On 3 September 2018, the International Union for Conservation of Nature (IUCN) featured an article written by David Johnson, Chief Executive, Margaret Pyke Trust, coordinator of the Population & Sustainability Network on population growth and the resulting pressure on the environment. You can read the original article on the IUCN's website here, or alternatively below.
One side effect of running an organisation with population in the name is that every week or so a friend will send me an article or video about population. Almost invariably there is mention of population being a "taboo" topic, or "the elephant in the room". This is in spite of the fact that I can't honestly recall ever speaking to a conservationist who disagreed that global population is a critical conservation issue, or was shy to say so. There is no taboo.
Given the United Nations projects that the global population will rise from 7.6 billion today to over 9.7 billion by 2050, this is hardly surprising. But there is something glib about merely stating the challenge. What needs to happen is that we move to consider the actions which form part of the effective response.
Evidence-based conservation has become standard practice. This includes the systematic assessment of information when designing conservation projects and determining policy. There is also a growing understanding of the need for cross-sector collaboration. So to move on from the mythical elephant in the room, the first step is to consider relevant health data.
The Convention on Biological Diversity (CBD), the United Nations Environment Programme (UNEP) and the World Health Organization (WHO) did exactly that in their 2015 joint report "Connecting global priorities: biodiversity and human health: a state of knowledge review". They acknowledge that population growth places increased demands on healthcare systems and can greatly increase pressures on natural resources. Most importantly, they acknowledge that population growth could be moderated by greater investments in family planning programmes.
They also explain that greater investment in the education of girls and women and improved access to and awareness of contraception would "not only improve human health and well-being directly, it would also help slow and reverse trends among countries with the highest projected growth rates and concomitant pressures on ecosystems".
In reaching their conclusions the CBD, UNEP and WHO relied on fertility and other reproductive health data. In the health sector that is elementary. It needs to become standard in conservation too. At the Trust I lead, we are working with an increasing number of conservation organisations which are considering not only population data, but also the evidence of how small improvements in family planning have significant changes on fertility rates, and consequently long-term population size and environmental pressures.
At the national policy level, there is often acceptance that population is a key conservation challenge. Nigeria's National Biodiversity Strategy and Action Plan (NBSAP), for instance, highlights high population growth as the first cause of biodiversity loss. Tanzania's NBSAP states, "Human activities due to rapidly growing population have been one of the major causes of habitat loss and degradation resulting into biodiversity decline in Tanzania." Nigeria and Tanzania are not alone in making such statements. Many developing nations, with relatively poor healthcare provision and higher fertility rates as a result, make this connection in their NBSAPs.
Poor rural communities often depend most directly on natural resources for their livelihoods, food, water, shelter and cultural practices. When localised human pressures on ecosystems exceed critical tipping points, both community health and environmental health suffer. Poor rural communities in developing nations also face the greatest barriers to the use of and access to reproductive health services.
This is why these areas are such a focus for health advocates. Barriers to family planning prevent women from choosing freely when and whether to have children, threaten family health, create challenges for girls who want to complete their education, and lead to higher levels of fertility and more rapid rates of population growth. These are issues which also directly impact on conservation, and there is often an overlap in areas with the greatest need for better reproductive health services and for conservation. Supporting policy developments that will improve reproductive health services is therefore something conservation organisations should consider including in their strategic priorities.
Conservation organisations can't provide clinical services. Similarly, family planning nurses can't implement conservation activities. But this does not mean family planning should be seen as solely a health issue. Conservationists are familiar with overlapping concerns. For example, we are confident to promote renewable energy as part of the response to the environmental threats of climate change, or to advocate for better public transport to improve air quality. These are not merely "energy" or "transport" issues. Conservation organisations do not need to run windfarms or provide train services to promote them as relevant to the environment.
Barriers to family planning increase pressures on ecosystems as well as families. Therefore, as a sector, we should find it as easy to promote universal access to family planning as we do renewable energy. After all, the Sustainable Development Goals call for both.
Some conservation and reproductive health organisations have already successfully joined forces. There is a growing movement of community-based Population, Health and Environment programmes which take a holistic approach to interconnected issues of poor health, family planning needs, food insecurity, poverty, and unsustainable use of natural resources. Project data shows this approach leads to greater uptake of family planning, better health and gender relations, and increased support for and participation in conservation activities by local communities. Such a multisector approach can also be more cost-effective, and generate more sustainable results.
The Inter-governmental Panel on Climate Change affirmed in a 2014 report that family planning can play a role in reducing climate change vulnerability and is a potential adaptation strategy. The book Drawdown: the most comprehensive plan ever proposed to reverse global warming, details the top 100 solutions that have the greatest potential to reduce emissions or sequester carbon from the atmosphere. Educating girls ranks sixth and family planning ranks seventh. The integration of family planning into conservation and climate change thinking has begun; we need this to become the norm.
The Margaret Pyke Trust has, from its inception, been at the forefront of developments in contraception. In the 1960s that meant opening and running what soon became the world's busiest family planning clinic. Today, the Trust is likely the only IUCN Member to have 50 years' family planning expertise. One of our strategic priorities is working with conservation organisations in developing countries to integrate family planning into conservation programmes. We do this because we believe all girls and women have the right to choose freely whether and when to have children, and how many. We also do this because we question whether many traditional conservation projects can have long term sustainable results if they don't.
On World Population Day in July 2019 we will be promoting the growing collective of visionary conservation organisations which are integrating family planning into their conservation activities by making appropriate partnerships, or supporting policy changes which will make this easier. The Trust is keen to work with other IUCN Members, as well as other conservation organisations worldwide, who understand that it is time we stopped talking about population growth and started to take action to remove barriers to family planning as an integral part of conservation policy.
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Fingermarks, also called Leopard Spot Effect, are actually a type of scratch-pattern left in the paint from using a product that's either too aggressive to be used by hand or not safe for clear coat paints.
The marks mimic the shape of your fingertips as they push down on some type of applicator pad while working a product over and into the paint. Finger Marks are inflicted into the paint anytime you stop the movement of the applicator pad for at this moment, as brief as it may be, there tends to be more pressure exerted to the fingertips at the point at which you either stopped moving your hand or changed the direction in which you're moving your hand.
This can happen if you apply products in a circular motion or in a back-and-forth, straight-line motion. It's not the direction of the motion, it's the stopping of motion or the change of direction of motion that the marks are imprinted into the paint.
It can also be caused by using too aggressive of a material for an applicator, for example a coarse cloth like cheesecloth a red shop rag.
In the last year I met a gentleman that purchased some Cheesecloth at a PBE store that said on the package it was safe for clear coat finishes, he used the cheesecloth to apply a hand applied polish to the paint on his brand new black Porsche and ended up instilling finger marks everywhere he rubbed the paint.
We all know you shouldn't apply paint care products to any type of paint using a red shop rag but I've met people that don't understand how easily clear coat paints are scratched and thus don't take the type of applicator or application material they're using into consideration. Automotive repair shops use a lot of shop rags, they're not always read, sometimes they gray or purple looking but they're all made out of fairly coarse material and it's not uncommon for some well-meaning mechanic to wipe or rub something off a car while it's in the shop for repair and in the process instill scratches and finger marks.
Old fashioned rubbing compounds in which the abrasives are hard and sharp will also cause Finger Marks when applied by hand. While most rubbing compounds have been reformulated to be safer than old style compounds that were commonly used on single stage paints before the 1980's, there are still zillions of cans of rubbing compound and polishing compound in garages and shops across the country that are used on clear coated cars because that's whats already out in the garage.
This is why you want to do a Test Spot to the paint anytime you're using products you're unfamiliar with or working on paint you've never worked on before.
If the results from your Test Spot look good, then simply duplicate your process over the rest of the car. If the results from your Test Spot don't look good... aren't you glad you only worked on a small section?
Always read the label of any scratch or swirl mark remover and make sure it states on the label that the product is safe for hand-application to clear coat finishes.
These were all instilled applying a product that was originally introduced as a hand applied swirl and scratch remover. It was never introduced to the market. It was given to me to test and I tested it on what started out as a flawless finish. It was applied using overlapping circular motions using a clean, soft foam applicator pad. The product was applied over the entire hood using a normal protocol of working small sections at a time, then when moving onto a new section overlapping a little into the previous section until the entire hood had been worked. All work was done in a garage with normal garage lighting and the finger marks did not show up until the car was pulled out into the sun for inspection.
While this test was done purposefully to document with pictures the problems caused by using products that are too abrasive for use by hand on clear coat finishes, it's always a best practice to only do a Test Spot when using a product for the first time, or working a a paint system you are unfamiliar with.
After the finger marks were documented, they were then removed using a DA Polisher and a clear coat safe swirl and scratch remover and the original swirl-free finish was restored to the demo car.
I myself have never put finger marks on my car before, but what I find odd is my cats have. and no they weren't trying to detail my car lol. they climb up on my car and I can see where their paws leave marks exactly like this, except in a shape of a cat paw. it's weird, and only my PC can remove it. anyone want some cats???
my 223 can take care of those... from a couple hundred yards away!!!
This Honda Pilot was buffed a lot, (it was our family grocery-getter), the paint on it buffed out like butter. I don't think I took an after picture as this was just another day-in-the-life for me. Take my word for it, the finish was restored to flawless before anyone one in the family knew it was used for testing.
Pics, or the recovery never happend!!
I try and evaulate my customers and then lay out the tools for the job, since the owner uses this as a daiy driver parked out 24/7 and doesnt really car to have a flawless finish, thats why a light duty cleaner wax was used.
But is it possible that even on the softest paints using the most gentle of products that fingermarks could be instilled?
Pics, or the recovery never happened!!
This was actually after purposefully swirling out the paint and then buffing with only M80 Speed Glaze. Then chemically stripping and then taking a sun shot for someone hung-up on fillers.
I posted these pictures because from time to time people will try rubbing their car's paint out by hand and end-up with a finish that has finger marks and most of the time the problem is either poor quality product or bad technique.
The bad technique is doing a quickie job not understanding how to use a compound and not understand you can only work small sections at at time when removing swirls out of clear coat paints by hand or machine.
So a point of reference is needed and I just happen have one in my "personal" collection of car pictures.
ugh my neighbor has like 5 outdoor cats that love to sleep on my car and i get those lovely little paw prints on my black stang as well. along with the scratches from their claws after i wax it because its slick. they have defenately made my paint job worse.


We all know how it feels like to have days where we don't feel like our best selves because of a rough night's sleep. If you keep on feeling this way even after a full night of sleep, you might be dealing with a potentially serious sleeping disorder called sleep apnea Fort Pierce. Did you know that over 22 million Americans are affected by this condition? In fact, about 80% of these cases are undiagnosed.
What is Sleep Apnea Fort Pierce?
Sleep apnea is a sleep disorder where your breathing starts and stops repeatedly. When your sleep is interrupted many times every night due to the lack of airflow, you won't reach your sleep cycle's restorative stages.
If you have sleep apnea, you won't be aware of the fact that your sleep is interrupted multiple times at night. However, one of the tell-tale signs that you may be suffering from this sleep disorder is when your partner or family members complain about your loud, disruptive, chronic heavy-duty snoring. Snoring is an indication that your airway is being blocked by your tongue or the collapsed muscles in your throat.
Another symptom would be waking up with a headache. Frequent pauses in breathing during sleep will mean less oxygen makes its way to your brain. When this happens, your blood vessels can widen and cause vascular headaches. Lastly, those who suffer sleep apnea often experience mood swings or feel irritable and depressed because of the lack of quality sleep.
Looking for Help with Sleep Apnea Fort Pierce?
Did you know that it's possible to effectively manage sleep apnea Fort Pierce with a mandibular advancement device? If you want to know more about how you can get a restful night's sleep, contact All Smiles Dentistry today to schedule an appointment.


text:
Christchurch murder: Teenage girl arrested as part of investigation into murder of 35-year-old Edward Reeve
Edward Reeve
(1822-1889) writer, museum curator and police magistrate
A 15-year-old girl has been arrested as part of an investigation in to the murder of a man in Dorset.
The 15-year-old was detained on suspicion of assisting an offender, after the death of Edward Reeve, 35, last week, Dorset Police said.
Mr Reeve's body was discovered at a property on Heath Road in Christchurch last week, prompting a murder investigation.
Two boys, both 16, from Christchurch and Bournemouth, are also in custody on suspicion of murder.
Police leading the investigation into the death are now appealing for CCTV footage from the Heath Road area in the days building up to the discovery of Mr Reeve's body.
Detective Inspector Simon Huxter, who is leading the investigation, said: "We have continued with our investigation and have been able to find out further details.
"I am grateful for all the information we have received from the public so far, but I am now appealing with regards to a wider time frame as previously given.
"Therefore, I urge residents and motorists in the area of Heath Road to please check any CCTV systems or dashcam footage from Wednesday 29 December 2021 to Tuesday 4 January 2022 and to report any suspicious activity they may have witnessed during this time period.
"As highlighted before, officers remain in the area carrying out enquiries and can be approached by members of the community with any information or concerns."
A thief has made off with 42,000 much-sought rapid antigen tests from a freight depot in Sydney, authorities said Wednesday.
Wife of critically ill Covid patient wins court battle to keep husband on ventilator
'Absent an order from the court, my husband will die,' Anne Quiner wrote to a Minnesota judge
'Winterwatch' 2022: When is it back? Where is it set? All you need to know
The nature show returns for another season.
Carer suffocated suspected burglar during citizen's arrest, court hears
Nathan Smith is accused of the manslaughter of Craig Wiltshire after he knelt on his back for nine minutes.
Winter Storm Creates Hazardous Road Conditions in Northern Ohio
A winter storm brought hazardous road conditions to Ohio on Monday, January 17, stranding vehicles in Wayne County, where up to a foot of snowfall was reported.Timelapse footage shared by the Orrville Police Department shows snowfall accumulation and conditions on North Main Street.Nearly 1,000 crews were at work to clear and treat roads across Ohio, according to the Ohio Department of Transportation. Credit: Orrville Police Department via Storyful
Woman rescued after car crashes through ice on frozen river
A woman has been rescued after her car smashed through the ice of a Canadian river. Locals used kayaks to retrieve the woman from her sinking vehicle on the Rideau River in Ottawa. Witnesses told local media a car driving on the frozen river crashed through the ice and into the cold water on Sunday.

redpajama_set_name: RedPajamaCommonCrawl


So this guy wants to make a Halloween costume. Here is his story.
That just made a very long day very much worth it. My Halloween costume seems lame in comparison.
words fail. oh thank you, tanster. thank you, thank you, thank you.
Wow. I was just gonna dress up as me…for Halloween. A scary version of me, but just me. This guy put me to shame.
I mean, that is is pretty cool!
It's two worlds colliding–cockeyed.com is one of my favorite websites…and The Office is my favorite show.
What a great idea, I wish I was handy enough to actually make something that cool.
Great stuff. Way too much free time.
A-mazing. I wonder if he is going to be sporting that at the convention, or if he just made it for Halloween? Either way, it's fantastic!
That so cool. But wouldn't it be easier to cut out three black holes of construction paper, stick it your shirt, and just go as "Three hole-punch Jim"? It's fast, simple, and no plaster necessary.
oh man… I hope that head bobbles in the end. That's amazing….
Wow, that's patently ridiculous and totally wonderful.
Is that supposed to be Dwight?
Oh my gosh, that's amazing! I can't wait to see how it turns out.
Well if that isn't a wee bit of brilliance. He gets all the candy.
But this guy's work is AMAZING! He should do an Angela one next!
I want to see a video of him actually walking in that thing! I wonder if other shows have fans that would go to such lengths to make a themed costume?
Seriously? Wouldn't it be easier/cheaper/less time consuming to just take the Jim route and go to the drugstore and Goodwill??
I mean, that's some serious talent there, but… wow.
Wow! That guy is awesome! I would freak if I saw him walking around my neighborhood with that costume on.
Wow, just…wow. Now that is dedication! And quite possibly the best Halloween costume EVER! Dwigt would be so proud to have a life size bobblehead.
Thanks for the nice comments! The costume is for a Contest Saturday night in Sacramento, CA. Wish me luck!
Sorry I will miss the convention! That sounds like more fun!
WOW!! I want to see it finished! I hope he takes pictures of the final product.
Oh gosh… that is awesome!
Wow. I'd like to see him try to navigate that thing through doorways.
I hope someone over at NBC has seen this – really hope Rainn has seen it. hahaha good luck at the contest! What could beat this!?
FYI, Rob usually winds up selling the stuff he builds like this on eBay once he's had his fun with it. You should also check out his site for non-Office things that are tremendous.
Yes! My first tip! I was so excited to see Rob's costume and when I realized it looked a lot like one of my favorite characters, I nearly choked. I knew you guys would appreciate the effort.
Neat! This guy's costumes are crazy! He did an awesome fandango paper bag that turned out really well, so I bet this year's Dwight will be his best yet… Can he turn his wife into Angela?
WOW!!! That is amazing! A true fan right there!!
I'm assuming this will be a bobblehead, but whatever it winds up being will be FANTASTIC. Such a lot of work!! If I worked that hard on a costume, I'd start a museum. Wonder what he does with stuff like that when he's done. Amazing!! Can't wait to see the rest!
The bobblehead is complete. Check out pages 6 and 7 to see the completed costume. It looks amazing. I hope Rob won that costume contest!
I, too, hope Rainn sees this. It's awesome!
My first thought was I could see this being used on the show. I thought Dwight would dress up as himself…but duh! Jim as Dwight is much better.
the site wanted to try it.
Fantastic. I love the creativity!
This was incredible! I am so jealous. Good job, this is amazing engineering and artistic work!
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text:
NESN's Elle Duncan set to join ESPN as SportsCenter anchor
2016By Joe Lucia on 04/27/2016 04/27/2016
Awful Announcing has learned that ESPN will be adding NESN anchor Elle Duncan as a SportsCenter anchor, and will soon announce the hiring.
Duncan has been with NESN since 2014, hosting both NESN Live and NESN Sports Today. Prior to her work in Boston, Duncan worked for Atlanta's NBC affiliate, WXIA, and was a sideline reporter for the Atlanta Hawks. Duncan also worked as an on-field reporter for SEC and ACC football on Fox Sports South.
When contacted by AA, Duncan confirmed that she'd be joining ESPN and talked about what she's looking forward to working for the company.
"I think for anybody who is a sports broadcaster, grew up in sports, watching sports, playing sports, ESPN is the penultimate [job]. The opportunity came about to join the team and for me, I'm all about learning and growing and wanting to get better, and I think that in order to do that, you surround yourself with the best in the business. ESPN is obviously filled with really talented people who essentially do what I'm hoping to one day get to, which is marry sports and their personalities and their opinions while also trying to be objective and a journalist.
"The ceiling is so high. I feel like you could work for ESPN for 15 years and not do everything that you could do at ESPN.
"As far as the timing goes, I spent a few great years at NESN learning and growing and getting better. When ESPN calls and you have an opportunity to join the Worldwide Leader, you take it."
Duncan also briefly discussed the necessary changes in her focus as a broadcaster going from a local outlet in NESN to a national one in ESPN.
"It's not only the reach, but it's your focus. My focus is pretty narrow [with NESN], but because I like sports, I stay on top of most things. My focus is on the four Boston teams, and in particular because of NESN, the Red Sox and Bruins. I think one of the difficulties is that your job never ends at ESPN. There's something happening at 1, 2 o'clock in the morning and again at 6, 7 o'clock in the morning and all throughout the day.
"I think that really widening that scope and staying on top of everything that they cover, whether that be your strong suits all the way to the sports you're not as knowledgeable about. For me, it's that – having to really widen what I've been focused on and thinking about national topics rather than regional ones."
About Joe Lucia
I hate your favorite team. I also sort of hate most of my favorite teams.
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text:
Voyages très extraordinaires de Saturnin Farandoul ("The Very extraordinary journeys of Saturnin Farandoul")  is a science-fiction novel by Albert Robida.

A parody of Jules Verne's Voyages extraordinaires, it was published in over one hundred installments between 1879 and 1880. It was first translated in English by Brian Stableford as The Adventures of Saturnin Farandoul.  The novel proved to be very successful in Italy, where it was adapted into a 1913 Italian silent film, Le avventure straordinarissime di Saturnino Farandola, directed and interpreted by Marcel Fabre  and into a RAI TV-series starring Mariano Rigillo and Daria Nicolodi in 1977.

A comic adaptation by Pier Lorenzo De Vita was published on Topolino in 1938 and 1940, and a three-parts sequel starring Donald Duck in the title role, still written by De Vita and drawn by Guido Martina, was published in 1959. The novel also inspired the Bonvi's comic series Marzolino Tarantola.

References

External links

 Voyages très extraordinaires de Saturnin Farandoul at Open Library

1879 novels
French novels adapted into films
Parody novels
French science fiction novels
French novels adapted into television shows

redpajama_set_name: RedPajamaWikipedia


text:
\section{Introduction.}\label{intro}
A situation frequently encountered in  dose-response studies is
identifying the minimum effective dose ($MED$). The $MED$ is defined
as the lowest dose such that the mean response is better than that
of a zero-dose control by a clinically significant difference.
Finding the MED is important since high doses often turn out to have
undesirable side effects.

Consider the one-way layout model
\begin{equation}
\label{model} Y_{ij}=\mu_i+\varepsilon_{ij}
\end{equation}
 for $i=0,...,k, j=1,...,n_i$, where $\mu_i$'s are the unknown response
means at different dose levels and $\varepsilon_{ij}\sim
N(0,\sigma^2)$ are the independent errors with an unknown variance.
The parameter space is
\begin{equation}
\label{para} H=\{\underline{\mu}=(\mu_0,\mu_1,...,\mu_k): \mu_1 \leq
...\leq \mu_k\}
\end{equation}
(here, for simplicity, we omit $\sigma$ in $H$), and the sufficient
statistics are the sample means, $\bar{Y}_{i}$, and the mean squared
error, denoted by $S^2$. Assume $i=0$ is the control group. One goal
is to find the smallest positive integer $N$ satisfying $\mu_{N}>
\mu_0+\delta$ for a clinically significant difference constant
$\delta \geq 0$. We call $N$ the minimum effective dose (MED).
Determination of the MED usually is done by step-down test
procedures, see Williams (1971), Ruberg (1989), Tamhane, Hochberg,
and Dunnett (1996),  Hsu and Berger (1999), and Hellmich and
Lehmacher (2005), among others. Tamhane, Hochberg and Dunnett (1996)
indeed proposed a step-up procedure SU1P to identify the MED. The
SU1P procedure is based on the step-up procedure of Dunnett and
Tamhane (1992), which controls the experimentwise error rate only
for balanced designs. However, Dunnett and Tamhane (1995)'s step-up
procedure for unbalanced designs case cannot control the
experimentwise error rate. Liu (1997) proposed a method of
calculating the critical values of the step-up procedure by Dunnett
and Tamhane (1995). The SU1P procedure does not make use of the
monotonicity, therefore its power should not be high. Intuitively it
seems that step-down procedures infer a larger dose as the MED.
Therefore, it is of interest to have a step-up procedure to use the
monotonicity to increase its power.

To derive a test of level-$\alpha$, one needs to find an appropriate
statistic and its least favorable distribution in the null
hypothesis. Thus a stochastic ordering for the test statistic is
needed. The desired statistic, the estimator of $\mu_i$, should be:
a) nondecreasing in $i$, b) and is also nondecreasing in each of
$\bar{Y}_j$'s. The PAVA algorithm generates the estimators that
achieve a). However, it is difficult to show b) directly for these
estimators using the PAVA. The PAVA was first proposed by Ayel,
Brunk, Ewing, Reid and Silverman (1955), and was introduced to
estimate the monotone proportions in independent binomial
experiments. Surprisingly, it has many applications in normal,
Poisson and multinomial distributions, etc. See more details in
Robertson, Wright and Dykstra (1988). The PAVA is an iterative
algorithm, each step is very simple to implement, however, it does
not have a closed form for the final estimator. Hence, it is hard to
establish analytic properties for the estimator. Notice these, a new
algorithm, the SDMMSA, is proposed to overcome the drawbacks. We
will show that the two algorithms yield the same estimators and each
estimator is a monotone function of each $\bar{Y}_i$. The second
fact is critical to determine the least favorable distribution in
the null hypothesis space.

The rest of the article is organized as follows. Section 2 provides
a new algorithm to generate estimators for $\mu_i$'s and discusses
their analytic properties. In particular,  a stochastic ordering for
the distributions of the proposed estimators is established. In
Section 3, one application of the stochastic ordering established in
Section 2 is given to identify the MED. A step-up multiple test
procedure that controls the experimentwise error rate in the strong
sense is provided by constructing a sequence of increasing rejection
regions of level-$\alpha$ for each null hypothesis in
$(\ref{null})$, and the proposed procedure is illustrated on a real
data set. Section 4 concludes with some discussion.

\section{A new algorithm to construct the estimator of $\mu_i$ and some analytic results.}

In this section, an estimator of $\mu_i$, denoted by $\hat{\mu}_i$,
for any integer $i\in [1,k]$ under $H$ is first constructed
iteratively. Then three facts are established: $\hat{\mu}_i$ is the
same as the estimator generated by the PAVA; $\hat{\mu}_i$ is the
mle under $H$, and the distribution of $\hat{\mu}_i$ is
stochastically non-decreasing in each $\mu_j$.

\subsection{A new iterative algorithm to construct  $\hat{\mu}_i$.}

Let
\begin{equation}
\label{combinedmean}  n_{i,j}=\sum_{h=i}^j n_h, \,\
\bar{Y}_{i,j}=\frac{\sum_{h=i}^j n_h\bar{Y}_h}{ n_{i,j}}, \,\
\forall 1\leq i\leq j\leq k
\end{equation}
be the sample size and the sample mean of a combined sample  of
treatments $i$ through $j$, respectively.

Step 1). We construct $\hat{\mu}_i$ starting from $i=k$ using the
data set $\{(\bar{Y}_i, n_i)\}_{i=1}^k$. Let
\begin{equation}
\label{step1set} A_1=\{j: \bar{Y}_{j,k}=max_{\{1\leq j'\leq
k\}}\{\bar{Y}_{j',k} \}   \}
\end{equation}
be a subset of $\{1,...,k\}$($A_1$ contains a single element with
probability one), and let
\begin{equation}
\label{step1index} i_1= min\{A_1\}.
\end{equation}
Then
\begin{equation}
\label{step1mle} \hat{\mu}_{i}\stackrel{def}{=}\bar{Y}_{i_1,k},\,\
\forall \,\ i\in [i_1,k].
\end{equation}
If $i_1=1$, then all $\hat{\mu}_{i}$'s are defined and stop;
otherwise go to the next step.
 Step 2). Note in this step $ i_1-1
\leq k-1$. Repeat Step 1 but using the data set
$\{(\bar{Y}_i,n_i)\}_{i=1}^{i_1-1}$. i.e., let
\begin{equation}
\label{step2set} A_2=\{j: \bar{Y}_{j,i_1-1}=max_{\{1\leq j'\leq
i_1-1\}}\{\bar{Y}_{j',i_1-1} \}   \}
\end{equation}
be a subset of $\{1,...,i_1-1\}$, and let
\begin{equation}
\label{step2index} i_2= min\{A_2\}.
\end{equation}
Then
\begin{equation}
\label{step2mle}
\hat{\mu}_{i}\stackrel{def}{=}\bar{Y}_{i_2,i_1-1},\,\ \forall \,\
i\in [i_2,i_1-1].
\end{equation}
If $i_2=1$, then all $\hat{\mu}_{i}$'s are defined and stop;
otherwise repeat this process for a number of times, say $h$ times,
until $i_h=1$. Such an integer $h$ exists, because $i_j$ strictly
decreases in $j$. Since $i_h=1$, then all $\hat{\mu}_{i}$'s are
defined and the construction on $\hat{\mu}_i$'s is complete. We name
this the step-down-maximum-mean-selection algorithm (SDMMSA).

\noindent{\bf Remark 1}.  There exists partition, $\cup_{u=1}^h
[i_{u},i_{u-1}-1]$, for $\{1,...,k\}$ with
$i_0-1\stackrel{def}{=}k$. Following the construction of
$\hat{\mu}_i$, each
 $\hat{\mu}_i$ is the sample mean of a combined sample of treatment(s) belonging to the interval in
 the partition that includes treatment $i$.
 Also
 $\hat{\mu}_i$ is constant in $i$ on each integer interval $[i_{u}, i_{u-1}-1]$ for $u=1,...h$, as shown in
  ($\ref{step1mle}$) and ($\ref{step2mle}$),
 and $\hat{\mu}_i$  is strictly increasing when $i$ moves from $[i_{u},
 i_{u-1}-1]$ to $[i_{u'}, i_{u'-1}-1]$ for $u>u'$, as shown in ($\ref{step1set}$) and ($\ref{step2set}$). Therefore,
 $\hat{\mu}_i$ is nondecreasing in $i$ for $i\in [1,k]$.
 ~\raisebox{.5ex}{\fbox{}}

\begin{lemma}
\label{mon} For partition $\cup_{u=1}^h [i_{u},i_{u-1}-1]$ given in
Remark 1, $\bar{Y}_{i_u-1}<\bar{Y}_{i_u}$ for any $u\in [1,h]$.
\end{lemma}

\noindent{\bf Proof}. Since the SDMMSA repeats itself in each step,
without loss of generality, we only need to prove Lemma~\ref{mon}
for $u=1$. i.e., $\bar{Y}_{i_1-1}<\bar{Y}_{i_1}$.

Suppose this is not true, i.e.,
 $\bar{Y}_{i_1} \leq \bar{Y}_{i_1-1}$.  Note
$\bar{Y}_{i_1,k} \geq \bar{Y}_{i_1+1,k}$ by the definition of $i_1$,
then
\begin{equation}
\label{contradiciton2}
 \bar{Y}_{i_1} \geq
\bar{Y}_{i_1+1,k}.
\end{equation} Similarly,  $\bar{Y}_{i_1-1} <
\bar{Y}_{i_1,k}$ is true due to $\bar{Y}_{i_1-1,k} <
\bar{Y}_{i_1,k}$.  Therefore, $\bar{Y}_{i_1} \leq \bar{Y}_{i_1-1} <
\bar{Y}_{i_1,k},$ which implies
$$ \bar{Y}_{i_1}<\bar{Y}_{i_1+1,k},$$ a contradiction to ($\ref{contradiciton2}$).
 ~\raisebox{.5ex}{\fbox{}}
\bigskip

\noindent{\bf Remark 2}. If $\bar{Y}_i$ is nondecreasing in
$i\in[1,k]$, then the partition, $\cup_{u=1}^h [i_{u},i_{u-1}-1]$,
for $\{1,...,k\}$ given in Remark 1 satisfies i) $\bar{Y}_i$  is
constant when $i\in[i_{u},
 i_{u-1}-1]$, and ii) is strictly increasing when $i$ moves from $[i_{u},
 i_{u-1}-1]$ to $[i_{u'}, i_{u'-1}-1]$ for $u>u'$. Therefore,
 $\hat{\mu}_i=\bar{Y}_i$  for $i\in [1,k]$.
 ~\raisebox{.5ex}{\fbox{}}

\bigskip

\noindent{\bf Example 1}. Consider the data in Table 1, taken from
Ruberg (1995). There are nine $(k=9)$ active dose groups and a zero
dose control group with six $(n_i=6, i=0,...,9)$ animals/group in
the experiment. Following Step 1, we obtain $i_1=9$ and then
$\hat{\mu}_9=\bar{Y}_9$; following Step 2, we obtain $i_2=6$, then
$\hat{\mu}_8=\hat{\mu}_7=\hat{\mu}_6$ and is equal to
$\bar{Y}_{6,8}=73.77$, the sample mean of the combined sample for
$i=8,7,6$. The construction of all $\hat{\mu}_i$'s ends at Step
7(=h) and their values are reported in Table 1. The partition given
in Remark 1 is now
$$[9]\cup [6,8] \cup [5]\cup [4]\cup [3]\cup [2]\cup [1],$$
with a notation of $[i]=[i,i]$.
 ~\raisebox{.5ex}{\fbox{}}

\bigskip
\subsection{The relationship between $\hat{\mu}_i$,   $\hat{\mu}_i^{mle}$ and $\hat{\mu}_i^{pava}$.}

So far, the estimator of $\mu_i$ under $H$ typically is obtained
following the pooled-adjacent-violators algorithm(PAVA, described
later), for example, see Barlow, Bartholomew, Bremner and Brunk
(1972), Robertson, Wright, and Dykstra (1988), and Silvapulle and
Sen (2005). Denote this estimator by $\hat{\mu}_i^{pava}$. Now we
show that $\hat{\mu}_i^{mle}=\hat{\mu}_i=\hat{\mu}_i^{pava}$ in
Theorem~\ref{sdmmsamle} and Theorem~\ref{mle} below.

\begin{theorem}
\label{sdmmsamle} Let $\hat{\mu}_i^{mle}$ be the maximum likelihood
estimator for $\mu_i$ under $H$. Then
$\hat{\mu}_i=\hat{\mu}_i^{mle}, \forall i \in [1,k]$. Therefore,
$\hat{\mu}_i=\hat{\mu}_i^{mle}, \forall i \in [1,k]$.
\end{theorem}

\noindent{\bf Proof of Theorem~\ref{sdmmsamle}}. Taking the log
transformation on the joint pdf of $Y_{ij}$, it is easy to see that
$\hat{\mu}_i^{mle}$ minimizes
$$f(\mu_1,...,\mu_k)=\sum_{i=1}^k n_i (\bar{Y}_i-\mu_i)^2=\sum_{u=1}^h
[\sum_{j=i_u}^{i_{u-1}-1} n_j
(\bar{Y}_j-\mu_j)^2]\stackrel{def}{=}\sum_{u=1}^h
f_i(\mu_1,...,\mu_k)$$
 under $H$, where the intervals $[i_{u},i_{u-1}-1]$ for $u=1,...,h$ are given
in Remark 1.

Now focus on each $f_i$. Without loss of generality, focus on $f_1$,
then
$$f_1(\mu_1,...,\mu_k)=\sum_{j=i_1}^k n_j (\bar{Y}_j-\mu_j)^2=\sum_{j=i_1}^k n_i [(\bar{Y}_j-\hat{\mu}_j)^2
+2(\bar{Y}_j-\hat{\mu}_j)(\hat{\mu}_j-\mu_j)+(\hat{\mu}_j-\mu_j)^2].$$
Rearrange the terms above and note $\hat{\mu}_j=\bar{Y}_{i_1,k}$ for
$j\in[i_1,k]$, then
$$f_1(\mu_1,...,\mu_k)=\{\sum_{j=i_1}^k n_j[
(\bar{Y}_j-\hat{\mu}_j)^2+(\hat{\mu}_j-\mu_j)^2]\} +2\sum_{j=i_1}^k
n_j(\bar{Y}_j-\hat{\mu}_j)(-\mu_j)\stackrel{def}{=}I_1+I_2.$$ It is
obvious that $I_1$ is minimized at $\mu_j=\hat{\mu}_j$ for
$j\in[i_1,k]$; for $I_2$, apply Abel's partial summation formula and
obtain
$$I_2=2\sum_{j=k}^{i_1} n_j(\hat{\mu}_j-\bar{Y}_j)\mu_j=2
\sum_{j=k}^{i_1+1}d_j(\mu_j-\mu_{j-1})+d_{i_1}\mu_{i_1}=2
\sum_{j=k}^{i_1+1}d_j(\mu_j-\mu_{j-1}),$$ where $d_j=\sum_{v=k}^j
n_v (\hat{\mu}_v-\bar{Y}_v)\geq 0$ and $d_{i_1}=0$ due to the
definition of $i_1$. Also note $\mu_j\geq \mu_{j-1}$. Thus $I_2$ is
nonnegative and achieves its minimum at $\mu_{i_1}=...=\mu_k$.
Therefore, combining $I_1$ and $I_2$, we conclude
$f_1(\mu_1,...,\mu_k)$ is minimized at $\mu_j=\hat{\mu}_j$ for
$j\in[i_1,k]$.

Repeat the same argument on $f_2$ through $f_h$, each $f_u$ is
minimized at  $\mu_j=\hat{\mu}_j$ for $j\in[i_u, i_{u-1}-1]$.
Lastly, note $\hat{\mu}_j$ nondecreasing, we conclude
$\hat{\mu}_j^{mle}=\hat{\mu}_j$ for any $j\in[1,k]$.
 ~\raisebox{.5ex}{\fbox{}}

\bigskip

\begin{theorem}
\label{mle} For any $i\in[1,k]$,
\begin{equation}
\label{mle1}\hat{\mu_i}(\bar{Y}_1,...,\bar{Y}_k)=\hat{\mu}_i^{pava}(\bar{Y}_1,...,\bar{Y}_k).
\end{equation}
\end{theorem}

For a data set of $\{(\bar{Y}_i,n_i)\}_{i=1}^k$, the PAVA proceeds
as follows:

Step 0-PAVA). If $\bar{Y}_i$ is nondecreasing in $i$ for
$i\in[1,k]$, then $\hat{\mu}_i^{pava}=\bar{Y}_i$; otherwise, go to
the next step.

Step 1-PAVA). Pick any consecutive pair $(\bar{Y}_j, \bar{Y}_{j+1})$
with $\bar{Y}_j>\bar{Y}_{j+1}$, let $j_l$ be the smallest integer so
that $\bar{Y}_i=\bar{Y}_j$ for $i\in [j_l,j]$ and let $j_u$ be the
largest integer so that $\bar{Y}_i=\bar{Y}_{j+1}$ for $i\in
[j+1,j_u]$. For each $i\in[j_l,j_u]$, replace each $\bar{Y}_i$ by
$\bar{Y}_{j_l,j_u}(=\frac{\sum_{i=j_l}^{j_u}n_i\bar{Y}_i}{\sum_{i=j_l}^{j_u}n_i})$,
and then obtain a new data set of $\{(a_i,n_i)\}_{i=1}^k$, where
$a_i=\bar{Y}_i$ for $i\not \in [j_l,j_u]$ and
$a_i=\bar{Y}_{j_l,j_u}$ for $i \in [j_l,j_u]$. Note two facts:
$\bar{Y}_i$ is non-increasing for $i\in[j_l,j_u]$, and  the number
of different $a_i$'s is strictly less than that of $\bar{Y}_i$'s.

Step 2-PAVA) Repeat this process on $\{(a_i,n_i)\}_{i=1}^k$ until
all $a_i$'s are nondecreasing. Then $\hat{\mu}_i^{pava}=a_i$. Since
the number of different $a_i$'s is strictly less than that in the
previous step, this algorithm has to stop in a finite steps.
 ~\raisebox{.5ex}{\fbox{}}
\bigskip

\noindent {\bf Proof of Theorem~\ref{mle}.} When $\bar{Y}_i$ is
nondecreasing in $i\in[1,k]$, then (\ref{mle1}) is true due to Step
0-PAVA) and Remark 2. When $\bar{Y}_j>\bar{Y}_{j+1}$ for some $j$,
let $a_i$ and $[j_l,j_u]$ be given in Step 1-PAVA). It suffices to
show
\begin{equation}
\label{mleya}
\hat{\mu}_i(\bar{Y}_1,...,\bar{Y}_k)=\hat{\mu}_i(a_1,...,a_k),
\end{equation}
for any $i\in[1,k]$. Let $\cup_{u=1}^h [i_{u},i_{u-1}-1]$ be the
partition of $[1,k]$ given in Remark 1 using data
$\{(\bar{Y}_i,n_i)\}_{i=1}^k$. The integer $j_u$ has to belong to
one of these intervals in the partition, say $[i_{u_j},i_{u_j}-1]$.
Since $\bar{Y}_i$ is non-increasing on $[j_l,j_u]$ as shown in Step
1-PAVA), by Lemma~\ref{mon}, $i_{u_j}\leq j_l$. Thus $[j_l,j_u]$ is
a subset of $[i_{u_j},i_{u_j}-1]$, an interval in the partition.
 Let $\cup_{u=1}^{h'} [i_{u}',i_{u-1}'-1]$ be the
partition of $[1,k]$ given in Remark 1 but using data
$\{(a_i,n_i)\}_{i=1}^k$. Therefore, $[j_l,j_u]$ also has to be a
subset of one of these intervals.

Case I). If $[j_l,j_u]\subset [i_1,k]$, i.e., $u_j=1$, consider
$$a_{i,k}\stackrel{def}{=}\frac{\sum_{u=i}^k n_u a_u}{\sum_{u=i}^k
n_u}$$ for $i\in[1,k]$. Note $a_{i,k}=\bar{Y}_{i,k}$ for $i\not \in
[j_l+1,j_u]$ and $i_1'\leq j_l$, then $i_1'=i_1$. Therefore,
$$\hat{\mu}_i(\bar{Y}_1,...,\bar{Y}_k)=\frac{\sum_{i=i_1}^k
n_i\bar{Y}_i}{\sum_{i=i_1}^k n_i}=\frac{\sum_{i=i_1'}^k
n_ia_i}{\sum_{i=i_1'}^k n_i}=\hat{\mu}_i(a_1,...,a_k),$$ for any
$i\in[i_1,k]$. For any $i<i_1$, since $\hat{\mu}_i$ only depends on
$\bar{Y}_1$ through $\bar{Y}_{i_1-1}$(or $a_1$ through $a_{i_1-1}$)
and $\bar{Y}_i=a_i$, we conclude (\ref{mleya}).

Case II). If $[j_l,j_u]\subset [i_2,i_1-1]$, i.e., $u_j=2$, we only
need to show $i_1'=i_1$. Then, similar to Case I) above,
(\ref{mleya}) is established. To prove $i_1'=i_1$, first note
$$a_{i_1,k} =\bar{Y}_{i_1,k}\geq \bar{Y}_{u,k}= a_{u,k}$$ for any $u\geq i_1$.
So
\begin{equation}
\label{i1'} i_1'\leq i_1
\end{equation}
 by the definition of
$i_1'$.

Suppose $i_1'<i_1$, we will construct a contradiction. Note
$i_1'\not \in (j_l, j_u]$ because $a_i$ is non-increasing on $(j_l,
j_u]$. Thus, $$\bar{Y}_{i_1',k}=a_{i_1',k}\geq
a_{i_1,k}=\bar{Y}_{i_1,k},$$ a contradiction with the definition of
$i_1$. Hence $i_1'\geq i_1$. Together with (\ref{i1'}), we conclude
$i_1'=i_1$.

For the other cases of $u_j=3,...,h$, similar to Case II), we can
show $i_u=i_u'$ for all $u=2,...,h$. Hence, $h'=h$ and two
partitions, $\cup_{u=1}^h [i_{u},i_{u-1}-1]$ and $\cup_{u=1}^{h'}
[i_{u}',i_{u-1}'-1]$ are identical. Also note that $[j_l,j_u]$ is
contained in one interval $[i_{u_j},i_{u_j-1}-1]$, (\ref{mleya}) is
established. ~\raisebox{.5ex}{\fbox{}}
\bigskip

\noindent{\bf Remark 3}. Although $\hat{\mu}_i$ and
$\hat{\mu}_i^{pava}$ generated by two algorithms are identical,
there are several advantages of the SDMMSA over the PAVA. First, it
is clear from the SDMMSA that $\hat{\mu}_i$ is uniquely defined, but
not clear for $\hat{\mu}_i^{pava}$ from the PAVA, since the latter
needs to show $\hat{\mu}_i^{pava}$ must be the same no matter where
to start the algorithm, which is not obvious at all. Secondly,
$\hat{\mu}_j$ has a closed form, $\bar{Y}_{i_j, i_{j-1}-1}$, if
$j\in [i_{j_0}, i_{j_0-1}-1]$ for some $j_0$, where $i_{j_0}$ is
given in Remark 1, while $\hat{\mu}_i^{pava}$ does not. This fact is
important for establishing the stochastic ordering of $\hat{\mu}_i$
as shown in the next section. Thirdly, it was mentioned, for
example, in Robertson, Wright and Dykstra (1988, p.10) that
$\hat{\mu}_i^{pava}$ is the mle. To the best knowledge of the
authors, no rigorous proof has been given. With $\hat{\mu}_i$, we
proved $\hat{\mu}_i=\hat{\mu}_i^{mle}$ and
$\hat{\mu}_i=\hat{\mu}_i^{pava}$. Thus
$\hat{\mu}_i^{pava}=\hat{\mu}_i^{mle}$. Lastly, regarding the
computation, the SDMMSA is easier to code than the PAVA since at
each step of the SDMMSA a certain number of the final estimators
($\hat{\mu}_i$) are defined.
 ~\raisebox{.5ex}{\fbox{}}

\bigskip

\subsection{A stochastic ordering of $\hat{\mu}_i$.}
We provide another major result in this paper which establishes a
stochastic ordering for each $\hat{\mu}_i$ in terms of each of
$\mu_j$'s. Let
\begin{equation}
\label{muik} \hat{\mu}_{i,k}=\hat{\mu}_i(\bar{Y}_1,...,\bar{Y}_k)
\end{equation}
be the estimator of $\mu_i$ obtained from the sample $\{(\bar{Y}_i,
n_i)\}_{i=1}^k$ following Steps 1 and 2 in Section 2.1.
 So the distribution of $\hat{\mu}_{i,k}$ depends on $\mu_1$ through $\mu_k$ and
$\sigma$.
\begin{theorem}
\label{order} For each i and j in $[1,k]$,  $\hat{\mu}_{i,k}$, as a
function of $\bar{Y}_j$,  is nondecreasing when the other
$\bar{Y}_{j'}$'s are held fixed. Therefore, $\hat{\mu}_{i,k}$ is
stochastically nondecreasing in $\mu_j$ when the other $\mu_{j'}$'s
are held fixed. i.e., $P(\hat{\mu}_{i,k}>x)$ is a nondecreasing
function of $\mu_j$ for any real number $x$.
\end{theorem}
\noindent {\bf Proof of Theorem~\ref{order}.} We will prove the
monotonicity of $\hat{\mu}_{i,k}$ in each $\bar{Y}_j$ by induction
on $k$.

 First for
the case of $k=1$, $\hat{\mu}_{1,1}=\bar{Y}_1$ is nondecreasing in
$\bar{Y}_1$.

Assume that, for the case of $k=m$, $\hat{\mu}_{i,m}$ is
nondecreasing in $\bar{Y}_j$ for any $i$ and $j$ no larger than $m$.
Now consider the case of $k=m+1$. Following Step 1, $i_1$ depends on
$k$, so write it as $i_1(k)$, i.e., obtain $i_1(k)$ using
$(\bar{Y}_1, n_1)$ through  $(\bar{Y}_k, n_k)$. Claim
\begin{equation}
\label{i1m} i_1(m+1)=\left\{
\begin{array}{ll}
m+1, & \mbox{if} \,\ \bar{Y}_{m+1}>\hat{\mu}_{i_1(m),m},\\
i_1(m),   & \mbox{if} \,\ \bar{Y}_{m+1}\in
(\hat{\mu}_{i_2(m),m},\hat{\mu}_{i_1(m),m}],
\\
...\\
i_j(m),   & \mbox{if} \,\ \bar{Y}_{m+1}\in
(\hat{\mu}_{i_{j+1}(m),m},\hat{\mu}_{i_j(m),m}],\\
....\\
 i_h(m), &  \mbox{if} \,\ \bar{Y}_{m+1} \leq
\hat{\mu}_{i_h(m),m}.
\end{array}
           \right.
\end{equation}

When $\bar{Y}_{m+1}>\hat{\mu}_{i_1(m),m}(=\hat{\mu}_{m,m})$, for any
$j\in[1,m]$, note
$$ \bar{Y}_{j,m+1}=
\frac{n_{j,m} \bar{Y}_{j,m}+n_{m+1}\bar{Y}_{m+1}}{
n_{j,m}+n_{m+1}}\leq \frac{ n_{j,m}
\hat{\mu}_{m,m}+n_{m+1}\bar{Y}_{m+1}}{
n_{j,m}+n_{m+1}}<\bar{Y}_{m+1},$$ then $i_1(m+1)=m+1$.

When $\bar{Y}_{m+1}\in
(\hat{\mu}_{i_2(m),m},\hat{\mu}_{i_1(m),m}]=(\bar{Y}_{i_2(m),i_1(m)-1},\bar{Y}_{i_1(m),m}]$.
i) For $j\in[i_1(m),m+1]$, note $n_{j,m+1}-n_{i_1(m),m+1}\leq 0$,
then
\begin{eqnarray*}
&&\bar{Y}_{i_1(m),m+1}-\bar{Y}_{j,m+1}\\&=&\frac{n_{j,m+1}(n_{i_1(m),m}\bar{Y}_{i_1(m),m}+n_{m+1}\bar{Y}_{m+1})
-n_{i_1(m),m+1}(n_{j,m}\bar{Y}_{j,m}+n_{m+1}\bar{Y}_{m+1})}{n_{i_1(m),m+1}n_{j,m+1}}\\
&=&\frac{
n_{j,m+1}n_{i_1(m),m}\bar{Y}_{i_1(m),m}-n_{i_1(m),m+1}n_{j,m}
\bar{Y}_{j,m}+
(n_{j,m+1}-n_{i_1(m),m+1})n_{m+1}\bar{Y}_{m+1}}{n_{i_1(m),m+1}n_{j,m+1}}\\
&\geq& \frac{
n_{j,m+1}n_{i_1(m),m}\bar{Y}_{i_1(m),m}-n_{i_1(m),m+1}n_{j,m}
\bar{Y}_{i_1(m),m}+
(n_{j,m+1}-n_{i_1(m),m+1})n_{m+1}\bar{Y}_{i_1(m),m}}{n_{i_1(m),m+1}n_{j,m+1}}\\
&=&0,\end{eqnarray*} and conclude $i_1(m+1)\leq i_1(m).$

ii) For $j\in[1, i_1(m)-1]$, note
$$\bar{Y}_{j,m+1}=\frac{n_{j,i_1(m)-1}\bar{Y}_{j,i_1(m)-1}+n_{i_1(m),m}\bar{Y}_{i_1(m),m}+n_{m+1}\bar{Y}_{m+1}}{
n_{j,m+1}},$$ and $\bar{Y}_{j,i_1(m)-1}\leq
\bar{Y}_{i_2(m),i_1(m)-1}<\bar{Y}_{m+1}$, then
\begin{eqnarray*}
&&\bar{Y}_{i_1(m),m+1}-\bar{Y}_{j,m+1}\\
&=&
\frac{n_{j,i_1(m)-1}n_{i_1(m),m}\bar{Y}_{i_1(m),m}+n_{j,i_1(m)-1}n_{m+1}\bar{Y}_{m+1}-
n_{j,i_1(m)-1}(n_{i_1(m),m}+n_{m+1})\bar{Y}_{j,i_1(m)-1}}{n_{i_1(m),m+1}n_{j,m+1}}\\
&>&0,
\end{eqnarray*}
and conclude $i_1(m+1)> i_1(m)-1$. Therefore, combining i) and ii)
we obtain $i_1(m+1)=i_1(m)$ when $\bar{Y}_{m+1}\in
(\hat{\mu}_{i_2(m),m},\hat{\mu}_{i_1(m),m}]$.

For the other cases of $\bar{Y}_{m+1}$, ($\ref{i1m}$) can be
established in a similar way. Therefore, we conclude that
$\hat{\mu}_{i,m+1}$ depends on $\{(\bar{Y}_j,n_j)\}_{j=1}^{m+1}$
through $\{(\hat{\mu}_{j,m},n_j)\}_{j=1}^{m}$ and
$(\bar{Y}_{m+1},n_{m+1})$. So write
\begin{equation}
\label{mui}\hat{\mu}_{i,m+1}=\hat{\mu}_{i,m+1}(\hat{\mu}_{1,m},...,\hat{\mu}_{m,m},\bar{Y}_{m+1}).
\end{equation}
Also write $\hat{\mu}_{i,m+1}$ as
\begin{equation}
\label{muiyj}\hat{\mu}_{i,m+1}=\hat{\mu}_{i,m+1}(\bar{Y}_j),
\end{equation}
since the other $\bar{Y}_{j'}$ are fixed. We will use any one of the
above two notations whenever it is convenient.
 For $y<y'$, we need to show the monotonicity below
\begin{equation}
\label{thm1mono} \hat{\mu}_{i,m+1}(y) \leq \hat{\mu}_{i,m+1}(y'),
\end{equation}
which establishes the theorem, in the following two cases.

Case 1: $j=m+1$. Since $\hat{\mu}_{j',m}$ does not involve
$\bar{Y}_{m+1}$ for all $j'\leq m$, ($\ref{thm1mono}$) changes to
\begin{equation}
\label{thm1mono1}
\hat{\mu}_{i,m+1}(\hat{\mu}_{1,m},...,\hat{\mu}_{m,m},y) \leq
\hat{\mu}_{i,m+1}(\hat{\mu}_{1,m},...,\hat{\mu}_{m,m},y'),
\end{equation}
which is established in Lemma~\ref{lemmacase1} by noting
$\hat{\mu}_{i,m}$ is nondecreasing in $i\leq m$. Case 2: $j<m+1$.
Since $\hat{\mu}_{j',m}(y)\leq \hat{\mu}_{j',m}(y')$
 for all $j'\leq m$ by the induction assumption on the case of $k=m$, ($\ref{thm1mono}$) changes to
\begin{equation}
\label{thm1mono2}
\hat{\mu}_{i,m+1}(\hat{\mu}_{1,m}(y),...,\hat{\mu}_{m,m}(y),\bar{Y}_{m+1})
\leq
\hat{\mu}_{i,m+1}(\hat{\mu}_{1,m}(y'),...,\hat{\mu}_{m,m}(y'),\bar{Y}_{m+1}),
\end{equation}
which is established in Lemma~\ref{lemmacase2}. Therefore, the proof
of the monotonicity of $\hat{\mu}_{i,k}$ in each $\bar{Y}_j$ is
complete.

Since $\bar{Y}_j$'s are independent random variables, and each is
stochastically increasing in $\mu_j$, $\hat{\mu}_{i,k}$, as a
nondecreasing function of each $\bar{Y}_j$, is also stochastically
nondecreasing in $\mu_j$. See, for example, Alam and Rizvi (1966) or
Lemma 2 in Wu and Wang (2007). The proof of Theorem~\ref{order} is
complete.
 ~\raisebox{.5ex}{\fbox{}}

\bigskip

\begin{lemma}
\label{lemmacase1} Let
$\hat{\mu}_{i,m+1}(\bar{Y}_1,...,\bar{Y}_m,y)$ be the estimator
following Steps 1 and 2 on a date set
$\{(\bar{Y}_v,n_v)\}_{v=1}^{m+1}$ with $\bar{Y}_{m+1}=y$. Then
$$\hat{\mu}_{i,m+1}(\bar{Y}_1,...,\bar{Y}_m,y)\leq \hat{\mu}_{i,m+1}(\bar{Y}_1,...,\bar{Y}_m,y')$$
if $\bar{Y}_v$ is nondecreasing in $v\in[1,m]$ and $y<y'$.
\end{lemma}

\noindent{\bf Proof of Lemma~\ref{lemmacase1}.} Now write $i_1(m+1)$
introduced in ($\ref{i1m}$) as $i_1(y)$.

First  note $i_1(y)\leq i_1(y')$, which follows ($\ref{i1m}$) and
$y<y'$.

Secondly, claim
\begin{equation}
\label{i1y} i_1(y)=min\{v\in[1, m+1]: \bar{Y}_v\geq
\bar{Y}_{v,m+1}\}\stackrel{\mbox{denoted by}}{=}A.
\end{equation}
Note
\begin{equation}
\label{yv}
\bar{Y}_{v,m+1}-\bar{Y}_{v+1,m+1}=\frac{n_v(\bar{Y}_v-\bar{Y}_{v,m+1})}{n_{v+1,m+1}}=\frac{n_v(\bar{Y}_v-\bar{Y}_{v+1,m+1})}{n_{v,m+1}}.
\end{equation}
Therefore, $\bar{Y}_{v,m+1}$ is nonincreasing in $v$ when $v\geq A$.
Hence $i_1(y)\leq A$. It is obvious  that
$\bar{Y}_{A-1,m+1}<\bar{Y}_{A,m+1}$ following the first equality in
($\ref{yv}$). Since $\bar{Y}_v$ nondecreasing in $v\in [1,m]$,
$\bar{Y}_{v,m+1}$ is nondecreasing in $v$ when $v\leq A-1$. Thus
($\ref{i1y}$) is established.

Thirdly, a) when $i< i_1(y)$, both
$\hat{\mu}_{i,m+1}(\bar{Y}_1,...,\bar{Y}_m,y)$ and
$\hat{\mu}_{i,m+1}(\bar{Y}_1,...,\bar{Y}_m,y')$ are constructed
based on $\{\bar{Y}_v\}_{v=1}^{i_1(y')-1}$, a subset of
$\{\bar{Y}_v\}_{v=1}^m$ which is nondecreasing in $v$. Thus
$\hat{\mu}_{i,m+1}(\bar{Y}_1,...,\bar{Y}_m,y)=\bar{Y}_i=
\hat{\mu}_{i,m+1}(\bar{Y}_1,...,\bar{Y}_m,y')$ following Remark 2.

b) When $i\in [i_1(y), i_1(y')-1]$,
$\hat{\mu}_{i,m+1}(\bar{Y}_1,...,\bar{Y}_m,y)=\bar{Y}_{i_1(y),m+1}$
with $\bar{Y}_{m+1}=y$; while
$\hat{\mu}_{i,m+1}(\bar{Y}_1,...,\bar{Y}_m,y')=\bar{Y}_i\geq
\bar{Y}_{i_1(y),m+1}$ following ($\ref{i1y}$) and $\bar{Y}_v$
nondecreasing in $v\in[1,m]$.

c) When $i\in [i_1(y'), m+1]$,
$\hat{\mu}_{i,m+1}(\bar{Y}_1,...,\bar{Y}_m,y)=\bar{Y}_{i_1(y),m+1}$
with $\bar{Y}_{m+1}=y$; while
$\hat{\mu}_{i,m+1}(\bar{Y}_1,...,\bar{Y}_m,y')=\bar{Y}_{i_1(y'),m+1}$
with $\bar{Y}_{m+1}=y'$. Then
$\hat{\mu}_{i,m+1}(\bar{Y}_1,...,\bar{Y}_m,y)<
\hat{\mu}_{i,m+1}(\bar{Y}_1,...,\bar{Y}_m,y')$ due to $i_1(y)\leq
i_1(y')$, $\bar{Y}_v$ nondecreasing in $v\in[1,m]$, $y<y'$ and
($\ref{i1y}$). The proof is complete.
 ~\raisebox{.5ex}{\fbox{}}

\bigskip

\begin{lemma}
\label{lemmacase2} Let
$\hat{\mu}_{i,m+1}(\bar{Y}_1,...,\bar{Y}_m,\bar{Y}_{m+1})$ be the
estimator following Steps 1 and 2 on a date set
$\{(\bar{Y}_v,n_v)\}_{v=1}^{m+1}$. Then
$$\hat{\mu}_{i,m+1}(\bar{Y}_1,...,\bar{Y}_m,\bar{Y}_{m+1})\leq
\hat{\mu}_{i,m+1}(\bar{Y}_1',...,\bar{Y}_m',\bar{Y}_{m+1})$$ if
$\bar{Y}_v$ and $\bar{Y}_v'$ are both nondecreasing  and $\bar{Y}_v
\leq \bar{Y}_v'$ for $v\in[1,m]$.
\end{lemma}

\noindent{\bf Proof of Lemma~\ref{lemmacase2}.} Claim
\begin{equation}
\label{muv}
\hat{\mu}_{i,m+1}(\bar{Y}_1,...,\bar{Y}_{v-1},\bar{Y}_v,\bar{Y}_{v+1}..,\bar{Y}_m,\bar{Y}_{m+1})\leq
\hat{\mu}_{i,m+1}(\bar{Y}_1,...,\bar{Y}_{v-1},\bar{Y}_v',\bar{Y}_{v+1}..,\bar{Y}_m,\bar{Y}_{m+1})
\end{equation}
for any $v\in[1,m]$ if $\bar{Y}_1\leq...\leq \bar{Y}_{v-1}\leq
\bar{Y}_v\leq \bar{Y}_v'\leq \bar{Y}_{v+1}\leq ...\leq \bar{Y}_m$.
If ($\ref{muv}$) is true, then
$$\hat{\mu}_{i,m+1}(\bar{Y}_1,...,\bar{Y}_m,\bar{Y}_{m+1})\leq
\hat{\mu}_{i,m+1}(\bar{Y}_1,...,\bar{Y}_{m-1},
\bar{Y}_m',\bar{Y}_{m+1})\leq ...\leq
\hat{\mu}_{i,m+1}(\bar{Y}_1',...,\bar{Y}_m',\bar{Y}_{m+1}).$$

To show ($\ref{muv}$), now write $i_1(m+1)$ introduced in
($\ref{i1m}$) as $i_1(\bar{Y}_v)$(note $i_1(y)$ introduced in the
beginning of the proof of Lemma~\ref{lemmacase1} has a different
argument $y=\bar{Y}_{m+1}$). Following ($\ref{i1y}$) and the second
equality of ($\ref{yv}$), $i_1(\bar{Y}_v)\leq i_1(\bar{Y}_v')$.
Similar to the proof of Lemma~\ref{lemmacase1}, ($\ref{muv}$) can be
shown in three cases a) $i< i_1(y)$, b) $i\in [i_1(y), i_1(y')-1]$
and c) $i\in [i_1(y'), m+1]$, and the proof is complete.
 ~\raisebox{.5ex}{\fbox{}}

\bigskip

In short, in this section, we proposed the SDMMSA to generate
estimators for monotone normal means, $\mu_i$, proved that the
SDMMSA and the PAVA are equivalent, both generate the mle's, and the
distribution of the proposed estimator is stochastically
nondecreasing when $\mu_i$ goes larger. The last is to be used to
derive a test to detect the MED in the response-dose study as shown
in the next section.

\section{A step-up testing procedure to detect the MED.}

Now return to the problem of finding the minimum effective
dose(MED). First, we formulate this as a multiple test problem by
proposing a sequence of decreasing null hypotheses. Then a general
result that identifies the least favorable distribution is provided.
Finally, a sequence of increasing rejection regions of
level-$\alpha$ is constructed.

\subsection{Motivation.}

Let
\begin{equation}
\label{trans} X_i=\bar{Y}_i-\bar{Y}_0\,\ \mbox{and} \,\
\eta_i=\mu_i-\mu_0\,\ \mbox{for}\,\ i=1,...,k.
\end{equation}
Since the $MED$ is to be found, one should start the search from
$i=1$ instead of $i=k$.  Therefore, a step-up procedure seems more
reasonable than a step-down one. To establish $N=1$, some authors
(see, for example Hsu and Berger (1999)) compare $min\{X_j:j\geq
1\}$ with $\delta$ and claim $N=1$ if $min\{X_j:j\geq 1\}-\delta$ is
large in the unit of $S$. Roughly speaking, they use $min\{X_j:
j\geq i\}$ to estimate $\eta_i$. This does not fully utilize the
assumption of the monotonicity on means. So we propose using the
maximum likelihood estimator of $\eta_i$, denoted by
$\hat{\eta}_i\stackrel{def}{=}\hat{\mu}_i-\bar{Y}_0$, as a test
statistic, where $\hat{\mu}_i$ is given by the SDMMSA in Section
2.1. If $\hat{\eta}_1-\delta$ is larger than a multiple of $S$, then
claim $N=1$ and stop; otherwise compare $\hat{\eta}_2-\delta$ with
$S$. Repeat this process until we find an $N$ so that
$\hat{\eta}_{N}-\delta$ is much larger than $S$. If no such $N$ can
be found, then the $MED$ does not exist.

To identify $N$(MED), let
\begin{equation} \label{null}
{\cal C}=\{H_{0i}=\{\eta_i \leq \delta\}: i\in [1,k]\}
\end{equation}
be the set of null hypotheses of interest in this section. For each
$i\geq 1$, the alternative $H_{Ai}$ claims $\eta_i>\delta$. If a
certain $H_{Ai}$ is established, then $N\leq i$ due to the
monotonicity in $\mu_i$'s for $i\geq 1$. Therefore, $N$ should be
equal to the smallest $i$ so that $H_{Ai}$ is true. For the strong
control of the experimentwise error rate, it is clear that $H_{0i'}$
is a subset of $H_{0i}$ if $i < i'$ due to the monotonicity(i.e.,
$H_{0i}$ is decreasing). Therefore, ${\cal C}$ itself is closed
under the operation of intersection. The closed test procedure
(Marcus, Peritz and Gabriel, 1976) can be applied on ${\cal C}$ to
construct a step-up testing procedure with the experimentwise error
rate controlled at $\alpha$ in the strong sense (see, for example,
Hochberg and Tamhane (1987) for a definition) as long as a
level-$\alpha$ test is constructed for each $H_{0i}$. Let $R_i$ be a
rejection region  for $H_{0i}$ for any $i$ between 1 and k.  In
order to strongly control the experimentwise error rate, as well as
being powerful, region $R_i$ should satisfy the following two
properties:

$*)$    $R_i$ is of level $\alpha$, i.e.,
$\mbox{sup}_{\underline{\mu}\in
H_{0i}}P_{\underline{\mu}}(R_i)=\alpha.$

$**)$    $R_i$ is increasing in $i$. i.e. $R_i\subset R_{i'}$ if
$i<i'$. Thus $R_i=\cap_{\forall H_{0i'}\subset H_{0i}}
R_{i'}=\cap_{i'=i}^k R_{i'}$.

 \noindent If these two are satisfied, then the multiple
tests, which assert $H_{Ai}$ if and only if $R_i$ occurs, strongly
control the experimentwise error rate at level $\alpha$, which is
the main result of this section.

\bigskip

\subsection{A general result.}

\begin{theorem}
\label{thmg} Let $T(t_1,...,t_k)$ and $g_i(t_1,...,t_k)$ for
$i=1,...,k$ be  non-decreasing function for any  $t_i$ when the
other $t_j$'s are held constant. Also
\begin{equation}
\label{linear} g_i(ct_1+d,...,ct_k+d)=cg_i(t_1,...,t_k)+d
\end{equation}
for any constants $c>0$ and $d$.
 Then
\begin{equation}
\label{f}
f(\eta_1,...,\eta_k,\sigma)\stackrel{def}{=}ET(\frac{g_1(\bar{Y}_1,...,\bar{Y}_k)-\bar{Y}_0-\delta}{S},...,\frac{g_k(\bar{Y}_1,...,\bar{Y}_k)-\bar{Y}_0-\delta}{S})
\end{equation}
is nondecreasing in each $\eta_i$ when the other $\eta_j$ and
$\sigma$ are held constants.
\end{theorem}

\bigskip

\noindent{\bf Proof of Theorem~\ref{thmg}.} Due to ($\ref{linear}$),
we assume $\bar{Y}_0$ has a mean 0 and $\bar{Y}_i$ has a mean
$\eta_i(=\mu_i-\mu_0)$. Let $\phi(x)$ be the pdf of $N(0,1)$ and
$g_{\nu}(y)$ be the pdf of a $\chi^2$-distribution with
$\nu=\sum_{i=0}^k n_i-(k+1)$ degrees of freedom. Then
$$f(\eta_1,...,\eta_k,\sigma)=\int \int ET(\frac{g_1-x\frac{\sigma}{\sqrt{n_0}}-\delta}{
\sqrt{\frac{\sigma^2 y}{\nu}}},
...,\frac{g_k-x\frac{\sigma}{\sqrt{n_0}}-\delta}{
\sqrt{\frac{\sigma^2 y}{\nu}}})\phi(x)g_{\nu}(y)dx dy.$$ For each
fixed $x$ and $y$, let
$$T_{x,y}(\bar{Y}_1,...,\bar{Y}_k)=T(\frac{g_1-x\frac{\sigma}{\sqrt{n_0}}-\delta}{
\sqrt{\frac{\sigma^2 y}{\nu}}},
...,\frac{g_k-x\frac{\sigma}{\sqrt{n_0}}-\delta}{
\sqrt{\frac{\sigma^2 y}{\nu}}}),$$ which is non-decreasing in each
$\bar{Y}_i$ due to the monotonicity of $T$ and $g_i$'s. Therefore,
the conditional distribution of $T_{x,y}$ for given $x$ and $y$ is
stochastically nondecreasing in each $\eta_i$(see Lemma 2 in Wu and
Wang (2007)). Hence its conditional expectation
\begin{equation}
\label{inte}
ET_{x,y}=ET(\frac{g_1-x\frac{\sigma}{\sqrt{n_0}}-\delta}{
\sqrt{\frac{\sigma^2 y}{\nu}}},
...,\frac{g_k-x\frac{\sigma}{\sqrt{n_0}}-\delta}{
\sqrt{\frac{\sigma^2 y}{\nu}}}) \end{equation}
 is nondecreasing in each
$\eta_i$. So is $f$, the integral of ($\ref{inte}$).
 ~\raisebox{.5ex}{\fbox{}}
\bigskip

\noindent{\bf Remark 3}. Each
$g_i(\bar{Y}_1,...,\bar{Y}_k)\stackrel{def}{=}\hat{\mu}_i$ satisfies
($\ref{linear}$) and is nondecreasing in each $\bar{Y}_j$. We will
use this to construct step-up tests in the next section.
~\raisebox{.5ex}{\fbox{}}

\bigskip

\noindent{\bf Remark 4}. If define
$g_i(\bar{Y}_1,...,\bar{Y}_k)=\bar{Y}_i$ for $i\in[1,k]$ and a
sequence of
\begin{equation}
\label{hb} T^{HB}_j=I_{\{min_{\{i\in[j,k]\}}\{
(\bar{Y}_i-\bar{Y}_0-\delta)/\sqrt{1/n_i+1/n_0}\}>t_{\alpha,\nu}\}},
\end{equation}
for $j\in[1,k]$, then $g_i$ and $T^{HB}_j$ satisfy the conditions of
Theorem~\ref{thmg}. Hsu and Berger's step-down tests (1999) claim
$N$, the MED, to be $j_0$ if $T^{HB}_{j_0}=1$ but
$T^{HB}_{j_0-1}=0$. ~\raisebox{.5ex}{\fbox{}}
\bigskip

\subsection{The construction of step-up tests}

We first construct a rejection region $R^I_i$ with level $\alpha$
for each individual $H_{0i}$.
\begin{lemma}
\label{lem1} For a constant $c$, let
\begin{equation} \label{rIi}
R^I_{i,c}=\{\frac{\hat{\mu}_i-\bar{Y}_0-\delta}{S}>c\}.
\end{equation}
Then
\begin{equation} \label{typeI}
sup_{\underline{\bfb{\mu}} \in H_{0i}}
P_{\underline{\bfb{\mu}}}(R^I_{i,c})=
P_{\underline{\bfb{\mu}}_i}(R^I_{i,c}),
\end{equation}
 where $\underline{\bfb{\mu}}_i=(\mu_0,\mu_1,...,\mu_k)$ with
$\mu_1=...=\mu_i=\mu_0+\delta$ and $\mu_{i+1}=...=\mu_k=+\infty$.
Therefore, for any $\alpha\in (0,1)$, $R^I_{i,c}$, with
$c=c_{i,\alpha}$, defines a level-$\alpha$ test for $H_{0i}$,
 where $c_{i,\alpha}$ is the solution of
\begin{equation}\label{cut}
 P_{\underline{\bfb{\mu} }_i}(R^I_{i,c})=\alpha.
\end{equation}
\end{lemma}

\bigskip

\noindent{\bf Proof of Lemma~\ref{lem1}.} Let $T=I_{R^I_{i,c}}$.
Then Lemma~\ref{lem1} follows Theorem~\ref{thmg}.
~\raisebox{.5ex}{\fbox{}}

\bigskip

\noindent{\bf Remark 5}.
$c_{1,\alpha}=t_{\alpha,\nu}\sqrt{1/n_1+1/n_0}$ due to
$\hat{\mu}_1=\bar{Y}_1$ when
$\underline{\bfb{\mu}}=\underline{\bfb{\mu}}_1$.
 ~\raisebox{.5ex}{\fbox{}}

\bigskip

Region $R^I_{i,c}$ satisfies property *), but not property **) in
Section 2. To obtain more powerful multiple tests, we propose
\begin{theorem}\label{thm1}
For any integer $i\in [1,k]$ and for a sequence of nonnegative
constants $c_1$ through $c_i$,  let
\begin{equation}
\label{rc} R_{c_1,...,c_i}=\cup_{j=1}^i R^I_{j,c_j}=
\cup_{j=1}^i\{\frac{\hat{\mu}_{j}-\bar{Y}_0-\delta}{S}>c_j\}.
\end{equation}
 Then
\begin{equation}
\label{max}
 sup_{\underline{\bfb{\mu}}\in
H_{0i}}P_{\underline{\bfb{\mu}}}(R_{c_1,...,c_i})=P_{\underline{\bfb{\mu}}_i}(R_{c_1,...,c_i}).
\end{equation}
Therefore, for any $\alpha\in (0,1)$, $R_{c_1,...,c_i}$, with
$c_1=c_{1,\alpha}$ given in Remark 5 and $c_i$ determined
iteratively by solving

\begin{equation}
\label{cutm} P_{\underline{\bfb{\mu} }_i}(R_{c_1,...,,c_i})=\alpha,
\end{equation}
for $i=2,...,k$, defines a level-$\alpha$ test for $H_{0i}$.
\end{theorem}

\noindent{\bf Proof of Theorem~\ref{thm1}.} Let
$T=I_{R_{c_1,...,c_i}}$. Then Theorem~\ref{thm1} follows
Theorem~\ref{thmg}. ~\raisebox{.5ex}{\fbox{}}

\bigskip

\begin{theorem}
\label{thm3} Consider all hypotheses in ${\cal C}$ in $(\ref{null})$
with the following testing procedure:
\begin{equation}
\label{multi} \mbox{assert  $H_{Ai}$ (or not $H_{0i}$) if
$R_i\stackrel{def}{=}R_{c_1,...,c_i}$ occurs} \end{equation} for any
fixed $\alpha\in (0,1)$. Then the experimentwise error rate is at
most $\alpha$. i.e., the probability of making at least one
incorrect assertion is at most $\alpha$.
\end{theorem}

\noindent{\bf Proof of Theorem~\ref{thm3}}. The proof is trivial if
one notices that $H_{0i}$ is decreasing in $i$, and $R_i$ is of
level-$\alpha$ and is increasing in $i$. Then Theorem~\ref{thm3}
follows the closed test procedure by Marcus, Peritz and Gabriel
(1976).
 ~\raisebox{.5ex}{\fbox{}}

\bigskip

\noindent{\bf Remark 6.} Region $R_i$ is increasing in $i$. Then
$R_i$ satisfies properties *) and **).
 ~\raisebox{.5ex}{\fbox{}}

\bigskip

\noindent{\bf Remark 7.} When the design is balanced, region $R_1$
contains the set of $\{T^{HB}_1=1\}$, on which Hsu and Berger's test
(1999) claims the MED=1. Therefore, the proposed test is uniformly
more powerful than Hsu and Berger's when the MED=1.
 ~\raisebox{.5ex}{\fbox{}}

\bigskip

\noindent{\bf Example 1(continued).}
 The sample standard deviation
$S=7.751$, $t_{0.05, 50}=1.676$. We compare the new step-up
procedure with the step-up procedure SU1P, the step-down Williams
procedure and step-down procedure SD1P in Tamhane et al (1996) and
the DR method in Hsu and Berger (1999). For illustration, $\delta
=6.5$. From Table 1 in Dunnett and Tamhane (1992) we have the
critical values for the step-up procedure SU1P $c_1=1.645,
c_2=1.933, c_3=2.071, c_4=2.165, c_5=2.237, c_6=2.294, c_7=2.342,
c_8=2.382$ (we treat $df=50$ as $df = \infty$). The SU1P procedure
infers $\widehat{MED} = 5$.
 The Williams procedure
has the $\bar{t}$ statistics: $\bar{t}_1=-1.810, \bar{t}_2=-0.961,
\bar{t}_3=0.313, \bar{t}_4=1.899, \bar{t}_5=5.788,
\bar{t}_6=\bar{t}_7=\bar{t}_8=9.334, \bar{t}_9=9.877$. The Williams
statistics are compared with the following critical values (taken
from Williams (1971)) in a step-down manner: $c_1=1.675, c_2=1.755,
c_3=1.780, c_4=1.790, c_5=1.795, c_6=1.800, c_7=1.805, c_8=1.805,
c_9=1.810.$. The Williams procedure infers $\widehat{MED} = 4$. By
simulation with 1,0000 repetition, the critical values of the new
statistic are $c_1=0.968, c_2=1.022, c_3=1.046, c_4=1.046,
c_5=1.034, c_6=1.043, c_7=1.044, c_8=1.047, c_9=1.030$,
respectively. Also $(\hat{\mu}_i-\bar{Y}_0-\delta)/S$ for
$i=1,2,3,4$ given in (\ref{rc}) are  -1.045,
     -0.555,
      0.181, and
       1.097, respectively. Thus the new step-up procedure concludes  $\widehat{MED} = 4$. So does
        Hsu and Berger (1999)'s DR method.

\section{Discussion.}

In this paper, we propose an alternative, SDMMSA, for the widely
used PAVA. Although the two are equivalent, the SDMMSA is important
by itself since it is easily coded and is well defined. Also a
stochastic ordering of the estimators for the monotone normal means
is established based on the SDMMSA. As one of its applications, a
step-up test procedure is proposed to identify the MED. It strongly
controls the experimentwise error rate, and is powerful to detect
the MED, especially when the true MED is small.

\section*{Acknowledgments.} We thank Professor Roger Berger for his helpful comments.

redpajama_set_name: RedPajamaArXiv


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Peaceful Passage Tribute. With elegance and grace this striking bouquet of fuchsia and purple roses sapphire lisianthus delicate dahlias plum Stargazer lilies lavender alstroemeria and magenta statice accented with tender greenery and displayed in a traditional white urn makes a timely and tasteful statement of your everlasting loyalty and love during trying times of personal loss. This consoling arrangement will bring a lovely touch of color and tranquility to any wake funeral or memorial service. It arrives with expedient same-day delivery instantly uplifting your dear ones. Style of container may vary. Product is not appropriate to send to a residence.
The Golden Garden Casket Spray offers an abundance of brilliant yellow blossoms with sunny yellow roses spray roses carnations chrysanthemums snapdragons and more. To add warmth and light to even the saddest of losses this arrangement is an excellent selection and it is available for same-day delivery when necessary by our highly trained florists.
Angel s Bounty Casket Spray. Each bloom in this lovely graceful casket arrangement is hand-selected and arranged by one of our local florists ready to be prominently displayed at a funeral service or burial. A bouquet of beautiful Stargazer lilies red roses coral carnations pale cream larkspur and fragrant hydrangeas are nestled amid soothing green ivy salal and aspidistra for a visual presentation that will bring peace and tranquility in your loved ones hour of need. This spray is traditionally ordered by an immediate family member and arrives with rapid same-day delivery to any location in the country. Product is not appropriate to send to a residence.
Rejoicing Spirit Bouquet. Give the gift of hope and courage in times of great sadness. This stunning bouquet of vibrant yellow roses golden Asiatic lilies Bells of Ireland and bright gladioli is thoughtfully arranged by our local florists and displayed amid lime green aspidistra leaves in a slender glass vase. Your fresh flowers will arrive with expedient same-day delivery ready to adorn an alter podium or tabletop at a funeral or memorial service or grace the home of grieving friends and family members letting them feel your endless support and love. Style of vase may vary.
The Our Hearts Speak to You Red Cross beautifully symbolizes the unyielding assistance that the Red Cross gives to individuals in need throughout the world. This gorgeous display is made up entirely of fresh red carnations to honor the memory of someone who served within the organization or relied on them for assistance at some point in their lives. Few arrangements can pay a more personal touching tribute at a final farewell service. Same-day delivery is offered as well.


He's bad, I know it.
UPDATE: To the person holding the "repent" sign -- I agree with your sentiments, but please, put down the poster board. You make us (Xians) all look silly.
UPDATE II: Watching him walk out of the courtroom, I think Jackson is on Xanax and ludes.
UPDATE III: If the verdict had been guilty, just what would that woman have done with the doves???
Folks, I have to admit, this is our legal system working properly, so to speak. They had no direct evidence and the accusers seriously lacked credibility.
However, our society is NOT working properly when a man who sleeps with underage children is venerated by screaming throngs.
In the best of worlds, we wouldn't have parents pimping thier children to creeps like this in hopes of scoring some huge payoff.
I think something happened here but the DA's office was stuck prosecuting for the wrong victim. . . .with the worst possible family. The defense just played the cards dealt.


text:
https://www.mysanantonio.com/news/texas/article/Mexico-extends-100-million-reforesting-program-14078903.php
Mexico extends $100 million reforesting program to Honduras
Updated 6:17 pm CDT, Monday, July 8, 2019
Migrant children eat a meal provided by United in Christ Ministry of Los Fresnos, Texas, a group who brings dinner every Thursday in rotation with other volunteers who organize to provide daily meals, at the Puerta Mexico international bridge, in Matamoros, Tamaulipas state, Mexico, Thursday, June 27, 2019. Mexican police, soldiers and National Guard have recently been raiding hotels, buses and trains to round up migrants, creating scenes that have caused an outcry in the United States, but in Mexico there has been little backlash despite the historical sympathy for the plight of migrants. less
Migrant children eat a meal provided by United in Christ Ministry of Los Fresnos, Texas, a group who brings dinner every Thursday in rotation with other volunteers who organize to provide daily meals, at the ... more
Photo: Rebecca Blackwell, AP
MEXICO CITY (AP) — Mexico said Monday it will extend a reforestation program to Honduras, and increase funding to $100 million to create jobs in Central America and stem migration from the region.
Mexico had already announced a $30 million tree-planting program for El Salvador and that money is presumably part of the $100 million total effort.
Marcelo Ebrard, Mexico's foreign relations secretary, said Honduras' president is expected to finalize details in a visit to Mexico this month.
Ebrard said creating jobs will do more to stop the flow of migrants than border enforcement measures.
It is unclear whether Guatemala will also join Mexico's reforestation program.
A similar plan to plant fruit and lumber trees is already being implemented in southern Mexico. It is one of the pet projects of President Andrés Manuel López Obrador.
Ebrard also said that 327 Central American migrants awaiting US asylum hearings have found jobs in northern Mexico and that companies had offered a total of 3,700 positions.
Under the Remain in Mexico program, 18,503 Central Americans who applied for asylum in the United States have been returned to Mexico so far this year to await resolution of their U.S. cases.
Critics claim the program, also known as the Migrant Protection Protocols, leaves migrants in Mexican border cities with no income or housing.
Ebrard said Mexican officials had overcome legal hurdles to allow migrants to work while they are waiting. Many of the jobs are at border assembly plants known as maquiladoras.
Ebrard said Mexico had also arranged with 16 border shelters to house at least some of the waiting migrants.
Mexico's labor ministry announced the "Juarez Initiative" to find jobs and provide shelter and support services for waiting migrants in the border cities of Tijuana, Mexicali and Ciudad Juarez. The federal government would pay for the actions.
Horacio Duarte Olivares, deputy secretary of employment, said the migrants wouldn't be taking jobs from Mexicans, but rather filling openings. The government would fast-track their applications for work permits, he said.
However, some asylum applicants appear to have given up. In early July, 66 Central Americans who had been returned to the border city of Ciudad Juarez to await the outcome of their U.S. asylum claims were bused back to their home countries.
Mexico has increasingly detained migrants in raids on buses, trucks, trains and motels, under pressure from the administration of U.S. President Donald Trump to prevent migrants from reaching the U.S. border.
On Monday, Mexico's federal Public Safety Department said federal police pulled 228 migrants off a freight truck that was supposedly carrying soft drinks in the southern border state of Chiapas. The driver and an assistant were arrested. The migrants came from Guatemala, El Salvador and Honduras.


text:
Der Vogelkopf ist ein  hoher Berg auf dem Gebiet der Gemeinde Wackersberg, südlich von Bad Tölz in Bayern.

Er kann als Bergwanderung von Arzbach aus erreicht werden, am Ende weglos. Der Gipfel selbst ist bewaldet und bietet wenig Aussicht.

Galerie

Einzelnachweise

Geographie (Wackersberg)
Berg im Landkreis Bad Tölz-Wolfratshausen
Bayerische Voralpen
Berg in den Alpen


text:
Population Pharmacokinetics and its Role in Anti-Tuberculosis Drug Development and Optimization of Treatment
Author(s): Eric Free Egelund, Aline Bergesch Barth, Charles Arthur Peloquin	Department of Pharmacotherapy and Translational Research, College of Pharmacy, P. O. Box 100486, University of Florida, Gainesville, FL 32610-0486, USA.
Tuberculosis (TB) has survived for millennia due to its unique physiochemical properties and its ability to persist in a "dormant" state. Because of these unique properties, the treatment of TB involves multiple drugs for half of a year or longer. Although the incidence of TB has declined in most developing countries, it remains poorly controlled in much of the developing world. This is partly driven by co-infection with HIV in these settings. Concern about this dual epidemic has renewed interest in anti-tuberculosis drugs, and this in turn has led to increasing amounts of pharmacokinetic data regarding TB drugs.
Population pharmacokinetics has proven useful for many disease states, providing a valuable tool with which to explore data and predict future events. It not only generates pharmacokinetic parameter estimates, but it includes patient characteristics (covariates) that may alter those parameters. We review the population pharmacokinetic studies undertaken with anti-TB medications. In addition to the primary medications used in TB treatment (rifampin, isoniazid, pyrazinamide, and ethambutol) we discuss the fluoroquinolones, various software packages, and the covariates that were found to be significant in each study.
Keywords: Population pharmacokinetics, rifampin, rifabutin, rifapentine, pyrazinamide, ethambutol, fluoroquinolones, epithelial lining fluid (ELF), alveolar cell concentrations, monotherapy
Title: Population Pharmacokinetics and its Role in Anti-Tuberculosis Drug Development and Optimization of Treatment
Author(s):Eric Free Egelund, Aline Bergesch Barth and Charles Arthur Peloquin
Affiliation:Department of Pharmacotherapy and Translational Research, College of Pharmacy, P. O. Box 100486, University of Florida, Gainesville, FL 32610-0486, USA.
Keywords:Population pharmacokinetics, rifampin, rifabutin, rifapentine, pyrazinamide, ethambutol, fluoroquinolones, epithelial lining fluid (ELF), alveolar cell concentrations, monotherapy
Abstract: Tuberculosis (TB) has survived for millennia due to its unique physiochemical properties and its ability to persist in a "dormant" state. Because of these unique properties, the treatment of TB involves multiple drugs for half of a year or longer. Although the incidence of TB has declined in most developing countries, it remains poorly controlled in much of the developing world. This is partly driven by co-infection with HIV in these settings. Concern about this dual epidemic has renewed interest in anti-tuberculosis drugs, and this in turn has led to increasing amounts of pharmacokinetic data regarding TB drugs.
Eric Free Egelund, Aline Bergesch Barth and Charles Arthur Peloquin, " Population Pharmacokinetics and its Role in Anti-Tuberculosis Drug Development and Optimization of Treatment", Current Pharmaceutical Design (2011) 17: 2889. https://doi.org/10.2174/138161211797470246
Identification of Proteins Interacting with Human SP110 During the Process of Viral Infections
Derivatives of 3-Isoxazolecarboxylic Acid Esters - A Potent and Selective Compound Class against Replicating and Nonreplicating Mycobacterium tuberculosis
Increased Expression of Forkhead Box Protein 3 Gene of Regulatory T Cells in Patients with Active Tuberculosis
Inflammation & Allergy - Drug Targets (Discontinued)
Methods for Diagnosis of Human Pulmonary Tuberculosis (Mini-Review)
Applied Clinical Research, Clinical Trials and Regulatory Affairs (Discontinued)
Artificial Neural Network Analysis of Pharmacokinetic and Toxicity Properties of Lead Molecules for Dengue Fever, Tuberculosis and Malaria
Current Computer-Aided Drug Design
Shikimate Kinase: A Potential Target for Development of Novel Antitubercular Agents
Characterization of Radiolabeled Dry Powder Leucine, a Constituent of Inhalable Capreomycin
Drug Delivery Letters
Liposomes as Potential Carrier System for Targeted Delivery of Polyene Antibiotics
Recent Patents on Inflammation & Allergy Drug Discovery (Discontinued)
Editorial [Tuberculosis Drugs and Drug Targets]
Infectious Disorders - Drug Targets
Recent Developments in Azole Compounds as Antitubercular Agent
Mini-Reviews in Organic Chemistry


text:
Kentucky's Capitol Building.
Thorns & Roses: Frankfort's War On Kentucky's Poor Is Unconscionable
By LEO Weekly
THORN: Frankfort's War On The Poor
Kentucky's lawmakers are really showing off their callousness, nay, straight out hatred, toward the poor this year. After floor rants about how nobody wants to work these days, the General Assembly approved a bill slashing unemployment benefits. Then, they voted to kill Kentucky's state of emergency, which will cause Kentucky to lose out on $50 million a month in federal SNAP benefits. All because they're too preoccupied with the one big man in the Governor's Mansion instead of the millions of little ones they represent.
ROSE: That's One Way To Be Remembered
Yvonne Miles was a character. You can tell from the obituary she wrote herself, which was published on the Courier Journal's website on March 10. "Well, it finally happened, I died," she wrote, "At 68 years of age, I stopped harassing people on 9-March, 2022." She also wrote that she chose to be cremated so she could "have the smokin' hot body" she always wanted. Yvonne, we didn't know you in life but we now realize we really missed out.
ROSE: A Worthy Honor
For the past two years, Southern Living Magazine's Southerner of the Year has been a famous musician. And while Dolly Parton (2020's winner) deserves all the accolades, the magazine's pick this year was special: Citizens of Mayfield, Kentucky, who survived devastating tornadoes this past December and who have been helping to repair the town.
ROSE: So… You're Saying There's A Chance?
If Kentucky legalizes medical marijuana, it will be up there with the 1876 Meat Shower in terms of surprise phenomenon. And yet, this year's bill is looking more and more possible. A key Kentucky senator, Whitney Westerfield, endorsed Rep. Jason Nemes' bill, after opposing his efforts previously. And during a recent House committee vote another Republican, Rep. Joe Fischer, made an about-face, voting to move the bill forward. Of course, Nemes had to make his bill extremely strict in order to appease his colleagues who are stuck in the past, but this softening of hearts has us feeling hopeful.
Published under News, Opinion | Tags: 2022 kentucky legislative session, funny obituary, kentucky tornaodes, mayfield kentucky, medical marijuana


Many of our customers are unable to visualize their industrial plant. The barrier like a wall or a hall roof hides a free overview. With our 3D animated explainer videos, we remove all the parts that obscure sight and show your system in a complete overview. We can visualize coherent processes in an easy-to-understand-way. The interaction of the individual components becomes apparent to your spectators. Showing your industrial plant in all their complexity underline your expertise and give an idea of the value of your business.
Due to the extensive models, 3D animated explainer videos are often quite expensive to produce.
If you have beautiful photos or graphics of your system, we can produce a 2D animated explainer video at a very affordable price.
Here we show an example of the visualization of an industrial plant. It is the representation of a paper mill. We created this image based on the data we used for a 3D animated video. The size of this image is big enough to print a poster.
Here is a selection of explainer videos about industrial plants. The selected examples are kept very short, so you can quickly get an overview of the variety of possibilities. Perhaps you too will find here a kind of presentation that could competently present your facility and your service.
Get in touch with us and let's talk about how we can present your industrial plant clearly and understandably.


Below are a few reasons I pick up a book to read!
1. I've read a book by the author before and loved their writing.
2. The book has been recommended to me by another avid reader that has the same taste I do in terms of genre, writing style, etc.
3. I like the book cover and/or book title.
4. The blurb about the book on its back cover makes me want to read it.
5. It's a book club pick.
6. I want to explore a new genre.
7. I'm interested in learning a new hobby or skill.
8. The book is required reading for a class.
9. I found the book at a Little Free Library.
10. I keep hearing a lot of positive buzz about a particular author.
Ooh nice set of reasons! I can definitely relate to a few of them as well!
Thank you for stopping by and leaving me a comment. Happy reading!!
I had cover and author I'd read before on my list this week too! I'll also pick up a book because the blurb is intriguing, I've been recommended it by a friend or if there's a lot of buzz around the author, particularly debuts.
Book club picks are huge for me too, though I probably influence those too much to say that's the ultimate reason I pick them up. 😂 Definitely reliant on a good blurb or a trusted recommendation as well.
Learning a new hobby or skill is a fantastic reason to pick up a book.
I love knitting, so have picked up a few books to help me as a newbie knitter!
A lot of these have been pretty popular today. But I like the idea of picking something up because you want to learn more about a subject.
All of these are great reasons, but I think the first four probably account for a huge amount of my new reads!
Those are great reasons to pick up a book. I love to read books from debut authors (especially if they're awesome people on Twitter), but I also like reading books by some of my favorites.
A very true and good list!
I adore Little Free Libraries and visit them @ every opportunity I can!! I find as many new ones as possible. Do you not have any Little Free Libraries near you? Happy reading!!
Little Free Libraries are the BEST! I very rarely find anything I want to read, but they are sprouting like dandelions in my region and I never tire of the thrill of peeking inside.
Yes they are the best!! I found two of them this past weekend in Sacramento.


text:
FROM php:5.6-apache

RUN docker-php-ext-configure pdo && \
    docker-php-ext-install pdo && \
    docker-php-ext-configure pdo_mysql &&  \
    docker-php-ext-install pdo_mysql

RUN a2enmod rewrite && a2enmod headers


Are you looking for new and exciting way to reach your audience with your message? Edd Fairman, the Wizard of Sorts, creates custom presentations that use comedy, magic, juggling, and technology to engage your audience in a novel, visual, and memorable way.
Receive your FREE consultation - Contact Edd Today!
Audience Bored by Another Power Point?
Trade Show Booth Lead Numbers Lacking?
Are People Avoiding your Team Building Activity?
What Can Edd Fairman Do For You?
Speaker, Motivator, and Mentalist Edd Fairman will leave your audience asking for more, and you looking like a winner.
Edd can provide a fully customized show that incorporates your company's values and message into the presentation. There is a reason that Ford, Harris Bank, and Four Seasons Hotels are among his extensive list of clients that just can't get enough of him and his friendly, clean shows.
It takes more than another PowerPoint Presentation to get your audience to remember your specific event or meeting's take away message. Edd will leave them engaged, entertained, and motivated.
Should a magician be in attendance at your next corporate event? Or more specifically, could Edd Fairman, Wizard of Sorts, make a real difference in getting the word out about your products and services? Think outside the box and consider all the benefits that come with bringing in someone for corporate entertainment ROI to handle your next company party or corporate event.
Who doesn't like to be entertained by someone like a magician? Many times employees show up to a corporate party or get together and expect that things are going to drag on. However, with the corporate mentalist "Wizard of Sorts" up at the front, interest is immediately peaked and there is no doubt that people are going to be paying attention. The same is true for trade shows, seminars, and conventions. Take a break from the traditional spokesperson and see how much attention is gained by adding a little bit of entertainment to the presentation.
There is a cost associated with Corporate Entertainment ROI. While that is expected, companies are often surprised to learn about the return that comes with the investment made to bring in someone new to deliver the message. In situations where potential customers are involved, Edd Fairman is sure to keep audiences captivated and find ways to make sure that people listen to the entire presentation. Many people will be in the audience, enjoying the entertainment, and recall their experience to others, later.
Having an entertainer help with corporate events doesn't mean that the presentation has nothing to do with a company's products and services. Instead, the "Wizard of Sorts" makes it not only possible, but essential to center the entertainment on a specific product or service, incorporating it into the act. While keeping everyone engaged, audiences are able to learn more about your company's message. This creates a unique experience. And, because your company's message about a product or service is so closely tied to the entertainment, this too will be remembered, outside of the wow factor.
Take advantage of the opportunity to shake things up at your next corporate event or presentation. Instead of settling for the same thing each year, try using the top corporate entertainment magician in Chicago to spice things up and get a little bit of added attention!


text:
Десятый Чемпионат Южной Америки был проведён в городе Сантьяго, столице  Чили. Он стартовал 12 октября 1926 года матчем  Чили  — Боливия, а завершился 3 ноября 1926 года матчем  Чили —  Парагвай.

Участники
  Аргентина
 Боливия
  Парагвай
  Уругвай
  Чили (организатор)

Составы команд

Матчи
12 октября, 1926

12 октября, 1926

17 октября, 1926

20 октября, 1926

23 октября, 1926

24 октября, 1926

28 октября, 1926

20 октября, 1926

1 ноября, 1926

3 ноября, 1926

Турнирная таблица

Чемпион

Лучшие бомбардиры

7 Голов
  Давид Арельяно

6 Голов
  Эктор Кастро
  Эктор Скароне

5 Голов
  Габино Соса

4 Гола
  Мануэль Рамирес

3 Гола
  Роберто Черро
  Бенхамин Дельгадо
  Пабло Рамирес

2 Гола
  Антонио Де Мигель
  Гильермо Субиабре
  Сеферино Рамирес
  Рене Борхас
  Сойло Сальдомбиде

Ссылки
 ЧЮА на rsssf.com

Кубок Америки по футболу
1926 год в футболе
Октябрь 1926 года
Ноябрь 1926 года
Чемпионаты Южной Америки в Чили
1920-е годы в Чили
Международные спортивные соревнования в Сантьяго
Международные соревнования по футболу в Чили


Graduated in Chemical Engineering from the Colorado School of Mines. I have worked as a petroleum engineer for 7 years.
Graduated from TCU with degree in Biology and Marketing. I have continued my education by receiving a masters in education.
Florida State University, Bachelor's of Science in Business Management with a Minor in Psychology. GPA - 3.35. Graduated in 2009.
University of Colorado at Boulder graduate in Spanish, Business, and Ethnic Studies.
BA, English; JD, Law; MA Cert. Education. Professional Secondary Language Arts Teaching License.


It is difficult for many families to have to juggle 1/2 kindergarten and find daycare arrangements and find employment that is flexible to accommidate both. That is why we offer a full day, every day kindergarten. On top of the convience of not worrying about daycare and work, there are many benefits for students who attend all day everyday kindergarten. It can be difficult for some students as full days require long periods of concentration, lots of work, lots of new information to absorb, and of coarse, no naps.
We have a stagered start for the kindergarten students. Half of the students will come on the first day and the rest will be on the second day. Everyone then comes together for the first time on the third day. This helps ease students into school life and helps them to become familiar with rules and routines in a small class setting with less distractions.
More information about Kindergarten can be found here.