* 1: This could represent the liver, given its large, somewhat irregular wedge shape, and it typically has a moderate to slightly high signal intensity on most MRI sequences.

* 2: This might be the gallbladder if it's located beneath the liver and appears as a small, streamlined or pear-shaped structure.

* 3: The pancreas is often more central in the abdomen, slightly flattened and elongated, and might correspond to this number.

* 4: The spleen is typically located to the left of the body (which would be on the right side of the image as we look at it), and it may appear as an oval or crescent-like structure with signal intensity slightly higher than that of the liver.

§.§ Tip 6: Appearance descriptors should be non-conflicting with any image annotations to avoid misinterpretations.

§.§.§ Case 1: Endoscopic image annotation recognition

2CASE 1

* Picture modal: Endoscopic picture

* Tasks: Laparoscopic cholecystectomy

* Reference Answer:

* Yellow Dot: Liver.

* Red Dot: Surgical grasper.

* Blue Dot: Gallbladder.

* Green Dot: Fats.

* Purple Dot: Intestine.

2CASE 1

Conflicted with annotation

[colback=gray!10,arc=0mm,auto outer arc]
Prompt: This is a picture of a cholecystectomy. In this picture, there are markers of different colors. Please tell me which tissues or instruments correspond to the annotations in the image. Intestinal tissue is typically pale pink, the liver is usually reddish-brown, the gallbladder is commonly grayish-green, and fat is generally pale yellow.

GPT-4V Response:
[colback=cyan!10,arc=0mm,auto outer arc]

* response 1:

* Blue Dot: This appears to be the gallbladder, which is typically of a grayish-green hue.

* Green Dot: This section might represent intestinal tissue, often presented in a pale pink color.

* Yellow Dot: This could likely be adipose tissue or fat, as it usually appears pale yellow.

* Red Dot: This is a yellow!20surgical instrument.

* Purple Dot: This section might represent the liver, typically seen in a reddish-brown hue.

* response 2:

* Blue Dot: Gallbladder, typically seen in a grayish-green hue.

* Green Dot: Intestinal tissue, often presented in a pale pink color.

* Yellow Dot: Adipose tissue, or fat, usually of a pale yellow appearance.

* Red Dot: This instrument is a surgical clamp, used for grasping or holding tissue in place.

* Purple Dot: Liver, often seen in a reddish-brown hue.

* response 3:

* Blue Dot: This appears to resemble the gallbladder, typically of a grayish-green hue.

* Green Dot: This section seems to be the intestinal tissue, seen in a pale pink hue.

* Yellow Dot: This could be adipose tissue, given its pale yellow appearance.

* Red Dot: This tool appears to be a yellow!20medical instrument used for surgical procedures.

* Purple Dot: This region likely represents the liver, with its reddish-brown appearance.

§.§ Tip 7: Clarifying the contextual relationships between sequential images enhances GPT-4v's analytical accuracy and detail recognition.

§.§.§ Case 1: Endoscopic video sequence analysis

2CASE 1

Without series number

With series number

* Picture modal: Endoscopy

* Tasks: Hysteroscopic polypectomy

* Reference Answer:

* Pic.1: Observe the surroundings, and locate the polyp.

* Pic.2: Excising the polyp.

* Pic.3: Further clearing of the polyp.

* Pic.4: Excision complete, removing the excised polyp.

2CASE 1

Control Group (without contextual relationships)

[colback=gray!10,arc=0mm,auto outer arc]
Prompt: These four photos are from a hysteroscopic polypectomy. Please analyze the procedures in these four images.

GPT-4V Response:
[colback=cyan!10,arc=0mm,auto outer arc]

* response 1:

* Picture 1: The image shows a polyp within the uterine cavity. A medical instrument (possibly a suction device or curette) has been introduced into the cavity, poised to perform a procedure on the polyp.

* Picture 2: The instrument is manipulating the polyp. It appears to be in the process of excision or curettage of the polyp.

* Picture 3: After the excision, portions of the polyp have been removed, but remnants still persist within the uterine cavity.

* Picture 4: There is continued clearing or treatment of the residuals inside the uterine cavity.

* response 2:

* Photo 1: Assisted by hysteroscopy, the physician is examining the endometrial lining. A distinct polyp is visible in this image, surrounded by some mild bleeding. Additionally, a surgical instrument (resembling forceps) is nearing the operational site.

* Photo 2: The surgical instrument approaches the polyp, preparing for excision. There's a slight increase in the bleeding surrounding the polyp.

* Photo 3: This is a pivotal moment in the excision procedure. The polyp, as evident from the image, has been grasped and is commencing detachment.

* Photo 4: The polyp has been successfully excised, and the surgical instrument is extracting it from the uterine cavity. Hemorrhage within the cavity has been managed.

* response 3:

* Photo 1: This image offers a perspective from the hysteroscope, showcasing the endometrial lining and a polyp. Hysteroscopic instruments are being employed to position and stabilize the polyp in preparation for its excision.

* Photo 2: In this image, the tool has approached the polyp and commenced its excision. There's a reddened region above the excision site, indicative of minor bleeding due to the surgical intervention.

* Photo 3: The image depicts a partially excised polyp. A section of the polyp has been cut off, with the remaining still attached to the uterine wall.

* Photo 4: This image presents the area post-polypectomy. There's a whitened zone on the endometrial lining, possibly a trace left after the polyp's removal.

2CASE 1

With contextual relationships

[colback=gray!10,arc=0mm,auto outer arc]
Prompt: These four photos are from a hysteroscopic polypectomy. Please analyze the procedures in these four images.

GPT-4V Response:
[colback=cyan!10,arc=0mm,auto outer arc]

* response 1:

* Picture 1: Initial Observation. At this stage, one can see the interior of the uterine cavity and a prominent polyp. At this point, the doctor might be looking for the best entry point and operational path.

* Picture 2: Beginning of Excision. Using medical tools, the doctor starts excising the polyp from its base. This is a critical step that requires careful operation to ensure the polyp is completely removed without damaging surrounding tissues.

* Picture 3: Continued Excision. At this stage, the polyp has been partially removed but still partially attached to the uterine wall. The doctor continues to operate carefully to ensure its complete removal.

* Picture 4: yellow!20Polypectomy Complete. At this stage, one can observe that the polyp has been fully removed and the interior of the uterine cavity looks relatively clean. Further cleaning or hemostasis might be performed.

* response 2:

* Picture 1: The first picture displays the scene within the uterine cavity. One can observe part of the polyp and the inner walls of the uterus. Surgical instruments have already entered the view, preparing to operate on the polyp.

* Picture 2: The second picture shows the process of the excision surgery. The surgical instrument seems to be grasping and partially excising the polyp.

* Picture 3: The third picture further displays the steps during surgery. From the picture, one can observe that the polyp has already been partially excised.

* Picture 4: The fourth picture showcases the situation inside the uterine cavity after the surgery. Most of the polyp seems to have been cleared, and surgical instruments might be conducting a final cleanup.

* response 3:

* Picture 1: The doctor might be using a hysteroscope to examine the condition of the endometrium and locate the polyp. In the picture, one can observe a surgical instrument capturing or securing the polyp for better operation.

* Picture 2: The picture appears to show that the polyp is already clamped by the surgical instrument and is being excised.

* Picture 3: The polyp has been partially excised, and one can see the excised polyp tissue and the remaining part.

* Picture 4: This picture displays the state of the endometrium after complete excision. The polyp has been fully removed, though there might be slight tissue remnants.

§.§ Tip 8: Splicing multiple images into one and providing a sequence order enhances task processing effectiveness compared to multiple simultaneous inputs.

§.§.§ Case 1: Endoscopic video sequence analysis

2CASE 1

Uploaded one by one via GPT-4V website. The upload system of GPT-4V may cause different sizes to be displayed, but the actual sizes are not affected

Uploaded after splicing

* Picture modal: Endoscopy

* Tasks: Hysteroscopic polypectomy

* Reference Answer:

* Pic.1: Observe the surroundings, and locate the polyp.

* Pic.2: Excising the polyp.

* Pic.3: Further clearing of the polyp.

* Pic.4: Excision complete, removing the excised polyp.

2CASE 1

Uploaded one by one via GPT4-v website

[colback=gray!10,arc=0mm,auto outer arc]
Prompt: These four photos are from a hysteroscopic polypectomy. Please analyze the procedures in these four images.

GPT-4V Response:
[colback=cyan!10,arc=0mm,auto outer arc]

* response 1:

* Picture 1: Showcases a slightly larger uterine polyp. One can observe the base of the polyp and a certain reddish area of bleeding.

* Picture 2: It seems to illustrate the doctor using a tool to grip or stabilize the polyp, preparing for excision.

* Picture 3: This image displays a close-up view of the polyp with signs of bleeding and potential inflammation on its surface.

* Picture 4: A tool approaches the polyp area, possibly in the process of excision or preparing for it.

* response 2:

* Picture 1: Displays the red tissue inside the uterus, which might be a polyp or another intrauterine tissue with hysteroscopic tools visible on the side.

* Picture 2: The tool seems to be in contact with or manipulating a certain tissue, possibly a step in the excision process.

* Picture 3: This image too appears to show the tool in contact with or manipulating tissue.

* Picture 4: The tool is near the tissue, suggesting that a part of the excision procedure might have been completed.

* response 3:

* Picture 1: Seems to display a polyp inside the uterine cavity. Polyps are growths of the endometrium, usually benign, but sometimes require excision to avoid complications or for further pathological analysis.

* Picture 2: This image shows that the doctor might be using surgical tools like forceps or scissors to excise the polyp.

* Picture 3: The view in this image might be of the uterine cavity after the excision of the polyp, showcasing certain bleeding or erythema, which is normal post-operatively.

* Picture 4: This image too shows the surgical area, looking like the uterine cavity post-polyp excision.

§.§ Tip 9: Providing comparative analysis opportunities, especially with temporal patient data, deepens lesion or condition assessments.

§.§.§ Case 1: CT scan tracking of pulmonary nodule

2CASE 1

Without comparison

With comparison

* Picture modal: CT scan

* Tasks: Pulmonary nodule growth tracking

* Reference Answer: adenocarcinoma

2CASE 1

Singe picture(without comparison) <ref>

[colback=gray!10,arc=0mm,auto outer arc]
Prompt: It took 6 years for the patient's nodule to grow to the size indicated by the arrow in the image. What disease is the patient most likely to have

GPT-4V Response:
[colback=cyan!10,arc=0mm,auto outer arc]

* response 1:

Some benign conditions that can present with slow-growing lung nodules include:

Inflammatory nodules (e.g., granulomatous infections like tuberculosis or fungal infections)

Rheumatoid nodules in patients with rheumatoid arthritis

Scar tissue or old infections

Some benign tumors

* response 2:

Slow-growing nodules in the lungs are most commonly benign. Benign pulmonary nodules can arise from a number of causes, including:

Granulomas: These are small areas of inflammation, often caused by past infections like tuberculosis or fungal infections.

Hamartomas: These are benign growths made up of a mixture of tissue types commonly found in the lungs.

Other benign tumors or cysts.

However, it's also essential to consider the possibility of a slow-growing malignancy, like certain types of lung cancer (e.g., adenocarcinoma or carcinoid tumors).

* response 3:

Given the slow growth rate over 6 years, one of the possibilities is a benign lung nodule. Benign lung nodules can be due to various causes, including:

Hamartoma: The most common type of benign lung tumor.

Granulomas: Caused by inflammation in the lung.

Infectious causes: Such as fungal infections or tuberculosis.

Other benign tumors or cysts.

2CASE 1

With comparison <ref>

[colback=gray!10,arc=0mm,auto outer arc]
Prompt: The patient's nodule has increased in size from left to right over 6 years, what kind of disease is the patient most likely to have

GPT-4V Response:
[colback=cyan!10,arc=0mm,auto outer arc]

* response 1:

* Primary lung cancer: The slow growth might suggest a subtype like adenocarcinoma, which can sometimes grow at a slower rate compared to other forms of lung cancer like small cell carcinoma. However, the speed of growth varies and cannot definitively diagnose the type.

* Infectious causes: Some infections can cause nodules that grow or change over time. Examples include fungal infections such as aspergillosis or coccidioidomycosis.

* Benign tumors: While benign tumors such as hamartomas or fibromas are typically slow-growing, they can still increase in size over time.

* Inflammatory conditions: Conditions such as rheumatoid nodules or granulomatous diseases can also present as lung nodules. The growth pattern might help differentiate between the different possibilities.

* Metastatic cancers: A lung nodule can be a metastasis from a primary tumor elsewhere in the body.

* response 2:

* Lung Cancer: Primary lung cancer is the most concerning possibility for a progressively enlarging nodule in the lung. It could be non-small cell lung cancer (e.g., adenocarcinoma, squamous cell carcinoma) or small cell lung cancer.

* Infectious Causes: Some infections can cause nodules in the lung that can grow over time. This includes bacterial infections like tuberculosis, fungal infections like aspergillosis or histoplasmosis, or parasitic infections like echinococcosis.

* Benign Tumors: Some benign lung tumors, such as hamartomas or fibromas, can also grow, although at a slower rate than malignant ones.

* Inflammatory Diseases: Conditions like granulomatosis with polyangiitis, rheumatoid nodules in the context of rheumatoid arthritis, and sarcoidosis can cause lung nodules.

* Metastases: If the patient has a history of cancer elsewhere in the body, it's possible that the nodule represents metastatic disease.

* response 3:

* Lung Cancer: One of the most concerning possibilities for a slow-growing pulmonary nodule is lung cancer, like adenocarcinoma. The risk is higher if the patient has a history of smoking, exposure to carcinogens, or a family history of lung cancer.

* Infectious Causes: Some infections, such as tuberculosis or fungal infections, can present as nodules that might change in size over time.

* Benign Tumors: Not all growing nodules are malignant. Some benign tumors can also grow over time.

* Inflammatory Diseases: Certain inflammatory conditions, like granulomas or rheumatoid nodules, can also appear as nodules in the lung.

§.§ Tip 10: Directing GPT-4v's focus to the interested areas will facilitate more targeted and relevant output.

§.§.§ Case 1: MRI cerebral scan

2CASE 1

* Picture modal: MRI

* Tasks: Cerebral lesions detection (after ischemic stroke)

* Reference Answer: A lesion is on the right frontal lobe (left part in image)

2CASE 1

Control Group

[colback=gray!10,arc=0mm,auto outer arc]
Prompt: Please examine this head MRI image and determine if it is normal (the information in this image is sufficient to make a judgment).

GPT-4V Response:
[colback=cyan!10,arc=0mm,auto outer arc]

* response 1:

The image displays a T1-weighted magnetic resonance imaging (MRI) scan of the head. It's imperative to note that I can only provide general observations based on this image, which should not be construed as a substitute for a professional medical opinion. From the given image:

Most cerebral structures appear to be within normal parameters.

Midline structures, such as the boundary between the cerebral ventricles and the cerebral hemispheres, seem to be symmetrical.

The dimensions of the third ventricle and lateral ventricles appear to be within the norm.

The image's resolution and contrast might constrain a more intricate analysis.

* response 2:

This represents a T1-weighted axial slice of a cranial MRI. From the scan:

The majority of the cerebral structures can be discerned, including the cerebral cortex, the ventricular system, and the butterfly-shaped white matter region in the brain's center.

At the current resolution, there are no evident neoplasms, hemorrhagic regions, or other manifest pathologies.

The size and morphology of the ventricles also appear to be within standard limits.

* response 3:

The image portrays an axial section of a cranial MRI. Upon initial examination:

The cerebrum's symmetry seems to be devoid of notable anomalies.

The ventricular system also appears unremarkable at first glance.

There should be no areas of abnormal intensity that could suggest the presence of lesions, tumors, or other abnormalities. The brain's gyri and sulci should follow a typical pattern without signs of displacement or compression.

2CASE 1

Directed to the interested area

[colback=gray!10,arc=0mm,auto outer arc]
Prompt: Please examine the upper left portion of this image and determine if it is normal (the information in this image is sufficient to make a judgment).

GPT-4V Response:
[colback=cyan!10,arc=0mm,auto outer arc]

* response 1:
Based on the provided MRI image, a hyperintense region is evident in the upper left quadrant, distinctly contrasting with the surrounding cerebral tissue. This region may signify a potential anomaly or lesion. However, the precise nature and clinical significance necessitate a more comprehensive medical evaluation for determination.

* response 2:
In the MRI image you've presented, there appears to be a hyperintense area in the upper left section, contrasting with the adjacent cerebral tissue. This could indicate the presence of a lesion or abnormality.

* response 3:
On the MRI image you provided, a hyperintense region is discernible in the upper left quadrant. Based on the image, this region seems to deviate from the typical cerebral tissue presentation. From this observation, the region may represent an aberration.

§.§.§ Case 2: PET scan for lung cancer detection

2CASE 2

* Picture modal: PET scan

* Tasks: Cancer detection

* Reference Answer: Lung cancer

2CASE 2

Control Group

[colback=gray!10,arc=0mm,auto outer arc]
Prompt: Please describe this PET scan and tell me what disease is most likely involved in this picture (only answer the most likely one)

GPT-4V Response:
[colback=cyan!10,arc=0mm,auto outer arc]

* response 1:
The scan shows several areas of increased tracer uptake, which are the bright spots in the image. The areas of high uptake of the tracer suggest a higher level of metabolic activity, which is commonly associated with cancer cells that are more active than normal cells.

Given the multiple bright spots of varying sizes throughout the body, this pattern may be indicative of a widespread metastatic disease, where cancer cells from an original tumor have spread to other parts of the body.

* response 2:
The scan shows several areas of high uptake, which are visible as bright spots. These bright spots indicate areas where a large amount of the tracer has accumulated, which often signifies a high level of metabolic activity that is typical of rapidly dividing cells, such as cancer cells. Given the multiple bright spots of varying sizes, which can be seen throughout the body, this pattern could suggest metastatic cancer, where the cancer has spread to multiple locations.

* response 3:
The bright areas in the head, chest, and abdomen could indicate areas of high metabolic activity, which is often associated with cancer, as cancer cells have a higher metabolic rate than normal cells. However, it's essential to consider the clinical context as these findings can also be seen in infection, inflammation, or benign tumors. Given the multiple areas of uptake, if this is indeed a pathological finding, it would suggest metastatic disease.

One of the most common diseases that could cause widespread increased uptake on a PET scan is metastatic cancer, where the primary cancer has spread to multiple locations throughout the body.

2CASE 2

Control Group

[colback=gray!10,arc=0mm,auto outer arc]
Prompt: Focusing on the chest, please describe this PET scan and tell me what disease is most likely involved in this picture (only answer the most likely one)

GPT-4V Response:
[colback=cyan!10,arc=0mm,auto outer arc]

* response 1:
Focusing on the chest area, there are at least two distinct areas of increased tracer uptake. One is located in the region that would correspond to the mediastinum, which is the central part of the thorax. The other is in the region that might correspond to the lung or pleura.

Given the presence of these bright spots in the chest, the most likely disease involved is lung cancer, possibly with mediastinal involvement which could suggest lymph node metastasis. However, other possibilities like lymphoma or infections could also present with increased uptake in these areas, but these are less likely based on the singular bright spots seen in the image. It is important to note that a definitive diagnosis would require correlation with the patient's clinical history, physical examination, and other diagnostic studies. A biopsy would be necessary to confirm the type of cancer or disease.

* response 2:
In the chest area of this scan, there are multiple areas that show increased tracer uptake (highlighted as bright spots). The presence of multiple bright spots can be associated with several conditions, but given that they are scattered throughout the chest, the most likely disease involved here is metastatic cancer. Metastases in the chest can arise from primary cancers located in the chest (like lung cancer) or from cancers that originated in other parts of the body and have spread to the chest.

* response 3:
In this particular PET scan image, there are multiple areas of increased radiotracer uptake (hypermetabolic activity) in the region of the chest. The most prominent hypermetabolic activity is noted centrally in the chest area, which could represent the mediastinum, the area between the lungs that contains the heart, great vessels, trachea, esophagus, and lymph nodes.

Given that there is focal, intense uptake in the chest without clear visualization of the surrounding anatomical structures, the most likely diagnosis in this scenario would be a mediastinal tumor. The most common type of mediastinal tumor that would show up on a PET scan is a lymphoma, especially if the hypermetabolic activity is associated with lymph nodes.

§ TEMPLATE 1

2CASE 1

* Picture modal:

* Tasks:

* Reference Answer:

§ TEMPLATE 2

2CASE 1

[colback=gray!10,arc=0mm,auto outer arc]
Prompt: TBD.

GPT-4V Response:
[colback=cyan!10,arc=0mm,auto outer arc]

* response 1:
The image

* response 2:
The image

* response 3:
The image

§ EXAMPLES OF CITATIONS, FIGURES, TABLES, REFERENCES

[8] [15, 16] and see [17].

The documentation for  may be found at

Of note is the command , which produces citations
appropriate for use in inline text.  For example,

Hasselmo, et al. (1995) investigated…

§.§ Figures

[10]
See Figure <ref>. Here is how you add footnotes. [Sample of the first footnote.]
[11]

[-.5cm]4cm4cm [-.5cm]4cm0cm
Sample figure caption.

§.§ Tables

[12]
See awesome Table <ref>.

Sample table title

Name         Description         Size ($\mu$m)

Dendrite     Input terminal      $\sim$100

Axon         Output terminal     $\sim$10

Soma         Cell body           up to $10^6$

§.§ Lists

* Lorem ipsum dolor sit amet

* consectetur adipiscing elit.

* Aliquam dignissim blandit est, in dictum tortor gravida eget. In ac rutrum magna.

[1]
Gpt-4 technical report, 2023.

[2]
Zhengyuan Yang, Linjie Li, Kevin Lin, Jianfeng Wang, Chung-Ching Lin, Zicheng Liu, and Lijuan Wang.
The dawn of lmms: Preliminary explorations with gpt-4v (ision).
arXiv preprint arXiv:2309.17421, 9, 2023.

[3]
Chaoyi Wu, Jiayu Lei, Qiaoyu Zheng, Weike Zhao, Weixiong Lin, Xiaoman Zhang, Xiao Zhou, Ziheng Zhao, Ya Zhang, Yanfeng Wang, and Weidi Xie.
Can gpt-4v(ision) serve medical applications? case studies on gpt-4v for multimodal medical diagnosis, 2023.

[4]
Zhichao Yang, Zonghai Yao, Mahbuba Tasmin, Parth Vashisht, Won Seok Jang, Beining Wang, Dan Berlowitz, and Hong Yu.
Performance of multimodal gpt-4v on usmle with image: Potential for imaging diagnostic support with explanations.
medRxiv, pages 2023–10, 2023.

[5]
Thomas Buckley, James A. Diao, Adam Rodman, and Arjun K. Manrai.
Accuracy of a vision-language model on challenging medical cases, 2023.

[6]
Tianyu Han, Lisa C Adams, Keno Bressem, Felix Busch, Luisa Huck, Sven Nebelung, and Daniel Truhn.
Comparative analysis of gpt-4vision, gpt-4 and open source llms in clinical diagnostic accuracy: A benchmark against human expertise.
medRxiv, pages 2023–11, 2023.

[7]
Gptv system card.
<https://cdn.openai.com/papers/GPTV_System_Card.pdf>, 2023.

[8]
Debesh Jha, Pia H. Smedsrud, Michael A. Riegler, Pål Halvorsen, Thomas de Lange, Dag Johansen, and Håvard D. Johansen.
Kvasir-seg: A segmented polyp dataset, 2019.

[9]
Salman Maqbool, Aqsa Riaz, Hasan Sajid, and Osman Hasan.
m2caiseg: Semantic segmentation of laparoscopic images using convolutional neural networks.
arXiv preprint arXiv:2008.10134, 2020.

[10]
Shaonan Zhong, Junyang Mo, and Zhantao Liu.
Autopet challenge 2022: Automatic segmentation of whole-body tumor lesion based on deep learning and fdg pet/ct, 2022.

[11]
Jakob Wasserthal, Hanns-Christian Breit, Manfred T. Meyer, Maurice Pradella, Daniel Hinck, Alexander W. Sauter, Tobias Heye, Daniel T. Boll, Joshy Cyriac, Shan Yang, Michael Bach, and Martin Segeroth.
Totalsegmentator: Robust segmentation of 104 anatomic structures in ct images.
Radiology: Artificial Intelligence, 5(5), September 2023.

[12]
Jun Ma, Yao Zhang, Song Gu, Cheng Zhu, Cheng Ge, Yichi Zhang, Xingle An, Congcong Wang, Qiyuan Wang, Xin Liu, Shucheng Cao, Qi Zhang, Shangqing Liu, Yunpeng Wang, Yuhui Li, Jian He, and Xiaoping Yang.
Abdomenct-1k: Is abdominal organ segmentation a solved problem?, 2021.

[13]
Ujjwal Baid and Satyam Ghodasara et al.
The rsna-asnr-miccai brats 2021 benchmark on brain tumor segmentation and radiogenomic classification, 2021.

[14]
Yuanfeng Ji, Haotian Bai, Jie Yang, Chongjian Ge, Ye Zhu, Ruimao Zhang, Zhen Li, Lingyan Zhang, Wanling Ma, Xiang Wan, and Ping Luo.
Amos: A large-scale abdominal multi-organ benchmark for versatile medical image segmentation, 2022.

[1]
Gpt-4 technical report, 2023.

[2]
Zhengyuan Yang, Linjie Li, Kevin Lin, Jianfeng Wang, Chung-Ching Lin, Zicheng Liu, and Lijuan Wang.
The dawn of lmms: Preliminary explorations with gpt-4v (ision).
arXiv preprint arXiv:2309.17421, 9, 2023.

[3]
Chaoyi Wu, Jiayu Lei, Qiaoyu Zheng, Weike Zhao, Weixiong Lin, Xiaoman Zhang, Xiao Zhou, Ziheng Zhao, Ya Zhang, Yanfeng Wang, and Weidi Xie.
Can gpt-4v(ision) serve medical applications? case studies on gpt-4v for multimodal medical diagnosis, 2023.

[4]
Zhichao Yang, Zonghai Yao, Mahbuba Tasmin, Parth Vashisht, Won Seok Jang, Beining Wang, Dan Berlowitz, and Hong Yu.
Performance of multimodal gpt-4v on usmle with image: Potential for imaging diagnostic support with explanations.
medRxiv, pages 2023–10, 2023.

[5]
Thomas Buckley, James A. Diao, Adam Rodman, and Arjun K. Manrai.
Accuracy of a vision-language model on challenging medical cases, 2023.

[6]
Tianyu Han, Lisa C Adams, Keno Bressem, Felix Busch, Luisa Huck, Sven Nebelung, and Daniel Truhn.
Comparative analysis of gpt-4vision, gpt-4 and open source llms in clinical diagnostic accuracy: A benchmark against human expertise.
medRxiv, pages 2023–11, 2023.

[7]
Gptv system card.
<https://cdn.openai.com/papers/GPTV_System_Card.pdf>, 2023.

[8]
Debesh Jha, Pia H. Smedsrud, Michael A. Riegler, Pål Halvorsen, Thomas de Lange, Dag Johansen, and Håvard D. Johansen.
Kvasir-seg: A segmented polyp dataset, 2019.

[9]
Salman Maqbool, Aqsa Riaz, Hasan Sajid, and Osman Hasan.
m2caiseg: Semantic segmentation of laparoscopic images using convolutional neural networks.
arXiv preprint arXiv:2008.10134, 2020.

[10]
Shaonan Zhong, Junyang Mo, and Zhantao Liu.
Autopet challenge 2022: Automatic segmentation of whole-body tumor lesion based on deep learning and fdg pet/ct, 2022.

[11]
Jakob Wasserthal, Hanns-Christian Breit, Manfred T. Meyer, Maurice Pradella, Daniel Hinck, Alexander W. Sauter, Tobias Heye, Daniel T. Boll, Joshy Cyriac, Shan Yang, Michael Bach, and Martin Segeroth.
Totalsegmentator: Robust segmentation of 104 anatomic structures in ct images.
Radiology: Artificial Intelligence, 5(5), September 2023.

[12]
Jun Ma, Yao Zhang, Song Gu, Cheng Zhu, Cheng Ge, Yichi Zhang, Xingle An, Congcong Wang, Qiyuan Wang, Xin Liu, Shucheng Cao, Qi Zhang, Shangqing Liu, Yunpeng Wang, Yuhui Li, Jian He, and Xiaoping Yang.
Abdomenct-1k: Is abdominal organ segmentation a solved problem?, 2021.

[13]
Ujjwal Baid and Satyam Ghodasara et al.
The rsna-asnr-miccai brats 2021 benchmark on brain tumor segmentation and radiogenomic classification, 2021.

[14]
Yuanfeng Ji, Haotian Bai, Jie Yang, Chongjian Ge, Ye Zhu, Ruimao Zhang, Zhen Li, Lingyan Zhang, Wanling Ma, Xiang Wan, and Ping Luo.
Amos: A large-scale abdominal multi-organ benchmark for versatile medical image segmentation, 2022.

is enabled.

[15]
George Kour and Raid Saabne.
Real-time segmentation of on-line handwritten arabic script.
In Frontiers in Handwriting Recognition (ICFHR), 2014 14th
International Conference on, pages 417–422. IEEE, 2014.

[16]
George Kour and Raid Saabne.
Fast classification of handwritten on-line arabic characters.
In Soft Computing and Pattern Recognition (SoCPaR), 2014 6th
International Conference of, pages 312–318. IEEE, 2014.

[17]
Guy Hadash, Einat Kermany, Boaz Carmeli, Ofer Lavi, George Kour, and Alon
Estimate and replace: A novel approach to integrating deep neural
networks with existing applications.
arXiv preprint arXiv:1804.09028, 2018.


# Kingman’s model with random mutation probabilities:
convergence and condensation I

Linglong Yuan111University of Liverpool, Department of Mathematical Sciences,
Peach Street, L69 7ZL, Liverpool, UK.
Email<EMAIL_ADDRESS>222 Xi’an Jiaotong-Liverpool University,
Department of Mathematical Sciences, Ren’ai Road, 215111, Suzhou, P.R.C.
China.

###### Abstract

For a one-locus haploid infinite population with discrete generations, the
celebrated Kingman’s model describes the evolution of fitness distributions
under the competition of selection and mutation, with a constant mutation
probability. This paper generalises Kingman’s model by using i.i.d. random
mutation probabilities, to reflect the influence of a random environment. The
weak convergence of fitness distributions to the globally stable equilibrium
is proved. The condensation occurs when almost surely a positive proportion of
the population travels to and condensates at the largest fitness value. The
condensation may occur when selection is more favoured than mutation. A
criterion for the occurrence of condensation is given.

###### Contents

  1. 1 Motivation and background
  2. 2 Models
  3. 3 Main results
  4. 4 Proofs
  5. A Analysis of $\ln\frac{h(1-b)}{\int x\mathcal{K}_{Q}(dx)}$ in Kingman’s model

Keywords and phrases. Population dynamics, Mutation-Selection balance, House
of Cards, Fitness distribution, Size-biased distribution, Distributional
equation.

AMS 2000 subject classifications. Primary 60F05; secondary 60G10, 92D15, 92D25

##  1 Motivation and background

Various biological forces interact with each other and drive the evolution of
population all together. One important competing pair consists of selection
and mutation. It was as early as 1937 when Haldane [14] put forward the
concept of mutation-selection balance. The mathematical foundation of this
subject was established by Crow and Kimura [7], Ewens [11], and Kingman [20].
For more details on this topic, we refer to Bürger [5, 6].

A simple setting is to consider a one-locus haploid infinite population with
discrete generations under selection and mutation. The locus is assumed to
have infinitely many possible alleles which have continuous effects on a
quantitative type. The continuum-of-alleles models were introduced by Crow and
Kimura [7] and Kimura [17] and are used frequently in quantitative genetics.

Kingman [18] suggested to explain the tendency that most mutations are
deleterious by the assumption of the independence of the gene before and after
mutation. This feature was named “House of Cards”, as the mutation destroys
the biochemical house of cards built up by evolution, by Kingman in [19] where
the most famous one-locus model was proposed. In this model, a population is
characterised by its type distribution, which is a probability measure on
$[0,1]$ and any $x\in[0,1]$ is a type value. In Kingman’s setting, an
individual with a larger type value is fitter, which means more productive. So
the type value can also be named fitness value. Kingman’s model can be seen as
the limit of a finite population model, see [13].

Bürger [4] generalised the selection mechanism by allowing the gene after
mutation to depend on that before and proved convergence in total variation.
The genetic variation of the equilibrium distribution was computed and
discussed. I proposed [22] a more general selection mechanism which can model
general macroscopic epistasis, with the other settings the same as in
Kingman’s model. This model was applied to the modelling of the Lenski
experiment (see [12] for a description of the experiment).

There exist also models on the balance of selection and mutation in the
setting of continuous generations. Bürger [3] provided an exact mathematical
analysis of Kimura’s continuum-of-alleles model, focusing on the equilibrium
genetic variation. Steinsaltz et al [21] proposed a multi-loci model using a
differential equation to study the ageing effect. Later on the recombination
was incorporated to the model [10]. Betz et al’s model [2] generalised a
continuous-time version of Kingman’s model and other models arising from
physics.

However to the author’s best knowledge, Kingman’s model has never been
generalised to a random version. In this paper we will assume that the
mutation probabilities of all generations form an i.i.d. sequence.
Biologically, we think of a stable random environment such that the mutation
probabilities vary on time but independently sampled from the same
distribution.

In Kingman’s model, condensation occurs if a certain positive proportion of
the population travels to and condensates at the largest fitness value. This
is due to the dominance of selection over mutation. In the random model
proposed in this paper, we also consider the convergence of (random) fitness
distributions to the equilibrium and the condensation phenomenon. Moreover,
Kingman’s model has been revisited recently in terms of the travelling wave of
mass to the largest fitness value [8]. The random model provides another
example for consideration in this direction.

##  2 Models

###  2.1 Kingman’s model with time-varying mutation probabilities

Consider a haploid population of infinite size and discrete generations under
the competition of selection and mutation. We use a sequence of probability
measures $(P_{n})=(P_{n})_{n\geq 0}$ on $[0,1]$ to describe the distribution
of fitness values in the $n$th generation. We can assume, more generally, that
the probability measures are supported on a finite interval, not necessarily
$[0,1]$. But since only fitness ratios will be relevant (see [19] or [22] for
a more explicit explanation), we adopt the setting of $[0,1]$, which was used
by Kingman [19], and which is equivalent to general finite supports.

Individuals in the $n$th generation are children of the $(n-1)$th generation.
First of all, the fitness distribution of children is initially $P_{n-1}$ (an
exact copy from parents). Then selection takes effect, such that the fitness
distribution is updated from $P_{n-1}$ to the size-biased distribution

$\frac{xP_{n-1}(dx)}{\int yP_{n-1}(dy)}.$

Here we use $\int$ to denote $\int_{0}^{1}.$ Basically the new population is
re-sampled from the existing population by using their fitness as a selective
criterion. Next, each individual mutates independent with the same mutation
probability which we denote by $b_{n}$ taking values in $[0,1).$ Each mutant
has the fitness value sampled independently from a common mutant distribution,
that we denote by $Q$, a probability measure on $[0,1]$. Then the resulting
distribution is the distribution of the $n$th generation

$P_{n}(dx)=(1-b_{n})\frac{xP_{n-1}(dx)}{\int yP_{n-1}(dy)}+b_{n}Q(dx).$ (1)

The fact that we exclude the case that $b_{n}$ equals $1$ is because in this
situation we have $P_{n}=Q$ which loses accumulated evolutionary changes. This
is not interesting neither biologically nor mathematically.

Expanding (1), we can also obtain

$P_{n}(dx)=\left(\prod_{l=0}^{n-1}\frac{1-b_{l+1}}{\int
yP_{l}(dy)}\right)x^{n}P_{0}(dx)+\sum_{j=1}^{n}\left(\prod_{l=j}^{n-1}\frac{1-b_{l+1}}{\int
yP_{l}(dy)}\right)b_{j}m_{n-j}Q^{n-j}(dx)$ (2)

where

$Q^{k}(dx):=\frac{x^{k}Q(dx)}{\int y^{k}Q(dy)},\quad m_{k}:=\int
x^{k}Q(dx),\quad\forall k\geq 0.$

In particular if $Q=\delta_{0},$ the Dirac measure on $\\{0\\}$,
$Q^{k}=\delta_{0}$ for any $k\geq 0$.

When all the $b_{n}$’s are equal to the same number $b\in[0,1),$ this is the
model introduced by Kingman [19]. In the general setting we allow the mutation
probabilities to be different. We call it Kingman’s model with time-varying
mutation probabilities or the general model for short.

We introduce a few more notations. Let $M$ be the space of (nonnegative) Borel
measures on $[0,1]$ and $M_{1}$ the subspace of $M$ consisting of probability
measures. Let $M,M_{1}$ be endowed with the topology of weak convergence. We
use $\stackrel{{\scriptstyle d}}{{\longrightarrow}}$ to denote weak
convergence. We say a sequence of measures $(u_{n})$ converges in total
variation to a measure $u$, denoted by $u_{n}\stackrel{{\scriptstyle
TV}}{{\longrightarrow}}u,$ if the total variation, $\sup_{B}|u_{n}(B)-u(B)|$
where the supremum is taken over all Borel sets, converges to $0$.

For any $u\in M_{1}$, define

$S_{u}:=\sup\\{x:u[x,1]>0\\}.$ (3)

So $S_{u}$ is interpreted as the largest fitness value in a population of
distribution $u.$ Define $h:=S_{P_{0}}$. It is not difficult to see that
$S_{P_{n}}=\max\\{S_{P_{0}},S_{Q}\\}$ if the equality holds for $n-1$ or
$0<b_{n}<1$. Since we are interested in asymptotics, it is thus without loss
of generality to assume that $h\geq S_{Q}$. Therefore $S_{Q}\leq h\leq 1.$

Note that the general model has parameters $(b_{n})_{n\geq 1},Q,P_{0},h$.
Kingman’s model shares the same parameters, but with $b_{n}$’s all equal to
$b.$ We call $(P_{n})$ the forward sequence or just the sequence. Although $h$
is determined by $P_{0}$, we still consider $h$ as a parameter as it will be
clear later that for Kingman’s model and the random model considered in this
paper, the limit of $(P_{n})$ depends on $P_{0}$ only through $h$. This is the
so-called global stability.

###  2.2 Convergence and condensation in Kingman’s model

Kingman [19] proved the convergence of $(P_{n})$ when all mutation
probabilities are equal, i.e., $b_{n}=b,\forall n\geq 1$.

######  Theorem 1 (Kingman’s Theorem, [19]).

1\. If $\int\frac{Q(dx)}{1-x/h}\geq b^{-1},$ then $(P_{n})$ converges in total
variation to

$\mathcal{K}(dx)=\frac{b\theta_{b}Q(dx)}{\theta_{b}-(1-b)x},$

with $\theta_{b}$, as a function of $b$, being the unique solution of

$\int\frac{b\theta_{b}Q(dx)}{\theta_{b}-(1-b)x}=1.$ (4)

2\. If $\int\frac{Q(dx)}{1-x/h}<b^{-1}$, then $(P_{n})$ converges weakly to

$\mathcal{K}(dx)=\frac{bQ(dx)}{1-x/h}+\Big{(}1-\int\frac{bQ(dy)}{1-y/h}\Big{)}\delta_{h}(dx).$

Note that $\mathcal{K}$ is uniquely determined by $b,Q,h$, but not the choice
of $P_{0}$. In this sense $\mathcal{K}$ is a globally stable equilibrium. For
simplicity, for any measure, say $\mu,$ its mass on a point $x$ is denoted by
$\mu(x)$ instead of $\mu(\\{x\\}).$ Then we say there is condensation at $h$
in Kingman’s model if $Q(h)=0$ but $\mathcal{K}(h)>0$. We call
$\mathcal{K}(h)$ the condensate size if $Q(h)=0$. In the case 1 above, there
is no condensation. The condition $\int\frac{Q(dx)}{1-x/h}\geq b^{-1}$ is
satisfied only if $b$ is big and/or $Q$ is fit (i.e., having more mass on
larger values). It means mutation is stronger against selection, so that the
limit does not depend on $P_{0}$ at all.

In the case 2, the condition $\int\frac{Q(dx)}{1-x/h}<b^{-1}$ implies
$Q(h)=0$, but we have that $\mathcal{K}(h)>0$. So there is condensation.
Contrarily to the first case, selection is more favoured so that the limit
depends on $P_{0}$ through $h$. If $P_{0}(h)=0$ (implying $S_{P_{n}}=h$ and
$P_{n}(h)=0$ for any $n$), a certain amount of mass
$\mathcal{K}(h)=\Big{(}1-\int\frac{bQ(dy)}{1-y/h}\Big{)}$ travels to the
largest fitness value $h$, by the force of selection.

Next we introduce the random model, which is the main object of study in this
paper.

###  2.3 Kingman’s model with random mutation probabilities

Let $(\beta_{n})_{n\geq 0}$ be an i.i.d. sequence of random variables in the
common probability space $(\Omega,\mathcal{F},\mathbb{P})$, taking values in
$[0,1)$ with common distribution $\mathcal{L}\in M_{1}$ supported on $[0,1)$.
The Kingman’s model with random mutation probabilities or simply the random
model is defined by the following dynamical system:

$P_{n}(dx)=(1-\beta_{n})\frac{xP_{n-1}(dx)}{\int
yP_{n-1}(dy)}+\beta_{n}Q(dx),\quad n\geq 1.$ (5)

The random model has parameters $(\beta_{n}),Q,P_{0},h$. It is a randomisation
of Kingman’s model, as we can set $\beta_{n}$’s to equal $b$ with probability
1.

We are interested in the convergence of $(P_{n})$ to the equilibrium and the
phenomenon of condensation. Since we are dealing with random probability
measures, i.e., random elements of $M_{1},$ let us recall the definition of
weak convergence in this context. Random (probability) measures $(\mu_{n})$
supported on $[0,1]$ converge weakly to a limit $\mu$ if and only if for any
continuous function $f$ on $[0,1]$ we have

$\int f(x)\mu_{n}(dx)\stackrel{{\scriptstyle d}}{{\longrightarrow}}\int
f(x)\mu(dx).$

We refer to [16] for a reference on random measures. The definition of weak
convergence for random measures stated in the follow-up paper [23] is
incorrect. But it does not affect anything there as the weak convergence
results are all proved in this paper.

As the sequence $(P_{n})$ is completely determined by $(\beta_{n}),Q,P_{0}$
and $h$, the only randomness arises from $(\beta_{n})$. In comparison to the
terminology in statistical physics, the weak limit of $(P_{n})$ is an annealed
limit, which is obtained given the law of $(\beta_{n})$. A quenched limit,
which is obtained by conditioning on $(\beta_{n})$, does not exist unless
$P_{0}=Q=\delta_{0}$. A simple reason for nonexistence is that $P_{n}$
contains $\beta_{n}Q$ which fluctuates persistently as $(\beta_{n})$ is
i.i.d.. However in Section 4.3 we will see that it is possible to obtain a
quenched limit if the evolution is seen backwards.

For the particular case that $Q=\delta_{0}$, we have

$P_{n}(dx)=(1-\beta_{n})\frac{x^{n}P_{0}(dx)}{\int
y^{n}P_{0}(dy)}+\beta_{n}\delta_{0}(dx).$

From this, it is easily deduced that the sequence $(P_{n})$ converges weakly
to the random element $(1-\beta)\delta_{h}+\beta\delta_{0}$, where $\beta$ is
a random variable with law $\mathcal{L}$, the common law of the $\beta_{n}$’s.
So we assume from now on $Q\neq\delta_{0}.$

##  3 Main results

###  3.1 Weak convergence

Recall that the sequence $(P_{n})$ in the random model has parameters
$(\beta_{n}),Q,P_{0}$ and $h$, with $h=S_{P_{0}}.$ Then $(P_{n})$ converges
weakly to a globally stable equilibrium, in the sense that the limit depends
on $P_{0}$ only through $h$. Recall $\beta$ is a random variable with law
$\mathcal{L}$, the common law of $\beta_{n}$’s.

######  Theorem 2.

For the random model (5), the sequence $(P_{n})$ converges weakly to a random
probability measure, denoted by $\mathcal{I}$, whose distribution depends on
$\mathcal{L},Q,h$ but not on the choice of $P_{0}$.

######  Remark 1.

In ([23], P.872), it is written that the distribution of $\mathcal{I}$ depends
on $\beta,Q,h$. The statement is true in the sense that the distribution of
$\mathcal{I}$ depends on $\beta$ via its distribution. Here we make it more
clear by replacing $\beta$ by $\mathcal{L}$.

######  Remark 2.

If we start with $P_{0}=\delta_{h}$ (recall that $h\in[S_{Q},1]$), then all
$P_{n}$’s are supported on $[0,S_{Q}]\cup\\{h\\}$, which entails that the
limit $\mathcal{I}$ is supported on the same set $[0,S_{Q}]\cup\\{h\\}$.
Moreover we have either $\mathcal{I}(h)>0$ a.s. or $\mathcal{I}(h)=0$ a.s. (a
justification is provided in Remark 11 in Section 4.4). In the latter case,
$\mathcal{I}$ is supported only on $[0,S_{Q}]$ and (the distribution of)
$\mathcal{I}$ does not depend on $h$ (see Theorem 3). Therefore, although we
say $h$ is a parameter of $\mathcal{I}$ but it should be understood in the
sense that $\mathcal{I}$ is weak limit of $(P_{n})$ with $h=S_{P_{0}}.$

The limit $\mathcal{I}$ is introduced in Section 4.3. But the proof of weak
convergence is deferred to a later stage, as it uses other main results such
as the condensation criterion for the random model.

###  3.2 Condensation criterion

The fact that either $\mathcal{I}(h)>0$ a.s. or $\mathcal{I}(h)=0$ a.s. allows
us to give the precise definition of condensation in line with that for
Kingman’s model, as follows:

######  Definition 1.

For the random model, we say there is condensation at the largest fitness
value $h$ if $Q$ assigns zero mass at $h$ (i.e., $Q(h)=0$) but the limiting
measure $\mathcal{I}$ assigns positive mass at it (i.e., $\mathcal{I}(h)>0$,
a.s.).

Next we give the condensation criterion. If $h=S_{Q}$, we write
$\mathcal{I}_{Q}$ for $\mathcal{I}$ and $\mathcal{K}_{Q}$ for $\mathcal{K}$.

######  Theorem 3 (Condensation criterion).

If there is no condensation at $h$, then $\mathcal{I}\stackrel{{\scriptstyle
d}}{{=}}\mathcal{I}_{Q}$. The condensation criterion for $\mathcal{I}$ at $h$
is as follows:

  1. 1.

If $h=S_{Q}$, then there is no condensation at $h$ if

$\mathbb{E}\left[\ln\frac{S_{Q}(1-\beta)}{\int
y\mathcal{I}_{Q}(dy)}\right]<0.$ (6)

  2. 2.

If $h>S_{Q}$, then there is no condensation at $h$ if and only if

$\mathbb{E}\left[\ln\frac{h(1-\beta)}{\int y\mathcal{I}_{Q}(dy)}\right]\leq
0.$ (7)

Here $\mathbb{E}\left[\ln\frac{1-\beta}{\int y\mathcal{I}_{Q}(dy)}\right]$ is
well defined, and takes values in $[-\infty,-\ln\int yQ(dy)]$, and depends
only on the marginal distributions of $\beta$ and $\mathcal{I}_{Q}.$

######  Remark 3.

In fact, if there is no condensation at $h$, then
$\mathcal{I},\mathcal{I}_{Q}$ are the same random probability measure, based
on the definition of $\mathcal{I}$ introduced at the end of Section 4.3. But
since here we do not have the definition yet, we write a weaker version
$\mathcal{I}\stackrel{{\scriptstyle d}}{{=}}\mathcal{I}_{Q}.$

By Remark 12 in Section 4.4, we can only have
$\mathbb{E}\left[\ln\frac{S_{Q}(1-\beta)}{\int
y\mathcal{I}_{Q}(dy)}\right]\leq 0$. About the occurrence of condensation in
the case where $h=S_{Q}$, the fact that we cannot say anything when
$\mathbb{E}\left[\ln\frac{S_{Q}(1-\beta)}{\int y\mathcal{I}_{Q}(dy)}\right]=0$
can be better understood in Kingman’s model, which is a special random model.
In this model, $\mathbb{E}\left[\ln\frac{S_{Q}(1-\beta)}{\int
y\mathcal{I}_{Q}(dy)}\right]=0$ becomes

$\ln\frac{S_{Q}(1-b)}{\int y\mathcal{K}_{Q}(dy)}=0.$

By some simple computations using Theorem 1, the above display is equivalent
to

$\int\frac{Q(dx)}{1-x/S_{Q}}\leq b^{-1}.$

But it covers cases with and without condensation. For full details please see
Appendix A, where the case $h>S_{Q}$ is also analysed.

We give some intuition why Theorem 3 holds true. Consider the unnormalised
variant of the dynamical system that is given by

$\overline{P}_{n}(dx)=(1-\beta_{n})x\overline{P}_{n-1}(dx)+\beta_{n}\left(\int
y\overline{P}_{n-1}(dy)\right)Q(dx)$ (8)

with $\overline{P}_{0}=P_{0}.$ By induction, it can be shown that

$\overline{P}_{n}=P_{n}\prod_{i=0}^{n-1}\int xP_{i}(dx),\forall n\geq 0.$ (9)

We can roughly think of the growth of $\overline{P}_{n}$ as contributed by two
parts, the initial $P_{0}$ and the subsequently arriving $Q$’s. If the initial
distribution is supported on $[0,S_{Q}]$, by Theorem 2, $P_{i}$ converges
weakly to $\mathcal{I}_{Q}$ as $i\to\infty.$ Then the part of
$\overline{P}_{n}$ contributed by the $Q$’s grows at rate
$\operatorname{gr}(Q):=\mathbb{E}[\ln\int x\mathcal{I}_{Q}]$ (see (9)). In
comparison the largest fitness value $h$ in $P_{0}$ can be assigned the growth
rate $\operatorname{gr}(h):=\mathbb{E}[\ln h(1-\beta)]$ (due to the term
$(1-\beta_{n})x\overline{P}_{n-1}(dx)$ in (8)). Then it is clear that the
occurrence of condensation is determined by the comparison of
$\operatorname{gr}(h)$ and $\operatorname{gr}(Q)$. However it is subtle when
$\operatorname{gr}(h)=\operatorname{gr}(Q)$: no condensation if $h>S_{Q}$ and
it is undetermined if $h=S_{Q}$.

In the follow-up paper [23], we provide a matrix representation for
$\mathcal{I}_{Q}$, so the condensation criterion can be written neatly
(Corollary 2, P.877). Moreover, using matrix analysis, we can compare the
fitness of equilibria from different models (Section 3.3-(3), P.878–879). The
challenging problem of finding a necessary and sufficient condition for the
occurrence of condensation in the case $h=S_{Q}$ has not been dealt with
anywhere and still remains open.

###  3.3 Invariant measure

We introduce the notion of invariant measure, which includes the limit
$\mathcal{I}.$ We will heavily use the invariant measures in the proofs.

######  Definition 2 (Invariant measure).

A random probability measure $\nu$ is invariant if it is supported on $[0,1]$
and satisfies

$\nu(dx)\stackrel{{\scriptstyle d}}{{=}}(1-\beta)\frac{x\nu(dx)}{\int
y\nu(dy)}+\beta Q(dx)$ (10)

where $\beta\text{ is independent of }\nu$.

Clearly $\mathcal{I}$ is an invariant measure, since it is the weak limit of
$(P_{n})$ defined by (5).

######  Theorem 4 (Compoundness of invariant measures).

For any invariant measure $\nu$, there exists a regular conditional
distribution of $\nu$ on $S_{\nu}$. Moreover, conditional on $S_{\nu}$,

$(\nu|S_{\nu})\stackrel{{\scriptstyle d}}{{=}}\mathcal{I},\quad\text{ almost
surely},$

where $\mathcal{I}$ is the random probability measure introduced in Theorem 2
with parameters $\mathcal{L},Q,h=S_{\nu}$ and satisfies
$\mathbb{P}(S_{\mathcal{I}}=S_{\nu}|S_{\nu})=1$, a.s..

######  Remark 4.

Remark 2 says that if there is no condensation at $h$, then $\mathcal{I}$ is
supported on $[0,S_{Q}]$. Since $\mathcal{I}$ is an invariant measure, the
above theorem entails that $\mathcal{I}\stackrel{{\scriptstyle
d}}{{=}}\mathcal{I}_{Q}.$ This assertion has been stated in Theorem 3.

Using the notion of invariant measures, we can solve a distributional equation
in the following example. For a survey on distributional equations, we refer
to Aldous and Bandyopadhyay [1].

Example 1. Consider a particular case: $Q$ is supported only on $\\{c\\}$ for
some $c\in(0,1)$, and $h\in(c,1)$. Let $\nu$ be an invariant measure supported
on $\\{c\\}\cup\\{h\\}$. Then $\nu$ can be written as
$\nu=X\delta_{c}+(1-X)\delta_{h}$ where $X$ is a random variable taking values
in $[0,1]$, and satisfies

$X\delta_{c}+(1-X)\delta_{h}\stackrel{{\scriptstyle
d}}{{=}}(1-\beta)\frac{cX\delta_{c}+h(1-X)\delta_{h}}{cX+h(1-X)}+\beta\delta_{c},$

where $\beta$ is independent of $X.$ The above display is equivalent to

$X\stackrel{{\scriptstyle d}}{{=}}\frac{c+(h\beta-c)(1-X)}{c+(h-c)(1-X)}.$

We are interested in a necessary and sufficient condition for the above
equation to have a solution $X$ with $0\leq X<1$ a.s. (i.e., $\nu(h)>0$ a.s.).
By Theorem 4, it is equivalently saying that there is condensation at $h$. By
Theorem 3, the necessary and sufficient condition is simply
$\mathbb{E}[\ln(h(1-\beta)/c)]>0.$ Moreover as such $\nu$ is unique (in
distribution), the solution $X$ is also unique (in distribution).

The paper is organised as follows. Section 4.1 and 4.2 provide necessary
preparations. Section 4.3 and 4.4 analyse the finite backward sequence, which
is the main tool used in this paper. Section 4.5 proves Theorem 3. Section 4.6
analyses the invariant measures, and the results obtained there will be used
in Section 4.7 to prove the weak convergence in Theorem 2. Section 4.8 is
dedicated to the proof of Theorem 4.

##  4 Proofs

###  4.1 Relations between measures

We introduce some notations to describe relations between measures.

1). For measures $u,v\in M$, we say $u$ is a component of $v$ on $[0,a]$
(resp. $[0,a)$), denoted by $u\leq_{a}v$ (resp. $\leq_{a-}$), if

$u(A)\leq v(A),\quad\text{ for any measurable set
}A\subset[0,a]\,\,(\text{resp. }[0,a)).$

For random measures $\mu,\nu\in M$, we write $\mu\leq_{a}^{d}\nu$ if there
exists a coupling $(\mu^{\prime},\nu^{\prime})$ with
$\mu^{\prime},\nu^{\prime}\in M$ such that

$\mu^{\prime}\leq_{a}\nu^{\prime}\text{ a.s. and
}\,\,\mu^{\prime}\stackrel{{\scriptstyle
d}}{{=}}\mu,\,\,\nu^{\prime}\stackrel{{\scriptstyle d}}{{=}}\nu.$ (11)

The relation $\mu\leq_{a-}^{d}\nu$ is defined in a similar way.

2). For measures $(u_{n})$ and $u$ in $M$, we introduce a notation

$u_{n}\leq_{a}\stackrel{{\scriptstyle TV}}{{\longrightarrow}}u$

which means that $u_{n}\leq_{a}u_{n+1}$ for any $n$, and $u_{n}$ converges in
total variation to $u.$ We define similarly $\leq_{a-}\stackrel{{\scriptstyle
TV}}{{\longrightarrow}}.$

3). For real-valued random variables $\xi,\eta,$ we write the well known
stochastic ordering $\xi\preceq\eta$, which holds if

$\mathbb{P}(\xi\leq x)\geq\mathbb{P}(\eta\leq
x),\quad\forall\,x\in\mathbb{R}.$

4). For any $u\in M_{1},$ let the distribution function of $u$ be

$D_{u}(x):=u([0,x]),\quad\forall x\in[0,1].$

For any $u,v\in M_{1},$ we use the same notation $\preceq$ of stochastic
ordering and write $u\preceq v$ if $D_{u}(x)\geq D_{v}(x)$ for any
$x\in[0,1]$. This definition is natural, as $\xi\preceq\eta$ is equivalent to
$u\preceq v$, if $u$ is the distribution of $\xi$ and $v$ is the distribution
of $\eta.$

######  Remark 5.

We make a comment between $\leq_{a-}$ and $\preceq.$ For two probability
measures $u,v\in M_{1}$, assume that $S_{u}=S_{v}=a$, then $u\leq_{a-}v$
implies that $v\preceq u.$ But the converse is not true.

######  Remark 6.

If we use notations $\leq_{a},\leq_{a-},\leq_{a}\stackrel{{\scriptstyle
TV}}{{\longrightarrow}},\leq_{a-}\stackrel{{\scriptstyle
TV}}{{\longrightarrow}},\preceq$ to describe the relations between random
measures, it should be understood that they hold in the almost sure sense, or
even pointwise sense (i.e., for every $\omega\in\Omega$) if possible.

Similarly if we use $\leq,<,\geq,>,=,\neq$ to compare random variables, it
should be understood in the almost sure sense, or pointwise sense.

###  4.2 Three sequences

To study the asymptotic behaviour of $(P_{n})$, we also introduce
$(P_{n}^{\prime}),(P_{n}^{\prime\prime})$ so that the three forward sequences
correspond respectively to

$((\beta_{n}),Q,P_{0},h),\quad((\beta_{n}),Q^{\prime},P_{0}^{\prime},h^{\prime}),\quad((\beta_{n}),Q^{\prime\prime},P_{0}^{\prime\prime},h^{\prime\prime}).$

The parameters of $(P_{n}^{\prime})$ and $(P_{n}^{\prime\prime})$ will be
specified when they are used. The two sequences will converge weakly when they
are used, and $(P_{n})$ is compared to them or one of them to show that
$(P_{n})$ also converges weakly. The first place where this technique is used
is in Section 4.6.

Using (2), we write

$P_{n}(dx)=\mathcal{M}_{n}(dx)+\mathcal{W}_{n}(dx)$ (12)

with

$\mathcal{M}_{n}(dx)=\Big{(}\prod_{l=0}^{n-1}\frac{1-\beta_{l+1}}{\int
yP_{l}(dy)}\Big{)}x^{n}P_{0}(dx)$

and

$\mathcal{W}_{n}(dx)=\sum_{j=1}^{n}\Big{(}\prod_{l=j}^{n-1}\frac{1-\beta_{l+1}}{\int
yP_{l}(dy)}\Big{)}b_{j}m_{n-j}Q^{n-j}(dx).$

Therefore $\mathcal{M}_{n}$ is the contribution to $P_{n}$ made by $P_{0}$,
while $\mathcal{W}_{n}$ is the contribution by the $Q$’s.

Similarly we introduce

$P_{n}^{\prime}(dx)=\mathcal{M}_{n}^{\prime}(dx)+\mathcal{W}_{n}^{\prime}(dx)$
(13)
$P_{n}^{\prime\prime}(dx)=\mathcal{M}_{n}^{\prime\prime}(dx)+\mathcal{W}_{n}^{\prime\prime}(dx)$
(14)

with
$\mathcal{M}_{n}^{\prime},\mathcal{W}_{n}^{\prime},\mathcal{M}_{n}^{\prime\prime},\mathcal{W}_{n}^{\prime\prime}$
defined correspondingly.

###  4.3 Introducing the finite backward sequences

####  4.3.1 The general model.

We introduce the finite backward sequence $(P_{j}^{n})=(P_{j}^{n})_{0\leq
j\leq n}$ for the general model which has parameters $n,(b_{j})_{1\leq j\leq
n},Q,P_{n}^{n},h$ with $S_{P_{n}^{n}}=h$:

$P_{j}^{n}(dx)=(1-b_{j+1})\frac{xP_{j+1}^{n}(dx)}{\int
yP_{j+1}^{n}(dy)}+b_{j+1}Q(dx),\quad\forall\,0\leq j\leq n-1.$ (15)

Here $h,Q$ are from the general model and $P_{n}^{n}$ can be any measure in
$M_{1}$ satisfying $S_{P_{n}^{n}}=h$. The $(b_{j})_{1\leq j\leq n}$ are the
first $n$ mutation probabilities in the general model. Here we use the index
$j$ to indicate that we are dealing with a finite backward sequence.

The sequence is backward in the sense that we use $b_{n}$ to generate
$P_{n-1}^{n}$ from $P_{n}^{n}$, and use $b_{n-1}$ to generate $P_{n-2}^{n}$
from $P_{n-1}^{n}$, etc. The $(b_{j})$ are used backwards and the
$(P_{j}^{n})$ are generated backwards. The advantage to take a backward
approach is that $(P_{j}^{n})$ converges as $n$ tends to infinity, in contrast
to the forward sequence.

######  Lemma 1.

In the general model, for the finite backward sequence with
$P_{n}^{n}=\delta_{h},$ $P_{j}^{n}$ converges in total variation to a limit,
denoted by $\mathcal{G}_{j}=\mathcal{G}_{j,h}$, as $n$ goes to infinity with
$j$ fixed, such that

$\mathcal{G}_{j-1}(dx)=(1-b_{j})\frac{x\mathcal{G}_{j}(dx)}{\int
y\mathcal{G}_{j}(dy)}+b_{j}Q(dx),\quad j\geq 1.$ (16)

As a consequence, $\mathcal{G}_{0}:[0,1)^{\infty}\to M_{1}$ is a measurable
function, with $\mathcal{G}_{j}=\mathcal{G}_{0}(b_{j+1},b_{j+2,\cdots})$
supported on $[0,S_{Q}]\cup\\{h\\}$ for any $j\geq 0.$

######  Remark 7.

We write $\mathcal{G}_{j,h}$ when $h$ has to be specified for clarity.
Otherwise we write $\mathcal{G}_{j}$. This logic applies to other terms which
will appear later.

######  Remark 8.

Note that, by (16), either $\mathcal{G}_{j}(h)$’s are all zero or all strictly
positive.

###### Proof.

We prove a stronger version below

$\text{ For any }j,\quad P_{j}^{n}\leq_{h-}\stackrel{{\scriptstyle
TV}}{{\longrightarrow}}\mathcal{G}_{j},\text{ as }n\to\infty.$ (17)

It suffices to show that

$P_{j}^{n}\leq_{h-}P_{j}^{n+1},$ (18)

as $P_{j}^{n}$’s are all supported on $[0,S_{Q}]\cup\\{h\\}$.

First of all, $P_{n}^{n}=\delta_{h}\leq_{h-}P_{n}^{n+1}$. Assume for some
$1\leq j\leq n$, we have $P_{j}^{n}\leq_{h-}P_{j}^{n+1}$. By definition

$P_{j-1}^{n}(dx)=(1-b_{j})\frac{xP_{j}^{n}(dx)}{\int
yP_{j}^{n}(dy)}+b_{j}Q(dx),\,\,P_{j-1}^{n+1}(dx)=(1-b_{j})\frac{xP_{j}^{n+1}(dx)}{\int
yP_{j}^{n+1}(dy)}+b_{j}Q(dx).$ (19)

Since $P_{j}^{n}\leq_{h-}P_{j}^{n+1}$ (hence $P_{j}^{n+1}\preceq P_{j}^{n}$,
see Remark 5), we have

$\int yP_{j}^{n+1}(dy)\leq\int yP_{j}^{n}(dy)$

and thus

$\frac{x}{\int yP_{j}^{n}(dy)}\leq\frac{x}{\int yP_{j}^{n+1}(dy)},\quad\forall
x\in[0,1].$

Together with $P_{j}^{n}\leq_{h-}P_{j}^{n+1}$ and (19), we get
$P_{j-1}^{n}\leq_{h-}P_{j-1}^{n+1}$. The induction shows that

$P_{j}^{n}\leq_{h-}P_{j}^{n+1},\text{ for any }0\leq j\leq n,n\geq 0.$ (20)

This completes the proof.∎∎

The monotonicity analysis in the above proof will be used many times in this
paper, as it applies to both backward and forward sequences. An immediate
application is the following: we can compare $(\mathcal{G}_{j})$ and
$(\mathcal{G}_{j}^{\prime})=(\mathcal{G}_{j,h^{\prime}})$ for $S_{Q}\leq
h<h^{\prime}\leq 1$ with the same $(b_{j}),Q$.

######  Corollary 1.

Let $(\mathcal{G}_{j})$ and $(\mathcal{G}_{j}^{\prime})$ be the above
sequences. Then we have

$\mathcal{G}_{j}^{\prime}\leq_{h-}\mathcal{G}_{j},\quad\mathcal{G}_{j}(h)\leq\mathcal{G}_{j}^{\prime}(h^{\prime}),\quad\forall
j\geq 0.$ (21)

Moreover we have the exact equalities in the above display for any
$h\in[S_{Q},h^{\prime}]$ if and only if
$\mathcal{G}_{0}^{\prime}(h^{\prime})=0$. In this case $(\mathcal{G}_{j})$ and
$(\mathcal{G}_{j}^{\prime})$ are all supported on $[0,S_{Q}]$, and both equal
to $(\mathcal{G}_{j,S_{Q}})$.

###### Proof.

Let $(P_{j}^{n})$ be the sequence in Lemma 1. Let $(P_{j,h^{\prime}}^{n})$ be
the variant of $(P_{j}^{n})$ with $P_{n}^{n}=\delta_{h^{\prime}}.$ By
following the same monotonicity analysis for proving (18), we obtain

$P_{j,h^{\prime}}^{n}\leq_{h-}P_{j}^{n},\quad P_{j}^{n}(h)\leq
P_{j,h^{\prime}}^{n}(h^{\prime}),\quad\forall 0\leq j\leq n.$

By Lemma 1, $P_{j,h^{\prime}}^{n}\stackrel{{\scriptstyle
TV}}{{\longrightarrow}}\mathcal{G}_{j}^{\prime}$ and
$P_{j}^{n}\stackrel{{\scriptstyle TV}}{{\longrightarrow}}\mathcal{G}_{j},$ as
$n\to\infty.$ Then we obtain (21).

Now let us prove the if-and-only-if statement. If
$\mathcal{G}_{0}^{\prime}(h^{\prime})=0$, then by (21),
$\mathcal{G}_{0}(h)=0$. Using Remark 8,
$\mathcal{G}_{j}^{\prime}(h^{\prime})=0,\mathcal{G}_{j}(h)=0$ for any $j$, and
so (21) holds with equalities. For the other direction, if
$\mathcal{G}_{0}^{\prime}(h^{\prime})>0$, then again by Remark 8,
$\mathcal{G}_{j}^{\prime}(h^{\prime})>0$ for any $j$. Using (21), it holds
that

$\int y\mathcal{G}_{j}^{\prime}(dy)>\int y\mathcal{G}_{j}(dy),\quad\forall j.$
(22)

Similar to (19)

$\mathcal{G}_{j-1}(dx)=(1-b_{j})\frac{x\mathcal{G}_{j}(dx)}{\int
y\mathcal{G}_{j}(dy)}+b_{j}Q(dx),\,\,\mathcal{G}^{\prime}_{j-1}(dx)=(1-b_{j})\frac{x\mathcal{G}^{\prime}_{j}(dx)}{\int
y\mathcal{G}^{\prime}_{j}(dy)}+b_{j}Q(dx).$

As (22) entails that $\frac{1-b_{j}}{\int
y\mathcal{G}_{j}(dy)}>\frac{1-b_{j}}{\int y\mathcal{G}^{\prime}_{j}(dy)}$, and
using again (21), we obtain
$\mathcal{G}^{\prime}_{j-1}\leq_{h-}\mathcal{G}_{j-1}$ but they are not equal
on $[0,h)$. Since they are probability measures, we have
$\mathcal{G}_{j-1}(h)<\mathcal{G}_{j-1}^{\prime}(h^{\prime})$ for any $j$.
Then the proof is complete.

If (21) holds with equalities, $(\mathcal{G}_{j})=(\mathcal{G}_{j}^{\prime})$
are all supported on $[0,S_{Q}]$. To show that they are equal to
$(\mathcal{G}_{j,S_{Q}})$, we only have to take $h=S_{Q}$ and apply the
equalities in (21). ∎

####  4.3.2 The random model.

Our goal of the paper is the random model, which is a randomised general
model. Since $(\mathcal{G}_{j})$ has parameters $(b_{j+1},b_{j+2},\cdots)$ and
$Q,h$, we can define

$\mathcal{I}_{j}=\mathcal{I}_{j,h}:=\mathcal{G}_{0}(\beta_{j+1},\beta_{j+2},\cdots).$

Therefore $\mathcal{I}_{j}$ is the quenched limit of the finite backward
sequences in the random model with $P_{n}^{n}=\delta_{h}$. Thanks to Lemma 1,
we have the following result.

######  Corollary 2.

The sequence $(\mathcal{I}_{j})=(\mathcal{I}_{j})_{j\geq 0}$ is stationary
ergodic and satisfies

$\mathcal{I}_{j-1}(dx)=(1-\beta_{j})\frac{x\mathcal{I}_{j}(dx)}{\int
y\mathcal{I}_{j}(dy)}+\beta_{j}Q(dx),\quad j\geq 1.$ (23)

######  Remark 9.

The equality (23) holds in the pointwise sense. In other words, given any
realisation of $(\beta_{j})$ (or equivalently, conditioning on $(\beta_{j})$),
the equality holds for any $j$ as in the general model. In the sequel, when we
present results regarding $(\mathcal{I}_{j})$, then conditioning on
$(\beta_{j})$ should be understood as in the pointwise sense. Sometimes we
omit saying either of them when the context is clear.

The proof of Corollary 2 requires the following lemma which is proved by Lemma
9.5 in [15].

######  Lemma 2.

Let $(S,\SS)$ and $(S^{\prime},\SS^{\prime})$ be measurable spaces. Let
$(\alpha_{j})\in S^{\infty}$ be a stationary ergodic sequence of random
variables. Let $f:S^{\infty}\to S^{\prime}$ be a measurable function. Then
$\left(f(\alpha_{j},\alpha_{j+1},\cdots)\right)$ is also stationary ergodic.

###### Proof of Corollary 2.

Since $(\beta_{j})$ is i.i.d., it is stationary ergodic. As $\mathcal{G}_{0}$
is a measurable function from $[0,1)^{\infty}$ to $M_{1}$, we apply Lemma 2 to
obtain that
$(\mathcal{I}_{j})=(\mathcal{G}_{0}(\beta_{j+1},\beta_{j+2},\cdots))$ is also
stationary ergodic. The recursive equation (23) is inherited from (16). ∎∎

Since $(\mathcal{I}_{j})$ is stationary ergodic, all $\mathcal{I}_{j}$’s have
the same distribution. We denote by
$\mathcal{I}:=\mathcal{I}_{0}=\mathcal{I}_{0,h}$ which is the weak limit
appeared in Theorem 2. The reason to drop off the index is to make it stand
out from the backward context, when it is appropriate to do so. The term
$\mathcal{I}_{Q}$ used in Theorem 3 is in fact $\mathcal{I}_{0,S_{Q}}.$

We comment further on the importance of finite backward sequences. Let
$(P_{n})$ be a forward sequence and $(P_{j}^{n})$ the finite backward sequence
with $P_{n}^{n}=P_{0}$, both in the random model with the same $(\beta_{j})$
and $Q$. Since $(\beta_{j})$ is i.i.d., we have

$(P_{0},P_{1},\cdots,P_{n})\stackrel{{\scriptstyle
d}}{{=}}(P_{n}^{n},P_{n-1}^{n},\cdots,P_{0}^{n}).$ (24)

So showing the weak convergence of $(P_{n})_{n\geq 0}$ is equivalent to
showing that of $(P_{0}^{n})_{n\geq 0}.$ But investigating the finite backward
sequences, via the general model, appears to be more convenient. In general a
dynamical system is easier to handle if we take a backward point of view, see
Diaconis and Freedman [9].

###  4.4 Finer analysis of the finite backward sequences

####  4.4.1 The general model.

We consider $(P_{j}^{n})$ with $P_{n}^{n}=\delta_{h}$, the one in Lemma 1.
Developing (15) we obtain

$\displaystyle P_{0}^{n}(dx)$
$\displaystyle=\left(\prod_{l=1}^{n}\frac{1-b_{l}}{\int
yP_{l}^{n}(dy)}\right)x^{n}P_{n}^{n}(dx)+\sum_{j=0}^{n-1}\left(\prod_{l=1}^{j}\frac{1-b_{l}}{\int
yP_{l}^{n}(dy)}\right)b_{j+1}m_{j}Q^{j}(dx)$ (25)
$\displaystyle=\left(\prod_{l=1}^{n}\frac{h(1-b_{l})}{\int
yP_{l}^{n}(dy)}\right)\delta_{h}(dx)+\sum_{j=0}^{n-1}\left(\prod_{l=1}^{j}\frac{1-b_{l}}{\int
yP_{l}^{n}(dy)}\right)b_{j+1}m_{j}Q^{j}(dx).$ (26)

We refer to (2) for the expansion of the forward sequence $(P_{n})$.

######  Proposition 1.

Let $(P_{j}^{n})$ be the finite backward sequence in the general model with
$P_{n}^{n}=\delta_{h}$. Then for the sequence $(\mathcal{G}_{j})$, we have

$\mathcal{G}_{0}(dx){=}G_{0}\delta_{h}(dx)+\sum_{j=0}^{\infty}\prod_{l=1}^{j}\frac{(1-b_{l})}{\int
y\mathcal{G}_{l}(dy)}b_{j+1}m_{j}Q^{j}(dx),$ (27)

where the second term on the right side of (26) converges to that of (27):

$\sum_{j=0}^{n-1}\left(\prod_{l=1}^{j}\frac{1-b_{l}}{\int
yP_{l}^{n}(dy)}\right)b_{j+1}m_{j}Q^{j}(dx)\leq_{S_{Q}-}\stackrel{{\scriptstyle
TV}}{{\longrightarrow}}\sum_{j=0}^{\infty}\prod_{l=1}^{j}\frac{(1-b_{l})}{\int
y\mathcal{G}_{l}(dy)}b_{j+1}m_{j}Q^{j}(dx)$ (28)

and the term $G_{0}=G_{0,h}$ satisfies the following assertions:

$\displaystyle\prod_{l=1}^{n}\frac{h(1-b_{l})}{\int yP_{l}^{n}(dy)}\text{
decreases in }n\text{ and converges to }G_{0},$ (29) $\displaystyle
G_{0}=1-\sum_{j=0}^{\infty}\prod_{l=1}^{j}\frac{(1-b_{l})}{\int
y\mathcal{G}_{l}(dy)}b_{j+1}m_{j}\in[0,1],$ (30) $\displaystyle
G_{0}=\mathcal{G}_{0}(h)\text{ , if }Q(h)=0,$ (31)
$\displaystyle\int\left(\frac{y}{h}\right)^{n}\mathcal{G}_{n}(dy)\prod_{l=1}^{n}\frac{h(1-b_{l})}{\int
y\mathcal{G}_{l}(dy)}\text{ decreases in }n\text{ and converges to
}G_{0},\text{ if }G_{0}>0.$ (32)

Moreover if we define $G_{j}$ for $\mathcal{G}_{j}$ similarly as $G_{0}$ for
$\mathcal{G}_{0}$, we have

$G_{j-1}=G_{j}\frac{h(1-b_{j})}{\int y\mathcal{G}_{j}(dy)},\quad\forall j\geq
1.$ (33)

As a consequence $G_{j}$’s are either all $0$ or all strictly positive.

###### Proof.

By (17), $\int yP_{l}^{n}(dy)$ increases in $n$ and converges to $\int
y\mathcal{G}_{l}(dy)$ as $n\to\infty$. Then using (26), we obtain (28).
Integrating on both sides of (26), we use (28) to deduce that

$\prod_{l=1}^{n}\frac{h(1-b_{l})}{\int
yP_{l}^{n}(dy)}=1-\int\sum_{j=0}^{n-1}\left(\prod_{l=1}^{j}\frac{1-b_{l}}{\int
yP_{l}^{n}(dy)}\right)b_{j+1}m_{j}Q^{j}(dx)$

decreases in $n$ and converges to the limit

$1-\int\sum_{j=0}^{\infty}\left(\prod_{l=1}^{j}\frac{1-b_{l}}{\int
y\mathcal{G}_{l}(dy)}\right)b_{j+1}m_{j}Q^{j}(dx)=:G_{0}.$

So (27), (29) and (30) are proved.

From (27) we observe (31). To show (32), we develop (16) as follows

$\displaystyle\mathcal{G}_{0}(dx)$
$\displaystyle=\left(\prod_{l=1}^{n}\frac{1-b_{l}}{\int
y\mathcal{G}_{l}(dy)}\right)x^{n}\mathcal{G}_{n}(dx)+\sum_{j=0}^{n-1}\left(\prod_{l=1}^{j}\frac{1-b_{l}}{\int
y\mathcal{G}_{l}(dy)}\right)b_{j+1}m_{j}Q^{j}(dx)$
$\displaystyle=\left(\int\left(\frac{y}{h}\right)^{n}\mathcal{G}_{n}(dy)\prod_{l=1}^{n}\frac{h(1-b_{l})}{\int
y\mathcal{G}_{l}(dy)}\right)\frac{x^{n}\mathcal{G}_{n}(dx)}{\int
y^{n}\mathcal{G}_{n}(dy)}+\sum_{j=0}^{n-1}\left(\prod_{l=1}^{j}\frac{1-b_{l}}{\int
y\mathcal{G}_{l}(dy)}\right)b_{j+1}m_{j}Q^{j}(dx).$

Combining the above display with (27) and (28), we obtain (32), and also that
$\frac{x^{n}\mathcal{G}_{n}(dx)}{\int y^{n}\mathcal{G}_{n}(dy)}$ converges
weakly to $\delta_{h}$. Finally, combining (16) and (27) leads to (33).∎∎

######  Remark 10.

The proposition implies that $(P_{0}^{n})$ with $P_{n}^{n}=\delta_{h}$ in the
random model converges in total variation to $\mathcal{I}=\mathcal{I}_{0}$,
pointwise. Then by (24), $(P_{n})$ in the random model with $P_{0}=\delta_{h}$
converges weakly to $\mathcal{I}.$ Therefore Theorem 2 is proved for the
particular case with $P_{0}=\delta_{h}.$ As will be clear later (Section 4.7),
a complete proof has to deal with different kinds of $P_{0}$. The one solved
here with $P_{0}=\delta_{h}$ is the simplest case.

####  4.4.2 The random model.

When carrying over the results of Proposition 1 to the random model, we change
the symbol $G$ to $I$, similar to the change from $\mathcal{G}$ to
$\mathcal{I}.$ For instance, we set
$I_{j}=G_{0}(\beta_{j+1},\beta_{j+2},\cdots)$ for any $j\geq 0$. Then we have
the following corollary.

######  Corollary 3.

The process $(I_{j})=(I_{j})_{j\geq 0}$ is stationary ergodic. Moreover
$\mathbb{P}(\\{I_{j}=0,\forall j\\})=\mathbb{P}(I_{0}=0)\in\\{0,1\\}$.

######  Remark 11.

If $Q(h)>0$, then it must be that $h=S_{Q}$ and
$\mathcal{I}(h)=\mathcal{I}_{0}(h)>0$ a.s.. If $Q(h)=0,$ then
$\mathcal{I}(h)=\mathcal{I}_{0}(h)=I_{0}.$ So applying Corollary (3), either
$\mathcal{I}(h)>0$ a.s., or $\mathcal{I}(h)=0$ a.s..

###### Proof.

By Proposition 1, $G_{0}=G_{0}(b_{1},b_{2},\cdots)$ is a measurable function
from $[0,1)^{\infty}$ to $[0,1]$. As $(\beta_{j})$ is i.i.d., we obtain that
$(I_{j})=(G_{0}(\beta_{j+1},\beta_{j+2},\cdots))$ is stationary ergodic,
thanks to Lemma 2.

By (33), for any $k$, $\\{I_{k}=0\\}=\\{I_{j}=0,\forall j\\}$. Note that
$\\{I_{j}=0,\forall j\\}$ is an invariant set in the sigma algebra generated
by $(I_{j})$. By ergodicity of $(I_{j})$, $\mathbb{P}(\\{I_{j}=0,\forall
j\\})=\mathbb{P}(I_{0}=0)\in\\{0,1\\}$. ∎∎

The following result provides us a tool to know more about $\mathcal{I}$ and
$Q$. Let $I=I_{0,h}$, and $I_{Q}=I_{0,S_{Q}}$. To summarise,
$\mathcal{I},I,\mathcal{I}_{Q},I_{Q}$ are identical in value to
$\mathcal{I}_{0,h},I_{0,h},\mathcal{I}_{0,S_{Q}},I_{0,S_{Q}}$ respectively.

######  Corollary 4.

The following statements about $\mathbb{E}\left[\ln\frac{1-\beta}{\int
y\mathcal{I}(dy)}\right]$ hold:

  1. 1).

$\mathbb{E}\left[\ln\frac{1-\beta}{\int y\mathcal{I}(dy)}\right]$ is well
defined, and takes values in $[-\infty,-\ln\int yQ(dy)]$, and depends only on
the marginal distributions of $\beta$ and $\mathcal{I}.$

  2. 2).

If $Q(h)=0$, then

$\mathbb{E}\left[\ln\frac{h(1-\beta)}{\int y\mathcal{I}(dy)}\right]\leq 0.$

  3. 3).

If $\mathcal{I}(h)>0$ a.s. and $Q(h)=0,$ then

$\mathbb{E}\left[\ln\frac{h(1-\beta)}{\int y\mathcal{I}(dy)}\right]=0.$

  4. 4).

If $h=S_{Q}$ and $Q(S_{Q})>0$, then

$\mathbb{E}\left[\ln\frac{S_{Q}(1-\beta)}{\int
y\mathcal{I}(dy)}\right]<0\text{ and }I=0,a.s..$

######  Remark 12.

If $h=S_{Q}$, we can only have $\mathbb{E}\left[\frac{S_{Q}(1-\beta)}{\int
y\mathcal{I}(dy)}\right]=\mathbb{E}\left[\frac{S_{Q}(1-\beta)}{\int
y\mathcal{I}_{Q}(dy)}\right]\leq 0.$

###### Proof.

1). By (27), $\mathcal{G}_{0}$ is a convex combination of probability measures
$\\{\delta_{h},Q,Q^{1},Q^{2},\cdots\\}$. As $Q^{j}\preceq
Q^{j+1}\preceq\delta_{h}$ for any $j\geq 0$, we have, in the pointwise sense

$Q\preceq\mathcal{I}=\mathcal{I}_{0}\preceq\delta_{h}.$ (34)

Then

$\ln\int yQ(dy)\leq\mathbb{E}\left[\ln\int y\mathcal{I}(dy)\right]\leq\ln h.$

So $\mathbb{E}\left[\ln\int y\mathcal{I}(dy)\right]$ is a finite term.
Consequently,

$\displaystyle\mathbb{E}\left[\ln\frac{1-\beta}{\int y\mathcal{I}(dy)}\right]$
$\displaystyle=\mathbb{E}\left[\ln(1-\beta)-\ln\int y\mathcal{I}(dy)\right]$
$\displaystyle=\mathbb{E}\left[\ln(1-\beta)\right]-\mathbb{E}\left[\ln\int
y\mathcal{I}(dy)\right]\in\left[-\infty,-\ln\int yQ(dy)\right].$

We observe that the above display depends only on the marginal distributions
of $\beta$ and $\mathcal{I}.$

2). Let $(P_{j}^{n})$ be the finite backward sequence in the random model with
$P_{n}^{n}=\delta_{h}$. By assumption, $Q(h)=0$. Adapting (26) into the random
model and taking the expectation of the mass on $h$ we obtain

$1\geq\mathbb{E}[P_{0}^{n}(h)]=\mathbb{E}\left[\left(\prod_{l=1}^{n}\frac{h(1-\beta_{l})}{\int
yP_{l}^{n}(dy)}\right)\right]\geq\exp\left(\sum_{l=1}^{n}\mathbb{E}\left[\ln\frac{h(1-\beta_{l})}{\int
yP_{l}^{n}(dy)}\right]\right)$

where the second inequality is due to Jensen’s inequality. By (17)

$\mathbb{E}\left[\ln\frac{h(1-\beta_{l})}{\int yP_{l}^{n}(dy)}\right]\text{
increases in $n$ and converges to
}\mathbb{E}\left[\ln\frac{h(1-\beta_{l})}{\int
y\mathcal{I}_{l}(dy)}\right]=\mathbb{E}\left[\ln\frac{h(1-\beta)}{\int
y\mathcal{I}(dy)}\right].$

Combining the above two displays, it must be that
$\mathbb{E}[\ln\frac{h(1-\beta)}{\int y\mathcal{I}(dy)}]\leq 0.$

3). Lemma 1 implies that there exists a measurable function
$T:[0,1)^{\infty}\mapsto(0,\infty)$ such that for any $j$,
$\frac{h(1-b_{j})}{\int y\mathcal{G}_{j}(dy)}=T(b_{j},b_{j+1},\cdots)$. By
Lemma 2,

$\left(\frac{h(1-\beta_{j})}{\int y\mathcal{I}_{j}(dy)}\right)\text{ is
stationary ergodic.}$

By $(\ref{k0k})$ and the fact that $\mathcal{I}(h)=\mathcal{I}_{0}(h)=I_{0}>0$
a.s. (because $Q(h)=0$ by assumption),

$\lim_{n\to\infty}(I_{0})^{1/n}=\lim_{n\to\infty}\exp\left(\frac{1}{n}\ln\int\left(\frac{y}{h}\right)^{n}\mathcal{I}_{n}(dy)+\frac{1}{n}\sum_{l=1}^{n}\ln\frac{h(1-\beta_{l})}{\int
y\mathcal{I}_{l}(dy)}\right)\stackrel{{\scriptstyle a.s.}}{{=}}1.$ (35)

As $(\mathcal{I}_{j})$ is stationary ergodic,
$\int\left(\frac{y}{h}\right)^{n}\mathcal{I}_{n}(dy)\in[I_{n},1]$ converges
weakly to $I_{0},$ which is strictly positive. Then

$\left[\frac{1}{n}\ln
I_{n},0\right]\ni\frac{1}{n}\ln\int\left(\frac{y}{h}\right)^{n}\mathcal{I}_{n}(dy)\stackrel{{\scriptstyle
d}}{{\longrightarrow}}0,\quad n\to\infty.$

Moreover, since $\left(\frac{h(1-\beta_{j})}{\int
y\mathcal{I}_{j}(dy)}\right)$ is stationary ergodic, we have

$\frac{1}{n}\sum_{l=1}^{n}\ln\frac{h(1-\beta_{l})}{\int
y\mathcal{I}_{l}(dy)}\stackrel{{\scriptstyle
a.s.}}{{\longrightarrow}}\mathbb{E}\left[\ln\frac{h(1-\beta)}{\int
y\mathcal{I}(dy)}\right],\quad n\to\infty.$

The above three displays yield

$1=\exp\left(\mathbb{E}\left[\ln\frac{h(1-\beta)}{\int
y\mathcal{I}(dy)}\right]\right)\,\,\text{ or equivalently
}\,\,\mathbb{E}\left[\ln\frac{h(1-\beta)}{\int y\mathcal{I}(dy)}\right]=0.$

4). We show $I=I_{0}=0$ a.s. by contradiction. Adapting (26) in the random
model

$P_{0}^{n}(dx)=\left(\prod_{l=1}^{n}\frac{S_{Q}(1-\beta_{l})}{\int
yP_{l}^{n}(dy)}\right)\delta_{S_{Q}}(dx)+\sum_{j=0}^{n-1}\left(\prod_{l=1}^{j}\frac{1-\beta_{l}}{\int
yP_{l}^{n}(dy)}\right)\beta_{j+1}m_{j}Q^{j}(dx).$

If $I_{0}>0$ a.s., we consider the mass on $S_{Q}$ in the above display. Note
that $m_{j}Q^{j}(S_{Q})=S_{Q}^{j}Q(S_{Q})$. Together with (29) we obtain

$1\geq P_{0}^{n}(S_{Q})\geq
Q(S_{Q})\sum_{j=0}^{n-1}\left(\prod_{l=1}^{j}\frac{S_{Q}(1-\beta_{l})}{\int
yP_{l}^{n}(dy)}\right)\beta_{j+1}\geq
Q(S_{Q})\sum_{j=0}^{n-1}I_{0}\beta_{j+1}\stackrel{{\scriptstyle
d}}{{\longrightarrow}}\infty.$

This is a contradiction. So $I_{0}=0,a.s..$ Note that by (34),
$\mathcal{I}(S_{Q})=\mathcal{I}_{0}(S_{Q})\geq Q(S_{Q})>0.$ Then we get
$\mathbb{E}\left[\ln\frac{h(1-\beta)}{\int y\mathcal{I}(dy)}\right]<0$ using
(35) and the arguments thereafter. ∎

∎

###  4.5 Proof of Theorem 3

###### Proof of Theorem 3.

The statement about $\mathbb{E}\left[\ln\frac{h(1-\beta)}{\int
y\mathcal{I}_{Q}(dy)}\right]$ concerns just a subcase of Corollary 4–1). So
this is proved.

If there is no condensation at $h$, then by Corollary 1,
$\mathcal{I}=\mathcal{I}_{0,h}=\mathcal{I}_{0,S_{Q}}=\mathcal{I}_{Q}$, then of
course $\mathcal{I}\stackrel{{\scriptstyle d}}{{=}}\mathcal{I}_{Q}.$

The first assertion in the condensation criterion holds due to Corollary 4–3).
We consider the second one. If there is condensation at $h$, then
$\mathcal{I}_{0,S_{Q}}\neq\mathcal{I}_{0,h}$. By Corollary 1,
$\mathcal{I}_{0,h}\leq_{S_{Q}-}\mathcal{I}_{0,S_{Q}}$ and
$\mathcal{I}_{0,S_{Q}}(S_{Q})\leq\mathcal{I}_{0,h}(h)$, which, together with
Corollary 4–3) entails that

$\mathbb{E}\left[\ln\frac{h(1-\beta)}{\int
y\mathcal{I}_{0,S_{Q}}(dy)}\right]>\mathbb{E}\left[\ln\frac{h(1-\beta)}{\int
y\mathcal{I}_{0,h}(dy)}\right]=0.$

The above inequality is strict because
$\mathcal{I}_{0,S_{Q}}\neq\mathcal{I}_{0,h}$.

If there is no condensation at $h$, then by Corollary 1,
$\mathcal{I}_{0,h}=\mathcal{I}_{0,S_{Q}}$. Using Corollary 4–2),

$\mathbb{E}\left[\ln\frac{h(1-\beta)}{\int
y\mathcal{I}_{0,S_{Q}}(dy)}\right]=\mathbb{E}\left[\ln\frac{h(1-\beta)}{\int
y\mathcal{I}_{0,h}(dy)}\right]\leq 0.$

∎∎

###  4.6 Some properties of invariant measures

In this section, we prove some results concerning invariant measures. But we
leave the proof of Theorem 4 to the end. Invariant measures will play
important roles in the proof of Theorem 2.

######  Lemma 3.

For any invariant measure $\nu$, $\mathbb{E}\left[\ln\frac{1-\beta}{\int
y\nu(dy)}\right]$ is well defined, and takes values in $[-\infty,-\ln\int
yQ(dy)]$, and depends only on the marginal distributions of $\beta$ and $\nu.$

###### Proof.

By the definition of invariant measure

$\displaystyle\mathbb{E}\left[\int y\nu(dy)\right]$
$\displaystyle=(1-\mathbb{E}[\beta])\mathbb{E}\left[\frac{\int
y^{2}\nu(dy)}{\int y\nu(dy)}\right]+\mathbb{E}[\beta]\int yQ(dy)$
$\displaystyle\geq(1-\mathbb{E}[\beta])\mathbb{E}\left[\int
y\nu(dy)\right]+\mathbb{E}[\beta]\int yQ(dy)$

where the inequality is due to the fact that $\int y^{2}\nu(dy)\geq(\int
y\nu(dy))^{2}.$ Then we obtain

$\int yQ(dy)\leq\mathbb{E}\left[\int y\nu(dy)\right]\leq 1.$

Proceeding similarly as in the proof of Corollary 4–1), we conclude that this
lemma holds. ∎∎

######  Corollary 5.

$\mathcal{I}_{Q}$ is the unique (in distribution) invariant measure supported
on $[0,S_{Q}]$.

###### Proof.

Let $\nu$ be any invariant measure on $[0,S_{Q}]$. We show that
$\nu\stackrel{{\scriptstyle d}}{{=}}\mathcal{I}_{Q}$. Note that
$S_{\nu}=S_{Q}$, a.s.. Let $(P_{n})$ and $(P_{n}^{\prime})$ be two forward
sequences as in Section 4.2 with

$Q=Q^{\prime};\quad h=h^{\prime}=S_{Q};\quad P_{0}\stackrel{{\scriptstyle
d}}{{=}}\nu,\quad P_{0}^{\prime}=\delta_{S_{Q}}$

and $P_{0}$ is independent of $(\beta_{n})$. The two sequences differ only in
the starting measures (satisfying $P_{0}\leq_{S_{Q}-}P_{0}^{\prime}$), with
other parameters identical. Since $\nu$ is invariant,
$P_{n}\stackrel{{\scriptstyle d}}{{=}}\nu$ for any $n\geq 0$. Using the
notations
$\mathcal{M}_{n},\mathcal{M}_{n}^{\prime},\mathcal{W}_{n},\mathcal{W}_{n}^{\prime}$
in Section 4.2, and by the monotonicity analysis as in the proof of Lemma 1,
we obtain in the pointwise sense,

$\int\mathcal{M}_{n}(dx)\leq\int\mathcal{M}_{n}^{\prime}(dx),\quad\mathcal{W}_{n}^{\prime}\leq_{S_{Q}}\mathcal{W}_{n}.$
(36)

If $I_{Q}=0$ a.s., by (29) in Proposition 1 and (24)

$\int\mathcal{M}_{n}^{\prime}(dx)\stackrel{{\scriptstyle
d}}{{\longrightarrow}}I_{Q}\stackrel{{\scriptstyle a.s.}}{{=}}0.$

Remark 10 says that
$P_{n}^{\prime}(=\mathcal{W}_{n}^{\prime}+\mathcal{M}_{n}^{\prime})\stackrel{{\scriptstyle
d}}{{\to}}\mathcal{I}_{Q}.$ Then

$\mathcal{W}^{\prime}_{n}{\stackrel{{\scriptstyle
d}}{{\longrightarrow}}}\mathcal{I}_{Q}.$

Thus applying (36) and $I_{Q}=0,$ a.s., we obtain

$\int\mathcal{M}_{n}(dx)\stackrel{{\scriptstyle
d}}{{\longrightarrow}}0,\quad\mathcal{W}_{n}{\stackrel{{\scriptstyle
d}}{{\longrightarrow}}}\mathcal{I}_{Q}.$

Consequently,

$P_{n}(=\mathcal{W}_{n}+\mathcal{M}_{n})\stackrel{{\scriptstyle
d}}{{\longrightarrow}}\mathcal{I}_{Q}.$

Since $\nu\stackrel{{\scriptstyle d}}{{=}}P_{n}$ for any $n$, we have
$\nu\stackrel{{\scriptstyle d}}{{=}}\mathcal{I}_{Q}.$

If $I_{Q}>0,a.s.,$ then by Corollary 4–4), $Q(S_{Q})=0$ and
$\mathcal{I}_{Q}(S_{Q})=I_{Q}>0,a.s.$. Then by Corollary 4–3), we have

$\mathbb{E}\left[\ln\frac{S_{Q}(1-\beta)}{\int
y\mathcal{I}_{Q}(dy)}\right]=0.$ (37)

Using again monotonicity analysis, in a pointwise sense,

$P_{n}^{\prime}\leq_{S_{Q}-}P_{n},\quad P_{n}(S_{Q})\leq
P_{n}^{\prime}(S_{Q}).$ (38)

As $P_{n}^{\prime}\stackrel{{\scriptstyle d}}{{\rightarrow}}\mathcal{I}_{Q}$,
and $P_{n}\stackrel{{\scriptstyle d}}{{=}}\nu$ for all $n$, the above display
entails that,

$\mathcal{I}_{Q}\leq_{S_{Q}-}^{d}\nu,\quad\nu(S_{Q})\preceq
I_{Q}=\mathcal{I}_{Q}(S_{Q}).$

Assume that $\nu$ is not equal to $\mathcal{I}_{Q}$ in distribution, then by
the above display and (37),

$\mathbb{E}\left[\ln\frac{S_{Q}(1-\beta)}{\int y\nu(dy)}\right]>0.$

The inequality implies that for $\varepsilon>0$ small enough, we have

$\mathbb{E}\left[\ln\frac{(S_{Q}-\varepsilon)(1-\beta)}{\int
y\nu(dy)}\right]>0.$

As $S_{\nu}=S_{Q}$ almost surely and $P_{0}\stackrel{{\scriptstyle
d}}{{=}}\nu$,
$\int(\frac{x}{S_{Q}-\varepsilon})^{n}P_{0}(dx)\stackrel{{\scriptstyle
d}}{{\rightarrow}}\infty,$ as $n\to\infty$. Using again the decomposition (12)
in Section 4.2, we get

$\displaystyle 1$ $\displaystyle=\mathbb{E}\left[\int
P_{0}(dx)\right]\geq\mathbb{E}\left[\int\mathcal{M}_{n}(dx)\right]$
$\displaystyle=\mathbb{E}\left[\exp\left(\ln\int\left(\frac{x}{S_{Q}-\varepsilon}\right)^{n}P_{0}(dx)+\sum_{l=0}^{n-1}\ln\frac{(S_{Q}-\varepsilon)(1-\beta_{l+1})}{\int
yP_{l}^{n}(dy)}\right)\right]$
$\displaystyle\geq\mathbb{E}\left[\exp\left(\sum_{l=0}^{n-1}\ln\frac{(S_{Q}-\varepsilon)(1-\beta_{l+1})}{\int
yP_{l}^{n}(dy)}\right)\right]\geq\exp\left(n\mathbb{E}\left[\frac{(S_{Q}-\varepsilon)(1-\beta)}{\int
y\nu(dy)}\right]\right)\stackrel{{\scriptstyle
n\to\infty}}{{\longrightarrow}}\infty$

where the third inequality is due to Jensen’s inequality. So this is a
contradiction, which means that $\nu$ is equal in distribution to
$\mathcal{I}_{Q}$. ∎∎

###  4.7 Proof of Theorem 2

Case 1. ${P_{0}=\delta_{h}}$.

###### Proof of Theorem 2.

This is shown in Remark 10.∎

∎

Case 2. ${I_{0,h}=0}$ a.s..

###### Proof of Theorem 2.

Let $(P_{n})_{n\geq 0}$, $(P^{\prime}_{n})_{n\geq 0}$ be two forward sequences
in Section 4.2 with

$Q=Q^{\prime},\quad h=h^{\prime},\quad P_{0}^{\prime}=\delta_{h}.$

So the two sequences differ only in the starting measures (satisfying
$P_{0}\leq_{h-}P_{0}^{\prime}$), with other parameters identical. Next it
suffices to follow the same procedure as in the proof of Corollary 5 for the
case $I_{Q}=0$ a.s.. The proof is omitted. ∎∎

Case 3. ${I_{0,h}>0}$ a.s. and $P_{0}(h)>0.$

First of all, we restate a result from ([22], p.10), where only $h=1$ is
considered. But it is easily generalised to any $h$. Recall also the
distribution function $D_{u}$ for $u\in M_{1}$, introduced in Section 4.1.

######  Lemma 4.

Let $u_{1},u_{2}\in M_{1}$ be any probability measures satisfying
$S_{u_{1}}=S_{u_{2}}=h$ and $u_{1}\leq_{h-}u_{2}$. If for some $\varepsilon>0$
there exists $a\in(0,h)$ such that $D_{u_{1}}(a)+\varepsilon\leq
D_{u_{2}}(a)$, then

$\int yu_{1}(dy)\geq c(a,\varepsilon)\int yu_{2}(dy)$

where $c(a,\varepsilon)=\frac{1}{1-\varepsilon(h-a)}>1.$

###### Proof of Theorem 2.

Let $(P_{n}),(P_{n}^{\prime})$ be the two forward sequences in the proof of
Case 2. Similarly as (38), conditionally on $(\beta_{n})$, we have

$P_{n}^{\prime}\leq_{h-}P_{n},\quad P_{n}(h)\leq P_{n}^{\prime}(h),$ (39)

implying

$\int yP_{j}^{\prime}(dy)\geq\int yP_{j}(dy),\quad\forall j\geq 0.$

For any $\varepsilon>0,a\in(0,h)$, let

$\kappa_{n}:=\\#\\{n:D_{P^{\prime}_{j}}(a)+\varepsilon\leq D_{P_{j}}(a),0\leq
j\leq n\\}.$

Note that by Proposition 1–4), $Q(h)=0$. So using (12) and (13) in Section 4.2

$P^{\prime}_{n}(h)=\prod_{l=0}^{n-1}\frac{h(1-\beta_{l+1})}{\int
yP^{\prime}_{l}(dy)},\quad
P_{n}(h)=\left(\prod_{l=0}^{n-1}\frac{h(1-\beta_{l+1})}{\int
yP_{l}(dy)}\right)P_{0}(h).$

Then by Lemma 4, we have $\prod_{l=0}^{n}\int yP_{l}^{\prime}(dy)\geq
c(a,\varepsilon)^{\kappa_{n}}\prod_{l=0}^{n}\int yP_{l}(dy).$ Therefore

$P^{\prime}_{n}(h)\leq\frac{1}{c(a,\varepsilon)^{\kappa_{n}}}\left(\prod_{l=0}^{n-1}\frac{h(1-\beta_{l+1})}{\int
yP_{l}(dy)}\right)=\frac{P_{n}(h)}{c(a,\varepsilon)^{\kappa_{n}}P_{0}(h)}\leq\frac{1}{c(a,\varepsilon)^{\kappa_{n}}P_{0}(h)}.$

But (29) of Proposition 1 and (24) entail that $P^{\prime}_{n}(h)$ converges
weakly to $I_{0,h}$ which is by assumption non-zero almost surely. Then
$\lim_{n\to\infty}\kappa_{n}<\infty$ a.s.. As $a,\varepsilon$ are arbitrary
numbers and by Case 1 of this theorem $P_{n}^{\prime}\stackrel{{\scriptstyle
d}}{{\longrightarrow}}\mathcal{I}_{0,h}$, we use (39) to conclude that $P_{n}$
also converges weakly to $\mathcal{I}_{0,h}.$ ∎∎

Case 4. ${I_{0,h}>0}$ a.s. and $P_{0}(h)=0.$

###### Proof of Theorem 2.

The idea is to use a tripling argument similarly as in the proof of Theorem 5
in [22]. For any $u\in M_{1}$ and any $a\in[0,1]$, define

$u^{a}=u_{[0,a)}+u([a,1])\delta_{a},\quad a<h$

where $u_{[0,a)}$ is the restriction of $u$ on $[0,a).$

We distinguish between $h>S_{Q}$ and $h=S_{Q}.$ For the former, let
$(P_{n}),(P_{n}^{\prime}),(P_{n}^{\prime\prime})$ be three forward sequences
in Section 4.2 with

$S_{Q}<h^{\prime\prime}<h=h^{\prime};\quad Q^{\prime}=Q,\quad
Q^{\prime\prime}=Q^{h^{\prime\prime}}=Q;\quad P_{0}^{\prime}=\delta_{h},\quad
P_{0}^{\prime\prime}=P_{0}^{h^{\prime\prime}}.$

So the three sequences differ in the starting measures including the largest
fitness values, but with the same $Q$ and $(\beta_{n})$. Since
$P_{0}^{\prime}(h^{\prime})=\delta_{h}(h)=1$ and
$0<P_{0}^{\prime\prime}(h^{\prime\prime})\leq 1$ and, we use Case 1 for
$(P_{n}^{\prime})$ and Case 2, Case 3 for $(P_{n}^{\prime\prime})$ to obtain
that,

$P_{n}^{\prime}\stackrel{{\scriptstyle
d}}{{\longrightarrow}}\mathcal{I}_{0,h},\quad
P_{n}^{\prime\prime}\stackrel{{\scriptstyle
d}}{{\longrightarrow}}\mathcal{I}_{0,h^{\prime\prime}}.$ (40)

Applying the monotonicity analysis, the following holds in the pointwise
sense:

$P^{\prime}_{n}\leq_{h-}P_{n}\leq_{h^{\prime\prime}-}P^{\prime\prime}_{n}.$
(41)

Letting $h^{\prime\prime}\to h$ and using Corollary 1, conditionally on
$(\beta_{j})$, $\mathcal{I}_{0,h^{\prime\prime}}$ converges weakly to a limit
in $M_{1}$, denoted by $\nu$. So $\nu$ is a (pointwise) weak limit of
$\mathcal{I}_{0,h^{\prime\prime}}$ as $h^{\prime\prime}\to h.$ We prove next
that $\nu=\mathcal{I}_{0,h}.$

Since $I_{0,h}>0,a.s.$ and $h>S_{Q}$, by Theorem 3,

$\mathbb{E}\left[\ln\frac{h(1-\beta)}{\int
y\mathcal{I}_{0,S_{Q}}(dy)}\right]>0.$

Then for $h^{\prime\prime}$ close enough to $h$, we also have

$\mathbb{E}\left[\ln\frac{h^{\prime\prime}(1-\beta)}{\int
y\mathcal{I}_{0,S_{Q}}(dy)}\right]>0.$

The above display entails that there is condensation at $h^{\prime\prime}$,
thanks to Theorem 3. Together with Corollary 4–3), we have

$\mathbb{E}\left[\ln\frac{h^{\prime\prime}(1-\beta)}{\int
y\mathcal{I}_{0,h^{\prime\prime}}(dy)}\right]=0,\quad
I_{0,h^{\prime\prime}}>0,\quad a.s.$ (42)

Since $\mathcal{I}_{0,h^{\prime\prime}}$ is an invariant measure, the limit
$\nu$ is still an invariant measure. Using (42) and Corollary 1, the pointwise
convergence of $\mathcal{I}_{0,h^{\prime\prime}}$ to $\nu$ as
$h^{\prime\prime}\to h$ entails

$\mathbb{E}\left[\ln\frac{h(1-\beta)}{\int
y\nu(dy)}\right]=0,\quad\nu(h)>0\,\,\,a.s..$ (43)

Using Corollary 1 again, in the pointwise sense

$\mathcal{I}_{0,h}\leq_{h^{\prime\prime}-}\mathcal{I}_{0,h^{\prime\prime}},\quad\mathcal{I}_{0,h^{\prime\prime}}(h^{\prime\prime})\leq\mathcal{I}_{0,h}(h),$

implying in the pointwise sense (since $\nu$ is a pointwise weak limit of
$\mathcal{I}_{0,h^{\prime\prime}}$),

$\mathcal{I}_{0,h}\leq_{h-}\nu,\quad\nu(h)\leq I_{0,h}.$ (44)

On the other hand, by assumption $I_{0,h}>0,a.s.$, so using Corollary 4–3),

$\mathbb{E}\left[\ln\frac{h(1-\beta)}{\int y\mathcal{I}_{0,h}(dy)}\right]=0.$

The above display with (43) and (44) entails that

$\nu=\mathcal{I}_{0,h},\quad\text{pointwise}.$ (45)

Therefore we proved that $\mathcal{I}_{0,h^{\prime\prime}}$ converges
pointwise to the weak limit $\mathcal{I}_{0,h}$ as $h^{\prime\prime}\to h.$

Now taking into account (40), for any continuous function $f,$

$\displaystyle\int f(x)P_{n}^{\prime\prime}(dx)$
$\displaystyle\stackrel{{\scriptstyle d}}{{\longrightarrow}}\int
f(x)\mathcal{I}_{0,h^{\prime\prime}}(dx)$
$\displaystyle\xrightarrow[h^{\prime\prime}\to h]{\text{pointwise}}\int
f(x)\mathcal{I}_{0,h}(dx)\stackrel{{\scriptstyle d}}{{\longleftarrow}}\int
f(x)P_{n}^{\prime}(dx).$ (46)

Note that using (41), for any bounded continuous increasing function $g$ we
have

$\int g(x)P_{n}^{\prime\prime}(dx)\preceq\int g(x)P_{n}(dx)\preceq\int
g(x)P_{n}^{\prime}(dx).$

Together with (4.7), we obtain

$\int g(x)P_{n}(dx)\stackrel{{\scriptstyle d}}{{\longrightarrow}}\int
g(x)\mathcal{I}_{0,h}(dx).$

Since by (41), $P_{n}^{\prime}\leq_{h-}P_{n}$ pointwise for any $n$, the above
display entails that $P_{n}$ converges weakly to $\mathcal{I}_{0,h}$, the same
weak limit of $(P_{n}^{\prime})$.

If $h=S_{Q}$, we follow the same procedure, except that to prove (45), we
require Corollary 5. ∎∎

###  4.8 Proof of Theorem 4

Firstly we prove two lemmas. Recall the definition of $S_{u}$ for $u\in
M_{1}$.

######  Lemma 5.

$S_{(\cdot)}$ is a continuous (hence measurable) function on $M_{1}$ with the
topology of the weak convergence.

###### Proof.

Assume that a sequence $(u_{n})$ converges weakly to $u$. If $S_{u_{n}}$ does
not converge to $S_{u}$, then there exists a subsequence $(u_{n_{k}})$ such
that $S_{u_{n_{k}}}$ converges to a limit $a$ with $a<S_{u}$ or $a>S_{u}$.
Without loss of generality, assume $a<S_{u}.$ We take a positive and
continuous function $f$ supported on $(\frac{a+S_{u}}{2},S_{u}]$ and then
$\int f(x)u(dx)>0.$ But $\int f(x)u_{n_{k}}(dx)$ converges to $0$. This is
against the weak convergence. The proof is then completed. ∎∎

The next lemma generalises Corollary 5.

######  Lemma 6.

For any invariant measure $\nu$ with $S_{\nu}=h$ a.s., we have
$\nu\stackrel{{\scriptstyle d}}{{=}}\mathcal{I}.$

###### Proof.

Let $(P_{n})$ be the forward sequence in the random model with
$P_{0}\stackrel{{\scriptstyle d}}{{=}}\nu$ and $P_{0}$ independent of
$(\beta_{n})$. By Theorem 2, conditionally on $P_{0}$, $P_{n}$ converges in
distribution to the same random measure $\mathcal{I}.$ Then unconditionally
$P_{n}\stackrel{{\scriptstyle d}}{{=}}\nu$ converges in distribution to
$\mathcal{I},$ implying $\nu\stackrel{{\scriptstyle d}}{{=}}\mathcal{I}$. ∎

###### Proof of Theorem 4.

Let $\nu$ be any invariant measure. By Definition (10), $S_{\nu}\in[S_{Q},1]$,
a.s.. By Lemma 5, $S_{\nu}$ is a random variable and then by Theorem 5.3 in
[15], there exists a regular conditional distribution of $\nu$ on $S_{\nu}$.

Conditioning on $S_{\nu}$ for both sides of (10), we see that $(\nu|S_{\nu})$
must be an invariant measure almost surely. By Lemma 6, conditionally on
$S_{\nu}$, we have $(\nu|S_{\nu})\stackrel{{\scriptstyle d}}{{=}}\mathcal{I}$
almost surely, where $\mathcal{I}$ is the random probability measure with
parameters $\mathcal{L},Q,h=S_{\nu}$ and satisfies
$\mathbb{P}(S_{\mathcal{I}}=S_{\nu}|S_{\nu})=1$, a.s.. Then the proof is
finished. ∎

∎

##  Appendix A Analysis of $\ln\frac{h(1-b)}{\int x\mathcal{K}_{Q}(dx)}$ in
Kingman’s model

We discuss respectively Theorem 1-1 (i.e., $\int\frac{Q(dx)}{1-x/h}\geq
b^{-1}$) and Theorem 1-2 (i.e., $\int\frac{Q(dx)}{1-x/h}<b^{-1}$). For the
former, let us compute $\int x\mathcal{K}_{Q}(dx)$ first.

$\displaystyle\int x\mathcal{K}_{Q}(dx)$
$\displaystyle=\int\frac{b\theta_{b}xQ(dx)}{\theta_{b}-(1-b)x}$
$\displaystyle=\int\frac{b\theta_{b}(x-\theta_{b}/(1-b))Q(dx)+b\theta_{b}^{2}/(1-b)Q(dx)}{\theta_{b}-(1-b)x}$
$\displaystyle=\frac{b\theta_{b}}{1-b}+\frac{b\theta_{b}^{2}}{1-b}\int\frac{Q(dx)}{\theta_{b}-(1-b)x}$
$\displaystyle=\theta_{b}$ (47)

where the last equality is due to the fact that $\theta_{b}$ is the solution
of the equation (4). The equation (4) also implies

$\int\frac{\theta_{b}Q(dx)}{\theta_{b}-(1-b)x}=\int\frac{Q(dx)}{1-(1-b)x/\theta_{b}}=b^{-1}.$

Recall $\int\frac{Q(dx)}{1-x/h}\geq b^{-1}.$ Then the above display entails
that

$\frac{1}{h}\geq\frac{1-b}{\theta_{b}}.$

Taking into account (A), we arrive at

$\frac{h(1-b)}{\int x\mathcal{K}_{Q}(dx)}\leq 1,\text{ or equivalently,
}\ln\frac{h(1-b)}{\int x\mathcal{K}_{Q}(dx)}\leq 0.$

The equality holds if and only if $\int\frac{Q(dx)}{1-x/h}=b^{-1}.$

For Theorem 1-2, we have

$\displaystyle\int x\mathcal{K}_{Q}(dx)$
$\displaystyle=\int\frac{bxQ(dx)}{1-x/S_{Q}}+\left(1-\int\frac{bQ(dx)}{1-x/S_{Q}}\right)S_{Q}$
$\displaystyle=S_{Q}+\int\frac{b(x-S_{Q})Q(dx)}{1-x/S_{Q}}$
$\displaystyle=(1-b)S_{Q}.$

Then we obtain

$\ln\frac{h(1-b)}{\int x\mathcal{K}_{Q}(dx)}=\ln\frac{h}{S_{Q}}\geq 0$

where the equality holds if and only if $h=S_{Q}$.

In conclusion, if $h>S_{Q},$ $\ln\frac{h(1-b)}{\int x\mathcal{K}_{Q}(dx)}\leq
0$ is equivalent to $\int\frac{Q(dx)}{1-x/h}\geq b^{-1}$ (non-condensation
case), and $\ln\frac{(1-b)S_{Q}}{\int x\mathcal{K}_{Q}(dx)}>0$ is equivalent
to $\int\frac{Q(dx)}{1-x/h}<b^{-1}$ (condensation case).

If $h=S_{Q}$, $\ln\frac{h(1-b)}{\int x\mathcal{K}_{Q}(dx)}=0$ is equivalent to
either $\int\frac{Q(dx)}{1-x/S_{Q}}=b^{-1}$ (non-condensation case) or
$\int\frac{Q(dx)}{1-x/S_{Q}}<b^{-1}$ (condensation case), and
$\ln\frac{S_{Q}(1-b)}{\int x\mathcal{K}_{Q}(dx)}<0$ is equivalent to
$\int\frac{Q(dx)}{1-x/S_{Q}}>b^{-1}$ (non-condensation case). The case
$\ln\frac{S_{Q}(1-b)}{\int x\mathcal{K}_{Q}(dx)}>0$ does not exist, which is
in line with Remark 12.

Therefore, if $h=S_{Q}$, knowing only $\ln\frac{h(1-b)}{\int
x\mathcal{K}_{Q}(dx)}=0$ cannot tell whether the condensation occurs or not.

## Acknowledgements

The author thanks Takis Konstantopoulos, Götz Kersting and Pascal Grange for
discussions. The author thanks the anonymous reviewers for their comments
which greatly improved the presentation of the paper, and for suggesting the
intuition for Theorem 3. The author acknowledges the support of the National
Natural Science Foundation of China (Youth Programme, Grant: 11801458).

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