\beta}_{0})J^{-1}(\boldsymbol{\beta}_{0})\right), $$
where J(β) is the Hessian matrix of logf(x;β), and I(β) is the Fisher Information matrix (per observation), each defined accordingly as:
$$ {\begin{aligned} J(\boldsymbol{\beta})\equiv\mathbb{E}\left[\begin{array}{cc} \frac{\partial^{2}\log f(x;\boldsymbol{\beta})}{\partial\gamma^{2}} & \frac{\partial^{2}\log f(x;\boldsymbol{\beta})}{\partial\gamma\partial\tau} \\ \frac{\partial^{2}\log f(x;\boldsymbol{\beta})}{\partial\gamma\partial\tau} & \frac{\partial^{2}\log f(x;\boldsymbol{\beta})}{\partial\tau^{2}} \\ \end{array}\right], \qquad I(\boldsymbol{\beta})\equiv\mathbb{E}\left[\begin{array}{cc} \left(\frac{\partial\log f(x;\boldsymbol{\beta})}{\partial\gamma}\right)^2 & \frac{\partial\log f(x;\boldsymbol{\beta})}{\partial\gamma}\frac{\partial\log f(x;\boldsymbol{\beta})}{\partial\tau} \\ \frac{\partial\log f(x;\boldsymbol{\beta})}{\partial\gamma}\frac{\partial\log f(x;\boldsymbol{\beta})}{\partial\tau} & \left(\frac{\partial\log f(x;\boldsymbol{\beta})}{\partial\tau}\right)^2 \end{array}\right]. \end{aligned}} $$
Case 3: All parameters unknown
To start, a nonparametric estimator of the survival (or reliability) function should be provided for each x(i), where x(i) is the i-th order statistic. The usual empirical survival function is \(\hat {S}(x_{(i)}) = (n-i)/n\), but we employ instead a common adjustment, \(\hat {S}(x_{(i)})=(n-i+1)/(n+1)\) to avoid the problematic situation of log(0) when i=n.
Now replace θ with the consistent estimate x(1), and equate empirical and population survival functions at x(i):
$$ 1-\hat{F}(x_{(i)})=\frac{n-i+1}{n+1} \approx \exp{\left(-\frac{(x_{(i)}-x_{(1)})^{\gamma}}{\tau x_{(i)}}\right)} = 1-F(x_{(i)}). $$
(A perhaps more common justification of (18) is to note the well-known property of uniform order statistics: \(\mathbb {E}[F(X_{(i)})]=i/(n+1)\).) Performing a log-log transformation of both sides then leads to:
$$y_{i}\equiv\log{\left(-\log{\left(\frac{n-i+1}{n+1}\right)}\right)}+\log(x_{(i)}) \approx \gamma \log{(x_{(i)}-x_{(1)})}-\log\tau\equiv\gamma z_{i}-\log\tau. $$
Denoting by yi the left-hand-side term of the above expression, and zi= log(x(i)−x(1)), we have, with the addition of the error term εi=yi−(a+bzi), the linear regression model
$$ y_i = a+{bz}_i+\varepsilon_i, \qquad i=2,\ldots,n, \quad\text{with }b=\gamma \text{ and } a=-\log\tau. $$
Obtaining the least squares estimates \(\hat {a}\) and \(\hat {b}\) for the regression parameters, yields the following starting values for the LDSWeibull model parameters:
$$ \theta^{(0)}=x_{(1)}, \qquad \gamma^{(0)}=\hat{b}, \qquad\tau^{(0)}=\exp{(-\hat{a})}. $$
Armed with these initial values, which are consistent by the next theorem, one can employ an efficient optimization algorithm to maximize (13) and obtain the MLEs \((\hat {\theta },\hat {\gamma },\hat {\tau })\). We note in passing that the procedure outlined above is nearly identical to the so-called "regression method" for estimating the parameters of the generalized extreme value distribution; see e.g., (Rinne (2009), Chapter 10).
(Central quantile limiting behavior) Note that consistent estimation of the right-hand-side of (18) subsumes the following limiting behavior for the integer 2≤i≤n appearing in the left-hand-side of (18):
$$ {\lim}_{n\rightarrow\infty}\frac{i}{n+1} = q_i,\qquad 0<q_i=F(\xi_i)<1, $$
where ξi is the population quantile corresponding to qi, and for notational expedience we omit the implicit dependence i≡i(n) in the limiting behavior of order statistics for the central quantile case (David and Nagaraja (2003), Chapter 10). (Note however that qi and ξi on the right-hand-side of (21) do not depend on n.)
Let β0 be in the restricted parameter range ΩC, and assume the central quantile limiting behavior in Remark 3. Then, the initial estimates β(0)=(θ(0),γ(0),τ(0))T given by (20) resulting from a random sample X1,…,Xn from the LDSWeibull (θ,γ,τ), are consistent for β0.
Simulation results
In this section we carry out a small simulation study to investigate the sampling properties of the MLEs for the three cases defined in "Estimation procedures and asymptotic results" section. To this end, Tables 1, 2, and 3 report the bias, variance, mean squared error (MSE), and coefficient of variation (CV) of the MLEs, empirically determined from 1000 simulated realizations.
Table 1 Summary statistics for the MLE of θ under Case 1: γ and τ are known
Table 2 Summary statistics for the MLEs of γ and τ under Case 2: θ is known
Table 3 Summary statistics for the MLEs of θ,γ and τ under Case 3: all parameters are unknown
We see a consistent decrease in all the metrics (bias, variance, MSE, CV) with increasing sample size, as expected. Interestingly, it appears that in general the parameter τ suffers from the most uncertainty, particularly noticeable in some large CV values at low samples.
Real data application
The motivating derivation of the LDSWeibull in Section 1 would behoove us to apply it to the results of a pullout test in order to infer the parameters of the underlying Weibull according to (8). In lack of such data, in this section we illustrate an application where the LDSWeibull (θ,γ,τ) provides a competitive fit to the Weibull (μ,β,σ).
To assess the prospective wind power at a given site, a distribution is often fit to the observed wind speeds. Although different locations tend to have different wind speed profiles, the Weibull has been found to closely mirror the actual distribution of hourly/ten-minute wind speeds at many locations (Masters 2013). In these cases the Weibull shape parameter β is often close to 2, and a Rayleigh distribution can therefore be used, offering a less accurate but simpler model.
The R package bReeze contains the data set "winddata", consisting of measured wind speed and direction at 10-min intervals collected by a meteorological mast, for a total of 36,548 consecutive observations on 17 variables. Of these variables, we selected winddata$v1_40m_max, which contains the maximum wind speed (m/s) over each 10-min interval recorded by the mast at a height of 40m above ground level. We divided up this long time series into 252 shorter time series of length n=144, each comprising the maximum wind speeds over a 24 h period (144 10-min intervals). The 6 (anomalous) wind speed values of zero were simply discarded before creating the resulting 252 time series data sets.
Parameters for the two distributions were estimated for each of these 252 data sets, and the differences in the attained maximized log-likelihood (LDSWeibull minus Weibull) recorded. No parameter restriction were placed on the LDSWeibull (θ,γ,τ), but for compatibility with the LDSWeibull and the reason mentioned in Remark 2, the parameter space for the Weibull (μ,β,σ) was restricted to μ>0, β≥1, and σ>0. Summary statistics for these log-likelihood differences are listed on Table 4. We can see that, although Weibull fits better approximately 75% of the time, the difference is typically very small.
Table 4 Summary statistics for the attained differences in the maximized log-likelihoods between LDSWeibull (θ,γ,τ) and Weibull (μ,β,σ) fits to each of the 252 daily time series data sets created from "winddata"
The top panels of Figs. 1 and 2 show some typical series where the differences in the log-likelihood are in excess of 5, and between 1 and 5, respectively. The corresponding LDS vs. Weibull marginal fits are displayed in the bottom panels as solid and dot-dash lines, respectively. The dashed KDE line tracking the shaded histogram corresponds to kernel density estimation. These plots typify two regimes: (i) generally calm days with a moderate burst of wind in Fig. 1, and (ii) a windy day with higher bursts (possibly as a consequence of a storm) in Fig. 2. The first regime is characterized by Weibull fits that coincide with an exponential (β=1), whereas the second is more of the Rayleigh type (β≈2).
Illustration of the LDS vs. Weibull marginal fits for selected daily time series of "winddata" where the difference in the attained maximized log-likelihood is in excess of 5. The dashed line tracking the shaded histogram corresponds to kernel density estimation (KDE, with the Sheather-Jones plug-in bandwidth)
Illustration of the LDS vs. Weibull marginal fits for selected daily time series of "winddata" where the difference in the attained maximized log-likelihood is between 1 and 5. The dashed line tracking the shaded histogram corresponds to kernel density estimation (KDE) with the Sheather-Jones plugin bandwith
Although we do not seek an exhaustive analysis here but merely an illustrative one, it is interesting to consider the question of goodness-of-fit. Anderson-Darling (AD) and Kolmogorov-Smirnov (KS) tests yield p-values lower than 10−4 in all the cases of Fig. 1, confirming the suspicion that neither distribution is sufficiently rich to capture this regime. The second regime of Fig. 2 is different however, as shown in Table 5. At the usual 5% significance level, the Weibull model only resoundly fits on Day 116, whereas the LDSWeibull fits in all but Day 26. In all of these examples, the distinctive feature is that the LDSWeibull model appears to be better able to resolve the peaks.
Table 5 Anderson-Darling (AD) and Kolmogorov-Smirnov (KS) goodness-of-fit test p-values for the fitted Weibull and LDSWeibull models of Fig. 2
Lemma 1
Let 0<y1≤⋯≤yn<∞ be an ordered sample of positive real numbers. Then, for any continuous function g(·),
$$U_{n}(r) \equiv \sum\limits_{i=1}^n \frac{c_{i}(r)^{r}}{\sum\nolimits_{j=1}^n c_{j}(r)^{r}}g(y_i) \longrightarrow g(y_n), \quad\text{as }r\rightarrow\infty, $$
provided that for some sufficiently large r∗, we have the ordering 0<c1(r)<⋯<cn(r)<∞ for all r≥r∗.
$$\begin{array}{@{}rcl@{}} U_{n}(r) &=& \sum\limits_i \frac{c_{i}^rg(y_i)}{\sum\nolimits_j c_{j}^{r}} = \sum \frac{g(y_i)}{\sum (c_j/c_i)^{r}} \\ &=& \sum\limits_{i=1}^n \frac{g(y_i)}{(c_1/c_i)^r+\cdots+(c_{i-1}/c_i)^r+1+(c_{i+1}/c_i)^r+\cdots+(c_{n}/c_i)^{r}}. \end{array} $$
Considering the terms in the denominator of the above summand, note that, for r≥r∗, since ci/cj<1 if i<j, and ci/cj>1 if i>j, we have that:
$${\lim}_{r\rightarrow\infty}\left(\frac{c_i}{c_j}\right)^r= \left\{\begin{array}{ll} 0, & \text{if }i<j, \\ \infty, & \text{if }i>j. \\ \end{array}\right. $$
Thus the first n−1 denominators of Un(r), corresponding to i=1,…,n−1, converge to ∞ as r→∞, while the last denominator converges to 1, which gives:
$${\lim}_{r\rightarrow\infty} U_{n}(r) = \frac{g(y_1)}{\infty} +\cdots+ \frac{g(y_{n-1})}{\infty} + \frac{g(y_n)}{1} = g(y_n). $$
Let Xn\,k,1≤k≤n, be a triangular array of random variables such that (i) plimn→∞Xn,k=Xk, and (ii) |Xn,k|≤|Y| a.s. for all n, with \(\mathbb {E}|Y|<\infty \). Then, it follows that:
$${\text{plim}_{n\rightarrow\infty}} \sum\limits_{k=1}^{n}X_{n,k} = \sum\limits_{k=1}^{\infty}{\text{plim}_{n\rightarrow\infty}} X_{n,k}=\sum\limits_{k=1}^{\infty}X_{k}. $$
By (Serfling (1980), Theorem §1.3.6), the hypothesized conditions on the sequence Xn,k imply that \(X_{n,k}{\xrightarrow {L_{1}}} X_{k}\), that is, \({\lim }_{n\rightarrow \infty }\mathbb {E}|X_{n,k}-X_{k}|=0\). Then, invoking the triangle inequality, we have, with the understanding that Xn,k=0 a.s. for k>n, that
$$\begin{array}{@{}rcl@{}} \left|\sum\limits_{k=1}^{n}X_{n,k}-\sum\limits_{k=1}^{\infty}X_{k}\right| &=& \left|\sum\limits_{k=1}^{n}\left(X_{n,k}-X_{k}\right)-\sum\limits_{k=n+1}^{\infty}X_{k}\right| \leq \sum\limits_{k=1}^{\infty}\left|X_{n,k}-X_{k}\right|+\sum\limits_{k=n+1}^{\infty}\left|X_{k}\right|, \end{array} $$
whence
$${\lim}_{n\rightarrow\infty}\mathbb{E}\left|\sum\limits_{k=1}^{n}X_{n,k}\,-\,\sum\limits_{k=1}^{\infty}X_{k}\right| \leq {\lim}_{n\rightarrow\infty}\sum\limits_{k=1}^{\infty}\mathbb{E}\left|X_{n,k}\,-\,X_{k}\right|+{\lim}_{n\rightarrow\infty}\sum\limits_{k=n+1}^{\infty}\mathbb{E}\left|Y\right| \,=\, \sum\limits_{k=1}^{\infty}0+0=0, $$
and therefore \(\sum \nolimits _{k=1}^{n}X_{n,k}{\xrightarrow {L_{1}}} \sum \nolimits _{k=1}^{\infty }X_{k} \left (\text {and}\ \sum \nolimits _{k=1}^{\infty }X_{n,k}{\xrightarrow {L_{1}}} \sum \nolimits _{k=1}^{\infty }X_{k}\right)\). The result now follows because convergence in the L1 norm implies convergence in probability (Serfling (1980), Theorem §1.3.2). □
Proof of Theorem 1
We will show that
$$ M(t)=\mathbb{E}\left(e^{tX}\right)={\int\limits_{\theta}^{\infty}\frac{1}{\tau x}e^{tx}(x-\theta)^{\gamma} \left[\frac{\gamma}{(x-\theta)}-\frac{1}{x}\right]e^{\frac{-(x-\theta)^{\gamma}}{x\tau}}\,dx}<\infty, $$
for γ>2,τ>0 and θ≥0, in a neighborhood 0<t<ε. (Note that it suffices to consider t>0 throughout since e−tx<etx, for t>0). Letting x=y+θ, the mgf becomes
$$ M(t)=\frac{e^{t\theta}}{\tau}\displaystyle{\int\limits_{0}^{\infty} {\left[\frac{y^{\gamma}}{y+\theta}\left(\frac{\gamma}{y}-\frac{1}{y+\theta}\right)\right]\, \exp\left\{ty-\frac{y^{\gamma}}{{(y+\theta)}\tau}\right\}\,dy}}, $$
and note that we only need to check convergence for θ near 0 and ∞. Hence we split the proof into the following 3 cases. Case θ=0. Although this corresponds to a Weibull distribution, for which existence of the mgf is well-known Rinne (2009), we outline here a new argument that will bound the mgf, and will subsequently be repeated with minor changes in the θ>0 case.
$$\begin{array}{*{20}l} M(t)&=\frac{1}{\tau}{\int\limits_{0}^{\infty}{e^{tx}x^{\gamma-1}\left[\frac{\gamma-1}{x}\right] e^{{\frac{-x^{\gamma-1}}{\tau}}}}\,dx}=\frac{\gamma-1}{\tau}{\int\limits_{0}^{\infty}\exp \left\{tx-\frac{1}{\tau}x^{\gamma-1}\right\}x^{\gamma-2}}\,dx\\ &=\frac{\gamma-1}{\tau}{\int\limits_{0}^{\infty}\exp\left\{tx-\frac{1}{2\tau}x^{\gamma-1} -\frac{1}{2\tau}x^{\gamma-1}\right\}x^{\gamma-2}\,dx}. \end{array} $$
Now, splitting the integral, we have
$$M(t)=\frac{\gamma-1}{\tau}\left[\underbrace{{\int\limits_{0}^{b}{e^{{x} (t-{\frac{1}{2\tau}x^{\gamma-2}})}{e^{-\frac{1}{2\tau}x^{\gamma-1}}}x^{\gamma-2}}\,dx}}_{A}+ \underbrace{{\int\limits_{b}^{\infty}{e^{{x}(t-{\frac{1}{2\tau}x^{\gamma-2}})} {e^{-\frac{1}{2\tau}x^{\gamma-1}}}x^{\gamma-2}}\,dx}}_{B}\right], $$
and we seek to bound each of the integrals A and B, for some b>0 sufficiently large. Since A constitutes the integral of a smooth function over a finite range, it follows immediately that A<∞. For B, since for x>b sufficiently large, γ>2, and any fixed t>0 and τ>0, we have xt<xγ−1/(2τ) which implies exp{x(t−τ−1xγ−2/2)}<1, this term can be dropped from the integrand of B. Performing the substitution y=xγ−1/(2τ), then leads to
$$0<B<{\int\limits_{b}^{\infty}{e^{\frac{-x^{\gamma-1}}{2\tau}}{x^{\gamma-2}}\,dx}}< {\int\limits_{0}^{\infty}{e^{\frac{-x^{\gamma-1}}{2\tau}}{x^{\gamma-2}}\,dx}} =\frac{2\tau}{\gamma-1}{\int\limits_{0}^{\infty}{e^{-y}\,dy}}<\infty. $$
Case θ>0. Starting from (23), we also separate the integral into two,
$$\begin{aligned} M(t)&=\frac{e^{t\theta}}{\tau}\underbrace{\int\limits_{0}^{b}{e^{ty-\frac{y^{\gamma}}{{2(y+\theta)}\tau}-\frac{y^{\gamma}}{{2(y+\theta)}\tau}} \left[\frac{y^{\gamma-1}}{(y+\theta)^{2}}({\gamma}{(y+\theta)}-y)\right]\,dy}}_{A}\\ &\qquad\qquad\qquad+\frac{e^{t\theta}}{\tau}\underbrace{{\int\limits_{b}^{\infty}{e^{ty-\frac{y^{\gamma}}{{2(y+\theta)}\tau}-\frac{y^{\gamma}}{{2(y+\theta)}\tau}} \left[\frac{y^{\gamma-1}}{(y+\theta)^{2}}({\gamma}{(y+\theta)}-y)\right]\,dy}}}_{B}, \end{aligned} $$
whence by a similar argument to the previous case, we have 0<A<∞. In B, note that when b is sufficiently large, we have, for y>b,
$$ty-\frac{y^{\gamma}}{{2(y+\theta)}\tau}=y\Bigg(t-\frac{y^{\gamma-1}}{{2y}{(1+\frac{\theta}{y})}{\tau}}\Bigg)<0, $$
whence we can omit the exponential of this term, as before, since the part of the integrand involving it would be bounded by e0=1. Thus,
$$\begin{aligned} B&<{\int\limits_{b}^{\infty}{\exp\left\{-\frac{y^{\gamma}}{{2y}{\left(1+\frac{\theta}{y}\right)\tau}}\right\} \left[\frac{y^{\gamma-1}}{{y^{2}}{\left(1+\frac{\theta}{y}\right)}^{2}}\left({\gamma}{y}{(1+\theta/y)}-{y}\right)\right]\,dy}}\\ &\qquad\qquad\qquad={\int\limits_{b}^{\infty}{\exp\left\{-\frac{y^{\gamma-1}}{{2}{\left(1+\frac{\theta}{y}\right)\tau}}\right\} \left[\frac{y^{\gamma-2}}{{\left(1+\frac{\theta}{y}\right)}^{2}}\left({\gamma}{(1+\theta/y)}-1\right)\right]\,dy}}, \end{aligned} $$
Now, since 2−2<(1+θ/y)−2<1−2 for y>b sufficiently large, we have
$$B<{\int\limits_{0}^{\infty}{y^{\gamma-2}}({2\gamma-1)}e^{-y^{\gamma-1}/(4\tau)}\,dy}\equiv C, $$
and performing the substitution x=yγ−1/(4τ), yields
$$C=\frac{4\tau(2\gamma-1)}{\gamma-1}{\int\limits_{0}^{\infty}e^{-x}\,dx}<\infty, $$
whence M(t)=A+B<A+C<∞. Case γ=2. To show γ=2 is sharp, let γ=2−ε, where 0<ε is small, t>0, and θ≥0. With these substitutions, reverting back to the τ=δγ parametrization, we have
$$M(t)={\int\limits_{\theta}^{\infty}{e^{tx-\frac{{(x-\theta)}^{2-\varepsilon}}{x\delta^{2-\varepsilon}}}} {\left[\frac{2-\varepsilon}{x-\theta}-\frac{1}{x}\right]}\,dx,} $$
whence, letting x=y+θ, we can successively refine the lower bound on M(t) as follows:
$$\begin{aligned} M(t)&=e^{t\theta}\int\limits_{0}^{\infty}\exp\left\{ty-\frac{y^{2-\varepsilon}}{(y+\theta)\delta^{2-\varepsilon}}\right\} \left[\frac{2-\varepsilon}{y}-\frac{1}{y+\theta}\right]\,dy\\ &>\int\limits_{0}^{\infty}\exp\left\{y\left(t-\frac{{y}^{1-\varepsilon}}{{(y+\theta)}\delta^{2-\varepsilon}}\right) \right\}\left[\frac{{(2-\varepsilon)}{(y+\theta)}-y}{y{(y+\theta)}}\right]\,dy\\ &>{\int\limits_{b}^{\infty}{e^{y(t-\frac{{y}^{-\varepsilon}}{\delta^{2-\varepsilon}})} {\left[\frac{{(2-\varepsilon)}{(y+\theta)}-(y+\theta)}{y{(y+\theta)}}\right]}\,dy}}\\ &>\exp\left\{\frac{-b^{-\varepsilon}}{\delta^{2-\varepsilon}}\right\}{\int\limits_{b}^{\infty}{e^{ty}{\frac{1}{y}}{(1-\varepsilon)}\,dy}}, \end{aligned} $$
for any 0<b<∞. Substituting x=ty, the mgf becomes
$${\begin{aligned} M(t)&\,=\,e^{\frac{-b^{-\varepsilon}}{\delta^{2-\varepsilon}}}{(1-\varepsilon)} {\int\limits_{tb}^{\infty}{e^{x}{\frac{t}{x}}{\frac{1}{t}}\,dx}}=e^{\frac{-b^{-\varepsilon}}{\delta^{2-\varepsilon}}} {(1-\varepsilon)}{\int\limits_{b}^{\infty}\frac{e^{x}}{x}\,dx}>{e^{b}}e^{\frac{-b^{-\varepsilon}} {\delta^{2-\varepsilon}}}{(1-\varepsilon)}{\int\limits_{b}^{\infty}\frac{1}{x}\,dx}\,=\,\infty. \end{aligned}} $$
Thus M(t) will not be finite in any neighborhood of t=0, whence γ=2 is a sharp upper bound on the divergence of the mgf.
We invoke (Serfling (1980), Theorem §4.2.2), and (Van der Vaart (1998), Theorem 5.41), where we must establish (i)–(v) as follows.
The absolute values of the two first order partials of f(x;β) are dominated by measurable functions in the vicinity of (γ0,τ0), whose integrals are finite. These derivatives are
$$\begin{array}{@{}rcl@{}} \frac{1}{f(x;\boldsymbol{\beta})}\frac{\partial f(x;\boldsymbol{\beta})}{\partial \gamma} &=& \frac{x}{x(\gamma-1)+\theta} + \frac{1-(x-\theta)^{\gamma}}{x\tau}\log(x-\theta), \\ \frac{1}{f(x;\boldsymbol{\beta})}\frac{\partial f(x;\boldsymbol{\beta})}{\partial \tau} &=& \frac{x\tau-(x-\theta)^{\gamma}}{x\tau^{2}}, \end{array} $$
whose absolute value is dominated by the function g1(x)=A(1+xB)(1+| logx|), for sufficiently large constants A and B, that is,
$$\left|\frac{\partial f(x;\boldsymbol{\beta})}{\partial\gamma}\right|\leq g_{1}(x)f(x;\boldsymbol{\beta}), \qquad\text{and}\qquad \left|\frac{\partial f(x;\boldsymbol{\beta})}{\partial\tau}\right|\leq g_{1}(x)f(x;\boldsymbol{\beta}), $$
whence \(\int g_{1}(x)f(x;\boldsymbol {\beta })dx=\mathbb {E} g_{1}(X)<\infty \).
The absolute values of the three 2nd order partials of f(x;β) are dominated by measurable functions in the vicinity of (γ0,τ0), whose integrals are finite. For example, the derivative with highest order terms is
$$\begin{array}{*{20}l} \frac{1}{f(x;\boldsymbol{\beta})}\frac{\partial^2 f(x;\boldsymbol{\beta})}{\partial \gamma^{2}} &= \left[\frac{2x}{x(\gamma-1)+\theta}-\frac{(x-\theta)^{\gamma}}{\tau(x(\gamma-1)+\theta)}\right]\log(x-\theta) \\ &\qquad\qquad\qquad\qquad+\left[1+\frac{(x-\theta)^{2\gamma}}{\tau^2x^{2}}-\frac{3(x-\theta)^{\gamma}}{x}\right]\log^{2}(x-\theta), \end{array} $$
whose absolute value is dominated by the function g2(x)=A(1+xB)(1+| logx|+| logx|2), for sufficiently large constants A and B, that is,
$$\left|\frac{\partial^2 f(x;\boldsymbol{\beta})}{\partial\gamma^{2}}\right|\leq g_{2}(x)f(x;\boldsymbol{\beta}), $$
whence \(\int g_{2}(x)f(x;\boldsymbol {\beta })dx=\mathbb {E} g_{2}(X)<\infty \). Tedious computations show that the remaining second order partials are likewise dominated by g2(x)f(x;β).
The absolute values of the four third order partials of logf(x;β) are dominated by integrable measurable functions in the vicinity of (γ0,τ0). The appropriate derivatives are:
$$\begin{array}{@{}rcl@{}} \frac{\partial^{3}\log f(x;\boldsymbol{\beta})}{\partial\gamma^{3}} &=& {\frac {2{x}^{3}}{ \left[ \left(\gamma-1 \right) x+\theta \right]^{3}}}-{\frac { \left(x-\theta \right)^{\gamma}}{x\tau }}\log^{3}\left(x-\theta \right), \\ \frac{\partial^{3}\log f(x;\boldsymbol{\beta})}{\partial\gamma^{2}\partial\tau} &=& {\frac { \left(x-\theta \right)^{\gamma} }{x{\tau}^{2}}}\log^{2}(x-\theta), \\ \frac{\partial^{3}\log f(x;\boldsymbol{\beta})}{\partial\gamma\partial\tau^{2}} &=& -{\frac { 2\left(x-\theta\right)^{\gamma}}{x{\tau}^{3}}}\log(x-\theta), \\ \frac{\partial^{3}\log f(x;\boldsymbol{\beta})}{\partial\tau^{3}} &=& {\frac {6\,\left(x-\theta \right)^{\gamma}-2\,x\tau}{x{\tau}^{4}}}, \end{array} $$
and we see that for (γ,τ) ranging over a sufficiently small neighborhood of (γ0,τ0), the absolute values of all of these are dominated by the general (integrable) function
$$ g_{3}(x)=A\left(1+x^{B}\right)\left(1+|\log x|+\cdots+|\log x|^{3}\right), $$
since for sufficiently large constants A and B, we have \(\mathbb {E} g_{3}(X)<\infty \).
The Hessian matrix,
$$J(\boldsymbol{\beta})=\mathbb{E}\left[\begin{array}{cc} -{\frac {{x}^{2}}{\left[\left(\gamma-1\right) x+\theta \right]^{2}}}-{\frac{\left(x-\theta \right)^{\gamma}}{x\tau}} \log^{2}\left(x-\theta \right) & {\frac{\left(x-\theta \right)^{\gamma}}{x{\tau}^{2}}}\log\left(x-\theta\right) \\ {\frac { \left(x-\theta \right)^{\gamma}}{x{\tau}^{2}}}\log\left(x-\theta\right) & {\frac{-2\left(x-\theta \right)^{\gamma}+x\tau}{x{\tau}^{3}}} \\ \end{array}\right], $$
exists and is nonsingular at (θ0,γ0,τ0). The existence part is verified by noting that, as in case (v) below, each term is finite due to the fact that it is dominated by the general integrable function (24). The matrix will be nonsingular if the first and second columns are linearly independent (a.s.). A glance at these terms reveals that it is impossible for one to be a multiple of the other (with positive probability).
The diagonal entries of I(β) are finite, I11(β)<∞ and I22(β)<∞, when evaluated at (θ0,γ0,τ0). Once again this follows similarly to case (iv) above by noting that the squares of each of the terms
$$\frac{\partial\log f(x;\boldsymbol{\beta})}{\partial\gamma} = \log\left(x-\theta \right) +{\frac {x}{ \left(\gamma-1 \right)x+\theta}}-{\frac{\left(x-\theta \right)^{\gamma}}{x\tau}}\log\left(x-\theta \right), $$
$$\frac{\partial\log f(x;\boldsymbol{\beta})}{\partial\tau} = {\frac{\left(x-\theta \right)^{\gamma}-x\tau}{x{\tau}^{2}}}, $$
are both dominated by the general integrable function (24). Note that the finiteness of the diagonals immediately implies that the off-diagonal term of I(β) is also finite.
With α=(a,b)T, write model (19) in vector/matrix form,
$$ \boldsymbol{y} = Z\boldsymbol{\alpha}+\boldsymbol{\varepsilon}, $$
and note that the (limit in probability of the) least squares estimates can be written as
$$\begin{array}{@{}rcl@{}} {\text{plim}_{n\rightarrow\infty}}\hat{\boldsymbol{\alpha}} &=& \boldsymbol{\alpha} + \left({\text{plim}_{n\rightarrow\infty}}\frac{1}{n-1}Z^{T} Z\right)^{-1}\left({\text{plim}_{n\rightarrow\infty}}\frac{1}{n-1}Z^{T}\boldsymbol{\varepsilon}\right) \\ &\equiv &\boldsymbol{\alpha} + \left({\text{plim}_{n\rightarrow\infty}} W\right)^{-1}\left({\text{plim}_{n\rightarrow\infty}}\boldsymbol{w}\right). \end{array} $$
To establish the required result, we will show that w=op(1), and W=Op(1). We will first derive the following basic results.
plimn→∞X(1)=θ. This follows easily by noting that
$$P(X_{(1)}>x) = \exp\left\{-\frac{(x-\theta)^{\gamma}}{x\tau/n} \right\}, $$
whence X(1)∼LDSWeibull(θ,γ,τ/n), so that for x>θ, P(X(1)>x)→e−∞=0, which implies P(X(1)≤x)→1 as n→∞; whereas P(X(1)>θ)=e−0=1, so that P(X(1)≤θ)=0.
plimn→∞X(i)=ξi=F−1(qi), for 2≤i≤n. This is a consequence of the asymptotic normality of X(i), which is a consistent estimate of the central quantile ξi. The asymptotic normality follows from the fact that F(·) is differentiable and f(ξi)>0 (David and Nagaraja (2003), Chapter 10).
From (i) and (ii), it follows immediately that plimn→∞yi= log[− log(1−qi)]+ logξi, and plimn→∞zi= log(ξi−θ).
In fact, since yi and zi are dominated by an integrable function (simply let x(i)↦x(n) in their corresponding definitions), implies the stronger L1 convergence:
$$y_i {\xrightarrow{L_{1}}} \log\left[-\log(1-q_i)\right]+\log\xi_{i}\equiv y_{i}^\ast, \qquad\text{and}\qquad z_i {\xrightarrow{L_{1}}} \log(\xi_i-\theta)\equiv z_{i}^\ast. $$
This follows because dominated convergence in probability implies L1 norm convergence (Lemma 2).
Now note that since
$$\xi_i = F^{-1}(q_i) \quad\Longleftrightarrow\quad \log\left[-\log(1-q_i)\right]+\log\xi_i = \gamma\log(\xi_i-\theta)-\log\tau, $$
yi∗ and zi∗ defined in (iv) satisfy the population regression equation: yi∗=a+bzi∗.
We can state the results in (iv)-(v) equivalently as:
$$\varepsilon_i=y_i-(a+{bz}_i) {\xrightarrow{L_{1}}} y_{i}^\ast-(a+{bz}_{i}^\ast)=0, $$
and since L1 norm convergence implies the weaker convergence in probability (see proof of Lemma 2), we have that plimn→∞εi=0.
Thus, from (iv)-(vi), we have that for any real numbers λ1 and λ2, \((\lambda _1+\lambda _2z_i)\varepsilon _{i}{\xrightarrow {L_{1}}} 0\), which also implies plimn→∞(λ1+λ2zi)εi=0.
To prove the first assertion (that plimn→∞w=0), invoke the Cramer-Wold device and Lemma 2 to see that, for any vector of reals λ=(λ1,λ2)T, and using the result in (vii),
$${\text{plim}_{n\rightarrow\infty}} \boldsymbol{\lambda}^{T} Z^{T}\boldsymbol{\varepsilon} = {\text{plim}_{n\rightarrow\infty}}\sum\limits_{i=2}^{n}(\lambda_1+\lambda_2z_i)\varepsilon_i = \sum\limits_{i=2}^{n}{\text{plim}_{n\rightarrow\infty}}(\lambda_1+\lambda_2z_i)\varepsilon_i = 0, $$
whence plimn→∞ZTε=0, and therefore
$$\boldsymbol{w} = \frac{1}{n-1}Z^{T}\boldsymbol{\varepsilon} = \left(\frac{1}{n-1}\right)\left(Z^{T}\boldsymbol{\varepsilon}\right) = o_{p}(1)o_{p}(1) = o_{p}(1). $$
To prove the second assertion (that W is bounded in probability), note that
$$W=\frac{1}{n-1}Z^{T} Z = \left[\begin{array}{cc} 1 & \frac{1}{n-1}\sum\nolimits_{i=2}^{n} z_{i} \\ \frac{1}{n-1}\sum\nolimits_{i=2}^{n} z_{i} & \frac{1}{n-1}\sum\nolimits_{i=2}^{n} z_{i}^{2} \\ \end{array}\right]. $$
An informal argument will now suffice. In the limit as n→∞, since the quantiles ξi are dense in the support of X, we have from (ii), and using the transformation u=F(x), that
$${\text{plim}_{n\rightarrow\infty}}\frac{1}{n}\sum\limits_{i=1}^{n} X_{(i)}= \lim\limits_{n\rightarrow\infty}\frac{1}{n}\sum\limits_{i=1}^{n}\xi_i=\int\nolimits_{0}^{1} F^{-1}(u)du = \mathbb{E} (X). $$
which generalizes immediately to
$${\text{plim}_{n\rightarrow\infty}}\frac{1}{n}\sum\limits_{i=1}^{n} g(X_{(i)})=\lim\limits_{n\rightarrow\infty}\frac{1}{n}\sum\limits_{i=1}^{n} g(\xi_i)=\int\nolimits_{0}^{1} g(F^{-1}(u))du = \mathbb{E} g(X), $$
for any integrable function g(·). Heuristically then, the fact that plimn→∞zi= log(ξi−θ) implies that
$$ {\begin{aligned} {\text{plim}_{n\rightarrow\infty}}\frac{1}{n-1}\sum\limits_{i=2}^{n} z_{i} = \mathbb{E}^{(T)}\log(X-\theta),\qquad\text{and}\qquad {\text{plim}_{n\rightarrow\infty}}\frac{1}{n-1}\sum\limits_{i=2}^{n} z_{i}^{2} = \mathbb{E}^{(T)}\log^{2}(X-\theta), \end{aligned}} $$
where \(\mathbb {E}^{(T)}\) denotes possible resulting truncation in the expectation operator in view of the fact that the summations begin at i=2 and may not span the entire support of the quantile function (see Remark 3). Now, since each of the sample averages in (26) is Op(1), we deduce that
$$ {\begin{aligned} {\text{plim}_{n\rightarrow\infty}}\left(\frac{1}{n-1}\sum\limits_{i=2}^{n} z_{i}\right)^{2} = [\mathbb{E}^{(T)}\log(X-\theta)]^{2}\neq\mathbb{E}^{(T)}\log^{2}(X-\theta)={\text{plim}_{n\rightarrow\infty}}\frac{1}{n-1}\sum\limits_{i=2}^{n} z_{i}^{2}. \end{aligned}} $$
whence we conclude that W is a.s. nonsingular and therefore Op(1).
The dataset analyzed in the current study is available in the CRAN R package bReeze repository [https://CRAN.R-project.org/package=bReeze].
Standard result for the cumulant generating function of any random variable, easily established by invoking Holder's Inequality.
cdf:
Cumulative distribution function
iid:
Independent and identically distributed
mgf:
Moment generating function
MLE:
maximum likelihood estimator
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Texas Tech University, Department of Mathematics & Statistics, Lubbock, 79409-1042, TX, USA
Roger W. Barnard, Chamila Perera, James G. Surles & A. Alexandre Trindade
Roger W. Barnard
Chamila Perera
James G. Surles
A. Alexandre Trindade
RWB developed the proof of Theorem 1, and spent considerable effort in trying to obtain analytical forms for several integrals that did not make it into the paper. CP worked closely with AAT to develop the results of "Formal definition, basic properties, and results" and "Estimation procedures and asymptotic results" sections, and was largely responsible for conducting the simulations in "Simulation results" section. JS was the main instigator of the paper, responsible for deriving the new distribution from first principles as described in the Introduction. He also spent considerable time looking for possible applications and datasets. AAT worked closely with CP to develop the results of "Formal definition, basic properties, and results" and "Estimation procedures and asymptotic results" sections, and was largely responsible for the real data analysis of "Real data application" section. All authors read and approved the final manuscript.
Correspondence to A. Alexandre Trindade.
Barnard, R.W., Perera, C., Surles, J.G. et al. The linearly decreasing stress Weibull (LDSWeibull): a new Weibull-like distribution. J Stat Distrib App 6, 11 (2019). https://doi.org/10.1186/s40488-019-0100-8
Accepted: 24 July 2019
Pullout test
Extreme values
Maximum likelihood estimate
Wind speed data


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text:
Q: Slide navigation menu from bottom I'm trying to implement a sliding menu, like in Prixing app. (The menu which lays under main layout and get visible when you take and drag main layout to the side)
But the trick is - i want it to appear on bottom of the screen, so I need to move the main layout to the top.
The second problem is that in tah layout I got a listview.
Please, I really need advices or maybe link to some usefull resources where I can understand how touch events are handled.
Thx!

A: i think you are try to do same as facebook app. i have found demo for it. so use this

A: there's really no magic on it.
To make it, move directly (in a button click or something):
you use a animation to change the LayoutParams.top from both fragments (the one on top (currently in view, and the one on the bottom, hidden under the screen).
On your XML layout make sure to set a fixed height for the bottom layout (in DIP) and during runtime animate the two layouts to move that amount of pixels.
To make it move with the user finger:
That will be tricky, because the listview MUST receive the the user input. Maybe if you call super on it.
But you can accomplish it by receiving OnTouchEvents:

*

*TOUCH_DOWN, you start the move (mark it with a boolean or something)

*TOUCH_MOVE, you apply the same change to the layout params that you did on the animation (1st option)

*TOUCH_UP or TOUCH_CANCEL, you stop the move (clear the flag).

hope it helps pointing you to the right direction.

redpajama_set_name: RedPajamaStackExchange


Difference between revisions of "Lloyd W. Chambers"
Chambers taught school from 1933 to 1941 in Topeka then became superintendent and joined the Navy as senior lieutenant in 1941.
After the war, he made equipment in wood for the magic fraternity. At one time he was with [[Floyd Thayer]] and [[Percy Abbott]].
Lloyd W. Chambers (1906-1960) was a semi-professional magician and owner of Chambers Manufacturing Co.
Early in his magic career, he performed at lodges and schools as Waldy Waldron.
Chambers was a contributor to Genii, Sphinx and the Linking Ring. He was a member of the I.B.M., P.C.A.M., S.A.M., and a charter member of the M.D.A.


Do you have a dominant PON ?
Do you have a dominant PON ??
If you have been asked to describe a dominant Pon, you might use the terms, "aggressive and assertive". Unfortunately this is a very common misperception. A dominant Pon tends to be very cool, calm and confident in just about every situation. This is why they are in fact, 'leaders'. Those Pons who fight and bicker among themselves are far too preoccupied with their own petty power struggles and anxieties, to be able to lead anybody anywhere. Over the years the word 'dominant' has taken on a negative image and is quite often used to describe Pons who lack sufficient training, are under socialized (therefore under confident), or extremely fearful, perhaps from being mishandled by owners. Some are aggressive and fearful because of medical conditions (eye and thyroid, allergy,etc. issues) that have gone undiagnosed.
1. Don't allow your dog to eat its meal until you (the top dog) have eaten first.
2. Don't allow your dog to leave the house (den) before you (top dog) have passed through the doorway first.
3. Don't allow your dog to climb onto the sofa or bed.
4. Don't allow your dog to climb your stairs, or to stare at you from the top of the staircase.
5. Don't allow your dog to look into your eyes.
Individual breed behavior needs to be considered when deciding if a particular dog is dominant or merely has breed characteristics. Sheepdogs follow a herding behavior pattern when dealing with both humans and animals. This behavior pattern always needs to be taken into account, regardless of circumstances. Further, I consider myself among those who believe that dogs are sentient beings, capable of making decisions, reviewing situations and learning for themselves. So bearing all this in mind, if you happen to hear someone repeating rigid rules on popular canine television shows, or read about them in doggy magazines, their opinions may be far from being entirely valid. Unfortunately many television networks and popular publications, do not check out the qualifications of their contributors and many of these "behaviorists" are in fact self proclaimed experts, nurses, and dog lovers who are merely regurgitating what they have read in popular dog psychology books, most of which have been written by people who are themselves passing along theories they have not scientifically documented. Owners need to carefully check the references of those who offer advice and try and determine credibility of theories they present. Being a good owner means thinking for one's self in all situations and taking nothing for granted.
Most Pons need a "job" and when allowed to participate in agility, conformation, therapy and other activities, turn their attention to these tasks and are more relaxed at home with their human families. These Pons have discovered how to be happy with their humans as well as being happy with their given tasks. Was this happiness achieved by humans asserting "pack leadership" or "alpha status" over the Pon's behavior ? Of course not ! The family Pon has had his life changed by offering him his genetically engineered job.
In summary, consider your Pon's genetic potential when trying to determine if your Pon is dominant or in fact, trying to fulfill his genetic role in life. A contented, well adjusted Pon is a busy Pon.


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text:
2018 gubernatorial race already heating up
11 Nov, 2015 by James Poulos
Still more than two years away, the race to succeed Gov. Jerry Brown in 2018 has already become one of the most closely watched competitions in California politics.
It has made adversaries of the state's two highest-profile personifications of its two biggest political power blocs — Lt. Gov. Gavin Newsom, hometown hero for the Bay Area's monied elite of self-styled progressives, and former Los Angeles Mayor Antonio Villaraigosa, still the darling of Southland Latinos who have long nursed a grudge against the San Francisco-Sacramento corridor's outsized influence on statewide politics.
Already, Newsom "has announced that he will take on the gun lobby with an initiative to, among other things, outlaw possession of large ammunition magazines and require background checks on any ammunition. He's also said he will back the legalizing of marijuana for general use, not just for medicinal purposes," according to the Los Angeles Times. "Villaraigosa's unofficial campaign is almost the polar opposite of Newsom's," the paper added. "The San Francisco mayor's gun-and-pot measures will probably be most popular among liberals along California's coast; Villaraigosa is spending his election run-up treading up and down the state's more conservative interior."
Strategy and scrutiny
Remarkably, however, Newsom and Villaraigosa have yet to clear the field of other candidates, several of whom have as strong a case — if not stronger — to run as Brown's rightful heir. In a recent Field poll reported by the San Jose Mercury News, respondents spread their support over enough possible contenders to leave Newsom and Villaraigosa almost tied. "The survey found that more than four in 10 likely voters say they'd back Villaraigosa, 62, and Newsom, 48, and a slightly smaller share say they'd endorse [current Los Angeles Mayor Eric] Garcetti, 44," noted the Mercury News. "It also shows that almost three in 10 voters say they'd support Democrats Tom Steyer, a billionaire environmentalist, and John Chiang, the state treasurer."
With Brown termed out, little surprise has accompanied the attention surrounding his party's scramble for a new generation of statewide leadership. Adding a twist, Republicans have remained remarkably uncompetitive at the gubernatorial level. "The prospect for any Republican gubernatorial candidate is dim in this heavily Democratic state," the Sacramento Bee observed. Visalia Republican Serenity Holden, who participated in the Field poll, told the Bee "I'm really kind of screwed."
But the plight of Holden's fellow Republicans has heightened the stakes for the race's contending Democrats and their party. The prospect of a Democrat-on-Democrat showdown would likely bring long-running tensions between the more corporatist and more ideological wings of the party.
Republicans in retreat
Perhaps with that dynamic in mind, the state GOP has gone out of its way to keep its powder dry in 2016. Party officials have even warned would-be candidates not to expect support if they jump into races where more moderate Democrats are likely to square off against hopefuls farther to their left. "Assemblywoman Kristin Olsen, R-Modesto, said party officials told her that if she runs against Sen. Cathleen Galgiani, D-Stockton, she'll be getting no help from them," the Los Angeles Times reported. "Neither state GOP chairman Jim Brulte nor Senate Republican leader Jean Fuller of Bakersfield would comment on the unusual decision to lay off a competitive race, especially one that could involve a legislative leader."
The party has established a precedent for this kind of unusual move. After Republican candidates were withdrawn from this year's special state Senate election, pro-business Democrat and former Brown advisor Steve Glazer triumphed over a big labor effort to push another candidate into office. In a recent interview with the Pleasanton Weekly, Glazer reaffirmed his centrist bona fides. "Certainly on public safety matters, I'm more conservative. On fiscal issues, I'm more conservative. On social issues, I'm very progressive. So you can't just put a broad label on it," he said, calling himself "the most conservative Democrat in the [state] Senate."
But the California GOP's primary season psych-outs may not scale upward to the gubernatorial level. Although Newsom has come much closer than Villaraigosa to embodying the kind of Democrat state Republicans might begrudgingly pull the lever for, Newsom's use of a base-pleasing agenda to advance his potent personal ambition has made him a far more threatening candidate to the GOP's long-term fortunes than either L.A.'s former mayor or its current one.
desmond 11 November, 2015, 18:47
When you look at that smile, you know he likes it in the @$S.
bob 12 November, 2015, 20:03
You think things are bad now in Colliefornia?
Just wait until pretty boy is governor and the Demonrats get back their 2/3rds majority in the legislature.
You will call the days of Brownie the good old days. How's that for depressing?
The Republocons are completely pathetic. We will not see a Republocon elected to a state wide office in our lifetimes.
LA Jones 13 November, 2015, 19:48
I sez Antonio stickin a hozd(_$do in Gavin. That smile just says mo mo mo…mmmm, that's goood. He dun likd hiz lipps.
Antonio VillaraigosaEric GarcettiGavin NewsomGov. Jerry BrownKristen Olsen
James Poulos
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text:
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The Latest: Massage therapist recalls Epstein's island
FILE - In this July 30, 2008 file photo, Jeffrey Epstein, center, appears in court in West Palm Beach, Fla. The wealthy financier pleaded not guilty in federal court in New York on Monday, July 8, 2019, to sex trafficking charges following his arrest over the weekend. Epstein will have to remain behind bars until his bail hearing on July 15. (Uma Sanghvi/Palm Beach Post via AP, File)
NEW YORK — The Latest on developments in the sex trafficking case against Jeffrey Epstein (all times local):
A massage therapist says she went to one of Jeffrey Epstein's private Caribbean islands to work dozens of times in the early 2000s and saw "nothing out of the ordinary."
She said she saw girls there on two occasions.
The woman spoke to The Associated Press on condition of being identified only by her initials, H.W., because she feared losing business.
She says she was never asked to do anything improper and didn't make anything of seeing the girls on Little St. James Island, Epstein's main retreat in the U.S. Virgin Islands. She was then in her 50s.
H.W. says one girl appeared to be 16 or 17 and rode an ATV. She glimpsed another girl hurrying from Epstein's house.
Epstein faces federal charges of abusing dozens of underage girls in New York and Florida in the early 2000s. He's pleaded not guilty.
Several employees who worked on Epstein's property have refused to talk because they signed non-disclosure agreements.
— By Danica Coto in San Juan, Puerto Rico
Financier Jeffrey Epstein's lawyers argue he should be held under house arrest with electronic monitoring pending trial despite prosecutors' claims he could flee to dodge sex trafficking charges.
The lawyers argued for bail in court papers Thursday in Manhattan federal court. A bail hearing for the 66-year-old is set for Monday.
Prosecutors argued in court this week for Epstein to be jailed ahead of trial. They say a trove of what seems to be nude pictures of underage girls was found in his mansion after his arrest on charges that he sexually exploited and abused dozens of underage girls from 2002 to 2005.
Epstein's lawyer says the new charges are "ancient history" and can't be brought because of a decade-old federal non-prosecution agreement when he pleaded guilty in Florida state court.
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