text:
Click to copyhttps://apnews.com/399b85de38354dd5bf9b20a09d7c0ab0
Couple finds love in rehabilitation center
By CATHY COOKJuly 12, 2019
PICAYUNE, Miss. (AP) — A couple that met as residents in the Picayune Rehabilitation and Healthcare Center tied the knot Thursday (July 4) after a four-month romance.
Cindy and Glenn Burlette met when Cindy was going to get coffee and Glenn was headed outside to smoke, Glenn said.
"It was love at first sight," Glenn said.
Cindy came to the center six months ago for rehabilitation to her hips, she said, while Glenn began his stay at the center last year after suffering a stroke, he said.
Both have been married before — Cindy for 35 years before being widowed and Glenn for 16 months 20 years ago.
Glenn has been lonely since his first marriage ended and was looking for love, he said. Cindy was lonely too, she said, but was surprised to find a new partner.
"You don't think that's ever going to happen," she said.
Cindy was not interested in even a boyfriend at first, Glenn said, but he kept bringing her coffee and buying her cigarettes.
Glenn chose the Fourth of July for the wedding date, so fireworks would go off on their wedding day, he said.
Cindy's nine grandkids and two great grandkids attended the wedding, as did Glenn's brother and sister-in-law.
Glenn bought wedding rings from another resident who sells jewelry, he said. However, the first rings went missing when Glenn went to get them resized, so he bought a second set. On the day of the wedding, the original wedding rings turned up, Cindy said.
The ceremony was not elaborate, but afterward the couple celebrated with a champagne toast and wedding cake. Cindy wore a knee length peach wedding dress.
"He cried like a baby," Cindy said, "I said why are you crying? He said, 'I'm happy.'"
The couple moved from their single rooms into the center's salon room Thursday, while they waited for their shared room to be set up, Cindy said.
Glenn's home burned down shortly after he became a resident in the center, he said.
"His brother is making arrangements to rebuild it and eventually we can move back into the house and try to live our own lives," Cindy said.
Both Glenn and Cindy are happier after getting married, they said.
Information from: Picayune Item, http://www.picayuneitem.com


text:
'There is a space to tell my story'
Marra B. Gad. Photo: Bobby Quillard.
By Masada Siegel, Special To The Observer
Independent film and television producer Marra B. Gad says she wrote The Color of Love: A Story of a Mixed Race Jewish Girl so that young, Jewish kids of color will know that there is beauty and pride in living exactly as they are.
"We are all made beautifully. B'tzelem Elokim, in the image of God," she says. "I never had anyone to look to who was like me. If I can be that for even one child, it would be a tremendous blessing."
Gad draws the reader into her world of growing up biracial in a Chicago Jewish family, and how she is treated by family and friends alike.
She'll talk about The Color of Love Nov. 20 at Wright Memorial Public Library in Oakwood as part of the JCC's Cultural Arts & Book Series.
In her book, Gad explains why she decides to help her racist, abusive, estranged Great-Aunt Nette after Nette develops Alzheimer's.
"I believe that racism, intolerance, antisemitism and anti-anything is taught, often from generation to generation — and that our children absolutely learn from the adults around them," Gad says. "Racism is often painted in large, broad strokes. But it is about far more than the large, clear stories. It is often a far more subtle thing. It lies in the people who assume that I am the hostess at the private club to which I belong in L.A., and not a member. Or in the people who assume that I am someone's nanny, rather than a member of the synagogue."
She describes in detail the intolerance she's experienced from people from all walks of life; the message Gad says she always receives is: "The way that you exist is wrong."
"For as long as humans have existed, there has been a dynamic at play wherein one person or group of people deems themselves better or more deserving than or superior to another, and it always leads to another person or a group of people getting stripped of being considered human by the former," Gad says. "It leads, as we well know, to violence, genocide, and clearly to slavery. We have seen it play out again and again and it is playing itself out now in every corner of the world."
She decided to tell her story now because when she was a child, she was the only brown face in the sea of white Jewish faces around. That's no longer the case, as the Jewish community continues to grow more diverse.
"It is the first time in my nearly 50 years that I have felt that there is a space to tell my story and that people might actually listen or be interested," she explains.
Photo: Bobby Quillard.
"We are finally having cultural conversations about intersectionality on all fronts and discussing racism and intolerance in ways that we have not during my lifetime because of the remarkable political climate currently at play in America and around the world. It felt like the right time to speak, and writing a book felt like the right way to begin what I hope will be an ongoing discussion about race, religion, tolerance, and love — in a time when so many choose not to be loving."
The JCC Cultural Arts & Book Series presents Marra B. Gad, 7 p.m., Wednesday, Nov. 20 at Wright Memorial Public Library, 1776 Far Hills Ave., Oakwood. The event is free and open to the public.
To read the complete November 2019 Dayton Jewish Observer, click here.
Tags:Marra B. GadMasada SiegelThe Color of Love
Basketball as backdrop to mid-century life in NY
Young entrepreneur designs wheelchair apparel in sister's memory
Reporting career to crime fiction
Novelist's mystical Jerusalem
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Masada's wedding wisdom

redpajama_set_name: RedPajamaCommonCrawl


text:
How Much Is a Still Worth?
beeradm
If you buy something through a link in our posts, we may get a small share of the sale.
Still is a device used to distill alcohol by heating it to the point of evaporation and then condensing the vapors into a separate container. Stills play an important role in producing many types of alcohol, including beer, whisky, and vodka. So, how much is a still worth?
Factors that Affect the Cost of a Still
The Type of Still
The Size of the Still
The Material of the Still
The Features of the Still
What Do You Get By Distilling Beer?
How Much Does a Still Produce?
A standard still can cost anywhere from 130 to 500 dollars if you are buying from the store or online. This is the price of a 5-gallon still, which is a common size for home alcohol distillation. If you want a larger still, you can expect to pay more because the manufacturer will charge more for the materials and labor.
If you are looking for a higher-quality still, the price will be significantly higher. For instance, 50-gallon stainless steel still can cost upwards of 2,000 dollars. This is because the materials are more expensive, and the construction is more complex. In most cases, spending this much money on a still is unnecessary unless you are distilling alcohol for commercial purposes.
Ultimately, the cost of a still depends on the size, quality, and features that you want. If you are looking for a basic still for home use, you can expect to pay a small amount. If you need a larger or higher-quality still, you will have to pay more.
A still's worth will can be affected by various factors that every distiller should consider before making a purchase. If you want to purchase your first still or upgrade from a smaller one, you will want to have the following factors in mind:
Stills come in different types, each with different purposes. Each type has its own set of pros and cons that you need to consider. You don't want to spend a lot of money on a still only to find out that it's not right for your needs. Remember that the type of still you choose will ultimately affect the cost. Here are the different types of stills that you need to know about:
Pot Still: A pot still is the most common type of still. It's also the simplest and most affordable type. The simplicity makes its price tag lower than other types of stills.
Reflux Still: A reflux still is a more complex type of still that's often used by commercial distillers. The complexity of its design increases the price. So, if you're looking for a still for home use, a pot still is a better option.
Column Still: A column still is the most complex type of still. It's also the most expensive type. The column still is often used in commercial alcohol production because of its high efficiency. This type is only recommended if you're sure that you need it.
The still size is another important factor to consider because it will affect the cost. If you want a small still for home use, you can expect to pay less than if you need a larger still for commercial use. The size of the still is often determined by the amount of alcohol you want to produce.
Most home brewers only need a small still because they're only brewing for personal consumption. If you want to produce more alcohol, you will need a larger still. The larger the still, the more expensive it will be.
The material of the still is another important factor that affects its cost. Most stills are made using stainless steel because they're durable and easy to clean. However, there are also stills made from other materials, such as copper.
In most cases, stills are made using copper because it's a good conductor of heat. This means that it can produce a higher-quality product. However, copper stills are more expensive than stainless steel stills. If you're on a budget, stainless steel is the better option.
Different stills come with different features. The more features a still has, the more expensive it will be. For instance, some stills come with a heating element. This feature allows you to control the temperature of the still, which is essential for distilling alcohol.
Other features that you might find in a still include a water pump, an automatic shut-off, and a pressure gauge. These features can make your distilling experience more convenient and efficient. However, they also come with a higher price tag.
If you distill beer, you will automatically get whisky. In a real sense, whisky is just distilled beer. The distilling beer removes the water content, leaving behind a more concentrated form of alcohol.
A still produces about 20% of your original liquid. This means that if you have 5 gallons of wort, you will end up with 1 gallon of whisky.
Still is an essential device for beer and whisky making. The still's size, quality, and features will determine its price. If you're on a budget, a small stainless steel pot still is a good option. For commercial use, a column still is the best choice.
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Full Coverage Car Insurance: What Is It?
We often have people call requesting a car insurance quote for "full coverage". However "full coverage" doesn't really exist.
The bank wants you to have comprehensive and collision because those coverages will pay to fix damage to your car so that you will keep making payments on the car loan. The state wants you to carry liability insurance because it helps pay for damage you cause in an accident. That is quite a bit of coverage but it may not be all that you need, which is why the term, "full coverage" may be misleading.
Comprehensive Coverage — Helps pay to repair your car if it's stolen or damaged by a covered cause, such as an animal collision, weather, a falling object, fire, or vandalism.
Collision Coverage — Helps pay for repairs to your car if you crash into a car or other object, such as a tree or a telephone pole.
Comprehensive and collision coverages are not cheap but a single claim can make it worth the cost. If replacing a stolen car or repairing your vehicle after a crash is more than you can afford or want to pay out of your own pocket, you will want to have comprehensive and collision coverage aka "full coverage".
Remember, also, that both comprehensive and collision coverages have a deductible. Most banks will allow you to reduce your insurance cost by having a $1,000 deductible but can you afford $1,000 to get your car out of the shop? Maybe a $500 deductible will only cost a few more dollars a month.
When buying a policy, "full coverage" means "the least amount of coverage I can pay for to make the bank happy and give me my car." To that same person, after a serious accident, "full coverage" means "everything is covered." Including towing, a rental car, medical bills, you name it.
Having "full coverage" car insurance doesn't mean you have "full protection no matter what". Depending on the seriousness of your accident and your own personal circumstances, having "full coverage" probably means you don't have enough protection.
Medical/Personal Injury Protection Coverage (PIP) — Helps pay for medical expenses for both you and your passengers regardless of who was at fault in a car accident. PIP can also pay if you are in someone else's car as a passenger or a pedestrian struck by a car.
Uninsured/Underinsured Motorist Coverage (UM/UIM) — With this coverage, your insurance company pays you as if they were the other driver's insurance company, when that driver either has no coverage or has less coverage than you do. This relieves you of the need to sue the other driver to recover your losses.
Customized Equipment & Parts Coverage — If you have a custom sound system, non-stock chrome rims, or other fancy tweaks to your car, "full coverage" won't help you replace or repair them after an accident — but this coverage can.
Rental Car Coverage — This can cover the cost of a rental car after a covered accident (usually it reimburses you).
Emergency Roadside Assistance Coverage (aka Towing) — Coverage if your car breaks down while on the road and you need towing or roadside service.
Gap coverage — If the insurance company determines that your car is totaled and you owe more on the car than it worth, gap coverage will pay off the remaining loan amount.
Since there is no such thing as "full coverage" when you ask an agent for "full coverage" it will may include more or less than what the next agent considers "full coverage" and that may not be the best car insurance coverage for you.
If you ask for a "full coverage" policy, you will still only have the coverages you choose and will only be covered up to the limits you select. If the damage you cause to the other person or object exceeds your liability limits, you could still be legally liable for the difference out of your own pocket (a good argument for higher limits).
Examine your own personal needs, your vehicle, and your budget to determine what coverages will work best for you then talk to your agent and get the amount of protection you need and that gives you peace of mind.
If you are shopping for car insurance and you're looking for maximum protection, consider all the coverage options above and look beyond what the misleading term "full coverage" includes.
A good rule of thumb: Buy as much coverage as you can comfortably afford to cover the risks you want to avoid. Towing is a nice coverage to have, but most of us could afford to pay for a tow every couple years. A rental car after an accident is nice, but if you have an extra car that you can use, why pay for the coverage?
You can call us or come into one of our offices in Kennewick, Pasco, or Yakima, or use our online quote page to build your car insurance policy with the coverages, options, and deductibles that make you feel "fully covered".


text:
class CParticleRenderIterator
{
friend class CParticleMgr;
friend class CParticleEffectBinding;
public:
	CParticleRenderIterator();

	// The sort key is used to sort the particles incrementally as they're rendered.
	// They only get sorted in the main rendered view (ie: not in reflections or monitors).
	// These return const because you should only modify particles during their simulation.
	const Particle* GetFirst();
	const Particle* GetNext( float sortKey );

	// Use this to render. This can return NULL, in which case you shouldn't render.
	// This being NULL is a carryover from when particles rendered and simulated together and
	// it should GO AWAY SOON!
	ParticleDraw* GetParticleDraw() const;

private:

	void TestFlushBatch();

private:
	// Set by CParticleMgr.
	CParticleEffectBinding *m_pEffectBinding;
	CEffectMaterial *m_pMaterial;
	ParticleDraw *m_pParticleDraw;
	CMeshBuilder *m_pMeshBuilder;
	IMesh *m_pMesh;
	bool m_bBucketSort;

	// Output after rendering.
	float m_MinZ;
	float m_MaxZ;
	float m_zCoords[MAX_TOTAL_PARTICLES];
	int m_nZCoords;

	Particle *m_pCur;
	bool m_bGotFirst;
	float m_flPrevZ;
	int m_nParticlesInCurrentBatch;
};

//
// Iterate the particles like this:
//
// Particle *pCur = pIterator->GetFirst();
// while ( pCur )
// {
//     ... simulate here.. call pIterator->RemoveParticle if you want the particle to go away
//     pCur = pIterator->GetNext();
// }
//
class CParticleSimulateIterator
{
friend class CParticleMgr;
friend class CParticleEffectBinding;
public:
	CParticleSimulateIterator();

	// Iterate through the particles, simulate them, and remove them if necessary.
	Particle* GetFirst();
	Particle* GetNext();
	float GetTimeDelta() const;

	void RemoveParticle( Particle *pParticle );
	void RemoveAllParticles();

private:
	CParticleEffectBinding *m_pEffectBinding;
	CEffectMaterial *m_pMaterial;
	float m_flTimeDelta;

	bool m_bGotFirst;
	Particle *m_pNextParticle;
};

// -------------------------------------------------------------------------------------------------------- //
// CParticleRenderIterator inlines
// -------------------------------------------------------------------------------------------------------- //

inline CParticleRenderIterator::CParticleRenderIterator()
{
	m_pCur = NULL;
	m_bGotFirst = false;
	m_flPrevZ = 0;
	m_nParticlesInCurrentBatch = 0;
	m_MinZ = 1e24;
	m_MaxZ = -1e24;
	m_nZCoords = 0;
}

inline const Particle* CParticleRenderIterator::GetFirst()
{
	Assert( !m_bGotFirst );
	m_bGotFirst = true;

	m_pCur = m_pMaterial->m_Particles.m_pNext;
	if ( m_pCur == &m_pMaterial->m_Particles )
		return NULL;

	m_pParticleDraw->m_pSubTexture = m_pCur->m_pSubTexture;
	return m_pCur;
}

inline void CParticleRenderIterator::TestFlushBatch()
{
	++m_nParticlesInCurrentBatch;
	if( m_nParticlesInCurrentBatch >= NUM_PARTICLES_PER_BATCH )
	{
		m_pMeshBuilder->End( false, true );
		m_pMeshBuilder->Begin( m_pMesh, MATERIAL_QUADS, NUM_PARTICLES_PER_BATCH * 4 );

		m_nParticlesInCurrentBatch = 0;
	}
}

inline const Particle* CParticleRenderIterator::GetNext( float sortKey )
{
	Assert( m_bGotFirst );
	Assert( m_pCur );

	TestFlushBatch();

	Particle *pNext = m_pCur->m_pNext;

	// Update the incremental sort.
	if( m_bBucketSort )
	{
		m_MinZ = min( sortKey, m_MinZ );
		m_MaxZ = max( sortKey, m_MaxZ );

		m_zCoords[m_nZCoords] = sortKey;
		++m_nZCoords;
	}
	else
	{
		// Swap with the previous particle (incremental sort)?
		if( m_pCur != m_pMaterial->m_Particles.m_pNext && m_flPrevZ > sortKey )
		{
			SwapParticles( m_pCur->m_pPrev, m_pCur );
		}
		else
		{
			m_flPrevZ = sortKey;
		}
	}

	m_pCur = pNext;
	if ( m_pCur == &m_pMaterial->m_Particles )
		return NULL;

	m_pParticleDraw->m_pSubTexture = m_pCur->m_pSubTexture;
	return m_pCur;
}

inline ParticleDraw* CParticleRenderIterator::GetParticleDraw() const
{
	return m_pParticleDraw;
}

// -------------------------------------------------------------------------------------------------------- //
// CParticleSimulateIterator inlines
// -------------------------------------------------------------------------------------------------------- //

inline CParticleSimulateIterator::CParticleSimulateIterator()
{
	m_pNextParticle = NULL;
#ifdef _DEBUG
	m_bGotFirst = false;
#endif
}

inline Particle* CParticleSimulateIterator::GetFirst()
{
#ifdef _DEBUG
	// Make sure they're either starting out fresh or that the previous guy iterated through all the particles.
	if ( m_bGotFirst )
	{
		Assert( m_pNextParticle == &m_pMaterial->m_Particles );
	}
#endif

	Particle *pRet = m_pMaterial->m_Particles.m_pNext;
	if ( pRet == &m_pMaterial->m_Particles )
		return NULL;

#ifdef _DEBUG
	m_bGotFirst = true;
#endif

	m_pNextParticle = pRet->m_pNext;
	return pRet;
}

inline Particle* CParticleSimulateIterator::GetNext()
{
	Particle *pRet = m_pNextParticle;

	if ( pRet == &m_pMaterial->m_Particles )
		return NULL;

	m_pNextParticle = pRet->m_pNext;
	return pRet;
}

inline void CParticleSimulateIterator::RemoveParticle( Particle *pParticle )
{
	m_pEffectBinding->RemoveParticle( pParticle );
}

inline void CParticleSimulateIterator::RemoveAllParticles()
{
	Particle *pParticle = GetFirst();
	while ( pParticle )
	{
		RemoveParticle( pParticle );
		pParticle = GetNext();
	}
}

inline float CParticleSimulateIterator::GetTimeDelta() const
{
	return m_flTimeDelta;
}

#endif // PARTICLE_ITERATORS_H

redpajama_set_name: RedPajamaGithub


text:
Fundraising and community outreach efforts of the Tanz Centre are guided by a dedicated and visionary group of individuals who have contributed their expertise to assist the Tanz scientists in achieving their mission to lead global research and education in neurodegenerative disease.
For many, their own experiences with neurodegenerative disease have inspired their contributions of time and energy to help understand and prevent Alzheimer's and other devastating brain diseases. We thank them for their tremendous support and guidance.
Stuart Tanz (Chair)
Stuart Tanz has served as the President and CEO of Retail Opportunity Investments Corporation (ROIC) since 2009. Prior to joining ROIC, Mr. Tanz was the Chief Executive Officer of United Income Properties Inc., a privately owned retail real estate development company in Southern California which developed, owned and operated various shopping centres in Southern California.
In addition to his philanthropic support, Mr. Tanz has been a member of the Tanz Centre Steering Committee since 2011. In 2014 he was appointed as the new Chair.
The Tanz Centre would like to acknowledge the dedicated efforts of the members of the External Advisory Committee, including:
Paul Godfrey
Heather Hunter
Mark Krembil
Geoff Matus
Joel Reitman
Lionel Schipper
Mark Tanz
Bernard Tanz
Russell Tanz


FRIDAY, Oct. 10, 2014 (HealthDay News) -- After surgery for prostate cancer, elevated levels of cholesterol and triglycerides may be linked with greater risk of the cancer's return, a new study suggests.
In a review of more than 800 men who had had their prostate gland removed, those with higher levels of these two blood fats were more likely to have their cancer come back, compared with men with normal levels.
"These findings suggest that normalization, or even partial normalization, of blood fats among men with high cholesterol and triglycerides may reduce the risk of prostate cancer recurrence," said lead researcher Emma Allott, a postdoctoral associate at Duke University School of Medicine. Cholesterol levels can be modified by diet or use of drugs called statins, she said.
Allott's team also found that increases in high-density lipoprotein (HDL), the "good" cholesterol, benefited some men. For every 10 milligrams per deciliter (mg/dL) increase in men with abnormal HDL, the risk for prostate cancer recurrence dropped by 39 percent, they said.
But Dr. Anthony D'Amico, chief of radiation oncology at Brigham and Women's Hospital in Boston, expressed a note of caution. "This is a study of association, not causality," he said, noting it doesn't prove that normal blood-fat levels prevent cancer recurrence.
D'Amico, who wasn't involved in the study, does not recommend that men take statins to protect themselves from prostate cancer based on this study or similar research.
A clinical trial that compares statins, such as Lipitor, with placebo (sham) drugs to specifically test whether or not they reduce the risk of prostate cancer is needed before these drugs can be recommended for prevention of cancer or its recurrence, D'Amico said.
His reluctance is based on disappointing results from trials that found no benefit from other associations that had looked promising.
"There might be ways to modify risk factors for prostate cancer recurrence that need to be studied," he said. "But this study doesn't prove that lowering cholesterol works."
The study was published Oct. 10 in Cancer Epidemiology, Biomarkers & Prevention.
Among the 843 men in the study, 325 had abnormal cholesterol levels, 263 had abnormal triglyceride levels, and 293 had a recurrence of prostate cancer, according to the report.
Men who had triglyceride levels of 150 mg/dL or higher had a 35 percent increased risk of their cancer coming back, compared with men with normal triglyceride levels, the researchers found.
Moreover, every 10 mg/dL increase in cholesterol above 200 mg/dL was linked with a 9 percent increase in risk for the return of prostate cancer, the researchers said.
"Understanding the role of high cholesterol as a modifiable risk factor for both heart disease and cancer, the most common causes of death, is of great importance," Allott said.
Approximately 45 percent of deaths worldwide are attributable to heart disease and cancer, and prostate cancer is the second most common cause of cancer death among American men, she added.


The EGGER Group again set new records in its annual results for 2017/2018. The sales increased during the business year 2017/2018 by +12.5% to reach EUR 2,683.8 million. The EBITDA increased by +22.6 % to reach EUR 445.8 million. The EBITDA margin exceeded previous years' levels by 16.6 %. The family business continues to enjoy a good credit rating thanks to a high equity ratio of 40.5 %. The quantity of rawboards, incl. timber, increased to 8.5 million m3 (+6.5 %), which means the full utilisation of all primary production capacities. On average last year, EGGER employed 8,765 employees group-wide.
Under the EGGER motto "Sustainable international growth", investments in property and intangible assets as well as acquisitions of EUR 483.8 million were posted during the business year 2017/2018 (previous year: EUR 259.2 million). Out of this amount, EUR 70.5 million was spent on maintenance investments and EUR 413.3 million on growth investments.
Over the coming years, EGGER will focus on completing the ongoing strategic development investments. On the market side, the Group Management counts for all European sales markets on a positive development of the overall economic situation. The framework conditions are seen as challenging due to the upcoming Brexit in the UK and the political and economic situation in Turkey as well as the currency instability in Argentina. Thanks to the commissioning of new capacities and the positive development in Europe and Russia while volumes are moved from weaker regions to alternative markets, EGGER expects turnover growth to continue across the Group for the coming business year 2018/2019.


Description: Lock-up unit within a line of similar units with the benefit of forecourt/parking area immediately to the front and shared access to visitor parking. Access by way of concertina metal door to a height of approximately 3.1m and a width of 3m. The property benefits from an eaves height of approximately 4m.


I'd never heard the term before last year when I made a new friend named Joe. Joe's a diabetes educator and therapist who works with families with diabetes. Joe was the first to acquaint me with the "diabetes wannabee" talking about a colleague of his whom he respects enormously who doesn't have diabetes but is so totally supportive you'd think she did. I found it a strange notion–after all why would anyone want to have diabetes? Then surprisingly soon after, I began wondering if my husband was exhibiting "diabetes wannabee" tendencies.
A recent blood test he had taken revealed a fasting glucose of 102 and he was immediately alarmed. The doctor reassured him that 2 points above the cut off was nothing to be concerned about, but it did concern him and living with me he had the opportunity to measure his blood glucose whenever he wanted so I gave him his own meter and lancing device.
He approached the task with what appeared to be unnatural gusto and glee deciding he would measure his glucose four times a day for three days. Like clockwork the first day at the appropriate hour he walked into the kitchen where we keep the meters, pulled out a test strip, inserted strip into meter, secured an appropriate finger and ample drop of blood and recorded his number. I would watch withholding a smile, yet after his third test I began to notice some uncomfortable feelings within myself: Was I jealous? Was he encroaching on my territory? Were we both now worthy of sympathy, not just me? I went so far as to wonder what would life be like if we both had diabetes? Midway through the second day, however, whatever novelty motivated his initial burst of enthusiasm wore off, the testing stopped, I got my fair share of, "I don't know how you do it!" and my "diabetes wannabee" was a "diabetes don'twannabeeanymore." Also, his numbers confirmed no diabetes.
I guess you could see having diabetes as membership in an exclusive club, like getting picked to play on a sports team without having had to wait embarrassedly while the team captain decided whether or not he'd pick you. Maybe that is the allure to the "diabetes wannabee," but whatever it be now that my "diabetes wannabee" has run off the field I get to reclaim the star position on this turf and I'm truly happy for a host of reasons that I'm the only one on the field.
← Remember when walking was something you did rather than measured?


text:
Q: Autolayout reseting views after changing frame in code I am trying to change some frames inside the code but I want autolayout to be respected.
I have this hierarchy

When I hit the search button on top right of navigation bar, I want the green view to go up 50px. It is working great but the tableView is not following the autolayout since its top is binded on the green view.
So when I hit the button I have a blank space between the navigation bar and the table view.
Any ideas how can I sort this out using autolayout?

A: When using auto layout, frame is dynamic based on constraints. Each time the view layout, it will recalculate by constraints and update view's frame. So instead of changing the frame of green view, you should change corresponding constraints.
Make a constraint reference in code call topSpace which is the constraint between green view and top layout guide and set its constant to -50 when user tap the button.

redpajama_set_name: RedPajamaStackExchange


Lecturing in America is a letter written by Arthur Conan Doyle first published in The Author in july 1895.
Sir, — I observed a passage in last month's Author which speaks of the gains to be made by lecturing in America. It is a subject upon Which there has been a great deal of exaggeration, and I think that a few words upon it may not be amiss — the more so as my name was coupled with the remarks.
Anyone who goes to America with the intention of seeing the place and the people, and counts on no more from his lectures than the payment of his expenses, will have a most enjoyable experience. He will come back with enlarged ideas, with a pleasant remembrance of hospitality received and with new friendships, which he will hope to retain until they are old ones.
But if he goes with the primary idea of making money he will be disappointed. Thackeray and Dickens made money, and when we have another Thackeray and Dickens they may do the same; but the British lecturer whose credentials are more modest will find that the margin left over, after his expenses are paid, is probably a less sum than he could have easily earned in his own study.
In the extract to which I refer from your American correspondence, the sum of 500 dollars a lecture is mentioned. This is nonsense. Taking an average a fifth part of it would be nearer the mark, which is no more than could be obtained from the better class provincial societies in Great Britain. For argument's sake, however, let us put the American average at 125 dollars. When the agent's commission of 15 per cent. and the high travelling and hotel expenses have been paid, the lecturer will probably have from 80 to 85 dollars clear. Allow him four lectures a week, and we have from 320 to 350 dollars as his gain. Two months of this will leave him something under 3000 dollars. From this he has to subtract his double passage-money, and about a month extra spent in the journey and preparations. If the balance will exceed what he would earn in the same period by his pen, it is then worth his while to go to America for money.
If any brother author should go, however, I strongly recommend him to put his affairs in the hands of my friend, Major J. B. Pond, in whom they will find a very sympathetic comrade as well as a keen business manager. My own trip to America was one of the most pleasant experiences of my life, but if it had been the wish to earn more than I could have done at home which had attracted me thither, I should certainly have been disappointed. This would be a merely personal and unimportant matter, were it not that the mention of exaggerated sums in your pages might mislead and cause disappointment to some of your readers.


The following scenarios offer examples of clergy who retire at normal or early retirement age. For illustrative purposes, these examples are based only on the formulas described in the section How Your Pension Benefit Is Calculated. They do not take into consideration the normal or optional forms of payment, which could change the amount of the benefit.
her Highest Average Compensation equals $72,000.
his Highest Average Compensation equals $70,000.
his Highest Average Compensation equals $80,000.


text:
\section{Introduction}

Non-commutative stochastic processes are continua of operators sitting inside non-commutative probability spaces.  Such processes can be used to model several phenomena in free probability and quantum information theory (see \cites{B1998, BKS1997}).  More recently, Voiculescu in \cite{V2014} introduced the notion of bi-free independence to simultaneously model the left and right actions of operators on reduced free product space.  Much of the terminology used in \cite{V2014} was based on the idea that the left action would model the past, the right action would model the future, and the theory of bi-free probability would act as a transition between the past and the future.  The goal of this paper is to realize this idea by using bi-free probability to model the transition between different times in non-commutative stochastic processes.

After reviewing the basics and transformations of bi-free probability in Section \ref{sec:bi-free} and some terminology on non-commutative stochastic processes in Section \ref{sec:stochastic}, Sections \ref{sec:self-adjoint} and \ref{sec:unitary} will examine the relation between bi-free probability and transition operators for self-adjoint and unitary non-commutative stochastic processes respectively.  Transition operators play a fundamental role in determining the progression from one point in a non-commutative stochastic process to another and dictate the expectations of the future onto the past.  It will be demonstrated that the transition operators of free Gaussian Markov processes are modelled by the bi-free Gaussian distributions (Example \ref{exam:bi-free-central-limit}) and free Poisson processes are modelled by bi-free Poisson distributions (Example \ref{exam:poisson}).  New formula to determine the transition operators for non-commutative stochastic process with freely independent increments will be examined (Theorems \ref{thm:free-additive-increments-Cauchy-transform-formula} and \ref{thm:free-multiplicative-increments-Cauchy-transform-formula}) and applied recovering some common examples.  Furthermore, the role of bi-free probability in studying non-commutative stochastic processes where the increments are not freely independent will be examined and a ``central limit'' type theorem will be developed for adding non-commutative stochastic processes (Theorem \ref{thm:central}).

\section{Bi-Free Probability Background}
\label{sec:bi-free}

In this section, we briefly recall some of the basic concepts in bi-free probability theory from \cites{CNS2015-2, GHM2016, HHW2018, HW2016, HW2018, S2016-2, S2018, V2014, V2016-2, BBGS2018, S2016-1, CNS2015-1, V2016-2}.

\begin{defn}[\cite{V2014}]
Let $({\mathcal{A}}, \varphi)$ be a non-commutative probability space (i.e. ${\mathcal{A}}$ a unital C$^*$-algebra and $\varphi : {\mathcal{A}} \to {\mathbb{C}}$ unital positive linear functional).  Pairs of unital $*$-subalgebras $(A_{\ell,1}, A_{r, 1})$ and $(A_{\ell,2}, A_{r, 2})$ of ${\mathcal{A}}$ are said to be \emph{bi-freely independent with respect to $\varphi$} if for $k \in \{1,2\}$ there exist Hilbert spaces $\H_k$, unit vectors $\xi_k \in \H_k$, and unital *-homomorphisms $\alpha_k : A_{\ell, k} \to \mathcal{B}(\H_k)$ and $\beta_k : A_{r,k} \to \mathcal{B}(\H_k)$ such that the following diagram commutes
\[
\begin{tikzpicture}[baseline]
\tikzset{
    myarrow/.style={->, >=latex', shorten >=1pt, thick},
}
			\node[above] at (0,1.5) {$A_{\ell,1} \ast A_{r,1} \ast A_{\ell,2} \ast A_{r,2}$};
			\node[above] at (4.5,1.5) {${\mathcal{A}}$};
			\node[above] at (7,1.5) {${\mathbb{C}}$};
			\node[below] at (0,0) {$\mathcal{B}(\H_1) \ast \mathcal{B}(\H_1) \ast \mathcal{B}(\H_2) \ast \mathcal{B}(\H_2)$};
			\node[below] at (7,0) {$\mathcal{B}(\H_1 \ast \H_2)$};
			\draw[myarrow] (0, 1.5) -- (0,0);
			\node[right] at (0,.75) {$\alpha_1 \ast \beta_1 \ast \alpha_2 \ast \beta_2$};
			\draw[myarrow] (2, 1.8) -- (3.95,1.8);
			\node[above] at (3,1.8) {$i$};
			\draw[myarrow] (2.75, -.3) -- (6,-.3);
			\node[above] at (4.35,-.3) {$\lambda_1 \ast \rho_1 \ast \lambda_2 \ast \rho_2$};
			\draw[myarrow] (5.05, 1.8) -- (6.75,1.8);
			\node[above] at (5.75, 1.8) {$\varphi$};
			\draw[myarrow] (7, 0) -- (7,1.5);
			\node[left] at (7, .75) {$\varphi_*$};
		    \end{tikzpicture}
\]
where $i$ is the inclusion map, $\H_1 \ast \H_2$ is the reduced free product of Hilbert spaces with respect to the vectors $\xi_1$ and $\xi_2$, $\varphi_*$ is the vector state on $\H_1 \ast \H_2$ induced by the vacuum vector, $\lambda_k$ is the left action of $\H_k$ on $\H_1 \ast \H_2$, $\rho_k$ is the right action of $\H_k$ on $\H_1 \ast \H_2$, and the free product of *-algebras are as normal.

A pair $(X, Y)$ is used to denote two operators $X, Y \in {\mathcal{A}}$ where $X$ is viewed as a left operator and $Y$ is viewed as a right operator in terms of the above definition.  Thus two pairs $(X_1, Y_1)$ and $(X_2, Y_2)$ are said to be bi-freely independent if $(*\text{-}\mathrm{alg}(X_1), *\text{-}\mathrm{alg}(Y_1))$ and $(*\text{-}\mathrm{alg}(X_2), *\text{-}\mathrm{alg}(Y_2))$ are bi-freely independent.
\end{defn}

It is elementary to see that if $(X_1, Y_1)$ and $(X_2, Y_2)$ are bi-freely independent with respect to $\varphi$, then $X_1$ and $X_2$ are freely independent with respect to  $\varphi$, $Y_1$ and $Y_2$ are freely independent with respect to $\varphi$, $X_1$ and $Y_2$ are classically independent with respect to $\varphi$ (i.e. $X_1$ and $Y_2$ commute in distribution and $\varphi(X_1^n Y_2^m) = \varphi(X_1^n) \varphi(Y_2^m)$ for all $n,m \in {\mathbb{N}}$), and $X_2$ and $Y_1$ are classically independent with respect to $\varphi$.  The converse does not hold (see \cite{S2016-2}).  However, in the commutative case, it is possible to deduce bi-free independence from free independence in certain situations.

\begin{thm}[\cite{CNS2015-1}*{Theorem 10.2.1}]\label{thm:bi-free-from-free}
Let $\H$ be a Hilbert space, let $\xi \in \H$ be a unit vector, and let $\varphi$ be the the vector space corresponding to $\xi$.  Suppose $(A_{\ell,1}, A_{r, 1})$ and $(A_{\ell,2}, A_{r, 2})$ are pairs of unital $*$-algebras in ${\mathcal{B}}(\H)$ such that
\begin{enumerate}
\item $A_{\ell, n}$ and $A_{r,m}$ commute for all $n, m \in \{1,2\}$, and
\item for each $T \in A_{r,k}$ there exists an $S \in A_{\ell,k}$ such that $T \xi = S \xi$ for all $k \in \{1,2\}$.
\end{enumerate}
Then $(A_{\ell,1}, A_{r, 1})$ and $(A_{\ell,2}, A_{r, 2})$ are bi-freely independent with respect to $\varphi$ if and only if $A_1$ and $A_2$ are freely independent with respect to $\varphi$.
\end{thm}

Similar to the theory of free cumulants via non-crossing partitions from \cite{S1994}, \cite{V2016-2} introduced the bi-free cumulants and \cite{CNS2015-2} extended the notion of bi-free cumulants using bi-non-crossing partitions.  As this paper will only make use of some elementary facts pertaining to the bi-free cumulants, we highlight that which is required here.

Given a pair $(Z_\ell, Z_r)$ in a non-commutative probability space $({\mathcal{A}}, \varphi)$, for all $n \in {\mathbb{N}}$ and maps $\chi : \{1,\ldots, n\} \to \{\ell, r\}$, there is a bi-free cumulant corresponding to $\chi$ denoted
\[
\kappa_\chi(Z_{\chi(1)}, \ldots, Z_{\chi(n)}).
\]
In the case that $Z_\ell$ and $Z_r$ commute, all joint moments of $(Z_\ell, Z_r)$ can be reduced to a moment of the form $\varphi(Z_\ell^n Z_r^m)$ for $n,m \in {\mathbb{N}}$ and thus \cite{CNS2015-2} implies only cumulants for $\chi : \{1,\ldots, n+m\} \to \{\ell, r\}$ where there exists a $k$ such that $\chi(k) = \ell$ if $k \leq n$ and $\chi(k) = r$ if $k > n$ matter in the moment-cumulant formula.  Consequently, for such $\chi$, we simplify notation to
\[
\kappa_\chi(Z_{\chi(1)}, \ldots, Z_{\chi(n)}) = \kappa_{n,m}(\underbrace{Z_{\ell}, \ldots, Z_{\ell}}_{n \text{ times}}, \underbrace{Z_r, \ldots, Z_r}_{m \text{ times}}) = \kappa_{n,m}(Z_\ell, Z_r).
\]
From this and the structure of bi-non-crossing partitions, it is elementary to show that
\begin{align}
\kappa_{n,m}(Z_\ell, Z_r) = \kappa_{n+m}(Z_\ell, \ldots, Z_\ell, Z_r, \ldots, Z_r) \label{eq:bi-free-to-free-cumu}
\end{align}
where $\kappa_{n+m}$ denotes the $(n+m)^{\text{th}}$ free cumulant function.  It was also shown in \cite{CNS2015-2} that if $(Z_\ell, Z_r)$ and $(Z'_\ell, Z'_r)$ are bi-freely independent, then mixed bi-free cumulants from the pairs vanish and thus
\begin{align*}
\kappa_{n,m}&(Z_\ell + Z'_\ell, Z_r+ Z'_r) = \kappa_{n,m}(Z_\ell,   Z_r) + \kappa_{n,m}(Z'_\ell,   Z'_r).
\end{align*}

\subsection*{Additive Transformations}
\label{subsec:add-trans}

Of great use in this paper are the bi-free transformations that are used to study commuting pairs of self-adjoint operators and commuting pairs of unitary operators. Since bi-free independence subsumes free independence, the transformations used in free probability are also of use.  We begin with the setting of self-adjoint operators and additive convolution (where free or bi-free independent objects are examined).

Let $X$ be a self-adjoint operator in a non-commutative probability space $({\mathcal{A}}, \varphi)$ and suppose the distribution measure of $X$ with respect to $\varphi$ is $\mu_X$.  The \emph{Cauchy transform of $X$ with respect to $\varphi$} is defined by
\[
G_X(z) = \varphi((z-X)^{-1}) = \frac{1}{z} + \sum_{n\geq 1} \frac{\varphi(X^n)}{z^{n+1}} = \int_{\mathbb{R}} \frac{1}{z-x} \, d\mu_X(x)
\]
for all $z \in {\mathbb{C}} \setminus {\mathbb{R}}$.  It is well known that if ${\mathbb{C}}^+$ and ${\mathbb{C}}^-$ denote the upper and lower half planes respectively, then $G_X({\mathbb{C}}^+) \subseteq {\mathbb{C}}^-$ and $G_X({\mathbb{C}}^-) \subseteq {\mathbb{C}}^+$.  Moreover, if $X$ has density $f_X$ so that $d\mu_X(x) = f_X(x) \, dx$, then $f_X$ can be recovered from $G_X$ via the formula
\begin{align}
f_X(x) = \lim_{\epsilon \searrow 0} -\frac{1}{\pi} \Im\left( G_X(x + i \epsilon)\right).  \label{eq:cauchy-inversion}
\end{align}

Recall the \emph{R-transform of $X$} is the power series that can be defined by
\[
R_X(z) = \sum_{n\geq 0} \kappa_{n+1}(X) z^n
\]
where $\kappa_{n}(X)$ is the $n^{\mathrm{th}}$ free cumulant of $X$.  Equivalently, if one defines
\[
K_X(z) = \frac{1}{z} + R_X(z),
\]
then $K_X$ is the unique function such that
\begin{align}
G_X(K_X(z)) = z = K_X(G_X(z)) \label{eq:cauchy-r-inversion}
\end{align}
for all $z \in {\mathbb{C}} \setminus {\mathbb{R}}$.  Thus $K_X({\mathbb{C}}^+) \subseteq {\mathbb{C}}^-$ and $K_X({\mathbb{C}}^-) \subseteq {\mathbb{C}}^+$.

One important use of the R-transform is that it completely determines the distribution of the free additive convolution of two self-adjoint operators via their individual distributions.  Indeed if $X$ and $X'$ are freely independent self-adjoint operators with respect to $\varphi$, then
\begin{align}
R_{X + X'}(z) = R_X(z) + R_{X'}(z). \label{eq:free-R}
\end{align}
Since $R_X$ and $R_{X'}$ are determined by $G_X$ and $G_{X'}$ respectively, and since $G_{X+X'}$ is determined by $R_{X + X'}$, we see that $G_{X+X'}$ is determined by $G_X$ and $G_{X'}$ (and thus $\mu_X$ and $\mu_{X'}$) so that $\mu_{X+X'}$ can be computed via (\ref{eq:cauchy-inversion}).

Related to computing $\mu_{X+X'}$ are the so-called subordination functions.  Indeed if
\[
\omega_X(z) = K_X(G_{X+X'}(z)) \quad\text{and}\quad \omega_{X'}(z) = K_{X'}(G_{X+X'}(z))
\]
then
\begin{align}
\omega_{X}(z) + \omega_{X'}(z) - z = \frac{1}{G_{X+X'}(z)} = \frac{1}{G_X(\omega_X(z))} = \frac{1}{G_{X'}(\omega_{X'}(z))}\label{eq:free-sub}
\end{align}
for all $z \in {\mathbb{C}}\setminus {\mathbb{R}}$ (see \cites{BB2007,BV1998}).

As per \cite{V2016-2}, given a pair of self-adjoint operators $(X, Y)$ in $({\mathcal{A}}, \varphi)$, the \emph{two-variable Green's function} is defined by
\[
G_{X,Y}(z,w) = \varphi((z-X)^{-1}(w-Y)^{-1}) = \frac{1}{zw} + \sum_{\substack{n,m \geq 0 \\ n+m \geq 1}}\frac{\varphi(X^n Y^m)}{z^{n+1} w^{n+1}}
\]
for all $(z,w) \in ({\mathbb{C}} \setminus {\mathbb{R}})^2$.  If $X$ and $Y$ commute and thus have joint distribution $\mu_{X,Y}$ on ${\mathbb{R}}^2$, then
\[
G_{X,Y}(z,w) = \int_{{\mathbb{R}}^2} \frac{1}{z-x} \frac{1}{w-y} \, d\mu_{X,Y}(x,y).
\]
Moreover \cite{HW2016} implies that if the pair $(X, Y)$ has density $f_{X,Y}$ so that $d\mu_{X,Y}(x) = f_{X,Y}(x) \, dx\, dy$, then $f_{X,Y}$ can be recovered from $G_{X,Y}$ via the formula
\begin{align}
f_{X,Y}(x,y) = \lim_{\epsilon \searrow 0}\frac{1}{\pi^2} \Im\left(\frac{G_{X,Y}(x+i\epsilon,y + i\epsilon) - G_{X,Y}(x+i\epsilon,y - i\epsilon)   }{2i}     \right). \label{eq:bi-cauchy-inversion}
\end{align}

As in \cite{V2016-2}, the \emph{bi-free partial R-transform of $(X,Y)$} is the power series that can be defined by
\begin{align*}
R_{X,Y}(z,w) &= 	\sum_{\substack{n,m \geq 0 \\ n+m \geq 1}} \kappa_{n,m}(X, Y) z^n w^m.
\end{align*}
As proved analytically in \cite{V2016-2} and combinatorially in \cite{S2016-2}, the relationship between the bi-free partial R-transform and the two-variable Green's function is given by
\[
R_{X,Y}(z,w) = 1 + z R_X(z) + w R_Y(w) - \frac{zw}{G_{X,Y}\left( K_X(z), K_Y(w)\right)}.
\]
As such the \emph{reduced bi-free partial R-transform of $(X,Y)$} is the power series defined by
\[
\tilde{R}_{X,Y}(z,w) = \sum_{n,m \geq 1} \kappa_{n,m}(X, Y) z^n w^m.
\]
Clearly
\[
R_{X,Y}(z,w)  = z R_X(z) + w R_Y(w) + \tilde{R}_{X,Y}(z,w)
\]
so that
\begin{align}
\tilde{R}_{X,Y}(z,w)  = 1 -\frac{zw}{G_{X,Y}\left( K_X(z), K_Y(w)\right)}.\label{eq:reduced-partial-R}
\end{align}

The reduced bi-free partial R-transform enables one to determine the distribution of the bi-free additive convolution of two pairs of commuting self-adjoint operators via their individual distributions.  Indeed if $(X,Y)$ and $(X',Y')$ are pairs of commuting self-adjoint operators that are bi-freely independent with respect to $\varphi$, then $(X + X', Y+Y')$ is a pair of commuting (note $X$ and $Y'$ need not commute but will commute in distribution which is enough) self-adjoint operators and thus
\begin{align}
R_{X + X', Y+Y'}(z,w) &= R_{X,Y}(z,w) + R_{X',Y'}(z,w) \text{ and} \nonumber \\
\tilde{R}_{X + X', Y+Y'}(z,w) &= \tilde{R}_{X,Y}(z,w) + \tilde{R}_{X',Y'}(z,w). \label{eq:bi-free-reduce-R-additive}
\end{align}
As per \cite{BBGS2018}, this implies that
\begin{align*}
&\frac{zw}{G_{X+X', Y+Y'}(K_{X+X'}(z),K_{Y+Y'}(w))} +1\\
&= \frac{zw}{G_{X,Y}(K_X(z), K_Y(w))}+ \frac{zw}{G_{X',Y'}(K_{X'}(z), K_{Y'}(w))}
\end{align*}
so that
\begin{align}
\frac{1}{G_{X+X', Y+Y'}(z,w)} &+ \frac{1}{G_{X+X'}(z) G_{Y+Y'}(w)} \nonumber  \\
&= \frac{1}{G_{X,Y}(\omega_X(z), \omega_Y(w))} + \frac{1}{G_{X',Y'}(\omega_{X'}(z), \omega_{Y'}(w))}. \label{eq:bi-free-addivite-subordination}
\end{align}
In particular, via (\ref{eq:bi-cauchy-inversion}), it is possible to describe $\mu_{X+X', Y+Y'}$ with knowledge of the individual distributions of $(X,Y)$ and $(X',Y')$.

\subsection*{Bi-Free Multiplicative Transformation}

To examine communing pairs of unitaries $(U, V)$ in a non-commutative probability space $({\mathcal{A}}, \varphi)$, it is first helpful to recall the free transforms for (bi-)free multiplicative convolution.  There are many approaches to the free transformations (see \cites{V1987, NS1997} for example), we will follow the structure of \cite{HW2018} for simple relation to the bi-free transformations.  Although the (bi-)free transforms generalize to non-unitary operators (see \cites{S2016-1, S2018, V2016-2} for the bi-free results), we will focus on such transforms for unitary operators only.

Let $U$ be a unitary operator in a non-commutative probability space $({\mathcal{A}}, \varphi)$ such that $\varphi(U) \neq 0$ and suppose $\mu_U$ is the distribution of $U$ with respect to $\varphi$.  The \emph{$\psi$-transform of $U$} is defined by
\begin{align}
\psi_U(z) = \varphi(zU (1-zU)^{-1})  = \int_{\mathbb{T}} \frac{zs}{1-zs} \, d\mu(s) \label{eq:psi}
\end{align}
for all $z \in {\mathbb{C}} \setminus {\mathbb{T}}$.

Note if $U$ has density $f_U$ so that $d\mu_U(s) = f_U(s) \, ds$ for $s \in {\mathbb{T}}$, then $f_U$ can be recovered from $\psi_U$.  Indeed
\begin{align}
\Re(2 \psi_U(z) + 1) = \int_{\mathbb{T}} \Re\left(\frac{1+zs}{1-zs}\right) \, d\mu(s) \label{eq:poisson}
\end{align}
is the Poisson integral of the measure $d\mu(\frac{1}{s})$ thereby yielding $\mu$ (see \cite{R1969}).

The \emph{$\eta$-transform of $U$} is defined by
\[
\eta_U(z) = \frac{\psi_U(z)}{1 + \psi_U(z)}
\]
for all $z \in {\mathbb{C}} \setminus {\mathbb{T}}$.  Note $\eta_U$ is holomorphic on its domain and $\eta'_U(0) = \varphi(U) \neq 0$.  Thus $\eta_U$ is invertible near the origin.  Moreover $\psi_U$ (and thus $\mu_U$) can be recovered from $\eta_U$ via the formula
\begin{align}
\psi^{-1}_U(z) = \eta^{-1}_U\left( \frac{z}{1+z}\right). \label{eq:free-S-inv-1}
\end{align}

The \emph{S-transform of $U$} is defined by
\begin{align}
S_U(z) = \frac{1}{z} \eta^{-1}_U(z).\label{eq:free-S-inv-2}
\end{align}
The importance of the S-transform is that it completely determines the distribution of the free multiplicative convolution of two untiary operators with respect to their individual distributions.  Indeed if $U$ and $U'$ are freely independent unitary operators with respect to $\varphi$, then
\begin{align}
S_{UU'}(z) = S_U(z) S_{U'}(z) = S_{U'U}(z).\label{eq:free-S-inv-3}
\end{align}
Since $\mu_U$ and $\mu_{U'}$ determine $S_U$ and $S_{U'}$ respectively, and since $S_{UU'}(z)$ determines $\eta^{-1}_{UU'}$ and thus $\psi_{UU'}$, the distribution of $UU'$ can be recovered via the above transformations and equation (\ref{eq:poisson}).

As per \cite{HW2018}, given a pair of commuting unitary operators $(U, V)$ in $({\mathcal{A}}, \varphi)$ with $\varphi(U) \neq 0$, $\varphi(V) \neq 0$, and joint distribution $\mu_{U, V}$, the \emph{two-variable $\psi$-transformation of $(U, V)$} is given by
\begin{align*}
\psi_{U,V}(z,w) &= \varphi\left((zU)(wV)(1-zU)^{-1}(1-wV)^{-1}\right) \\
&= \int_{{\mathbb{T}}^2} \frac{zs}{1-zs} \frac{wt}{1-wt} \, d\mu_{U,V}(s,t)
\end{align*}
for all $(z,w) \in ({\mathbb{C}} \setminus {\mathbb{T}})^2$.  By \cite{HW2018}, if
\begin{align}
g_{U,V}(z,w) &= 4 \psi_{U,V}(z,w) + 2(\psi_{U}(z) + \psi_{V}(w)) + 1 \nonumber\\
&= \int_{{\mathbb{T}}^2} \frac{1+zs}{1-zs} \frac{1+wt}{1-wt} d\mu_{U,V}(s,t) \label{eq:g-for-U-inversion}
\end{align}
for all $(z,w) \in ({\mathbb{C}} \setminus {\mathbb{T}})^2$, then
\begin{align}
\Re\left( \frac{g(z,w) - g(\frac{1}{\overline{z}},w)}{2}\right) = \int_{{\mathbb{T}}^2} \Re\left(\frac{1+zs}{1-zs}\right) \Re\left( \frac{1+wt}{1-wt}\right) \, d\mu(s,t) \label{eq:bi-free-poisson}
\end{align}
will recover the Poisson integral of the measure $d\mu\left(1/s, 1/t\right)$ and thus $\mu$  (see \cite{R1969}).

By defining
\begin{align}
H_{U,V}(z,w) &= \psi_{U,V}(z,w) + \psi_{U}(z) +  \psi_{V}(w) + 1 \nonumber \\
&= \int_{{\mathbb{T}}^2} \frac{1}{(1-zs)(1-wt)} \, d\mu(s,t) \label{eq:H-definition}
\end{align}
which is well-defined in a neighbourhood of $(0,0)$, the \emph{opposite bi-free partial S-transform of $(U,V)$} was defined in \cite{HW2018} to be
\begin{align}
S^{\mathrm{op}}_{U,V}(z,w) = \frac{w(z+1)}{z(w+1)} \left(\frac{H_{U,V}(\psi^{-1}_{U}(z), \psi^{-1}_{V}(w)) - (w+1)}{H_{U,V}(\psi^{-1}_{U}(z), \psi^{-1}_{V}(w)) - (z+1)}\right). \label{eq:op-S-via-H}
\end{align}
The opposite bi-free partial S-transform is analytic in a neighbourhood of $(0,0)$.

An alternative formulation of the opposite bi-free partial S-transform was provided in \cite{S2018}*{Proposition 2.5} (also see \cite{S2018}*{Remark 2.6}).  Indeed by defining
\[
K_{U,V}(z,w) = \sum_{n,m\geq 1} \kappa_{n,m}(U, V) z^n w^m,
\]
we have
\begin{align}
S^{\mathrm{op}}_{U,V}(z,w)  = \frac{1 + \frac{1}{z} K_{U,V}(z S_U(z), w S_V(w))}{1 + \frac{1}{w} K_{U,V}(z S_U(z), w S_V(w))} \label{eq:op-S-transform-via-cummulants}
\end{align}
in a neighbourhood of $(0,0)$.

If $(U', V')$ is another pair of commuting unitary operators in $({\mathcal{A}}, \varphi)$ with $\varphi(U') \neq 0$ and $\varphi(V') \neq 0$ that is bi-free from $(U, V)$, then \cite{HW2018} proved analytically and \cite{S2018} proved combinatorially that
\begin{align}
S^{\mathrm{op}}_{UU', V'V}(z,w) = S^{\mathrm{op}}_{U,V}(z,w)S^{\mathrm{op}}_{U', V'}(z,w). \label{eq:op-S}
\end{align}
This formula in conjunction with equation (\ref{eq:bi-free-poisson}) enables one to compute $\mu_{UU', V'V}$ in terms of $\mu_{U,V}$ and $\mu_{U', V'}$.  Indeed since $U$ and $U'$ are freely independent and $V$ and $V'$ are freely independent, it is possible to compute $\mu_{UU'}$ and $\mu_{V'V}$ as above.  As equation (\ref{eq:op-S}) enables one to compute $S^{\mathrm{op}}_{(UU', V'V)}(z,w)$ and thus $H_{U,V}(z,w)$, it is possible to compute $\psi_{U,V}$ and thus $\mu_{U,V}$ via equation (\ref{eq:bi-free-poisson}).  Thus the algorithm for determining this bi-free multiplicative convolution is more complicated then the bi-free additive convolution.

It should be pointed out that the above is considered the `opposite' bi-free partial S-transform due to historical development.  Indeed \cite{V2016-3} proved analytically and \cite{S2016-1} proved combinatorially the existence of the bi-free partial S-transform $S_{U,V}(z,w)$ such that
\[
S_{UU', VV'}(z,w) = S_{U,V}(z,w)S_{U', V'}(z,w).
\]
It was shown in \cite{HW2018} that there is a relation between these two partial S-transforms provided unitary operators are being considered with the opposite bi-free partial S-transform being more useful there and in this paper.

\section{Non-Commutative Stochastic Processes Background}
\label{sec:stochastic}

In this section, we will introduce the concepts and terminology on non-commutative stochastic processes required in the subsequent sections.

\begin{defn}
Let $({\mathcal{A}}, \varphi)$ be a non-commutative probability space.  A \emph{non-commutative stochastic process} is a collection $(Z_t)_{t \in T}$ of elements in ${\mathfrak{M}}$.  The index set $T$ is considered a time parameter.  A non-commutative stochastic process $(Z_t)_{t \in T}$ is said to be
\begin{itemize}
\item \emph{self-adjoint} if $Z_t$ is self-adjoint for all $t \in T$.
\item \emph{unitary} if $Z_t$ is unitary for all $t \in T$.
\end{itemize}
\end{defn}

Of particular interest in free probability are the following types of non-commutative processes.

\begin{defn}[\cite{B1998}]
A self-adjoint non-commutative stochastic process $(X_t)_{t \in T}$ in a non-commutative probability space $({\mathcal{A}}, \varphi)$ is said to have \emph{freely additive increments} if for all $t_1 < t_2 < \cdots < t_n$ in $T$, the operators $X_{t_1}, X_{t_2} - X_{t_1}, \ldots, X_{t_n} - X_{t_n-1}$ are freely independent.
\end{defn}

\begin{defn}[\cite{B1998}]
A unitary non-commutative stochastic process $(U_t)_{t \in T}$ in a non-commutative probability space $({\mathcal{A}}, \varphi)$ is said to have \emph{(left) multiplicatively free increments} if for all $t_1 < t_2 < \cdots < t_n$ in $T$, the operators $U_{t_1}, U_{t_2}U^{-1}_{t_1}, \ldots, U_{t_n}U^{-1}_{t_{n-1}}$ are freely independent.
\end{defn}

This paper will mainly focus on the transitioning between different operators in a non-commutative stochastic process.  In particular, given a non-commutative stochastic process $(Z_t)_{t \in T}$, we will consider $Z_\ell$ and $Z_r$ where $\ell,r \in T$ are such that $\ell \leq r$.  In the bi-free terminology, `left operators' will be used to model the past whereas `right operators' will be used to model the future.  To study such transitions, the following terminology is useful.

\begin{defn}
Let $(Z_t)_{t \in T}$ be a non-commutative stochastic process in tracial von Neumann algebra $({\mathfrak{M}}, \tau)$.  Denote
\begin{align*}
{\mathfrak{M}}_{t]} &= W^*(\{Z_\ell \, \mid \, \ell \leq t\}) \subseteq {\mathfrak{M}}\\
{\mathfrak{M}}_{[t]} &= W^*(\{Z_t\}) \subseteq {\mathfrak{M}}\\
{\mathfrak{M}}_{[t} &= W^*(\{Z_r \, \mid \, r \geq t\}) \subseteq {\mathfrak{M}}.
\end{align*}
It is said that $(Z_t)_{t \in T}$ is a \emph{Markov process} if for all $\ell,r \in T$ with $\ell \leq r$ we have
\[
E_{{\mathfrak{M}}_{\ell]}}(A) \in {\mathfrak{M}}_{[\ell]} \quad \text{for all }A \in {\mathfrak{M}}_{[r]}.
\]
\end{defn}

\begin{defn}
Let $(Z_t)_{t \in T}$ be a self-adjoint or unitary non-commutative stochastic process in a tracial von Neumann algebra $({\mathfrak{M}}, \tau)$.  For $t \in T$, let $\mu_t$ be the distribution of $X_t$.  Note $W^*(Z_t)$ is isomorphic to $L_\infty(\mu_t)$.

For $\ell,r \in T$ with $\ell \leq r$, an operator $K_{\ell,r} : L_\infty(\mu_r) \to L_\infty(\mu_\ell)$ determined by
\[
E_{{\mathfrak{M}}_{[\ell]}}(h(Z_r)) = (K_{\ell,r}(h)) (Z_\ell)
\]
for all $h \in L_\infty(\mu_r)$ is called a \emph{transition operator} of the process $(Z_t)_{t \in T}$.
\end{defn}

If $(Z_t)_{t \in T}$ is a Markov process then all of the transition operators exist (see \cite{BKS1997} for example) and
\[
E_{{\mathfrak{M}}_{\ell]}}(h(Z_r)) = E_{{\mathfrak{M}}_{[\ell]}}(h(Z_r)) = (K_{\ell,r}(h)) (Z_\ell)
\]
for all $\ell \leq r$ and $h \in L_\infty(\mu_r)$.  Furthermore, \cite{B1998} showed the existence of transition operators for all self-adjoint non-commutative stochastic processes with additively free increments and for all all unitary non-commutative stochastic processes with multiplicatively free increments.

\section{Self-Adjoint Non-Commutative Stochastic Processes}
\label{sec:self-adjoint}

In this section we will examine the relation between bi-free probability theory and the transition operators of self-adjoint non-commutative stochastic processes and, more generally, the expectation of a von Neumann algebra generated by a single self-adjoint operator onto another von Neumann algebra generated by a single self-adjoint operator.  Our initial discussions will also consider unitary operators which will be examined further in the next section.

Let $({\mathfrak{M}}, \tau)$ be a tracial von Neumann algebra and let $X_\ell, X_r \in {\mathfrak{M}}$ be self-adjoint (respectively unitary) operators.  If $\mu_\ell$ and $\mu_r$ are the distribution measures of $X_\ell$ and $X_r$ with respect to $\tau$ respectively, then $W^*(X_\ell) \cong L_\infty(\mu_\ell)$ and $W^*(X_r) \cong L_\infty(\mu_r)$.  If $E : {\mathfrak{M}} \to W^*(X_\ell) $ is the trace-preserving conditional expectation of ${\mathfrak{M}}$ onto $W^*(X_\ell)$, then, by standard von Neumann algebra theory, for all $S \in {\mathfrak{M}}$ the value of $E(S)$ is determined by the values of
\[
\tau(E(S) X_\ell^n) = \tau(SX_\ell^n) = \tau(X_\ell^n S)
\]
for all $n \in {\mathbb{N}} \cup \{0\}$ (respectively $n \in {\mathbb{Z}}$). Thus, the restriction of $E$ to $W^*(X_r) \cong L_\infty(\mu_r)$ yields a transition operator $K_{\ell,r} : L_\infty(\mu_r) \to L_\infty(\mu_\ell)$ determined by
\[
E(h(X_r)) = (K_{\ell,r}(h)) (X_\ell)
\]
for all $h \in L_\infty(\mu_r)$.

Consider the GNS Hilbert space $L_2({\mathfrak{M}}, \tau)$ generated by $({\mathfrak{M}}, \tau)$, let $\xi = 1_{\mathfrak{M}} \in L_2({\mathfrak{M}}, \tau)$, and for $S \in {\mathfrak{M}}$ let $L(S)$ and $R(S)$ denote the left and right actions of $S$ on $L_2({\mathfrak{M}}, \tau)$ respectively.  Note for all $n,m \in {\mathbb{N}} \cup \{0\}$ (respectively $n,m \in {\mathbb{Z}}$) that
\[
\tau(X_\ell^n X_r^m) = \langle L(X_\ell)^n R(X_r)^n \xi, \xi\rangle.
\]
As this vector state is positive and $L(X_\ell)$ and $R(X_r)$ commute, there exists a probability measure $\mu_{\ell,r}$ such that if $\Omega = {\mathbb{R}}$ (respectively $\Omega = {\mathbb{T}}$) then
\[
\int_{\Omega^2} x^n y^m \, d\mu_{\ell,r}(x,y) = \tau(X_\ell^n X_r^m)
\]
for all $n,m \in {\mathbb{N}} \cup \{0\}$ (respectively $n \in {\mathbb{Z}}$).

Clearly $\mu_{\ell, r}$ completely determines the transition operator $K_{\ell, r}$.  In particular, if $\mu_{\ell,r}$ is absolutely continuous with respect to the two-dimensional Lebesgue measure and thus can be written as $f_{\ell,r}(x,y) \, dx \, dy$,  then
\[
f_\ell(x) = \int_\Omega f_{\ell,r}(x,y) \, dy
\]
is the density function of $X_\ell$ and the transition operator $K_{\ell,r} : L_\infty(\mu_r) \to L_\infty(\mu_\ell)$ is obtained via
\[
(K_{\ell,r}(h))(x) = \int_\Omega h(y) k_{\ell,r}(x,dy)
\]
where
\begin{align}
k_{\ell,r}(x, dy) = \frac{f_{\ell,r}(x,y)}{f_\ell(x)}  \, dy.\label{eq:transition-op-from-2-d-distribution}
\end{align}

Using bi-free probability theory and the above, the transition operators of some common non-commutative stochastic processes become easy to describe and, interestingly, are related to the bi-free generalizations of such objects.

\begin{exam}\label{exam:bi-free-central-limit}
Let $\H$ be a real Hilbert space, let $\H_{\mathbb{C}}$ be the complexification of $\H$, let ${\mathcal{F}}(\H_{\mathbb{C}})$ denoted the Fock space associated to $\H_{\mathbb{C}}$, and let $\tau : {\mathcal{B}}({\mathcal{F}}(\H_{\mathbb{C}})) \to {\mathbb{C}}$ denote the vacuum vector state.  For an index set $T$, let $(f_t)_{t \in T}$ be a set of vectors in $\H$.  As per \cite{BKS1997}, a \emph{free (centred) Gaussian Markov process} is the self-adjoint non-commutative stochastic process $(X_t)_{t \in T}$ where
\[
X_t = l(f_t) + l^*(f_t)
\]
for all $t \in T$ where $l$ and $l^*$ are the left creation and annihilation  operators on ${\mathcal{F}}(\H_{\mathbb{C}})$ respectively.  Moreover \cite{BKS1997}*{Remark 3.4} implies that free Gaussian Markov processes depend only on the covariance function $c : T \times T \to {\mathbb{R}}$ defined by
\[
c(\ell, r) = \langle f_\ell, f_r\rangle
\]
and does not depend on the choice of $\H$ nor the choice of $(f_t)_{t \in T}$.  Some examples of free Gaussian Markov processes include the free Brownian motion whose covariance function is given by $c(\ell, r) = \min(\ell, r)$ with $T = [0, \infty)$, the free Brownian bridge whose covariance function is given by $c(\ell, r) = \ell(1-r)$ for $\ell \leq r$ with $T = [0,1]$, and the free Ornstein-Uhlenbeck process whose covariance function is given by $c(\ell, r) = e^{-|\ell-r|}$ with $T = {\mathbb{R}}$.

In \cite{BKS1997}*{Theorem 4.6} (also see \cite{B1998}*{5.3}) the transition operators for the free (centred) Gaussian Markov process were described.  Using standard bi-free probability theory, we obtain a connection between the transition operators of these free Gaussian Markov processes and the bi-free Gaussian distributions.  Indeed for $\ell, r \in T$ with $\ell < r$, note the the joint distribution of $L(X_\ell)$ and $R(X_r)$ is precisely the distribution of the pair
\[
(l(f_\ell) + l^*(f_\ell), r(f_r) + r^*(f_r))
\]
where $r$ and $r^*$ are the right creation and annihilation operators on ${\mathcal{F}}(\H_{\mathbb{C}})$ respectively.  This pair is a centred self-adjoint bi-free central limit pair by \cite{V2014}*{Theorem 7.6} with bi-free partial and reduced bi-free partial R-transforms given by
\begin{align*}
R_{X_\ell, X_r}(z,w) &= c(\ell, \ell) z^2 + c(\ell, r) zw + c(r,r) w^2\\
\tilde{R}_{X_\ell, X_r}(z,w) &= c(\ell, r) zw.
\end{align*}

The distribution $\mu_{\ell, r}$ was computed in \cite{HW2016}*{Example 3.4} (also see the correction comment in \cite{HW2018}*{Comment 1}) in the case that $c(\ell, \ell) = 1 = c(r,r)$ and $c(\ell, r) = c$.  For completeness and to demonstrate the technology from Section \ref{subsec:add-trans}, we will provide a proof of the formula for the density of $\mu_{\ell, r}$ in the general case.

By equation (\ref{eq:reduced-partial-R}) we know that
\[
G_{X_\ell, X_r}(K_{X_\ell}(z), K_{X_r}(w)) = \frac{zw}{1-c(\ell, r)zw }.
\]
Recall $X_\ell$ and $X_r$ are semicircular operators with spectrum $[-2\sqrt{c(\ell, \ell)} ,2\sqrt{c(\ell, \ell)}]$ and $[-2\sqrt{c(r, r)}, 2\sqrt{c(r, r)}]$ respectively.  Hence the distribution $\mu_{\ell, r}$ is supported on the Cartesian product of these two intervals.  Since
\[
G_{X_\ell}(z) =
\frac{z - \sqrt{z^2 - 4c(\ell,\ell)}}{2 c(\ell,\ell)}  \quad\text{and}\quad G_{X_r}(w) =
\frac{w - \sqrt{w^2 - 4c(r,r)}}{2 c(r,r)}
\]
for $\Im(z), \Im(w) > 0$, since $G_{X_r}(\overline{w}) = \overline{G_{X_r}(w)}$, and since equation (\ref{eq:cauchy-r-inversion}) enables us to compute $G_{(X_\ell, X_r)}(z,w)$ by replacing $z$ and $w$ with $G_{X_\ell}(z)$ and $G_{X_r}(w)$ respectively, we see by equation (\ref{eq:bi-cauchy-inversion}) with $a = c(\ell,\ell)$, $b = c(r,r)$, and $c = c(\ell, r)$  that for all $(x,y)$ in the support of $\mu_{\ell, r}$,
\begin{align*}
&f_{\ell,r}(x,y) \\
&= \frac{1}{\pi^2}   \Im\left( \frac{1}{2i}\left[ \frac{\left(\frac{x- i\sqrt{4 a - x^2}}{2a} \right)\left(\frac{y- i\sqrt{4 b - y^2}}{2b}\right) }{1 - c \left(\frac{x- i\sqrt{4 a - x^2}}{2a} \right) \left(\frac{y- i\sqrt{4 b - y^2}}{2b}   \right)} -  \frac{\left(\frac{x- i\sqrt{4 a - x^2}}{2a} \right)\left(\frac{y+ i\sqrt{4 b - y^2}}{2b}\right) }{1 - c \left(\frac{x- i\sqrt{4 a - x^2}}{2a} \right) \left(\frac{y+ i\sqrt{4 b - y^2}}{2b}   \right)}   \right] \right)\\
&= \frac{1}{2\pi^2}\left(\frac{ -\frac{xy}{4ab} + \frac{\sqrt{4a-x^2}\sqrt{4b-y^2}}{4ab} + \frac{c}{ab}}{\left(1 + \frac{c^2}{ab}\right) - \frac{c}{2 a b} xy + \frac{c}{2 a b} \sqrt{4a - x^2}\sqrt{4b - y^2}} - \frac{-\frac{xy}{4ab} -\frac{\sqrt{4a-x^2}\sqrt{4b-y^2}}{4ab} + \frac{c}{ab}    }{ \left(1 + \frac{c^2}{ab}\right) - \frac{c}{2 a b} xy - \frac{c}{2 a b} \sqrt{4a - x^2}\sqrt{4b - y^2} }              \right)\\
&= \frac{1}{4\pi^2ab} \left(\frac{ \left(1 - \frac{c^2}{ab}  \right)\sqrt{4a-x^2} \sqrt{4b-y^2}  }{\left(1- \frac{c^2}{ab}\right)^2 - \frac{c}{ab}\left(1 + \frac{c^2}{ab}\right) xy + \frac{c^2}{a^2 b^2} \left( bx^2 + ay^2\right)  }    \right).
\end{align*}
To simplify this expression, note if we let $\lambda_t = \sqrt{c(t,t)}$ for $t \in \{\ell, r\}$ and
\[
\lambda_{\ell,r} = \frac{c(\ell,r)}{\lambda_\ell \lambda_r},
\]
then
\[
f_{\ell,r}(x,y) = \frac{1}{4\pi^2 \lambda_\ell^2 \lambda_r^2} \left(\frac{ \left(1 - \lambda_{\ell, r}^2 \right)\sqrt{4\lambda_\ell^2-x^2} \sqrt{4\lambda^2_r-y^2}  }{\left(1- \lambda_{\ell, r}^2\right)^2 - \lambda_{\ell, r}\left(1 + \lambda_{\ell, r}^2\right) \left( \frac{x}{\lambda_\ell}\right) \left(\frac{y}{\lambda_r}\right)  + \lambda_{\ell, r}^2 \left(\left(\frac{x}{\lambda_\ell}   \right)^2 +\left( \frac{y}{\lambda_r}  \right)^2         \right)  }    \right).
\]
Therefore, since $X_\ell$ is a centred semicircular operator with variance $c(\ell, \ell) = \lambda_\ell^2$ so that
\[
f_{\ell}(x) = \frac{1}{2\pi \lambda_\ell^2} \sqrt{4 \lambda_\ell^2 - x^2}
\]
for all $x \in [-2\lambda_\ell, 2\lambda_\ell]$,  we obtain by equation (\ref{eq:transition-op-from-2-d-distribution}) that
\[
k_{\ell,r}(x, dy) = \frac{1}{2\pi \lambda_r^2} \frac{(1-\lambda_{\ell,r}^2) \sqrt{4 \lambda_r^2 - y^2} \, dy}{(1-\lambda_{\ell,r}^2)^2 - \lambda_{\ell,r}(1 + \lambda_{\ell,r}^2) \left(\frac{x}{\lambda_\ell}\right)\left(\frac{y}{\lambda_r}\right) + \lambda_{\ell,r}^2\left(  \left(\frac{x}{\lambda_\ell}\right)^2 + \left(\frac{y}{\lambda_r}\right)^2\right)}
\]
in agreement with \cite{BKS1997}*{Theorem 4.6}.

Note that although the above focused on the free Gaussian Markov processes, the same bi-free computations work for any non-commutative stochastic process consisting of semicircular operators.
\end{exam}

\begin{exam}\label{exam:poisson}
Recall the following example from \cite{VDN1992}.  Let $({\mathfrak{M}}, \tau)$ be a tracial von Neumann algebra and let $I \mapsto P_I$ be a projection valued process; that is, this map is normal, projection valued, if $I, J \subseteq [0,1]$ are disjoint then $P_I P_J = 0$ and $P_I + P_J = P_{I\cup J}$, and $\tau(P_I) = |I|$ for all $I \subseteq [0,1]$ where $|I|$ denotes the Lebesgue measure of $I$.  If $S$ is a centred semicircular operator in ${\mathfrak{M}}$ of variance 1 and free from $\{P_I \, \mid \, I \subseteq [0,1]\}$, then $I \mapsto SP_IS$ is called a \emph{free Poisson process}.

For all $t \in [0,1]$ let $X_t = SP_{[0,t)} S$ which yields the self-adjoint non-commutative process $(X_t)_{t \in[0,1]}$.  Then
\[
\kappa_m(X_t) = t
\]
for all $t \in [0,1]$. Furthermore, for $\ell,r \in [0,1]$ with $\ell \leq r$, we obtain that
\[
\kappa_{n+m}(X_\ell, \ldots, X_\ell, X_r, \ldots, X_r) = \begin{cases}
\ell &\text{if } n > 0 \\
r & \text{if $n=0$ and $m > 0$}
\end{cases}
\]
since $X_r = X_\ell + SP_{[\ell, r)}S$ and since $SP_{[\ell, r)}S$ is freely independent from $X_\ell$ by \cite{NS1996}*{Theorem 1.6}.  Thus, equation (\ref{eq:bi-free-to-free-cumu}) implies that the ordered bi-free cumulants for the bi-free compound Poisson distribution with rate $\lambda = r$ and jump size $\nu = \frac{\ell}{r} \delta_{(1,0)} + \frac{r-\ell}{r}\delta_{(1,1)}$ where $\delta_{(x,y)}$ is the Kronecker delta measure at $(x,y)$ (see \cite{GHM2016}).   Thus the distribution of $(L(X_\ell), R(X_r))$ is the bi-free compound Poisson distribution and thus the transition operator from $X_r$ to $X_\ell$ can be computed.  Note by \cite{HHW2018}*{Corollary 7.6} that such distributions are not full.
\end{exam}

Note the free Poisson process is an example of a self-adjoint non-commutative stochastic process with freely additive increments.  Using technology from bi-free probability, an approach to study such processes that differs from that in \cite{B1998}*{Theorem 3.1} is obtained:

\begin{thm}\label{thm:free-additive-increments-Cauchy-transform-formula}
Let $X$ and $Y$ be freely independent self-adjoint operators in a tracial von Neumann algebra $({\mathfrak{M}}, \tau)$.  Then
\[
G_{L(X), R(X+Y)}(z,w) = - \frac{G_X(z) - G_{X+Y}(w)}{z - K_X(G_{X+Y}(w))}.
\]
Therefore, since equation (\ref{eq:free-sub}) allows one to compute $G_{X+Y}(w)$ based on the distributions of $X$ and $Y$, equation (\ref{eq:bi-cauchy-inversion}) enables one to compute the distribution of the pair $(L(X), R(X+Y))$ and thus the transition operator of $X+Y$ onto $X$.

In particular if $(X_t)_{t \in T}$ is a self-adjoint non-commutative stochastic process with freely additive increments, the above holds for $X = X_\ell$ and $Y = X_r - X_\ell$ for all $\ell < r$ and thus $G_{L(X_\ell), R(X_r)}$ is computable via the individual transformations of $X_\ell$ and $X_r$.
\end{thm}
\begin{proof}
Since $X$ and $Y$ are freely independent, Theorem \ref{thm:bi-free-from-free} implies that the pairs
\[
(L(X), R(X)) \quad\text{and}\quad (L(0), R(Y))
\]
are bi-freely independent with respect to $\tau$.   Hence equation (\ref{eq:bi-free-reduce-R-additive}) along with bi-free cumulants of constants vanishing (see \cite{CNS2015-1}*{Proposition 6.4.1}) imply that
\begin{align*}
\widetilde{R}_{L(X), R(X+Y)}(z,w) &= \widetilde{R}_{L(X), R(X)}(z,w) + \widetilde{R}_{L(0), R(Y)}(z,w) \\
&= \widetilde{R}_{L(X), R(X)}(z,w) + 0 \\
&= \sum_{n,m\geq 1} \kappa_{n,m}(L(X), R(X)) z^n w^m \\
&= \sum_{n,m\geq 1} \kappa_{n+m}(X) z^n w^m \\
&= \sum_{k \geq 2} \kappa_k(X)  zw\frac{z^{k-1} - w^{k-1}}{z-w} \\
&= zw \sum_{k \geq 1} \kappa_{k+1}(X) \frac{z^{k} - w^{k}}{z-w} \\
&= zw \frac{R_{X}(z) - R_{X}(w)}{z-w}.
\end{align*}
Alternatively, since $X$ and $Y$ are freely independent, for $n \geq 1$ and $m \geq 1$,
\begin{align*}
\kappa_{n+m}(\underbrace{X, \ldots, X}_{n \text{ times}}, \underbrace{X+Y, \ldots X+Y}_{m \text{ times}}) &= \kappa_{n+m}(X, \ldots, X) + \Sigma
\end{align*}
where $\Sigma$ is a sum of free cumulants containing $n$ copies of $X$ and $m$ operators that are either $X$ or $Y$ with not all being $X$, and thus vanishes as $X$ and $Y$ are freely independent.  Therefore, by equation (\ref{eq:bi-free-to-free-cumu}),
\[
\kappa_{n,m}(X,X+Y) = \begin{cases}
\kappa_{n+m}(X) & \text{if }n > 0 \\
\kappa_{n+m}(Y) & \text{if }n = 0
\end{cases}
\]
and the above reduced bi-free partial R-transform can be computed from this.

Using equation (\ref{eq:reduced-partial-R}) we obtain that
\begin{align*}
-\frac{zw}{G_{L(X),R(X+Y)}\left( K_{X}(z), K_{X+Y}(w)\right)} &= zw \frac{R_{X}(z) - R_{X}(w)}{z-w} - 1 \\
&= zw\frac{(R_{X}(z) + \frac{1}{z} ) -   (R_{X}(w) + \frac{1}{w})}{z-w}
\end{align*}
so
\[
G_{L(X),R(X+Y)}\left( K_{X}(z), K_{X+Y}(w)\right) = - \frac{z-w}{K_{X}(z) -   K_X(w)}.
\]
Therefore equation (\ref{eq:cauchy-r-inversion}) yields the desired equation.
\end{proof}

Similar to  \cite{B1998}*{Theorem 3.1}, Theorem \ref{thm:free-additive-increments-Cauchy-transform-formula} can be used to compute the transition operators of many self-adjoint non-commutative stochastic processes:
\begin{exam}
The \emph{free Cauchy process} is the self-adjoint non-commutative stochastic process with freely additive increments where
\[
\mu_t(dx) = \frac{1}{\pi} \frac{t}{x^2 + t^2} \, dx
\]
for all $t \in [0, \infty)$. To compute the transition operators, note for all $t \in [0, \infty)$ that
\[
G_{X_t}(z) = \begin{cases}
\frac{1}{z + it} & \text{if }\Im(z) > 0 \\
\frac{1}{z-it} & \text{if } \Im(z) < 0
\end{cases}
\]
so that
\[
K_{X_t}(z) = \begin{cases}
\frac{1}{z} + it & \text{if }\Im(z) > 0 \\
\frac{1}{z}-it & \text{if } \Im(z) < 0
\end{cases}.
\]
Thus if $l,r \in [0,\infty)$ are such that $l < r$, then for all $z,w \in {\mathbb{C}}$ with $\Im(z) > 0$ and $\Im(w) > 0$ we have by Theorem \ref{thm:free-additive-increments-Cauchy-transform-formula} that
\begin{align*}
G_{L(X_\ell), R(X_r)}(z,w) &= - \frac{\frac{1}{z+i\ell} - \frac{1}{w+ir}}{z - \left(\frac{1}{\frac{1}{w + ir}} + i\ell\right)} = \frac{1}{z+i\ell}\frac{1}{w + ir},
\end{align*}
and for all $z,w \in {\mathbb{C}}$ with $\Im(z) > 0$ and $\Im(w) < 0$ we have by Theorem \ref{thm:free-additive-increments-Cauchy-transform-formula} that
\begin{align*}
G_{L(X_\ell), R(X_r)}(z,w) &= - \frac{\frac{1}{z+i\ell} - \frac{1}{w-ir}}{z - \left(\frac{1}{\frac{1}{w - ir}} - i\ell\right)} \\
&= \frac{(z-w) + i(\ell + r)}{(z+i\ell)(w-ir)((z-w) + i(r-\ell))}.
\end{align*}
Hence equation (\ref{eq:bi-cauchy-inversion}) implies that
\begin{align*}
f_{\ell,r}(x,y)
&= \frac{1}{\pi^2}   \Im\left(\frac{1}{2i} \frac{1}{x + i\ell} \left[\frac{1}{y + i r} - \frac{(x - y) + i (\ell+r)}{(y - ir)((x-y) + i (r-\ell)}  \right]     \right)\\
&= \frac{1}{\pi^2} \frac{\ell}{x^2 + \ell^2} \frac{r-\ell}{(x-y)^2 + (r-\ell)^2}.
\end{align*}
Therefore, since
\[
f_\ell(x) = \frac{1}{\pi} \frac{\ell}{x^2 + \ell^2},
\]
we obtain by equation (\ref{eq:transition-op-from-2-d-distribution}) that
\[
k_{\ell,r}(x, dy) = \frac{f_{\ell,r}(x,y)}{f_\ell(x)} \, dy = \frac{1}{\pi} \frac{r-\ell}{(x-y)^2 + (r-\ell)^2} dy
\]
in agreement with \cite{B1998}*{5.1}.
\end{exam}

One benefit of the bi-free approach to non-commutative stochastic processes over \cite{B1998} is the ability to understand a free additive convolution of non-commutative stochastic processes without freely additive increments.  This is accomplished via the following.

\begin{thm}\label{thm:free-convolution-self-adjoint-stochastic-processes}
Let $X_1, X_2, Y_1, Y_2$ be self-adjoint operators in a tracial von Neumann algebra $({\mathfrak{M}}, \tau)$ such that $\mathrm{alg}(\{X_1, Y_1\})$ and $\mathrm{alg}(\{X_2, Y_2\})$ are freely independent.  Thus $G_{X_1 + X_2}(z)$ and $G_{Y_1 + Y_2}(w)$ can be computed via the process described after equation (\ref{eq:free-R}).  With
\[
\omega_{X_k}(z) = K_{X_k}(G_{X_1 + X_2}(z)) \quad\text{and}\quad \omega_{Y_k}(w) = K_{Y_k}(G_{Y_1 + Y_2}(w))
\]
for $k = 1,2$, we have
\begin{align*}
\frac{1}{G_{X_1+X_2, Y_1+Y_2}(z,w)} &+ \frac{1}{G_{X_1+X_2}(z) G_{Y_1+Y_2}(w)} \\
&= \frac{1}{G_{X_1,Y_1}(\omega_{X_1}(z), \omega_{Y_1}(w))} + \frac{1}{G_{X_2,Y_2}(\omega_{X_2}(z), \omega_{Y_2}(w))}.
\end{align*}
Thus equation (\ref{eq:bi-cauchy-inversion}) can be used to compute the transition operator of $Y_1 + Y_2$ onto $X_1 + X_2$.
\end{thm}
\begin{proof}
Since $\mathrm{alg}(\{X_1, Y_1\})$ and $\mathrm{alg}(\{X_2, Y_2\})$ are freely independent, Theorem \ref{thm:bi-free-from-free} implies that
\begin{align*}
&(\mathrm{alg}(\{L(X_1), L(Y_1)\}), \mathrm{alg}(\{R(X_1), R(Y_1)\})) \quad\text{and}\quad \\
& (\mathrm{alg}(\{L(X_2), L(Y_2)\}), \mathrm{alg}(\{R(X_2), R(Y_2)\}))
\end{align*}
are bi-freely independent with respect to $\tau$.   Hence
\[
(L(X_1),  R(Y_1)) \quad\text{and}\quad ( L(X_2) ,  R(Y_2))
\]
are bi-freely independent with respect to $\tau$.  Thus $(L(X_1 + X_2), R(Y_1, Y_2))$ is the additive bi-free convolution of $(L(X_1),  R(Y_1))$ and $( L(X_2) ,  R(Y_2))$ so equation (\ref{eq:bi-free-addivite-subordination}) applies thereby yielding the result.
\end{proof}

Theorem \ref{thm:free-convolution-self-adjoint-stochastic-processes} is particularly useful to compute the transition operators of a free additive convolution of two self-adjoint non-commutative stochastic processes where the transition operators are known, even if those non-commutative stochastic processes do not have freely additive increments.  Indeed suppose $(X_t)_{t \in T}$ is a self-adjoint non-commutative stochastic process in $({\mathfrak{M}}, \tau_{\mathfrak{M}})$ and $(Y_t)_{t \in T}$  is a self-adjoint non-commutative stochastic process in $({\mathfrak{N}}, \tau_{\mathfrak{N}})$.  By consider the reduced free product $({\mathfrak{M}} \ast {\mathfrak{N}}, \tau_{\mathfrak{M}} \ast \tau_{\mathfrak{N}})$, we can consider $(X_t)_{t \in T}$ and $(Y_t)_{t \in T}$ in the same tracial von Neumann algebra in such a way that $X_t$ and $Y_s$ are freely independent for all $t,s \in T$. The self-adjoint non-commutative stochastic process $(X_t + Y_t)_{t \in T}$ then falls under the purview of Theorem \ref{thm:free-convolution-self-adjoint-stochastic-processes} to compute the transition operators.

As examples of the above are at least as complicated as computing the distributions of free additive convolutions, which often need computer assistance, we only provide the following example.

\begin{exam}
In the above exposition, suppose $X_t$ and $Y_t$ are centred semicircular operators for all $t \in T$.  For $\ell, r \in T$ with $\ell < r$, Example \ref{exam:bi-free-central-limit} yields the transition operator of $X_r$ onto $X_\ell$ (respectively $Y_r$ onto $Y_\ell$) where $c(t,t) = \tau_{\mathfrak{M}}(X_t^2)$ (respectively $c(t,t) = \tau_{\mathfrak{N}}(Y_t^2)$) for $t \in \{\ell, r\}$ and $c(\ell, r) = \tau_{\mathfrak{M}}(X_\ell X_r)$ (respectively $c(\ell, r) = \tau_{\mathfrak{N}}(Y_\ell Y_r)$).  As the (bi-)free additive convolution of such operators is of the same form, the transition operator of $X_r + Y_r$ onto $X_\ell + Y_\ell$ is given by the transition operators in Example \ref{exam:bi-free-central-limit} with $c(t,t) = \tau_{\mathfrak{M}}(X_t^2) + \tau_{\mathfrak{N}}(Y_t^2)$ for $t \in \{\ell, r\}$ and $c(\ell, r) = \tau_{\mathfrak{M}}(X_\ell X_r) + \tau_{\mathfrak{N}}(Y_\ell Y_r)$.
\end{exam}

Furthermore, the algebraic bi-free central limit theorem (\cite{V2014}*{Theorem 7.9}) immediately yield the following for limits of free additive convolutions of non-commutative stochastic processes.  In particular, the transition operators of such non-commutative stochastic processes `tend' to those from Example \ref{exam:bi-free-central-limit}.

\begin{thm}\label{thm:central}
Let $\{(X_{n,t})_{t \in T}\}_{n \in {\mathbb{N}}}$ be a sequence of self-adjoint non-commutative stochastic processes in a tracial von Neumann algebra $({\mathfrak{M}}, \tau)$ such that for all $\ell, r \in T$ with $\ell < r$, $\mathrm{alg}(\{X_{n,\ell}\}_{n \in {\mathbb{N}}})$ and $\mathrm{alg}(\{X_{n,r}\}_{n \in {\mathbb{N}}})$ are freely independent.  Suppose further that
\begin{enumerate}
\item $\tau(X_{n,t}) = 0$ for all $n \in {\mathbb{N}}$ and $t \in T$,
\item $\sup_{n \in {\mathbb{N}}} \left|\tau(X_{n,t_1} \cdots X_{n, t_k})\right| < \infty$ for all $\ell, r \in T$, for all $k \in {\mathbb{N}}$, and for all $t_1, \ldots, t_k \in \{\ell, r\}$, and
\item $\lim_{N \to \infty} \frac{1}{N} \sum^N_{n=1} \tau(X_{n,\ell} X_{n,r}) = c(\ell, r)$ for all $\ell, r \in T$.
\end{enumerate}
By letting $S_{N, t} = \frac{1}{N} \sum^N_{n=1} X_{n,t}$ for all $t \in T$ and $N \in {\mathbb{N}}$ and letting $E_{N,t} : {\mathfrak{M}} \to W^*(S_{N, t})$ denote the conditional expectation onto $W^*(S_{N, t})$, we have for all $\ell, r \in T$ with $\ell < r$ and $m \in {\mathbb{N}}$ that
\[
\lim_{N \to \infty} \tau(E_{N,\ell}(S_{N,r}^m)) = \int_{{\mathbb{R}}^2} y^m f_{\ell, r}(x,y) \, dx \, dy
\]
where $f_{\ell, r}$ is as in Example \ref{exam:bi-free-central-limit}.
\end{thm}

Of course, a stronger analytic version of the bi-free central limit theorem could immediately yield stronger information about the transition operators of such processes.

\section{Unitary Stochastic Processes and Bi-Free Probability}
\label{sec:unitary}

The same ideas used in the previous section  can be transferred to describe the transition operators for unitary operators in von Neumann algebra using bi-free probability theory and the S-transforms.  We begin with the following.

\begin{thm}\label{thm:free-multiplicative-increments-Cauchy-transform-formula}
Let $U$ and $V$ be freely independent unitary operators in a tracial von Neumann algebra $({\mathfrak{M}}, \tau)$ such that $\tau(U) \neq 0$ and $\tau(V) \neq 0$.  Then
\[
H_{L(U),R(VU)}(z,w) = 1 + \frac{z \psi_U(z) - \psi^{-1}_U(\psi_{VU}(w)) \psi_{VU}(w)}{z-  \psi^{-1}_U(\psi_{VU}(w))}.
\]
Therefore, since equations (\ref{eq:free-S-inv-1}), (\ref{eq:free-S-inv-2}), and (\ref{eq:free-S-inv-3}) allows one to compute $\psi_{VU}$ based on the distributions of $U$ and $V$, it is possible to compute $\psi_{L(U), R(VU)}$, $\psi_U$, and $\psi_{VU}$.  Thus $g_{L(U),R(VU)}$ is computable by equation (\ref{eq:g-for-U-inversion}) so the distribution of $(L(U), R(VU))$ and the transition operator of $VU$ onto $U$ are computable via equation (\ref{eq:poisson}).

In particular if $(U_t)_{t \in T}$ is a unitary non-commutative stochastic process with (left) multiplicatively free increments, the above holds for $U = U_\ell$ and $V = U_r U_\ell^{-1}$ for all $\ell < r$  and thus $H_{L(U_\ell), R(U_r)}$ is computable via the individual transformations of $U_\ell$ and $U_r$.
\end{thm}
\begin{proof}
Since $U$ and $V$ are freely independent, Theorem \ref{thm:bi-free-from-free} implies that the pairs
\[
(L(U), R(U)) \quad\text{and}\quad ( L(V),  R(V))
\]
are bi-freely independent with respect to $\tau$.  Hence equation (\ref{eq:op-S}) implies that
\begin{align*}
S^{\mathrm{op}}_{L(U),R(VU)}(z,w) &= S^{\mathrm{op}}_{L(1)L(U), R(U)R(V)}(z,w)  \\
&= S^{\mathrm{op}}_{L(1), R(V)}(z,w)  S^{\mathrm{op}}_{L(U), R(U)}(z,w)
\end{align*}
However, by equation (\ref{eq:op-S-transform-via-cummulants}) together with the fact that  bi-free cumulants of constants vanishing (see \cite{CNS2015-1}*{Proposition 6.4.1}), we obtain that
\[
S^{\mathrm{op}}_{L(1), R(V)}(z,w)  = 1
\]
so that
\begin{align*}
S^{\mathrm{op}}_{L(U),R(VU)}(z,w) &= S^{\mathrm{op}}_{L(U), R(U)}(z,w).
\end{align*}
Hence equation (\ref{eq:op-S-via-H}) implies that
\[
H_{L(U),R(VU)}(\psi^{-1}_U(z), \psi^{-1}_{VU}(w)) = H_{L(U),R(U)}(\psi^{-1}_U(z), \psi^{-1}_{U}(w))
\]
and thus
\[
H_{L(U),R(VU)}(z,w) = H_{L(U),R(U)}(z, \psi^{-1}_{U}(\psi_{VU}(w))).
\]
Note if $\mu_U$ is the distribution of $U$, then distribution of $(L(U), R(U))$ is supported on
\[
D = \{(s,s) \, \mid \, s \in {\mathbb{T}}\}
\]
with $d\mu_{L(U),R(U))}(s,s) = d \mu_U(s)$.  Hence equations (\ref{eq:psi}) and (\ref{eq:H-definition}) imply that
\begin{align*}
H_{(U,U)}(z,w) &= \int_{{\mathbb{T}}} \frac{1}{(1-zs)(1-ws)} d\mu_U(s) \\
&= \int_{\mathbb{T}} 1 + \frac{z^2s}{(z-w)(1-zs)} - \frac{w^2s}{(z-w)(1-ws)} \, d\mu_U(s) \\
&= 1 + \frac{z \psi_U(z) - w \psi_U(w)}{z-w}
\end{align*}
so the result follows.
\end{proof}

\begin{exam}
In \cite{B1998}*{5.2}, a unitary multiplicative analogue of the free Levy process $(U_t)_{t \geq 0}$ was examined where
\[
\mu_t(ds) = \frac{1 - e^{-2t}}{|s- e^{-t}|^2}  \, ds
\]
and thus
\[
\psi_{U_t}(z) = \frac{z}{e^t-z}
\]
for all $z \in {\mathbb{C}}$ with $|z| < 1$.  Thus if $l,r \in [0,\infty)$ are such that $l < r$, then for all $z,w \in {\mathbb{C}} \setminus {\mathbb{T}}$ we have by Theorem \ref{thm:free-multiplicative-increments-Cauchy-transform-formula} that
\begin{align*}
H_{L(U_\ell), R(U_r)}(z,w) &= 1 + \frac{z \psi_{U_\ell}(z) - e^{\ell-r} w \psi_{U_r}(w)}{z-e^{\ell-r} w}.
\end{align*}
Note that
\[
\psi_{U_\ell}\left(\frac{1}{\overline{z}}\right) = -1 - \overline{\psi_{U_\ell}(z)}.
\]
Therefore, since
\[
g_{L(U_\ell), R(U_r)}(z,w) = 4 H_{L(U_\ell), R(U_r)}(z,w) - 2 \psi_{U_\ell}(z) - 2 \psi_{U_r}(w) - 3,
\]
equation (\ref{eq:bi-free-poisson}) implies for all $s,t \in {\mathbb{T}}$ that the density of $d\mu\left(\frac{1}{s}, \frac{1}{t}\right)$ is given by
\begin{align*}
&2 \Re\left(\frac{s \psi_{U_\ell}(s) - e^{\ell-r} t \psi_{U_r}(t)}{s - e^{\ell-r} t} - \frac{-s -s \overline{\psi_{U_\ell}(s)}  - e^{\ell-r}  t \psi_{U_r}\left(t\right)}{s - e^{\ell-r} t} \right) - \Re\left( \psi_{U_\ell}(s)+ \overline{\psi_{U_\ell}(s)} + 1\right)\\
&= \Re\left(\frac{2s}{s-e^{\ell-r} t}     -   1 \right) \left( \psi_{U_\ell}(s)+ \overline{\psi_{U_\ell}(s)} + 1\right)\\
&= \Re\left(\frac{(s+e^{\ell-r}t)(\overline{s} - e^{\ell-r} \overline{t}) }{|s-e^{\ell-r} t|^2}\right) \Re\left( 2\psi_{U_\ell}(s)+ 1\right)\\
&= \frac{1-e^{2(\ell-r)}}{|s - e^{\ell-r} t|^2} \frac{e^{2\ell} - 1}{|e^\ell-s|^2}.
\end{align*}
Thus, by density of $d\mu(s,t)$ is given by
\begin{align*}
\frac{1-e^{2(\ell-r)}}{|\frac{1}{s} - e^{\ell-r} \frac{1}{t}|^2} \frac{e^{2\ell} - 1}{|e^\ell-\frac{1}{s}|^2}&=\frac{1-e^{2(\ell-r)}}{|s^{-1}t - e^{\ell-r} |^2} \frac{1-e^{-2\ell} }{|e^{-\ell}-s|^2}.
\end{align*}
Thus, by dividing by the density of $\mu_\ell$, we obtain that
\[
k_{\ell, r}(s,dt) = \frac{1-e^{2(\ell-r)}}{|s^{-1}t - e^{\ell-r} |^2}  dt,
\]
which agrees with \cite{B1998}*{5.2}.
\end{exam}

Again one benefit of the bi-free approach to non-commutative stochastic processes over \cite{B1998} is the ability to understand a free multiplicative convolution of non-commutative stochastic processes without freely multiplicative increments.  This is accomplished via the following.

\begin{thm}\label{thm:free-convolution-unitary-stochastic-processes}
Let $U_1, U_2, V_1, V_2$ be unitary operators with non-vanishing first moments in a tracial von Neumann algebra $({\mathfrak{M}}, \tau)$ such that $*\text{-}\mathrm{alg}(\{U_1, V_1\})$ and $*\text{-}\mathrm{alg}(\{U_2, V_2\})$ are freely independent so $\psi_{U_1 U_2}(z)$ and $\psi_{V_1V_2}(w)$ can be computed via the process described after equation (\ref{eq:free-S-inv-3}).  The pair
\[
(L(U_1U_2), R(V_1V_2)) = (L(U_1)L(U_2), R(V_2)R(V_1))
\]
is the opposite bi-free multiplicative convolution of $(L(U_1), R(V_1))$ and $(L(U_2), R(V_2))$.  Therefore $\psi_{U_1 U_2, V_1 V_2}(z,w)$ can be computed via the process described after equation (\ref{eq:op-S}).  Hence equations (\ref{eq:bi-free-poisson}) and (\ref{eq:transition-op-from-2-d-distribution}) can be used to compute the transition operator of $V_1V_2$ onto $U_1U_2$.
\end{thm}
\begin{proof}
Since  $\{U_1, V_1\}$ and $\{U_2, V_2\}$ are freely independent, Theorem \ref{thm:bi-free-from-free} implies that
\begin{align*}
&(*\text{-}\mathrm{alg}(\{L(U_1), L(V_1)\}), *\text{-}\mathrm{alg}(\{R(U_1), R(V_1)\})) \quad\text{and}\quad \\
& (*\text{-}\mathrm{alg}(\{L(U_2), L(V_2)\}), *\text{-}\mathrm{alg}(\{R(U_2), R(V_2)\}))
\end{align*}
are bi-freely independent with respect to $\tau$.   Hence
\[
( L(U_1),   R(V_1)) \quad\text{and}\quad ( L(U_2),   R(V_2))
\]
are bi-freely independent with respect to $\tau$.  Thus
\[
(L(U_1U_2), R(V_1V_2)) = (L(U_1)L(U_2), R(V_2)R(V_1))
\]
is the opposite bi-free multiplicative convolution of $(L(U_1), R(V_1))$ and $(L(U_2), R(V_2))$ thereby yielding the result.
\end{proof}

redpajama_set_name: RedPajamaArXiv