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Time to think small? Gregory Sterzel Tue Nov 22 09:30:00 EST 2022 Time to think small? http://www.federatedinvestors.com/texPool/static/images/texpool/texpool-logo-amp.png http://www.federatedinvestors.com/texPool/daf\images\insights\article\magnifying-glass-forest-small.jpg 11-21-2022 November 21 2022 11-22-2022 November 22 2022
Time to think small?
Perhaps. But the focus should be on quality.
Published 11-22-2022 November 22 2022
Gregory Sterzel, CFA, CAIA
Client Portfolio Manager
A tough year for equities has been even harder on small-cap stocks. No great surprise there. Because of their lack of scale relative to larger companies, smaller companies tend to be more sensitive to changes in the economy—to the upside when growth is accelerating and the downside when it's decelerating. And because smaller companies typically carry more debt than larger companies, much of it variable, they're inclined to be more vulnerable to changes in rates. So, in the first half of the year when real GDP contracted and the Fed embarked on its most aggressive rate-hike cycle since the early 1980s, small caps as measured by the Russell 2000 underperformed larger companies as measured by the S&P 500.
The past month, however, has seen smaller companies take the lead. Again, no great surprise. The fourth quarter got off to a relatively strong start—the Atlanta Fed currently has Q4 GDP tracking at 4.2% annualized rate, its best pace since last year. Similarly, futures suggest the Fed is closing in on an end point for its rate-hike cycle. At this writing, the yield on the 10-year Treasury is down 50 basis points from its October cycle peak of 4.25%. Stronger growth and lower rates represent tailwinds for small caps—they've outperformed large caps in all six of the six recoveries since 1980, often times meaningfully, Strategas Research says. After the dot.com bubble ended, for example, small caps outpaced large caps by more than 40%, cumulative, from October 2002 to October 2007.
To be sure, one month does not a trend make. There are many reasons growth could slow abruptly in the year ahead—that in fact is the consensus view of the Federated Hermes Macroeconomic Policy Committee. And while the Fed may be nearing a terminal rate, all indications are it will be in no rush to start easing. Futures currently are pricing a peak policy rate of 5.25% through summer. So, caution is merited. That's why when we think small caps, we're thinking quality companies that have lower debt, stronger balance sheets and a history of earnings relative to the overall small-cap universe. Since its inception on July 1, 2007, the Russell 2000 Quality Factor Index has outperformed, on a rolling 60-month basis, the broader Russell 2000 Index more than 80% of the time through Q3 2022, according to Morningstar Direct. This would suggest that, in volatile times, quality matters.
Tags Equity . Markets/Economy . Active Management .
Past performance is no guarantee of future results.
Bond prices are sensitive to changes in interest rates, and a rise in interest rates can cause a decline in their prices.
Diversification and asset allocation do not assure a profit nor protect against loss.
Gross Domestic Product (GDP) is a broad measure of the economy that measures the retail value of goods and services produced in a country.
Russell 2000® Index: Measures the performance of the 2,000 smallest companies in the Russell 3000 Index, which represents approximately 8% of the total market capitalization of the Russell 3000 Index. Investments cannot be made directly in an index.
Russell 2000 Quality Factor Index: A composite of the broader Russell 2000 Index that focuses on constituents with a history of profitability, efficiency, earnings quality and leverage. Indexes are unmanaged and investments cannot be made in an index.
Small company stocks may be less liquid and subject to greater price volatility than large capitalization stocks.
S&P 500 Index: An unmanaged capitalization-weighted index of 500 stocks designated to measure performance of the broad domestic economy through changes in the aggregate market value of 500 stocks representing all major industries. Indexes are unmanaged and investments cannot be made in an index.
Stocks are subject to risks and fluctuate in value.
Federated Advisory Services Company


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The TortureCast
Tag Archives: draft
by TortureCast | June 10, 2019 · 4:38 PM
Episode #169: MadBum vs. MadMax
Madison Bumgarner barks at Max Muncy shortly after Muncy deposited a 92 MPH fastball into McCovey Cove. Bum didn't like the brief admiring pause that Muncy gave at the plate, but honestly, we didn't think there was much to be upset about. Photo by CBS Sports.
You can download the episode here, or stream it below!
https://torturecast.podomatic.com/enclosure/2019-06-10T16_28_24-07_00.mp3
The Giants just can't seem to push through in any particular week, and last week was no exception. Although they did win the first game of each series against the Mets and Dodgers, they proceeded to lose the next two games and each series. They also mustered a horrendous 2 for 24 with runners in scoring position against the Dodgers, glaringly one time unable to score with the bases loaded and no outs and again with runners on the corners with no outs in their 1-0 loss on Sunday. Speaking of Sunday, anyone see that Muncy splash hit and the subsequent reaction by MadBum? Yeah, Eric and Chad have thoughts on that.
We also review the draft, this week in Giants history, and who's hot and who's not. We'll be back in two weeks for episode 170.
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Filed under Episodes
Tagged as blog, bruce bochy, draft, Hunter Bishop, Joey Bart, Los Angeles Dodgers, Madison Bumgarner, Max Muncy, New York Mets, podcast, San Francisco Giants, Todd Frazier
by TortureCast | October 2, 2017 · 11:34 PM
Episode #130: Free Fallin'
Matt Cain, we are going to miss you. Photo by Brad Mangin
What a horrendous day. Just coming off the news from Puerto Rico, then comes the Vegas shooting and now Tom Petty. Let's talk about baseball, shall we?
Chad, Eric and Willie reflect on the end of the 2017 season for the Giants, including what went wrong, how they may address it, Pablo's season-ending walk-off and how that affects the draft, and of course, Matt Cain's last start. Chad talks about covering Cain's last start and Eric shares a poignant moment he had with his father when they watched Matt Cain and Barry Bonds beat the Cubs in Chicago.
We also dwell upon the impending playoffs and offer up our predictions. Chad and Willie both predict Houston and Washington making it to the World Series with Houston winning it all, while Eric thinks the Dodgers will hoist the Tiffany trophy after defeating the Astros. (GOD I hope Eric is wrong)
That means it'll be the Twins vs the Rockies in the World Series, of course.
We'll be back in a week or two to reflect on the playoffs, then again after the World Series before we head into the hot stove season.
Tagged as draft, Houston Astros, Los Angeles Dodgers, Major League Baseball, Matt Cain, MLB Playoffs, pablo sandoval, San Francisco Giants, Tom Petty
by TortureCast | July 31, 2013 · 2:58 PM
Wilson Signs With LA: Giants Freak Out By Not Trading Anyone
AAAAHAHWISALKJASDOIJ LKJAWE@#$ASDJK~~!!!
We all knew it was a distinct possibility.
We just didn't want to believe it was possible.
Brian Wilson is a Dodger.
My initial gut reaction (Ben and Willie do not necessarily share my opinions on this) was angst, disgust, betrayal. As witnessed by the twitterxplosion, I'd say many Giants fans felt the same way.
At first, that is.
But, when you break it down and look at the known facts (and not everything is known), I do believe it was a business decision, and nothing more. Let's look at his motivating factors for the bearded one signing with our arch-rivals:
IT WAS THE ONLY CONTRACT OFFER HE RECEIVED: Initially it was reported that he received 3 offers, but as of what is known now, only the Dodgers gave him a formal offer on paper. The man is unemployed and looking for work. His loyalty to the Giants and their fans does not supersede a job offer. On the surface, I don't blame him. He was quoted by TMZ as saying, "I just play baseball. You know, if there are 30 teams out there and 29 teams don't want me, what am I going to do, say I'm not going to play baseball?" He's also probably concerned about what he can turn this into in 2014, and this is a springboard to it, maybe his only one.
The Dodgers are in first place: Ok, it took me 5 minutes to type that sentence out. But, even though the Pirates and Giants were in "talks" with him, he knows the Giants aren't going any where this year, and perhaps he felt it was more of a "sure thing" with the Dodgers anyway. Just look at how the Pirates fell flat on their face last year. Although, that beard would have been a true "pirate" look.
He lives in Hollywood: Ok, he lives in the neighborhood. He also loves California and apparently doesn't want to move anywhere else.
Ned Colletti: Ned Colletti, the GM for the Dodgers, drafted Wilson when he was assistant GM with SF in 2003, and this was after his first Tommy John surgery. Now he's signed him after his second TJ. There's some loyalty there, perhaps, or perhaps that's why he was the only GM to offer Wilson a contract.
It's not out of spite: Asked how Giants fans will feel about seeing him in Dodger blue, Wilson said, "I've got much love for San Francisco. We had a good time. But there's nothing I can do. They don't want me back so it's all good. I'll just play baseball hopefully here with the major league team and see if I can help them win." This doesn't sound like spite to me. And, after firing up his Twitter account among other news sources, he's fully aware of the backlash amongst Giants fans.
Brian Wilson speaks with TMZ today in LA
Standing Pat
Well, as of 1 pm Pacific today, it's official: The San Francisco Giants did not make any trades. Hunter Pence raised his hands and yelled, "YES!" right at the stroke of the deadline. Javier Lopez was reportedly nervous up until then, but was relieved when it passed. It says something about their teammates and the organization when they would rather be on a last place team with these teammates, rather than going to a contender. I believe this is a great sign that they will want to re-sign for next year and they can regroup and have a Mulligan on their season.
Also, don't forget to listen to our latest podcast here.
Now, let's see if the Giants can recuperate some respectability the rest of the way.
@chadk21
Tagged as Brian Wilson, dodgers, draft, hollywood, hunter pence, javier lopez, la, ned colletti, San Francisco Giants, Tim Lincecum, tmz, trade deadline
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Q: How to display the text in one line in wpf textblock I'm a newbie with wpf , what i want to display the text in one line in wpf textblock.
eg.:
<TextBlock
    Text ="asfasfasfa
    asdasdasd"
</TextBlock>

TextBlock display it in two lines default,
but i want it in only one line like this"asafsf asfafaf". I mean show all the text in one line even there are more than one lines in the text
   what should i do?

A: Instead of this:
            <TextBlock Text="Hello
                How Are
                You??"/>

Use this:
            <TextBlock>
                Hello
                How Are
                You??
            </TextBlock>

or this:
            <TextBlock>
                <Run>Hello</Run>
                <Run>How Are</Run>
                <Run>You??</Run>
            </TextBlock>

or set Text property in code behind like this :
(In XAML)
            <TextBlock x:Name="MyTextBlock"/>

(In code - c#)
            MyTextBlock.Text = "Hello How Are You??"

Code-behind approach has an advantage that you can format your text before setting it.
Example: If the text is retrieved from a file and you want to remove any carriage-return new-line characters you can do it this way:
 string textFromFile = System.IO.File.ReadAllText(@"Path\To\Text\File.txt");
 MyTextBlock.Text = textFromFile.Replace("\n","").Replace("\r","");

A: Use a Converter:
    <TextBlock Text={Binding Path=TextPropertyName,
Converter={StaticResource SingleLineTextConverter}}

SingleLineTextConverter.cs:
public class SingleLineTextConverter : IValueConverter
{
    public object Convert(object value, Type targetType, object parameter, System.Globalization.CultureInfo culture)
    {
        string s = (string)value;
        s = s.Replace(Environment.NewLine, " ");
        return s;
    }

    public object ConvertBack(object value, Type targetType, object parameter, System.Globalization.CultureInfo culture)
    {
        throw new NotImplementedException();
    }
}

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Home Worldwide North America DNA Kit Traces Man's Ancestry Back 55 Generations In What Could Be...
DNA Kit Traces Man's Ancestry Back 55 Generations In What Could Be The Oldest Known Lineage In America
When Darrell "Dusty" Crawford was in school in Montana, he was taught that his ancestors, the Blackfeet Nation, were some of the first people to step foot in America. Judging by his DNA analysis, his teachers might have been on to something.
On the wish of his dying brother, Crawford took a home DNA test to trace his family's ancestor. Remarkably, the DNA company claims to have traced his ancestry back at least 55 generations with 99 percent accuracy, as reported by Great Falls Tribune.
For a person of strong Native American ancestry, that's pretty impressive. The company, CRI Genetics, says this is the furthest back they've ever been able to trace anyone's ancestry in the Americas, as the tests typically only give information about a recent handful of generations before it gets too diluted to trace.
Crawford had an unusually high percent of Native American DNA in his results, around 83 percent, with small amounts of European, East Asian, and South Asian ancestry mixed in. Remarkably, up to 73 percent of his Native American heritage was from the same lineage. The more mixed a person's heritage, the harder it is to trace, so with high percentages like these, it narrows down the field considerably.
His mitochondrial DNA can be pinned down to the haplogroup B2, one of five haplogroups found among the indigenous peoples of the Americas. Mitochondrial DNA (mtDNA) is inherited through the maternal line, passed down from mother to child, and the woman who founded this B2 line lived sometime around 17,000 years ago. As far as we can tell, one of the earliest mtDNA haplogroups to reach the Americas was B2.
However, here's where it gets a bit strange. The B2 lineage appears to have sprung up in the middle of America, near modern-day Arizona, but nowhere near Alaska or Canada, the main entry route into North America by ancient people. The most widely-accepted theory of how humans populated the Americas is that they crossed the Bering Strait, a land bridge that once existed between russia and Alaska, towards the end of the Ice Age at least 16,500 years ago (although some researchers believe it was actually earlier than this). On the other hand, haplogroup B2 suggests a dIfferent story. While it originated in Arizona, it isn't found in native Alaskan populations and it's relatively uncommon in Canada. This hints that his ancestors may have actually migrated from the PacIfic to South America and traveled upwards to modern-day North America.
"Its path from the Americas is somewhat of a mystery as there are no frequencies of the haplogroup in either Alaska or Canada. Today this Native American line is found only in the Americas, with a strong frequency peak on the eastern coast of North America," CRI Genetics said, according to the Great Falls Tribune.
Of course, Crawford's DNA is just one person's, so it would be premature to reach any grand conclusions. Nevertheless, Crawford says he encourages anyone who has Native American ancestry to have a DNA test as it could provide some invaluable insights into our understanding of the history of people in America.
Related: North America
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Why I Wrote Little Leaders: Bold Women in Black History
Published on: 2 Mawrth 2018 Author: Vashti Harrison
Early in February 2017 I had an idea. I wanted to create a drawing challenge for myself, in the vein of "Mer-May" and "Ink-tober," but one that felt like it was all my own. Using Black History Month in the United States as a jumping off point I would draw a famous figure every day and post it along with a short bio to my social media.
I was thinking a lot about the sentiment behind Black History Month. When Carter G. Woodson started Negro History Week in 1926 he wanted to celebrate the stories that were often neglected throughout history. I felt inspired to use Black History Month as an opportunity to focus on black women in particular, whose stories have been doubly neglected.
I wanted to learn more about famous and lesser-known figures and share their stories with others. However, when I started the project, I didn't expect how deeply connected I would feel to their stories - stories of hard work, dedication, courage through adversity, love for craft and love for family. Many of these women didn't have a choice but to be bold, and I just felt overwhelmingly thankful for them paving the way for others and for me.
Suddenly these incredible figures in history were not just names on a page but humans with passions and dreams and fears and struggles. I wanted to convey that in this book and dug deep to find those human connections in each one of their stories. Once I knew that this project would become a book, I tried to be very thoughtful in selecting the list of women. It was important to me to showcase a diverse list of names and fields of study, to celebrate the many contributions black women have made to history and to share with young readers the many possibilities that are out there for them.
Topics: 5-8 years, 9-12 years, Non-fiction, Historical, Picture book, Around the world, Librarian, Parent/carer, Publisher/bookseller, Teacher (primary), Writer, Press, Diversity (BAME), Features
More inspiring women & girls
Here's a list of some of our favourite stories with inspiring heroines for older girls and young teens.
More amazing non-fiction
Perfect for enquiring young minds, these engaging non-fiction books approach subjects from biography to nature and science to history in a variety of interesting ways.
Follow Vashti on Twitter
Hey Philly! Join me (and a ton of amazing authors and illustrators) for the African American Childrens Book Fair this Saturday Feb 1! https://t.co/W0qzX3RDeE https://t.co/yKBZVth0ja 12 hours ago
I had a dream last night that was animated by Glen Keane 1 day ago
😭 https://t.co/oooTpFSVDI 2 days ago


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Why aren't mothers the main protagonists of books or movies?
March 23, 2016 trishmercer 1 Comment
Have you ever noticed mothers are not the main character . . . in anything?
Try this: name every Disney movie where the mother is absent and/or dead.
Are you done yet? There's a lot, isn't there?
And when a mother is present in a book/movie, what's her role?
Supporting.
Try this: name a book or movie where a mom is the principle protagonist.
Yeah, none sprang to my mind, either.
Recently I ran into a series of articles about "Strong Female Characters," and how they tend to collapse into a few categories:
they're heartless witches, manipulative and demeaning; or,
they're remarkably tough—like a female male—until the male shows up to aid/rescue/eclipse her; or,
the females vanish from the story because they really didn't play much of a role anyway, only to show up at the end as an afterthought.
When analyzing the stories I love, I realized to my dismay that most of the females fall into one of those three categories.
For example, "How to Train Your Dragon 2." I liked that movie, until my husband pointed out that not only is the mom rather useless after we meet her, but Hiccup disobeys his father, gets him killed, then still gets to become the chief. Dave thought all of that was pretty unfair. Then again, he's the father of five sons.
But dang it, he was right, especially about Valka. She's powerful, but useless, and never really mothered her son, aside from giving him some pat advice when he needed it.
At least she fares better than most female/mother figures in literature and movies. Consider "The Emperor's New Groove," "Cinderella," or any other Disney movie—
Except for "Mulan." Mulan is brilliant, in every way. She not only has BOTH parents (rare in Disney), but even a grandmother, and Mulan does the rescuing, several times. (And, thankfully, Disney allowed Merida to "Brave"ly keep both of her parents, although the mother needs saving later.)
Still, even in "Mulan" the mother is in the background, only as a supporting role.
Not that there's anything wrong with the supporting role; it's vital, in everything. Where would any of us be without those who support us?
By why not make mother The Hero?
That thought was part of my unconscious when I created Mahrree for my series "Forest at the Edge." When we first meet her in The Forest at the Edge of the World, she's a single woman, albeit motherly in that she teaches school and worries immensely for her students. Then we have the obligatory romance (because hey, I've got some Strong Male Characters, too) and then she becomes a mother.
Sometime around there, while drafting the series, I realized that I was writing a main character that I hadn't encountered before: a mother actively trying to improve the world for her children. (Aside from sappy made-for-cable movies or disease-of-the-week tales.)
Everything she says and does is in effort to reform her government, allow her children options in their future, and make sure school isn't ridiculously boring.
"Mom Wants to Make the World a Better Place for Her Children". It's a universal theme, so why isn't it more prevalent in what we read and watch?
Why have young mothers, middle-aged women, and even grandmothers not been the movers and drivers of a plot? They are when they're in a family, and everyone knows it. And I don't mean in a snarky, antagonistic way; enough of evil step-mothers.
I have friends who see themselves as primarily mothers, but are also involved in politics, providing aid for abused women and children, volunteering to preserve the environment, fighting pornography, promoting families, helping their schools, and improving the health and finances of those in need.
Maybe writers and Hollywood don't see that as exciting as clichéd "Save the World" themes, but these women really are saving the world, more than any fictional superhero will.
I'd love to see more thoughtful mothers, experienced grandmothers, and confident women playing bigger roles in the narratives of the world, because they do that everywhere else.
Terry Pratchett gets close at times, with his witches who aren't evil but keep the balance in the world, and in their villages. Nanny Ogg, who has more than a dozen children and oodles of grandchildren, is occasionally a major player in the witches books, but even then, Granny Weatherwax, who was never a mother, prevails as the main protagonist.
Mahrree Peto Shin isn't a character I created, as I erroneous stated earlier; she manifested herself to me through one shocking, powerful image that I won't reveal entirely here, because it happens in book 8 which will come out in probably another 3-4 years. (Book 5 will be out before this summer.)
But that image—of a woman facing an enormous threat, alone—riveted me. I wondered for months how she came to that position, and she slowly revealed her story to me, until I had enough to begin writing it.
(As an aside, I'm not psychotic; lots of writers have had this experience of characters demanding that they write their story. I can see how the ancient Greeks believed they had muses speaking to them, and my daughter gave me a mug which reads, "Writer's Block: when your imaginary friends refuse to talk to you." It's true, you know.)
The "Forest at the Edge" series isn't wholly about Mahrree, but about her family. She plays a major role in that family, and all the trouble she causes is primarily because she's a worried mother.
The mother-perspective is powerful, more so than the perspective of a president, or a general, or a king, or any other hero. Those characters may lead countries or armies, but I doubt they could ever genuinely love them as much as a mother loves her children. References comparing mothers to bears and their cubs aren't accidental.
Imagine Harry Potter being told from the perspective of Molly Weasley, who had seven children in peril, along with Harry who she tried to mother.
Or Hunger Games, from the perspective of President Coin, who lost her child.
Or even "Brave," from the perspective Queen Elinor, who tries to help her daughter find a husband, then has to be saved from her unintentionally turning her into a bear.
How engrossing, heart-breaking, and hopeful would these perspectives be?
Maybe . . . maybe they'd be too much.
Too close to reality. Too gut-wrenching.
Maybe that's why it's rare to find a mother-perspective in literature, because all of us have mothers, most of us appreciate them, and deep down we're worried about what they're really thinking.
Maybe many of us are still suffering from Bambi's Mother Syndrome, the idea of losing mother, or seeing things from her perspective, which may be too intense. (Although Disney loves to make children miserable with mother-loss. No adult movie franchise kills off mothers as much as Disney.)
As moms, I think we frequently have a hard time facing our greatest fears for our families and their future.
But I think it's time we tried. I submit there isn't a stronger force in nature or in humanity than that of a mother trying to protect and provide for her children.
What fantastic stories that premise could ignite! Let's see them!
Book 5 Teaser–Mahrree's big mouth
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← The Concludinator: can't really fight him–A Really Bad Book
Book 5 Teaser–I take comfort in my delusions! →
One thought on "Why aren't mothers the main protagonists of books or movies?"
Madison Pearce says:
Thought of 1 and kind of a 2nd:
Chocolat has a mom as a main character.
And The Sound of Music; Maria is kind of like a mom.


Raymond Perez works as a lease operator in Artesia, NM. Earlier this week he spotted a blue flame coming out of the enardo valve on a production tank. Enardo valves vent-off gas in the tanks and can ignite if struck by lightning. Being that these are steel tanks storing flammable liquids it was very important that Raymond spotted it. He prevented the tank battery from exploding or causing injury by calling the fire department to spray it with foam.
Do you have a co-worker that goes above and beyond? Submit their story on the High Performance Culture blog on the Danos intranet. Just log on and tell us how they go above and beyond!


text:
Meitav () is a military unit created in May 2006 by merging the Bakum (בסיס קליטה ומיון, Bsis Klita UMiyun lit. Reception and Sorting Base) and the Minhag (מנהל הגיוס, Minhal HaGiyus, lit. Recruitment Administration). In practice, it is the unit that oversees new Israel Defense Forces recruits and sorts them into their respective corps or basic training bases, similar to USMEPCOM in the U.S. military. Other responsibilities include (but are not limited to) initial allocation of basic gear and discharging soldiers at the end of their service.

The Reception and Sorting Base, located in Tel HaShomer is still commonly referred to as the Bakum and its new name, Meitav, as the unit's name, is virtually unknown outside the army. A possible reason for this is that meitav is the name of the military unit that resides on the base and is responsible for the different sorting tasks, but not the name of its location.

Meitav's headquarters are located at the Tel HaShomer military base, and its current commander as of March 2023 is Colonel Alon Matzliah.

In 2010, Meitav was the third-largest IDF unit, with 1,026 soldiers.

History
With the founding of the IDF, Meitav was called Kelet () and was under the jurisdiction of the manpower branch of the Ministry of Defense. It was located in Tzrifin. The Kelet unit, which inducted soldiers, was separate from the sorting unit/base, which was located in the Kiryat Meir base in Tel Aviv. In the 1950s, the two units were merged into the Bakum, and moved to Tel HaShomer.

In 1966, the jurisdiction of the Bakum passed from the Defense Ministry to the army.

Recruitment Offices
Under Meitav's responsibility is the management of 5 regional recruitment offices across the country (Tel Hashomer, Beer Sheva, Tiberias, Haifa and Jerusalem.
The recruitment offices are in charge of the sorting process for service, including Tzav Rishon, medical exams, interviews and tests.

References

External links
 Meitav website

Military installations of Israel
Military units and formations of Israel

redpajama_set_name: RedPajamaWikipedia


This was fun to watch. Fox vs squirrel.
There was a squirrel near the tree when the fox came around. The squirrel jumped onto the tree and was about seven feet from the ground. The fox rounded the tree a couple times eyeing up the odds while the squirrel flagged with its tail. Soon the fox jumped about three feet up into the crotch of the tree. At that instant the squirrel rushed straight up to where we see it (top box). A moment or so later the fox decided to leave (bottom box).
Back yard - looking over towards the neighbors yard (the dog might be out).
About twenty minutes after the fox came through another came up the driveway. Now I'm wondering if it could be the same fox. As best as I can tell this fox has less white on the tip of its tail.
Each stopped at this item in the yard. I should see what it could be (it was an old wash cloth).
Head'n down the road (their common path).


text:
The Marshall Building,
London School of Ecomomics
London School of Ecomnomics and Political Science (LSE)
Dry Lining / Ceilings
I&S Construction Ltd
18,000m²
BREEAM Excellent
© Peter Smith
Introducing The Marshall Building
A £145m investment to improve the student and staff experience at LSE.
The new building, which opened its doors in January 2022, demonstrates the School's ambition to create a world class estate and university quarter in the heart of London.
Designed by Dublin-based Grafton Architects, the large, multi-purpose building occupies a pivotal position at the southern corner of Lincoln's Inn Fields.
Award Winning Starting Point
LSE has been ranked as the top university in London in the Complete University Guide 2023 for the 11th consecutive year recognising the School's commitment to world-leading research and education. So, it was no surprise that it wanted to partner with the best architects to help mastermind the new Marshall Building.
In April 2016, LSE announced that award winning Dublin based practice Grafton Architects had been selected as the winners of the international architectural competition to design The Marshall Building at 44 Lincoln's Inn Fields.
Located at the south-west corner of Lincoln's Inn Fields, The Marshall Building is LSE's largest building with ten storeys plus two basement levels. It accommodates the Departments of Accounting, Finance and Management, as well as the Marshall Institute of Philanthropy and Social Enterprise, the Systemic Risk Centre and the Financial Markets Group.
At approximately 18,000m² it provides students with innovative lecture theatres, adaptable teaching rooms, flexible study spaces and a base hub where students can spend time on campus when not in lectures and classes.
Given the scale of the project and the size of the Sports Hall and Squash Courts it was vital to use an impact resistant ceiling which was also acoustically absorbent to meet reverberation requirements. Economy was also an important factor. In the highly serviced areas it was critical to have a ceiling that was readily demountable.
We had experience working with Knauf Ceiling Solutions on Institut Mines Telecom in Paris and the Kingston Town House. We really like HERADESIGN® for its aesthetics and performance and have used previously. Getting the perforation in the ceiling panels right was very important to the overall look of the ceiling design and we were really pleased with the flexibility of the service available to get the dots exactly as we wanted them.
David Healy / Grafton Architects
HERADESIGN® was chosen for its acoustic performance, aesthetics, economy and lowcarbon. The impact resistant type was also specified for the sports areas.
HERADESIGN® Creative is a brilliantly innovative and versatile product, valued by specifiers for its high performance and aesthetic wood wool properties. It can be matched to any interior colour scheme and is available in a wide selection of shapes, face patterns, edge details and sizes. It can be fixed to walls or suspended as ceilings, baffles and rafts. Manufactured from robust sustainable materials, HERADESIGN® offers high impact resistance and superior fire resistance.
Knauf Ceiling Solutions METAL R-H 200 panels provide a sophisticated open and contemporary look. They were chosen for areas where more service access was required but where a consistent aesthetic was also important. The metal panels were also used as 'floating canopies' so a high perimeter was specified to give a finished look at all sides.
Working during the pandemic was not without its challenges but the team at Knauf Ceiling Solutions were great to work with. Installation of HERADESIGN® and the metal ceilings was pretty simple and we had the help we needed from the Knauf Ceiling Solutions team as required. Over all we were really pleased with the finished result.
Glenn Giddens / I&S Construction Ltd
UK The Marshall Building, London School of Economics, London pdf | 1 MB


text:
\section{Introduction}\label{sec:intro}

Optimal investment is a tremendously rich source of mathematical
challenges in stochastic control theory. The key driver in this
problem is the tradeoff between risk and return. Thus, information on
the investment opportunities is playing a role which is as important
for the mathematical theory as it is in practice where investors go at
great lengths to secure even the slightest advantage in knowledge.  So
it is no wonder that insider information has been widely studied in
the literature; see, for instance,
\cite{PK:96,APS:98,I:03,ABS:03,ADI:06} where an investor obtains extra
information about the stock price evolution at some fixed point in
time. By contrast to these studies, the present paper takes a more
dynamic view on information gathering and affords the investor the
opportunity to \emph{continually} peek $\Delta$ units of time into the
future.  Closest to such an investor in reality may be high-frequency
traders (``frontrunners'') that get access to order flow information
earlier or are able process it faster than their competition. To the
best of our knowledge, this paper is the first continuous-time
stochastic control paper with such a feature, apart from the optimal
stopping problem of \cite{BZ:16}.

Of course, perfect knowledge about future stock prices easily lets
optimal investment problems degenerate and so it is of great interest
to understand how market mechanisms may curb an investor's ability to
take advantage of this extra information. A most satisfactory approach
from an economic point of view is the equilibrium approach due to Kyle
\cite{Kyle:85} where the insider knows the terminal stock price right
from the start and internalizes the impact of her orders on market
prices. Generalizations of this approach are challenging; see
\cite{BackBaruch:04}, \cite{Cetin:18}, \cite{BackEtal:21} and the
references therein.  For dynamic information advantages in this
context, we refer to \cite{CampiCetinDanilova:11,
  CampiCetinDanilova:13} who consider an insider receiving a dynamic
signal on, respectively, the terminal asset price or the traded firm's
default time. These models, however, do not get close to addressing
the intrinsically infinite-dimensional information structure of our
peek-ahead setting. Fortunately, also the much simpler market impact
model of \cite{AlmgrenChriss:01} that just imposes quadratic
transaction costs for the investor turns out to be sufficient friction
to make the optimal investment problem viable. In an insider model
where additional information is obtained just once,
\cite{AnkirchnerEtAl:2016} use such a friction for optimal portfolio
liquidation. A combination between Kyle's equilibrium setting and
quadratic price impact costs is solved in \cite{Donnelly:21}. With
peek-ahead information as in the present paper, \cite{DZ:21} study
super--replication, albeit in a discretized version of the Bachelier
model.

It is in the continuous-time Bachelier model that the present paper
provides its main result, namely the explicit optimal investment
strategy for an exponential utility maximizer who knows about future
prices $\Delta$ time units before they materialize in the market, but
cannot freely take advantage of her extra knowledge due to quadratic
transaction costs. The optimal policy turns out to be a combination of
two trading incentives. On the one hand, there is the urge to trade
towards the optimal frictionless position given by the well-known
Merton ratio. On the other hand, there is the desire to take advantage
of the next stock price moves and this contributes to the optimal
turnover rate through an explicitly given average of stock prices over
the window of length $\Delta$ on which our investor has extra
information.

Due to its peek-ahead feature, our optimal control problem can be
viewed as a contribution to pathwise stochastic control. A closely
related work is \cite{BM:07} where the authors studied a hidden
stochastic volatility model with a controller who has full information
on the extra noise.  The theory of delayed or partial information also
shares the infinite-dimensional pathwise control issues we need to
address here; see the recent papers \cite{SZ:19,S:19} and the
references therein. Finally, our control theoretic setting is also
related to models discussed in the monograph \cite{Gozzi:17}.

Instead of dynamic programming (which would be challenging in this
infinite memory setting; cf. \cite{Gozzi:17}), our methodology is
based on duality. For the case of exponential utility and quadratic
transaction costs, this theory is developed with flexible information
flow in great generality in an essentially self-contained appendix. It
shows that the primal optimal control is determined by the conditional
expectation of the terminal stock price under the dual optimal
probability measure. For the Brownian framework that we focus on in
the main body of the paper, we derive a particularly convenient
representation of the dual target functional which leads to
\emph{deterministic} variational problems. These problems can be
solved explicitly, and results from the theory of Gaussian Volterra
integral equations (\cite{H:68,HH:93}) allow us both to construct the
solution to the dual problem and to compute the primal optimal
strategy. These Gaussian Volterra integral equations also occur in
\cite{Cheridito:01} albeit in the rather different context of (no)
arbitrage criteria in fractionally perturbed financial models.

In Section~\ref{sec:mainresult} we specify our model and formulate and
interpret our main result.  Section~\ref{sec:proof} contains the proof
of the main result and the appendix~\ref{appendix} presents the
duality results necessary for these developments.

\section{Problem Formulation and Main Result}\label{sec:mainresult}
We consider an investor who knows about market movements some time
before they happen, but cannot arbitrarily exploit them due to market
frictions. Specifically, apart from a riskless savings account bearing
zero interest (for simplicity), the investor has the opportunity to
trade in a risky asset with Bachelier price dynamics
\begin{align}
S_t=s_0+\mu t+\sigma W_t, \quad t \geq 0,
\end{align}
where $s_0 \in \mathbb R$ is the initial asset price,
$\mu\in\mathbb{R}$ is the constant drift, $\sigma>0$ is the
constant volatility and $W$ is a one-dimensional Brownian motion
supported on a complete probability space
$(\Omega, \mathcal{F}, \mathbb P)$. Rather than having access to just
the natural augmented filtration $(\mathcal{F}^S_t)_{t \geq 0}$ for
making investment decisions, we assume that our investor can peek
$\Delta \in [0,\infty)$ time units into the future, and so her information
flow is given by the filtration
$$
\mathcal G^{\Delta}_t:=\mathcal F^S_{t+\Delta}, \quad t\geq 0.
$$
\begin{rem}
  As suggested by an anonymous referee, one could more generally
  consider a non-decreasing time shift $\tau:[0,\infty) \to [0,\infty)$ with
  $\tau(t) \geq t$ to model time-varying ability to peek ahead. To
  keep the exposition here as simple as possible, we leave this
  extension of our model as a topic for future research.
\end{rem}
Taking advantage of the inside information is impeded by the
investor's adverse market impact. Following
\cite{AlmgrenChriss:01}, we model this impact in a temporary linear
form and, thus, when at time $t$ the investor turns over her position $\Phi_t$ at
the rate $\phi_t=\dot{\Phi}_t$ the execution price is $S_t + \frac{\Lambda}{2}
\phi_t$ for some constant $\Lambda >0$. As a result, the profits and
losses from trading are given by
\begin{align}\label{eq:pnl}
V^{\Phi_0,\phi}_T:=\Phi_0(S_T-S_0)+\int_{0}^T \phi_t(S_T-S_t)dt-\frac{\Lambda}{2} \int_{0}^T \phi^2_t dt,
\end{align}
where, for convenience, we assume that the investor marks to market her position
$\Phi_T = \Phi_0+\int_0^T \phi_t dt$ in the risky asset that she has acquired by time
$T>0$.

Fixing a time horizon $T>0$, the natural class of admissible strategies
is then
\begin{align}
\mathcal A^\Delta:=\left\{\phi=(\phi_t)_{t\in [0,T]}: \ \phi \text{ is } \
  \mathcal G^{\Delta}\text{-optional with } \int_{0}^T \phi^2_t dt<\infty
  \ \text{ a.s.}\right \}.
\end{align}
The investor's preferences are described by an exponential utility function
$$u(x):=-\exp(-\alpha x), \quad x\in\mathbb R,$$
with constant absolute risk aversion parameter $\alpha>0$, and her goal
is thus to
\begin{align}\label{problem}
\text{Maximize } \mathbb E\left[u(V^{\Phi_0,\phi}_T)\right]=\mathbb
  E\left[-\exp\left(-\alpha V^{\Phi_0,\phi}_T\right)\right]\text{ over
  } {\phi\in\mathcal A^\Delta}.
\end{align}

The paper's main result is the following solution to this optimization
problem:
\begin{thm}\label{thm2.1}
  In the utility maximization problem~\eqref{problem}, the investor's
  optimal turnover rate $\hat{\phi}_t$ at time $t \in [0,T]$ depends
  on the risk-liquidity ratio
\begin{align}
  \label{eq:22}
  \rho:=\frac{\alpha \sigma^2}{\Lambda},
\end{align}
  on the position $\hat{\Phi}_t=\Phi_0+\int_0^t \hat{\phi}_sds$
  acquired so far and the privileged information on the next stock prices
  $(S_{t+s})_{s \in [0,\Delta]}$ in the feedback
  form
  \begin{align}\label{eq:feedback}
\hat{\phi}_t=&
               \frac{1}{\Lambda}\left(\bar{S}^\Delta_t-S_t\right
                          +\frac{\Upsilon^\Delta(T-t)}{\Delta}
                            \left(\frac{\mu}{\alpha\sigma^2} -\hat{\Phi}_t\right),
  \end{align}
  where $\bar{S}^\Delta$ is the stock price average given by
  \begin{align}
    \label{eq:24}
    \bar{S}^\Delta_t := \left(1-\Upsilon^\Delta(T-t)\right)S_{(t+\Delta) \wedge T}
+\Upsilon^\Delta(T-t)\frac{1}{\Delta}
\int_0^{\Delta} S_{t+s}ds
  \end{align}
 with $\Upsilon^\Delta(\tau)=\Delta\sqrt{\rho}\tanh(\sqrt{\rho}(\tau-\Delta)^{+})/(1+\Delta\sqrt{\rho}\tanh(\sqrt{\rho}(\tau-\Delta)^+))$.
The maximal utility this policy generates is
  \begin{align}\label{eq:primalvalue}
  \max_{\phi \in \mathcal A^\Delta} \mathbb  E&\left[-\exp\left(-\alpha\sigma
                                                V^{\Phi_0,\phi}_T\right)\right] =\\\nonumber
    &-\exp\left(
      \frac{\alpha\Lambda\sqrt{\rho}}{2\coth\left(\sqrt{\rho}T\right)}{\left(\Phi_0-\frac{\mu}{\alpha
      \sigma^2}\right)^2}-\frac{1}{2}\frac{\mu^2}{\sigma^2}T\right)\\\nonumber&\qquad\cdot \exp\left(-\frac{1}{2}\int_{0}^T \frac{(s\wedge\Delta)\rho}{1+(s\wedge\Delta) \sqrt{\rho}\tanh\left(\sqrt{\rho}(T-s)\right)}ds\right).
  \end{align}
\end{thm}

Our feedback description~\eqref{eq:feedback} can be interpreted as
follows: First, without privileged information, i.e. for $\Delta=0$,
we have $\bar{S}^\Delta_t=S_t$ and, therefore, the first term
in~\eqref{eq:feedback} vanishes leaving us with the optimal policy
\begin{align}
         \hat{\phi}_t=\sqrt{\rho}\tanh(\sqrt{\rho}(T-t))
                            \left(\frac{\mu}{\alpha\sigma^2}
  -\hat\Phi_t\right), \quad t \in [0,T].
\end{align}
So the uninformed agent will trade towards the optimal position
$\mu/(\alpha\sigma^2)$ well known from the frictionless Merton
problem. Due to the impact costs, she does so with finite urgency
$\sqrt{\rho}\tanh(\sqrt{\rho}(T-t))$. With a long time to go, this
urgency is essentially $\sqrt{\rho}$ and thus dictated by the
risk/liquidity ration $\rho=\alpha \sigma^2/\Lambda$;
as $t$ approaches the time horizon $T$, the urgency vanishes because,
towards the end, position improvements have an ever shorter time to
yield risk premia but the investor still has to pay the same impact
costs that obtain at the start of trading.  \footnote{We will prove
  this result along the way to our main result with future knowledge
  $\Delta>0$. Let us note though that, for $\Delta=0$, a closely
  related result is obtained by dynamic programming techniques in
  \cite{Schied_2007} who, in contrast to our setting, impose a
  liquidation constraint $\hat{\Phi}_T=0$ and assume $\mu=0$.}

 For the informed agent, i.e. for $\Delta>0$, the desire to be close to
the Merton ratio persists, but the urgency reduces to
\begin{align}
\frac{\Upsilon^\Delta(T-t)}{\Delta}=\frac{\sqrt{\rho}\tanh(\sqrt{\rho}(T-t-\Delta)^{+})}{1+\Delta\sqrt{\rho}\tanh(\sqrt{\rho}(T-t-\Delta)^+)},
\end{align}
 leaving ``some air'' to take advantage of the knowledge on future
price movements. This is done by averaging out in~\eqref{eq:24} the
latest relevant stock price available to the investor,
$S_{(t+\Delta)\wedge T},$ with the mean stock price
$\frac{1}{\Delta} \int_0^{\Delta} S_{t+s}ds$ to be realized over the
next $\Delta$ time units in an effort to assess the earnings potential
over today's stock price~$S_t$. Put into relation with the impact costs $\Lambda$, this yields
the second contribution $(\bar{S}^\Delta_t-S_t)/\Lambda$ to the optimal turnover rate. The weight that this
assessment of earnings assigns to the average stock prices is given by
$\Upsilon^\Delta(T-t) \in [0,1]$; it is about
$\Delta \sqrt{\rho}/(1+\Delta\sqrt{\rho})$ when there is still a lot
of time to go, but vanishes completely as soon as $T-t\leq \Delta$,
i.e. as soon as full knowledge of stock price movements over the
relevant time span $[0,T]$ is attained. In this terminal regime also
the ambition to be close to the Merton ratio is wiped out and the
investor just chases the earning potential $S_T-S_t$ from the stock,
of course still in a tradeoff against the liquidity costs $\Lambda$;
this latter effect is also immediate from separate, pointwise
optimization over $\phi_t$ in the representation~\eqref{eq:pnl} of
profits and losses (which leads to $\phi_t^*=(S_T-S_t)/\Lambda$, $t
\in [0,T]$, an admissible
strategy as soon as $S_T$ becomes known).
\begin{figure}
\includegraphics[width=0.3\textwidth]{stockaverageupsilon}
\includegraphics[width=0.3\textwidth]{earningstracking}
\includegraphics[width=0.3\textwidth]{positions}
\caption{The first of these illustrations shows an evolution of the stock price
  $S$ (blue), the corresponding average $S^\Delta$
  (orange) along with the underlying weight $\Upsilon^\Delta$; the second shows
  the resulting trading rates due to ``frontrunning'' (grey) and
  due to tracking the Merton portfolio (black); the third display
  shows the ensuing stock position $\Phi$ (red) together with the
  Merton ration $\mu/(\alpha \sigma^2$ (light red). Parameters where
  chosen as $s_0=0$, $\mu=.1$, $\sigma=.3$, $T=10$, $\Delta=1$,
  $\alpha=.03$, $\Phi_0=0$, $\Lambda = .01$.}
\end{figure}

The monetary value of being able to peek ahead by $\Delta$ is best described by
the certainty equivalent
  \begin{align}\label{eq:7}
c(\Delta) &= -\frac{1}{\alpha} \log\frac{ \max_{\phi \in \mathcal A^{\Delta}} \mathbb  E\left[-\exp\left(-\alpha
                                                V^{\Phi_0,\phi}_T\right)\right]}{ \max_{\phi \in \mathcal A^{0}} \mathbb  E\left[-\exp\left(-\alpha
                                                V^{\Phi_0,\phi}_T\right)\right]}\\
    &=\frac{1}{2\alpha}  \int_{0}^T
      \frac{(s\wedge\Delta)\rho}{1+(s\wedge\Delta)
      \sqrt{\rho}\tanh\left(\sqrt{\rho}(T-s)\right)}ds
  \end{align}
determined by comparing the utility attainable for an informed investor
(with admissible strategy set $\mathcal A^\Delta$) and an uninformed one (who
is confined to strategies from the smaller class $\mathcal A^0$)\footnote{The
      integral in~\eqref{eq:7} can be computed explicitly, but the resulting
  formulae turn out to be not more informative than the above integral
  and are therefore omitted.}.
Interestingly, the certainty equivalent does not depend on
the stock's risk premium $\mu$, but is determined by the
risk/liquidity ratio $\rho=\alpha \sigma^2/\Lambda$, the investor's
time horizon $T$ and the time units $\Delta$ she can look
ahead. Except for a period of length $\Delta$ and with a lot of time
to go, it accrues at about the rate
$\Delta \rho/(2(1+\Delta \sqrt{\rho}))$ which increases with $\Delta$
to the upper bound $\sqrt{\rho}/2$, revealing again the curb frictions
put on the earning potential of even extreme information advantages.

The proof of Theorem~\ref{thm2.1} is carried out in the next
section. It is obtained by solving the dual problem to
\begin{align}
  \intertext{$\qquad$Minimze}\label{eq:dualproblem}
  \mathbb E_{\mathbb Q}\left[\Phi_0(S_T-S_0) +\frac{1}{2\Lambda}\int_{0}^T\left|
\mathbb E_{\mathbb Q}\left[S_T\middle|\mathcal
  G^{\Delta}_t\right]-S_t\right|^2 dt\right]+\frac{1}{\alpha}
  \mathbb E_{\mathbb Q}\left[\log \frac{d\mathbb Q}{d\mathbb P}\right]
  \intertext{$\qquad$over $\mathbb Q \sim \mathbb P$ with finite relative entropy
$\mathbb E_{\mathbb Q}\left[\log \frac{d\mathbb Q}{d\mathbb
    P}\right]<\infty$.}
\end{align}
The corresponding duality theory holds true beyond the Brownian
framework specified here and is developed in a self-contained manner
in the Appendix~\ref{appendix} as a second key contribution of our
paper

\section{Proof of Theorem~\ref{thm2.1}}\label{sec:proof}
Let us first note that it suffices to treat the case
\begin{align}\label{eq:wlog}
  S=W\text{, i.e., without loss of generality $s_0=0$, $\sigma=1$, $\mu=0$.}
\end{align}
Indeed, by passing from $\alpha$, $\Lambda$, $\mu$ to, respectively,
$\alpha'=\alpha\sigma$, $\Lambda'=\Lambda/\sigma$, $\mu'=\mu/\sigma$,
the utility with $\sigma'=1$ obtained from a given strategy will
coincide with the one obtained from this strategy under the original
parameters. Moreover, rewriting the expected utility under
$\mathbb P' \sim \mathbb P$ with density
$$\frac{d{\hat {\mathbb P'}}}{d\mathbb P}{|\mathcal F^W_T}:=\exp\left(-\mu'
  W_T-\frac{1}{2}\mu'^2 T\right),$$ under which
$W'_t=W_t+\mu' t$, $t \geq 0$, is a driftless Brownian motion, the expected
utilities under $\mathbb P$ coincide, up to the factor $\exp\left(\frac{1}{2}\mu'^2 T\right)$, with those under $\mathbb P'$ if we start with
$\Phi_0'=\Phi_0-{\mu}/({\alpha \sigma^2})$ rather than $\Phi_0$
risky assets.

The proof of Theorem \ref{thm2.1} will be accomplished via the dual
problem whose properties are summarized in the following proposition
which is an immediate consequence of the general duality results
presented in Appendix~\ref{appendix}.
\begin{prop}\label{prop2.1}
Denoting by $\mathcal Q$ the set of all probability measures $\mathbb Q\sim\mathbb P$ with finite
entropy
$$\mathbb E_{\mathbb Q}\left[\log\left(\frac{d\mathbb Q}{d\mathbb P}\right) \right]<\infty$$
relative to $\mathbb P$, we have
\begin{align}
\max_{\phi\in\mathcal A}&\left\{-\frac{1}{\alpha}\log\mathbb E\left[\exp\left(-\alpha V^{\Phi_0,\phi}_T\right)\right]\right\}\\
&=\min_{\mathbb Q\in\mathcal Q}\mathbb E_{\mathbb Q}\left[\Phi_0(S_T-S_0)+\frac{1}{\alpha}\log\left(\frac{d\mathbb Q}{d\mathbb P}\right) +\frac{1}{2\Lambda}\int_{0}^T\left|
\mathbb E_{\mathbb Q}(S_T|\mathcal G^{\Delta}_t)-S_t\right|^2 dt\right].\label{eq:1}
\end{align}
Furthermore, the minimizer $\hat{\mathbb Q}$ for the dual
problem is unique  and yields via
\begin{equation}\label{portfolio}
\hat\phi_t:=\frac{\mathbb E_{\hat{\mathbb Q}}\left[S_T|\mathcal G^{\Delta}_t\right]-S_t}{\Lambda}, \quad t\in [0,T],
\end{equation}
the unique optimal portfolio for the primal problem.
\end{prop}
\begin{proof}Follows from Proposition~\ref{propapp} below with the
  choice $\mathcal{G}_{t}:=\mathcal{G}_{t}^{\Delta}$ after
noting that $S_{t}/\sqrt{t}$ is standard Gaussian and so
$$
\sup_{t\in [0,T]}\mathbb E[\exp(a S_{t}^{2})]\leq \mathbb E[\exp(a S_{T}^{2})]<\infty,
$$
clearly holds for some small enough $a>0$.
\end{proof}

In order to solve the utility maximization problem it therefore
suffices to find the minimizer $\hat{\mathbb Q}$ of the dual problem
and work out the conditional expectation in~\eqref{portfolio}. This is
the path we will follow for the rest of this section. In a first step
we derive a particularly convenient representation for the target
functional of our dual problem:

\begin{lem}\label{lem:1}
  The dual infimum in~\eqref{eq:1} coincides with the one taken over
  all $\mathbb Q \in \mathcal Q$ whose densities take the form
  \begin{align}
\frac{d\mathbb Q}{d\mathbb P} = \exp\left(-\int_0^T\theta_t
    dW_t-\frac{1}{2}\int_0^T\theta_t^2dt\right) \label{eq:standarddensity}
  \end{align}
  for some bounded and adapted $\theta$ changing values only at finitely
  many deterministic times. For such $\mathbb Q$ the induced value~\eqref{eq:dualproblem} for
  the dual problem can be written as
  \begin{align}
    &\mathbb E_{\mathbb Q}\left[\Phi_0(S_T-S_0)+\frac{1}{\alpha}\log\left(\frac{d\mathbb Q}{d\mathbb P}\right) +\frac{1}{2\Lambda}\int_{0}^T\left|
\mathbb E_{\mathbb Q}\left[S_T\middle|\mathcal G^{\Delta}_t\right]-S_t\right|^2
      dt\right]\\
    =&-\Phi_0\int_{0}^T a_t dt+\frac{1}{2\alpha}\int_{0}^T a^2_t
      dt+\frac{1}{2\Lambda}\int_{0}^T \left(\int_t^T a_udu\right)^2 dt\\
&+\int_{0}^T  \mathbb E_{\mathbb Q}\left[\frac{1}{2\alpha}\int_{s}^{T}
                                                                             l^2_{t,s} dt+
\frac{1}{2\Lambda}
\int_{s}^{T}\left(\int_t^T l_{u,s}du\right)^2dt
+\frac{s\wedge\Delta}{2\Lambda}\left(1-\int_s^Tl_{u,s}du\right)^2\right]ds
  \end{align}
  where, for $t\in[0,T]$, $a_t$ and $l_{t,.}$ are
  determined by the It\^o-representations
  \begin{align}\label{eq:3}
     \theta_t = a_t + \int_0^t l_{t,s} dW^{\mathbb Q}_s
  \end{align}
  with respect to the $\mathbb Q$-Brownian motion $W^{\mathbb
    Q}_s=W_s+\int_0^s \theta_r dr$, $s \geq 0$.
\end{lem}
\begin{proof}
  For any $\mathbb Q \in \mathcal Q$ the martingale representation
  property of Brownian motion gives us a predictable $\theta$ with
  $\mathbb E_{\mathbb Q}[\log(d\mathbb Q/d\mathbb P)]=\mathbb
  E_{\mathbb Q}[\int_0^T\theta^2_sds]/2<\infty$ such that the density
  $d\mathbb Q/d\mathbb P$ takes the
  form~\eqref{eq:standarddensity}. Using this density to rewrite the
  dual target functional as an expectation under $\mathbb P$, we can
  follow standard density arguments to see that the infimum over
  $\mathbb Q \in \mathcal Q$ can be realized by considering the
  $\mathbb Q$ induced via~\eqref{eq:standarddensity} by simple
  $\theta$ as described in the lemma's formulation. As a consequence,
  the It\^o representations of $\theta_t$ in~\eqref{eq:3} can be
  chosen in such a way that the resulting $(a_t,l_{t,.})$ are also
  measurable in $t$: in fact they only change when $\theta$ changes
  its value, i.e., at finitely many deterministic times.  This joint
  measurability will allow us below to freely apply Fubini's theorem.

  Let us rewrite the dual target functional in terms of $a$ and $l$. In
  terms of $\theta$ and the $\mathbb Q$-Brownian motion
  $W^{\mathbb Q}$, it reads
  \begin{align}\label{eq:firstrepdual}
\mathbb E_{\mathbb Q}&\left[\Phi_0(S_T-S_0)+\frac{1}{\alpha}\log\left(\frac{d\mathbb Q}{d\mathbb P}\right) +\frac{1}{2\Lambda}\int_{0}^T\left(
\mathbb E_{\mathbb Q}\left[S_T\middle|\mathcal G^{\Delta}_t\right]-S_t\right)^2 dt\right]\\
&=\mathbb E_{\mathbb Q}\Bigg[-\Phi_0\int_{0}^T \theta_t dt+\frac{1}{2\alpha}\int_{0}^T \theta^2_u du\\
&\qquad\qquad+\frac{1}{2\Lambda}\int_{0}^T
\left(W^{\mathbb Q}_{(t+\Delta)\wedge T}-W^{\mathbb Q}_t-\mathbb E_{\mathbb Q}\left[\int_{t}^T\theta_u du\middle|\mathcal G^{\Delta}_t\right]\right)^2 dt\Bigg].
 \end{align}
From It\^o's isometry and Fubini's theorem we obtain
\begin{align}\label{3.3}
\mathbb E_{\mathbb Q}\left[\int_{0}^T \theta^2_u du\right]=\int_{0}^T a^2_t dt+\int_{0}^T\int_{s}^{T}\mathbb E_{\mathbb Q}\left[l^2_{t,s} \right]dt\, ds.
\end{align}
Again by Fubini's theorem it follows that
\begin{align}
\mathbb E_{\mathbb Q}\left[\int_{t}^T \theta_udu\middle|\mathcal G^{\Delta}_t\right]&=
\int_{t}^T a_u du+\mathbb E_{\mathbb Q}\left[\int_{0}^T\int_{t\vee s}^T l_{u,s}du\, dW^{\mathbb Q}_s\middle|\mathcal G^{\Delta}_t\right]\\
&=\int_{t}^T a_u du+\int_{0}^{(t+\Delta)\wedge T} \int_{t\vee s}^T l_{u,s} du\, dW^{\mathbb Q}_s
\end{align}
for any $t\in [0,T]$, where the last equality follows from the
martingale property of stochastic integrals. Thus, another application of It\^o's isometry yields
\begin{align}
\mathbb E_{\mathbb Q}&\left[\left(W^{\mathbb Q}_{(t+\Delta)\wedge T}-W^{\mathbb Q}_t-\mathbb E_{\mathbb Q}\left[\int_{t}^T \theta_u du\middle|\mathcal G_t^\Delta\right]\right)^2 \right]\\
&=\left(\int_{t}^T a_u du\right)^2+ \mathbb E_{\mathbb Q}\left[\int_{0}^t \left(\int_{t}^T l_{u,s} du\right)^2ds \right]\\
&\qquad
+\mathbb E_{\mathbb Q}\left[\int_{t}^{(t+\Delta)\wedge T} \left(1-\int_{s}^T l_{u,s} du\right)^2ds \right].
\end{align}
Plugging this together with~\eqref{3.3} into~\eqref{eq:firstrepdual}
and using Fubini's theorem then provides us with the claimed formula
for our dual target value:
\begin{align}
\mathbb E_{\mathbb Q}&\left[-\Phi_0(S_T-S_0)+\frac{1}{\alpha}\log\left(\frac{d\mathbb Q}{d\mathbb P}\right) +\frac{1}{2\Lambda}\int_{0}^T\left(
\mathbb E_{\mathbb Q}\left[S_T\middle|\mathcal G^{\Delta}_t\right]-S_t\right)^2 dt\right]\\
=&-\Phi_0\int_{0}^T a_t dt+\frac{1}{2\alpha}\int_{0}^T a^2_t
      dt+\frac{1}{2\Lambda}\int_{0}^T \left(\int_t^T a_udu\right)^2 dt\\
&+\int_{0}^T \mathbb E_{\mathbb Q}\left[\frac{1}{2\alpha}\int_{s}^{T}
                                                                             l^2_{t,s} dt+
\frac{1}{2\Lambda}
\int_{s}^{T}\left(\int_t^T l_{u,s}du\right)^2dt\right.\\&\qquad\qquad\qquad\qquad\qquad\qquad\qquad\left.
+\frac{s\wedge\Delta}{2\Lambda}\left(1-\int_s^Tl_{u,s}du\right)^2\right]ds.
\end{align}
\end{proof}
The crucial point of the above representation is that by taking the
minimum separately over $a$ and over $l_{.,s}$ for each $s \in [0,T]$
we obtain \emph{deterministic} variational problems that can be solved
explicitly (see the next Lemma~\ref{lem:2}) \emph{and} this
deterministic minimum yields a lower-bound for the dual target value
that, using some Gaussian process theory, will ultimately be shown to
actually coincide with it (see Lemma~\ref{lem:3} below).

\begin{lem}\label{lem:2}
   Recall our notation $\rho=\alpha\sigma^2/\Lambda=\alpha/\Lambda$
   (because $\sigma=1$; cf.\ \eqref{eq:wlog}).
  \begin{enumerate}
  \item[(i)] The minimum of the functional
    \begin{align}
-\Phi_0\int_{0}^T a_t dt+\frac{1}{2\alpha}\int_{0}^T a^2_t
      dt+\frac{1}{2\Lambda}\int_{0}^T \left(\int_t^T a_udu\right)^2 dt
    \end{align}
    over $a \in L^2([0,T],dt)$ is attained for $\hat{a}\Phi_0$ where
    \begin{align}\label{eq:4}
      \hat{a}_t = \frac{\alpha
      \cosh(\sqrt{\rho}(T-t))}{\cosh(\sqrt{\rho} T)}, \quad t \in [0,T].
    \end{align}
    The resulting minimum value is $-\hat{A}_T\Phi_0^2$ where
    \begin{align}\label{eq:hatA}
      \hat{A}_T=\Lambda\sqrt{\rho}\tanh(\sqrt{\rho} T)/2.
    \end{align}
 \item[(ii)] For any $s \in [0,T]$, the minimum of the functional
    \begin{align}
\frac{1}{2\alpha}\int_{s}^{T}
                                                                             l^2_{t} dt+
\frac{1}{2\Lambda}
\int_{s}^{T}\left(\int_t^T l_{u}du\right)^2dt
+\frac{s\wedge\Delta}{2\Lambda}\left(1-\int_s^Tl_{u}du\right)^2
    \end{align}
   over $l \in L^2([s,T],dt)$ is attained at
    \begin{align}\label{eq:5}
       \hat{l}_{t,s} =  \frac{ \rho (s \wedge \Delta)
      \cosh(\sqrt{\rho}(T-t))}{\cosh(\sqrt{\rho}(T-s))+ \sqrt{\rho}
      (s \wedge \Delta) \sinh(\sqrt{\rho}(T-s))}, \quad t\in[s,T].
    \end{align}
    The corresponding minimum value is
    \begin{align}\label{eq:hatL}
      \hat{L}_s=\frac{1}{2\Lambda}\frac{s \wedge  \Delta}{1+(s\wedge \Delta)\sqrt{\rho} \tanh(\sqrt{\rho}(T-s))}.
     \end{align}
    \end{enumerate}
  \end{lem}
\begin{proof}
  We start with (ii).  The uniqueness follows from strict convexity of
  the functional to be minimized over $l \in L^2([s,T],dt)$. To write
  this as a standard variational problem, put
  $H(u,v):=\frac{1}{2\Lambda}u^2+ \frac{1}{2\alpha} v^2$ for
  $ u,v\in\mathbb R$, reparametrize $l$ via $g(t) = \int_t^T l_u du$,
  $t \in [s,T]$, and
  consider, for any $s \in [0,T]$ and any $\Theta\in\mathbb R$, the
  problem to minimize $\int_{s}^T H(g_t,\dot{g}_t) dt$ over
  $g \in C^1[s,T]$ subject to the constraints $g(s)=\Theta$, $g(T)=0$.

The optimization problem is convex and so it has a unique solution which has to satisfy the
Euler–Lagrange equation (for details see Section~1 in \cite{GF:63})
$$\frac{d}{dt}\frac{\partial H}{\partial \dot{g}}=\frac{\partial H}{\partial{g}}.$$
Thus, the optimizer is the unique solution of the linear ODE
$$\ddot{g}=\rho g, \ \ g(s)=\Theta, \ \ g(T)=0,$$
namely
$$g^{\Theta,s}(t):=\frac{\Theta\sinh\left(\sqrt \rho(T-t)\right)}
{\sinh\left(\sqrt{\rho}(T-s)\right)}, \ \ t\in [s,T].$$

Next, observe that for the function $g(t):=\int_{t}^T i_u du$, $t\in [s,T]$ we have
$\dot{g}=-l$ where $\dot{g}$ is the weak derivative of $g$, and so,
$$\frac{1}{2\alpha}\int_{s}^T \psi^2_t dt+\frac{1}{2\Lambda}\int_{s}^T\left(\int_{t}^{T}\psi_u du\right)^2dt=\int_{s}^T H(g_t,\dot{g}_t) dt.$$
Thus, from simple density arguments (needed since $g$ is not necessarily smooth) we obtain that
\begin{align}
\inf_{l\in L^2 ([s,T],dt)}&\left\{\frac{1}{2\alpha}\int_{s}^{T}l^2_{t} dt+
\frac{1}{2\Lambda}
\int_{s}^{T}\left(\int_t^T l_{u}du\right)^2dt
+\frac{s\wedge\Delta}{2\Lambda}\left(1-\int_s^T l_{u}du\right)^2\right\}
\\
&=\inf_{\Theta\in\mathbb R}\left\{\int_{s}^T H(g^{\Theta,s}_t,\dot{g}^{\Theta,s}_t) dt+\frac{s\wedge\Delta}{2\Lambda}(1-\Theta)^2\right\}\\
&=\frac{1}{2\Lambda}\inf_{\Theta\in\mathbb R}
\left\{\frac {\coth\left(\sqrt{\rho}(T-s)\right)}{\sqrt{\rho}}\Theta^2+(s\wedge\Delta) (1-\Theta)^2\right\}
\end{align}
where the last equality follows from simple computations.

Finally, the minimum of the above quadratic pattern (in $\Theta$) is atained at
$$\Theta^{*}=-\frac{ (s\wedge\Delta)\sqrt{\rho}}{\coth\left(\sqrt{\rho}(T-s)\right)+(s\wedge\Delta)\sqrt{\rho}}.$$
This gives \eqref{eq:5}--\eqref{eq:hatL}.

The proof of~(i) is almost the same as the of~(ii), but slightly simpler.
Observe that
\begin{align}
\inf_{a\in L^2([0,T],dt)}&\left\{-\Phi_0\int_{0}^T a_t dt+\frac{1}{2\alpha}\int_{0}^T a^2_t
      dt+\frac{1}{2\Lambda}\int_{0}^T \left(\int_t^T a_udu\right)^2 dt\right\}\\
&=\inf_{\Theta\in\mathbb R}\left\{-\Phi_0 \Theta+\int_{0}^T H(g^{\Theta,0}_t,\dot{g}^{\Theta,0}_t) dt\right\}\\
&=\inf_{\Theta\in\mathbb R}\left\{-\Phi_0 \Theta+\frac {\coth\left(\sqrt{\rho}T\right)}{2\sqrt{\rho}\Lambda}\Theta^2\right\}.
\end{align}
The minimum of the above quadratic pattern (in $\Theta$) is attained in
$$\tilde{\Theta}^{*}=\Phi_0\sqrt{\rho}\Lambda \tanh\left(\sqrt{\rho}T\right).$$
This gives \eqref{eq:4}--\eqref{eq:hatA}.
\end{proof}

The previous two lemmas suggest a way to construct a candidate for the
solution to the dual problem: Find $\hat{\mathbb Q} \sim \mathbb P$
whose density is given by
 \begin{align} \label{eq:2}
\frac{d\hat{\mathbb Q}}{d\mathbb P} = \exp\left(-\int_0^T\hat\theta_t
    dW_t-\frac{1}{2}\int_0^T\hat\theta_t^2dt\right)
  \end{align}
with
\begin{align}
  \label{eq:hattheta}
  \hat{\theta}_s =\hat{a}_s\Phi_0+ \int_0^s\hat{l}_{s,r}
  d\hat{W}^{\hat{\mathbb Q}}_r, \quad s \in [0,T].
\end{align}
For the associated Brownian motion
$\hat{W}=W^{\hat{\mathbb Q}}=W+\int_0^. \hat{\theta}_r dr$ this implies the
Volterra-type integral equation
\begin{align}\label{eq:Volterra}
  W_t+\int_0^t \hat{a}_s  \Phi_0 ds  = \hat{W}_t-\int_0^t \int_0^s\hat{l}_{s,r}
  d\hat{W}_rds, \quad t \in [0,T].
\end{align}
Integral equations of this type occur in~\cite{H:68,HH:93}; see also \cite{Cheridito:01}. By
considering $W+\int_0^. \hat{a}_r \Phi_0 dr$ as a Brownian motion with
respect to some probability measure which is equivalent to
$\mathbb P$, we can apply the results from Section~6.4 in \cite{HH:93}
(in particular see Theorem 6.3 and its proof there). We obtain that
\eqref{eq:Volterra} has a unique solution given by
\begin{align}
  \label{eq:hatW}
\hat{W}_t& = W_t+\Phi_0\int_0^t \hat{a}_s ds
-\int_{0}^t\int_{0}^s \hatk_{s,r} \left(dW_r+\Phi_0\hat{a}_r dr\right) ds \nonumber\\
&=W_t-\int_{0}^t\int_{0}^s \hatk_{s,r} dW_r ds+\Phi_0\left(\int_{0}^t\hat{a}_s ds-\int_{0}^t\int_{0}^s \hatk_{s,r}\hat a_r dr ds\right)
\end{align}
where $\hat{k}$
is the associated resolvent kernel characterized by the equation
\begin{align}
   \label{eq:6}
\hat{k}_{t,s}+\hat{l}_{t,s}=\int_{s}^t
   \hat{l}_{t,u}\hat{k}_{u,s}du, \quad 0 \leq s \leq t \leq T.
\end{align}
Moreover, $\hat{W}$ is a Brownian motion with respect to $\hat{\mathbb Q}$ which is well defined by~\eqref{eq:2}.

As our $\hat{l}$ is separable multiplicatively, \eqref{eq:6}
can be reduced to a linear ODE from which we compute the explicit
solution
\begin{align}\label{eq:hatkappa}
 \hat{k}_{t,s}=-\exp\left(\int_s^t
  \hat{l}_{u,u}du\right)\hat{l}_{t,s}, \quad 0 \leq s \leq t \leq T .
\end{align}

We are now in a position to solve the dual problem:
\begin{lem}\label{lem:3}
The dual infimum~\eqref{eq:1} is attained by $\hat{\mathbb Q} \sim
\mathbb P$ with density
\begin{align}
  \label{eq:9}
  \frac{d\hat{\mathbb Q}}{d\mathbb P} = \exp\left(-\int_0^T
  \hat{\theta}_tdW_t-\frac{1}{2}\int_0^T \hat{\theta}_t^2dt\right)
\end{align}
for $\hat{\theta}$ constructed in~\eqref{eq:hattheta} with $\hat{W}$ as
given by~\eqref{eq:hatW}; this $\hat{W}$ coincides with the
$\hat{\mathbb Q}$-Brownian motion induced by the $\mathbb P$-Brownian
motion $W$ via Girsanov's theorem. The value of the dual problem is
\begin{align}\label{eq:18}
-\frac{\Lambda\Phi^2_0 \sqrt{\rho}}{2\coth\left(\sqrt{\rho}T\right)}+\int_{0}^T \frac{1}{2\Lambda}\frac{(s\wedge\Delta)}{1+(s\wedge\Delta) \sqrt{\rho}\tanh\left(\sqrt{\rho}(T-s)\right)}ds.
  \end{align}
\end{lem}
\begin{proof}
  The construction of $\hat{\mathbb Q}$, $\hat{W}$ and $\hat{\theta}$
  has already been established by the preceding discussion. It is
  readily checked that $\hat{\mathbb Q}$ has finite entropy relative
  to $\mathbb P$ and so $\hat{\mathbb Q} \in \mathcal Q$. Note that
  $\hat{W}$ and $W$ generate the same filtration because
  of~\eqref{eq:Volterra} and~\eqref{eq:hatW} and so we can follow the
  same reasoning as in the proof of Lemma~3.2 to obtain its
  representation for the dual target functional also for
  $\hat{\mathbb Q}$. Recalling the minimizing properties of $\hat{a}$ and
  $\hat{l}_{.,s}$, $s \in [0,T]$, it then follows that
  $\hat{\mathbb Q}$ solves the dual problem with value~\eqref{eq:18}.
\end{proof}
By~\eqref{eq:1} the above value~\eqref{eq:18} of the dual problem
already yields the claimed value~\eqref{eq:primalvalue} for our primal
utility maximization problem.  For the completion of the proof of
Theorem~\ref{thm2.1} it therefore remains to work out the optimal
turnover policy $\hat{\phi}$. Due to its dual
description~\eqref{portfolio}, it suffices to compute
$\mathbb E_{\hat{\mathbb Q}}\left[S_T|\mathcal G^{\Delta}_t\right]=
\mathbb E_{\hat{\mathbb Q}}\left[W_T|\mathcal
  F_{t+\Delta}\right]$. Recalling the Volterra-type
equation~\eqref{eq:Volterra} and using Fubini's theorem we can write
\begin{align}
  \label{eq:10}
     W_T&= \hat{W}_T- \int_0^T \left(\hat{a}_u\Phi_0+ \int_0^u\hat{l}_{u,s}
          d\hat{W}_s\right)du \\
  &= \int_0^T \left(1-\int_s^T
    \hat{l}_{u,s}du\right)d\hat{W}_s-\int_0^T\hat{a}_u du\Phi_0.
\end{align}
Thus, for any $t \in [0,T]$, we find
\begin{align}
  \label{eq:11}
  &\mathbb E_{\hat{\mathbb Q}}\left[S_T\middle|\mathcal G_t^\Delta\right] \\
  &=
\int_0^{(t+\Delta)  \wedge T} \left(1-\int_s^T \hat{l}_{u,s}du\right)d\hat{W}_s-
  \int_0^T\hat{a}_u du\Phi_0 \\
&= \int_0^{(t+\Delta)  \wedge T} \left(1-\int_s^T
                                                                \hat{l}_{u,s}du\right)\left(dW_s-\int_{0}^s \hatk_{s,r}
                                                                d W_r ds\right)\\
 &\qquad +
  \Phi_0 \left(\int_0^{{(t+\Delta)
  \wedge T}}\left(1-\int_s^T \hat{l}_{u,s}du\right)\left(\hat{a}_sds-\int_0^s\hat{k}_{s,r}\hat{a}_rdr\,ds\right)-\int_0^T\hat{a}_u du\right)
\end{align}
where in the second step we used~\eqref{eq:hatW} to get an expression
in terms of the original input to our problem $W$ rather than
$\hat{W}$. The structure of this expression suggests to consider for
$X=W$ and $X=\int_0^. \hat{a}_s ds$ the integral operator
\begin{align}
  \mathcal I_{t}^{T}(X) := \int_0^{(t+\Delta)\wedge T} \left(1-\int_s^T
                                                                \hat{l}^T_{u,s}du\right)\left(dX_s-\int_{0}^s \hatk^T_{s,r}
                                                                d X_r ds\right)
\end{align}
for continuous paths $X$. Notice that the $dX$-integrals can be
defined through integration by parts which reveals in particular that
$\mathcal I^T_t(X)$ depends continuously on $X$; notice also that we
used the notation $\hat{l}^T$ and $\hat{k}^T$ in lieu of $l$
and $k$ to emphasize that these kernels depend on the time
horizon $T$. In conjunction with~\eqref{portfolio} and $S_t=W_t$, this
provides us with a (somewhat) explicit `open loop' expression of the
optimal turnover policy:
\begin{align} \label{eq:openloop}
  \hat{\phi}_t = \frac{1}{\Lambda}
  \left(\mathcal{I}_{t}^{T}\left(W\right)-W_
 +\Phi_0\left(\mathcal{I}_{t}^{T}\left(\int_0^.\hat{a}^T_sds\right)-\int_0^T\hat{a}^T_udu\right)\right),
\end{align}
where, again, $\hat{a}^T$ is used to recall that $\hat{a}$
of~\eqref{eq:4} depends on $T$.

To establish the policy's more informative feedback description given in
Theorem~\ref{thm2.1}, we note next that dynamic programming holds for
our problem:

\begin{lem}\label{lem:policy}
  The optimal policy
  $\hat{\phi}$ of~\eqref{eq:openloop} can alternatively be described in the form
  \begin{align}\nonumber
    \hat{\phi}_t = \frac{1}{\Lambda}
  \Bigg(&\mathcal{I}_{0}^{T-t}\left(W_{t+.}-W_t\right)-W_t\\         \label{eq:13}
&\qquad+\hat{\Phi}_t\left(\mathcal{I}_{0}^{T-t}\left(\int_0^.\hat{a}^{T-t}_sds\right)-\int_0^{T-t}\hat{a}^{T-t}_udu\right)\Bigg)
  \end{align}
   where $\hat{\Phi}_t=\Phi_0+\int_0^t \hat{\phi}_sds$ for $t \in [0,T]$.
 \end{lem}
\begin{proof}
  The righthand side of~\eqref{eq:openloop} gives us for each time
  horizon $T$ a continuous functional
  $\Psi^T:\mathbb R\times C[0,T]\rightarrow C[0,T]$ such that for any
  initial stock position $\Phi_0 \in \mathbb R$ and any stock price
  evolution $W$, $\Psi^T(\Phi_0,W|_{[0,T]})$ is the correspondingly
  optimal strategy $\hat{\phi}$ for the utility maximization problem.

Assume by contradiction that the statement of our lemma does not hold.
Then, by continuity of sample paths of $\hat{\phi}$, there exists $t_0\in [0,T]$ such
that with positive probability $\hat{\phi}_{t_0}$ does not coincide with the
righthand side of~\eqref{eq:13}.  Now consider the strategy $\tilde{\phi}$
that coincides with
$\hat{\phi}$ up to time $t_0$ when it switches to
$$\tilde\phi_t:=\Psi^{T-t_0}\left(\hat\Phi_{t_0},W_{.+t_0}-W_{t_0}\right)_{t-t_0},
\quad t \in [t_0,T].$$
For any strategy $\phi$, we can write the contribution over the interval $[t_0,T]$ to the resulting terminal wealth as
\begin{align}
  V^{\Phi_0,{\phi}}_T - V^{\Phi_0,{\phi}}_{t_0}
  =\hat\Phi_{t_0}(S_T-S_{t_0})+\int_{t_0}^T{\Phi}_t(S_T-S_t)dt-\frac{\Lambda}{2}\int_{t_0}^T{\phi}^2_tdt=:V_{[t_0,T]}^{\Phi_{t_0},,\phi},
\end{align}
where $\Phi_{t_0}:=\Phi_0+\int_0^{t_0}\phi_t dt$. Of course,
$\tilde{\Phi}_{t_0}:=\Phi_0+\int_0^{t_0}\tilde{\phi}_t
dt=\hat{\Phi}_{t_0}$. So, by the Markov property of Brownian motion
and choice of $\tilde{\phi}$ as the unique optimal policy as of time
$t_0$, this allows us to observe that
\begin{align}
  \mathbb{E}\left[-\exp\left(-\alpha V_{[t_0,T]}^{\tilde{\Phi}_{t_0},\tilde{\phi}}\right)\middle|\mathcal
                                                             G^\Delta_{t_0}\right]
 \geq  \mathbb{E}\left[-\exp\left(-\alpha V_{[t_0,T]}^{\hat{\Phi}_{t_0},\hat{\phi}}\right)\middle|\mathcal
                                                             G^\Delta_{t_0}\right],
\end{align}
with ``$>$'' holding on $\{\hat{\phi}_{t_0} \not= \tilde{\phi}_{t_0}\}$
(i.e.\ where~\eqref{eq:13} is violated) because continuity of
$\hat{\phi}$ and $\tilde{\phi}$ ensures that they will differ on an
open interval once they differ at all. Since by assumption this
happens with positive probability,  it follows for the unconditional expected
utility from $\tilde{\phi}$ that
\begin{align}
\mathbb{E} \left[-\exp(-\alpha
  V^{\Phi_0,\tilde{\phi}}_T)\right]&=\mathbb{E} \left[\exp(-\alpha V_{t_0}^{\Phi_0,\tilde{\phi}})  \mathbb{E}\left[-\exp\left(-V_{[t_0,T]}^{\tilde{\Phi}_{t_0},\tilde\phi}\right)\middle|\mathcal
                                                             G^\Delta_{t_0}\right]\right]\\&>\mathbb{E} \left[\exp(-\alpha V_{t_0}^{\Phi_0,\tilde{\phi}})  \mathbb{E}\left[-\exp\left(-V_{[t_0,T]}^{\hat{\Phi}_{t_0},\hat\phi}\right)\middle|\mathcal
                                                             G^\Delta_{t_0}\right]\right]\\&=\mathbb{E} \left[-\exp(-\alpha
  V^{\Phi_0,\hat{\phi}}_T)\right],\label{eq:26}
\end{align}
contradicting the optimality of $\hat{\phi}$.
\end{proof}

 As a consequence of this dynamic programming result, it
 suffices to verify our feedback policy
 description~\eqref{eq:feedback} for time $t=0$:

 \begin{lem}\label{lem:policy1}
   The optimal initial turnover rate is
 \begin{align}\label{eq:17}
\hat{\phi}_0=
&\frac{1}{1
+\Delta\sqrt{\rho}\tanh(\sqrt{\rho}(T-\Delta)^{+})}\frac{S_{\Delta \wedge T}}{\Lambda}
\\&+\frac{\sqrt{\rho}}{\coth(\sqrt{\rho}(T-\Delta)^{+}) +\Delta\sqrt{\rho}}
\int_0^{\Delta \wedge T} \frac{S_{s}}{\Lambda}ds-\frac{S_0}{\Lambda}\nonumber\\
                          &+\frac{\sqrt{\rho}}{\coth(\sqrt{\rho}(T-\Delta)^{+}) +\Delta\sqrt{\rho}}
                            \left(\frac{\mu}{\alpha\sigma^2}-\Phi_0\right).\nonumber
  \end{align}
 \end{lem}
 \begin{proof}
   In view of~\eqref{eq:openloop}, we need to compute for $X=W$ and
   $X=\int_0^.\hat{a}_udu$ the operator
 \begin{align}
   \mathcal{I}_{0}^T\left(X\right) &=
                        \int_0^{\Delta \wedge T} \left(1-\int_s^T
                                                                \hat{l}_{t,s}dt\right)\left(dX_s-\int_{0}^s \hatk_{s,r}
                                                                d X_r
                        ds\right)\\\label{eq:15}
   &= \int_0^{\Delta \wedge T} \left(1-\int_s^T  \hat{l}_{t,s}dt\right)dX_s-I
 \end{align}
 with
 \begin{align}
  I&:=\int_0^{\Delta \wedge T} \left(1-\int_s^T\hat{l}_{t,s}dt\right)
                             \int_{0}^s \hatk_{s,r} d X_r ds\\  \label{eq:16}
                           &= \int_0^{\Delta \wedge T}
                             \left(\int_r^{\Delta\wedge T}
                                \hat{k}_{s,r}ds
                             -\int_r^T\int_r^{t\wedge \Delta}
                             \hat{l}_{t,s}\hat{k}_{s,r}ds \ dt
                             \right)  dX_r
 \end{align}
 where the last equality is due to Fubini's theorem. For
 $t \in [r,\Delta \wedge T]$, the kernel identity~\eqref{eq:6} shows
 that the second $ds$-integral in~\eqref{eq:16} gives
 $\hat{k}_{t,r}+\hat{l}_{t,r}$. For $t \in (\Delta \wedge T,T]$,
 we note that $\Delta<T$ and we let $n_t$ denote the numerator in the
 definition of $\hat{l}_{t,.}$ in~\eqref{eq:5} to write
 $\hat{l}_{t,r}=\hat{l}_{\Delta,r} n_t/n_{\Delta}$. It follows by another use of the kernel identity~\eqref{eq:6}
 that for such $t$ the second $ds$-integral above amounts to
 \begin{align}
   \label{eq:19}
   \int_r^{\Delta} \hat{l}_{t,s}\hat{k}_{s,r}ds & =
     \int_r^{\Delta} \hat{l}_{\Delta,s}\hat{k}_{s,r}ds
                                                       \frac{n_t}{n_{\Delta}}
                                                     = \left(\hat{k}_{\Delta,r}+\hat{l}_{\Delta,r}\right) \frac{n_t}{n_{\Delta }}
                                                     \\ & =\hat{l}_{t,r}-\exp\left(\int_r^{\Delta}\hat{l}_{u,u}du\right)\hat{l}_{t,r}
 \end{align}
 where we used~\eqref{eq:hatkappa} in the final step. Plugging all
 this into~\eqref{eq:16}, we see that the contribution to the
 $dt$-integral from $[r,\Delta \wedge T]$ is partially cancelled by
 the first $ds$-integral there, leaving us with
 \begin{align}
   \label{eq:20}
     I=\int_0^{\Delta \wedge T}\left(-\int_r^T
   \hat{l}_{t,r}dt+\int_{\Delta \wedge T}^T\exp\left(\int_r^{\Delta}\hat{l}_{u,u}du\right)\hat{l}_{t,r}dt\right)dX_r.
 \end{align}
 Inserting this into~\eqref{eq:15}, we see a cancellation of integrals
 over $\hat{l}$ and arrive at
 \begin{align}
   \mathcal{I}_{0}^T\left(X\right) &=
                        \int_0^{\Delta \wedge T} \left(1-\int_{\Delta
                        \wedge
                        T}^T\exp\left(\int_s^{\Delta}\hat{l}_{u,u}du\right)\hat{l}_{t,s}dt\right)dX_s\\\label{eq:21}
   &=\int_0^{\Delta \wedge T} \left(1-f_T(s)\right)dX_s
 \end{align}
 where in view of \eqref{eq:5} we have
  \begin{align}\label{eq:fT}
    f_{T}(s) := &\exp\left(\int_s^{\Delta \wedge T}
                         \frac{u\rho}{1+u\sqrt{\rho}\tanh(\sqrt{\rho}(T-u))}du\right) \\
    & \qquad \cdot \frac{s\sqrt{\rho}\sinh(\sqrt{\rho}(T-\Delta)^+)}{\cosh(\sqrt{\rho}(\tau-s)) +s\sqrt{\rho}\sinh(\sqrt{\rho}(T-s))}\\
    &=\frac{s\sqrt{\rho}\sinh(\sqrt{\rho}(T-\Delta)^+)}{\cosh(\sqrt{\rho}(T-\Delta)^{+}) +\Delta\sqrt{\rho}\sinh(\sqrt{\rho}(T-\Delta)^{+})}.
  \end{align}
 Now we apply~\eqref{eq:21} to $X=W$ and $X=\int_.^T\hat{a}_udu$ to
 rewrite the open loop description~\eqref{eq:openloop} of
 $\hat{\phi}_0$ as
 \begin{align}
   \hat{\phi}_0 =&
\frac{W_{\Delta \wedge T}}{\Lambda}    (1-f_T(\Delta \wedge T))
                            +\int_0^{\Delta \wedge T} \frac{W_{s}}{\Lambda}  f'_T(s)ds\\
                          &-
                            \Phi_0\left(
                            \int_{\Delta \wedge
                            T}^T\frac{\hat{a}_u}{\Lambda}du    (1-f_T(\Delta \wedge T))+\int_0^{\Delta\wedge T}\int_{s}^T\frac{\hat{a}_u}{\Lambda}du f'_T(s)ds\right).
 \end{align}
We conclude the claimed representation for the optimal policy~\eqref{eq:17} by inserting~\eqref{eq:fT} and
 \begin{align}
   \label{eq:23}
   \int_{s}^T\frac{\hat{a}_u}{\Lambda}du =
   \frac{\sqrt{\rho}\sinh(\sqrt{\rho}(T-s))}{\cosh(\sqrt{\rho}T)}, \quad s \in [0,T],
 \end{align}
in the above formula for $\hat{\phi}_0$.

As a final step, we need to recall the simplifying steps from the
beginning of this chapter where we reduced everything to the case
$S=W$ underpinning our calculations so far. Reversing these steps then
leads to the formulae given in the present lemma which work for the
general case required in our main theorem.
\end{proof}

redpajama_set_name: RedPajamaArXiv


text:
Q: wso2 api management 2.0 production cluster deployment architecture We are designing wso2 apim 2.0 cluster deployment architecture, from 2.0 clustering, we just found distribution deployment description, so

*

*Is the wso2 apim gateway 2.0 cluster same with 1.10?

*Any recommendation for system CPU/MEMORY requirement?

*wso2 apim 2.0 has traffic manager, how to cluster it?

*For store/publisher, how to cluster them?

*When integration with APIM Analytics, do we just need to config gateway workers?

A: 1) There are configuration changes, and also new configs, in gateways. But gateway clustering has not been changed.
2) See Installation Prerequisites and Tuning Performance
3) There is no public document yet. It'll be there soon. Until now, you can cluster it in the same way gateways are clustered.
4) Store and publisher should be in the same cluster.
5) Gateway workers, store and publisher
Update:
About Dep-Sync: In gateway worker-manager setup, we only publish API synapse files to manager node, and then use dep-sync to copy those APIs to gateway workers. You can use anything that can sync artifacts between nodes instead of SVN based dep-sync. (eg. rsync)


text:
Accenture Announces Technology Leadership Appointments
-- Paul Daugherty to lead technology architecture, Gavin Michael to lead technology innovation --
NEW YORK; June 29, 2011 – Accenture (NYSE: ACN) announced two senior leadership appointments within its Technology organization, effective July 1, naming Paul Daugherty chief technology architect and Gavin Michael chief technology innovation officer. Both executives will report to Kevin Campbell, group chief executive – Technology.
As chief technology architect, Daugherty, 47, will have broad responsibility for expanding Accenture's business in new and emerging technologies and managing the growth and development of Accenture's specialized team of technology architects. He will continue to lead the company's Advanced Systems & Technology delivery platform and provide leadership for the company's digital, web, social media, mobility and cloud computing areas, collaborating with leaders across the company.
As chief technology innovation officer, Michael, 46, will be responsible for leading Accenture's innovation agenda including the company's technology vision, research and development agenda, and technology alliances, strategy and execution. He will also lead the Accenture Technology Labs, Accenture's technology R&D organization that focuses on how to apply new and emerging technologies to solve business issues.
"Paul and Gavin are very experienced and highly regarded innovators and technology architects, and we are very fortunate to have such strong leaders driving Accenture's technology innovation and architecture teams," Campbell said. "As Accenture's clients look to us for the technology vision and practical know-how we offer as the pace of innovation accelerates, Accenture is well positioned to help them meet the challenges they face in applying the latest technologies to drive business performance."
Daugherty and Michael will assume the management responsibilities held by Don Rippert, Accenture's chief technology officer and managing director – Technology, who is retiring from Accenture after 30 years to become chief executive of Basho Technologies, an open source software company focused on big data.
"I want to thank Don Rippert for his many years of service at Accenture," said Campbell. "His passion for driving Accenture's technology vision and his contributions to the growth of our business has been significant. We wish him nothing but the best in the future."
Accenture is a global management consulting, technology services and outsourcing company, with more than 223,000 people serving clients in more than 120 countries. Combining unparalleled experience, comprehensive capabilities across all industries and business functions, and extensive research on the world's most successful companies, Accenture collaborates with clients to help them become high-performance businesses and governments. The company generated net revenues of US $21.6 billion for the fiscal year ended Aug. 31, 2010. Its home page is www.accenture.com.
Alex Pachetti
alex.pachetti@accenture.com
Peter Soh
Peter.y.soh@accenture.com
Global Media & Industry Analyst Relations


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