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Endocrinol Metab > Volume 36(4); 2021 > Article
Kim, Kim, Lim, Kim, and Park: The Concept of Economic Evaluation and Its Application in Thyroid Cancer Research

### Abstract

Economic evaluation is a type of comparative analysis between interventions in terms of both their resource use and health outcomes. Due to the good prognosis of thyroid cancer (TC), the socioeconomic burden of TC patients post-diagnosis is increasing. Therefore, economic evaluation studies focusing on TC are recommended. This study aimed to describe the concept and methods of economic evaluation and reviewed previous TC studies. Several previous studies compared the costs of interventions or evaluated recurrence, complications, or quality of life as measures of their effectiveness. Regarding costs, most studies focused on direct costs and applied hypothetical models. Cost-minimization analysis should be distinguished from simple cost analysis. Furthermore, due to the universality of the term “cost-effectiveness analysis” (CEA), several studies have not distinguished CEA from cost-utility analysis; this point needs to be considered in future research. Cost-benefit analyses have not been conducted in previous TC research. Since TC has a high survival rate and good prognosis, the need for economic evaluations has recently been pointed out. Therefore, correct concepts and methods are needed to obtain clear economic evaluation results. On this basis, it will be possible to provide appropriate guidelines for TC treatment and management in the future.

### INTRODUCTION

The incidence of thyroid cancer has markedly increased worldwide in recent decades [1]. In South Korea (hereafter, Korea), the incidence rate of thyroid cancer increased sharply by more than 10 times between 2000 and 2012 (age-standardized incidence rate [ASIR] based on the Korean population in 2000: 7.2 per 100,000 in 2000; 74.8 per 100,000 in 2012), decreased from 2012 to 2015, and then slightly increased again after 2015 [2]. Although the incidence of thyroid cancer in Korea has changed to a pattern of a slow increase since 2015, the incidence in 2018 is still the highest in the world, twice that of the United States and three times that of Canada [3].
Thyroid cancer has a 5-year survival rate of almost 100%, irrespective of whether surgery is performed [2]. Furthermore, it has been suggested that the rapid increase of the ASIR of thyroid cancer in Korea may reflect over-diagnosis and over-screening [4]. Collectively, these points call into question the value of thyroid cancer screening. However, there are limitations in making an appropriate choice because no comparative results have been reported. It is also necessary to select appropriate medical services in a resource-limited environment. Economic evaluation methodologies are useful for resolving such issues [5].
With its system of universal coverage through the National Health Insurance Service, Korea is facing a variety of issues, such as providing adequate services to the public, guaranteeing people’s right to make choices for their own health, and ensuring efficient compensation for providers [6]. Since thyroid cancer is a disease with a favorable prognosis, the problems in Korea also encompass challenges in screening, diagnosis, treatment, and long-term health care for thyroid cancer, suggesting that an appropriate allocation of limited resources is necessary. Thus, economic evaluation methodology is particularly important because efficient reimbursement in the National Health Insurance Service and people’s right to make choices for their own health are both important issues. Therefore, it is vitally important for thyroid cancer research to understand and apply economic evaluation methodology.
The purpose of this study was to introduce the methods of economic evaluation, key elements, and key indicators. To present economic evaluation methodology in an easy-to-understand way for thyroid cancer researchers, we briefly introduce several representative examples of previous thyroid cancer studies using economic evaluation methods. We also present several applicable examples from studies currently in progress at the Korean Thyroid Association.

### METHODS

This review focused on illustrating the concept and methods of economic evaluation with examples from thyroid cancer research. First, in order to explain the concept and methods of economic evaluation in an easy-to-understand way for thyroid cancer researchers, textbooks on economic evaluation were selected and summarized based on keywords and main methods [7,8]. Existing thyroid cancer studies applying the economic evaluation methodology were searched up to May 31, 2021 using a combination of keywords: (“Thyroid Cancer” & [“Cost analysis”, or “Cost-minimization analysis” or “Cost-effectiveness analysis” or “Cost-utility analysis”]) in MEDLINE (PubMed) and Google Scholar. A search was also performed with the same keywords in the Cochrane Database. The resulting 1,323 papers were all searched using the keywords, and after excluding irrelevant papers such as inappropriate literature based on the abstracts and titles, non-English papers, and papers with no access to the full text, 49 papers remained. Among them, we selected examples of straightforward interventions that are easy to understand for thyroid cancer researchers. Four thyroid cancer researchers selected each topic, and one to two examples of each method were selected through discussion and consensus.

### Concepts and main methods of economic evaluation

Economic evaluation is a type of comparative analysis between health interventions in terms of both their resource use (costs) and health outcomes [9,10]. Its goal is usually to allocate resources with sufficient efficiency that the maximum outcomes can be expected in a complicated health care environment [9,10].
Based on the methods and outcomes, economic evaluations can be classified into four types: cost-minimization analysis (CMA), cost-effectiveness analysis (CEA), cost-utility analysis (CUA), or cost-benefit analysis (CBA) (Table 1) [11]. In these types of economic evaluations, costs are measured in the same way, but the outcomes (cost, effectiveness, utility, and benefit) and their measurement methods are different.
CMA is a method of selecting a more economical intervention by comparing the cost of the interventions when the clinical effects (e.g., in terms of therapeutic effects between generics and patent medicines, safety and tolerability, and quality of life) of each intervention is the same as a prerequisite [12]. It calculates the additional cost of the intervention compared to the comparative intervention for one thyroid cancer case. An assumption of this approach is that the intervention and the comparative intervention have the same effect. A positive number indicates additional cost and a negative number indicates reduced cost. CMA has the benefit of simple and easy interpretation, and it is commonly used to compare generics and patent medicines as standard medications. However, CMA is inappropriate when there are differences in the effects of the intervention and comparison or in the measures and units of the effects.
If the clinical outcome and input cost of the targeted intervention and the comparative intervention are different, we calculate the cost difference (numerator) of the two interventions per the difference in outcomes (denominator) of the two interventions, which is the additional cost of the targeted intervention per a 1-unit increment of the outcome. Here, the result of each intervention is clinical effectiveness for CEA and quality of life for the CUA. The incremental cost-effectiveness ratio (ICER), which identifies the incremental (decreased) cost per 1-unit increased effectiveness, is used in the CEA framework [13]. CEA is the main decision-making tool in economic evaluations, and the effectiveness of interventions is measured in “natural units,” such as the recurrence rate, complication-free survival, complication incidence, or life-years [7]. For decision-making, it is necessary to compare the estimated ICER with the maximum acceptable ICER, which represents the maximum amount that society is willing to pay for an additional unit of health outcome [14]. The main advantage of CEA is that measuring outcomes in natural units simplifies the analysis, making it intuitive and facilitating easier communication among physicians. However, CEA faces limitations when comparing different diseases or projects with different measurement units. As another limitation, CEA overlooks quality of life; therefore, CUA has been suggested as an alternative method for analysis [8]. Similar to CEA, CUA uses the incremental cost-utility ratio (ICUR) as the outcome. The ICUR is defined as the incremental (decreased) cost per 1-unit of increased utility (quality of life or quality-adjusted life-year [QALY] due to thyroid cancer treatment or complications). Additionally, some previous studies dealt with quality of life or QALY from cancer treatment and complications as measures of utility [15-17].
CBA is a method of converting all possible alternatives into monetary values and comparing the benefits per cost (in monetary terms) of the alternatives. CBA compares the benefits of two alternatives according to their costs. To choose between alternatives, the costs should be assessed for both alternatives. If the costs of both alternatives are the same, the alternative with the larger benefit would be selected [18]. It is a method of measuring the opportunity cost of investing resources for their efficient allocation. CBA estimates the benefit per cost of a specific intervention, reported as the benefit-cost ratio (BCR). A BCR is greater than 1 indicates that the net present value of the project exceeds zero [19]. The BCR is commonly used in commercial transactions, business or policy decisions, and project investments [20]. When evaluating the economic value of interventions for medical health services, CBA is rarely used because it is difficult to assign a monetary value to clinical results [19]. Therefore, in the latter context, it is recommended to use CEA or CUA.
Of particular note, in economic evaluation, the main economic evaluation method should be selected before constructing the analysis model. The model can be a decision tree or Markov model (Supplemental Fig. S1). A decision tree model consists of decision nodes (squares), chance nodes (circles), probabilities, and outcome nodes (triangles), as shown in Supplemental Fig. S1A. A decision node is defined as a point where the decision-maker makes a choice. A chance node is defined as a probabilistic event beyond the control of the decision-maker. In addition, each chance node has probability branches, the sum of which should be 1. The probability can be estimated from actual data or a literature review. A Markov model is conducted based on the health status of a study population during an observation period. It has a constant rate of health transition for each health status over time, and it accumulates the cost or quality of life while individuals stay in that health status. The health statuses used in a Markov model should be mutually exclusive, and the model should have as simple a design as possible. Additionally, as shown in Supplemental Fig. S1B, there is a probability of moving to and from each state, which is defined as the transition probability. The sum of the transition probabilities for moving from each health status to other statuses should also be 1 (e.g., probability of moving from mild status to mild status, 0.37; from mild to moderate, 0.49; from mild to severe, 0.11; and from mild to death, 0.03; 0.37+0.49+0.11+0.03=1).

### Costs and outcomes as key elements in economic evaluation

#### Costs

One of the main challenges in economic evaluation analyses is to decide which costs should be included and how these costs should be measured and valued. Costs can be divided into direct, indirect, and intangible costs (Fig. 1) [21]. Direct costs can be divided into medical and non-medical costs. Direct medical costs are further divided into fixed costs, which must be paid irrespective of treatment (e.g., facility management and building leases), and variable costs, which vary depending on the amount of health care resources used (e.g., inpatient, outpatient, and pharmaceutical costs). Non-medical costs encompass transportation for outpatient, inpatient, and social services [22]. Because of individuals’ personal interest in their health status, it is also necessary to consider health supplements and alternative medicine within the category of complementary and alternative medicine (CAM) [23]. Although CAM is not included as a medical cost, it is relevant for a detailed accounting of non-medical costs. Indirect costs include loss of productivity, the value of the output lost due to a patient’s death (converted to monetary value), and inpatient time loss due to a patient’s morbidity or injury. Intangible costs are those incurred for psychological damages such as a patient’s anxiety and pain (converted to monetary value) [22]. Intangible costs are treated as a type of outcome in CEA and CUA, but as costs in CBA.
The value of current costs and future costs cannot be compared equally. Due to the time preference for the present, the value of future costs (the future monetary value) is lower than that of current costs (the current monetary value) [24]. The discount rate is used to correct for the future value. If the discount rate is 3% and considered 5-year of period the following formula can be applied: [present value]=[future value]×1/(1+ 0.03)5. The discount rate is applied in the process of assessing costs and benefits at different times. The discount rate varies depending on the economic status of each country and its health care system, and it should be considered in economic evaluations.

#### Outcomes

Outcome indicators vary depending on the types of economic evaluation. They can be divided into (1) clinical outcomes, (2) psychosocial outcomes, and (3) economic outcomes. Clinical outcomes consist of complication-free survival, morbidity, mortality, or complications resulting from interventions for thyroid cancer recurrence. Psychosocial outcomes consist of utility-based quality of life metrics (e.g., QALY). In particular, QALY is a generic measure of the outcome of medical interventions, including both the quality and the quantity of life [25]. The duration of time an individual is likely to spend in a health condition is weighted by a utility score from standard valuations (Fig. 2). Health-related quality of life is generally expressed on a numerical scale ranging from 0 (death) to 1 (perfect health), and it assumes that a year lived in perfect health is worth 1 QALY [25]. QALY is usually used for CUA. Lastly, economic outcomes include productivity loss, such as absence from work and work changes.
In CEA, clinical outcomes are mainly used as indicators of effectiveness, whereas in CUA, psychosocial results that have been adjusted for quality of life are mainly used as indicators of utility. Economic outcomes are generally included as indirect costs when estimating costs. In CBA, the benefit of an investment is estimated by considering the increase or decrease of all the above-mentioned results after the implementation of a specific program.

### Examples of economic evaluation in previous thyroid cancer studies

#### Cost minimization analysis

A previous study conducted in 2014 in Hong Kong sought to determine whether total thyroidectomy or total thyroidectomy with prophylactic central neck dissection was more cost-saving for nodal-negative (cN0) papillary thyroid cancer (PTC) (Table 2) [26]. A hypothetical model was applied based on a decision tree design for 20 years. The reference case was a non-pregnant female patient aged 50 years with a unifocal intrathyroidal 1.5-cm cN0 PTC. For cost estimation, the study applied an institutional perspective, considering only direct costs. For medical services provided within the national or state health care system, direct cost information was estimated from the government website or its publications. As cost items, workup procedures, complications, hospitalizations, and drugs related to the operation were included and derived from the government database. The probability of death based on reoperation, complications, or non-thyroidal causes following initial surgery was estimated from the literature. A 3% discount rate was applied for cumulative cost calculation, a sensitivity analysis was conducted, and the increased cost per patient was presented as economic evaluation indicators. The CMA analysis suggested that total thyroidectomy alone is less costly than total thyroidectomy with prophylactic central neck dissection in nodal-negative PTC ($6,702.81 vs.$10,062.35), with a consistent finding also reported for 20-year cumulative costs ($19,888.36 vs.$22,760.86).

#### Cost-effectiveness analysis

Many previous economic evaluation studies compared interventions to assess which management strategy (e.g., different surgery types) is more appropriate in thyroid cancer patients (Table 2) [27,28]. The two studies selected as examples of CEA compared two thyroid cancer surgical methods by calculating the additional costs per 1-unit increment of the clinical effectiveness of each surgical method relative to each other [27,28]. In a 2007 study, surgical complications, cause-specific mortality, the survival rate, and the recurrence rate were defined as effectiveness metrics for hemi-thyroidectomy and total thyroidectomy (Table 2) [27]. To assess the effectiveness of each intervention, a Markov model was constructed with a 20-year follow-up period. The probability of effectiveness according to the intervention was derived from the literature. For cost estimation, that study applied direct costs, such as diagnostic examinations including ultrasonography or fine-needle aspiration, surgery, and non-surgical treatment based on the New York state government and Medicare cost databases. The value of death resulting from unintentional impairment was also considered. The discount rate was defined as 6%, and the sensitivity analysis suggested acceptable thresholds for various parameters. The researchers concluded that total thyroidectomy was cost-effective compared to hemi-thyroidectomy based on the ICER. Total thyroidectomy was predominantly cost-effective based on overall survival (ICER: total thyroidectomy, $13,979; hemi-thyroidectomy,$15,184), recurrence-free survival (ICER: total thyroidectomy, $15,464; hemi-thyroidectomy,$20,005) compared to hemi-thyroidectomy. However, it showed heterogeneity according to the institution in long-term treatment outcomes.
Another study focused on two intervention strategies for bilateral neck dissection (BND): simultaneous and two-stage operations (Table 2) [28]. Patients with bilateral PTC from 2006 to 2012 who did not have a prior history of surgery or radiation exposure comprised the study population. Their pathology records were identified from postoperative clinical data. Surgical and medical expenses including neck dissection, ultrasonography, computed tomography, fine-needle aspiration, and outpatient, inpatient, and postoperative treatment charged by the hospital were considered as medical expenses. As indirect costs, accommodation, meals, and transportation fees were included. Disease-free status was defined as serum thyroglobulin levels of 1 ng/mL after recombinant human thyroid-stimulating hormone (TSH) therapy, negative neck ultrasound findings, and undetectable thyroglobulin antibody. A discount rate was not considered, and sensitivity and threshold analyses were not performed. The simultaneous BND strategy (cost-effectiveness ratio, defined as the cost per disease-free survivor: $4,423.34) was more cost-effective than the two-stage BND strategy (cost-effectiveness ratio:$8,158.25). However, two-stage BND was recommended in cases with invasion of the internal jugular vein.

#### Cost-utility analysis

CUA is a type of CEA used to compare the cost of a target intervention and a comparative intervention for thyroid cancer and the effect on the health status of patients who receive the intervention to determine the cost per 1-unit increment of the effect on health status. A life-year at perfect health status (theoretical life-year) is considered as 1 QALY and a life-year at death is 0 QALY [25]. In thyroid cancer, the meaning of “perfect health” is the status of no complications or no recurrence after primary treatment of thyroid cancer or having the best quality of life after maintenance treatment. The former two scales can estimate QALY as a value distributed between 1 QALY in the optimal state and 0 QALY in the death state using complications or recurrence-free years in the population. When using a health-related quality of life questionnaire such as the 36-item Short-Form Health Survey, the optimal quality of life is 1 QALY and the worst quality of life is 0 QALY [7,8]. QALY values between 0 and 1 can be estimated through the distribution of scores in each questionnaire and the validation of weights in a randomly selected sample population [7,8,25]. However, the conversion method depends on whether QALY is measured as a value at a single point in time or a cumulative value over several years; therefore, it is necessary to clarify how the results will be operationalized in advance.
A study published in 2020 conducted CUA for fiberoptic laryngoscopy prior to total thyroidectomy in low-risk thyroid cancer patients (T2N0M0 PTC) based on a decision model for 30 years (Table 2) [29]. The study only considered direct costs, which were converted into 2019 direct costs according to public expense data (Medicare) and the literature with a 3% discount rate. The probabilities and health utility values (ipsilateral and contralateral vocal fold paralysis, as well as hypoparathyroidism) in the model were derived from the literature including complication-free survival years and recurrence-free survival years according to the intervention. The threshold for willingness-to-pay was defined as 100,000($)/QALY. Given the sensitivity analysis, cost-acceptability curve, and time horizon according to each parameter in CUA, fiberoptic laryngoscopy prior to total thyroidectomy was found to be cost-effective for low-risk thyroid cancer patients ($45,193/QALY) and acceptable based on the willingness-to-pay threshold. As a sensitivity analysis, laryngoscopy was found to be cost-effective in 90.9% of cases, and the intervention was cost-effective if the incidence of vocal fold paralysis was at least 0.57%.
Another topic of interest is the immediate biopsy and active surveillance of thyroid nodules at 1-cm levels (Table 2) [30]. Therefore, a Markov model was constructed for CUA between immediate biopsy and active surveillance, and the reference case was a 40-year-old woman with 1-cm nodules in 2018. As inclusion criteria, the study considered patients without clinical or imaging evidence of cervical nodal metastases, a history of radiation exposure, or a family history of thyroid cancer. The outcome was defined as the performance and surgical outcomes of fine-needle aspiration, and the probability was derived from the literature. If the probability could not be identified from the literature, it was replaced by expert opinion. The effectiveness of each intervention was measured in QALYs. The time period was 50 years, but the interval of the ultrasound examinations was specifically set according to the nodule condition. For costs, medical and hospitalization fees were considered from the Medicare database or obtained from the literature if they could not be identified from the database, and a 3% discount rate was applied. The threshold value for willingness-to-pay was defined as 100,000/QALY. Through sensitivity analyses, the dominance point between active surveillance and biopsy was identified, as was the change in ICUR according to the malignancy probability. As a conclusion, active surveillance with ultrasonography was more cost-effective than an immediate biopsy (incremental cost,1,829; QALY, 0.016), with consistent dominance when the probability of malignancy is less than 84% and if the cost of ultrasonography does not exceed $793. However, if active surveillance caused considerable anxiety, immediate biopsy was more effective; therefore, quality of life should be considered for each intervention. #### Cost-benefit analysis CBA is generally used as a preliminary feasibility assessment of financial projects or as an analysis comparing the cost-benefit ratio between interventions. For example, rapid-transit railway enterprise and new town developments are domains where CBA has been widely performed. A single intervention can be judged as an economical project if the BCR is greater than 1 or if the net benefit is greater than 0. However, no previous CBA studies have been conducted on thyroid cancer. ### DISCUSSION In this study, the concept and methods of economic evaluation were summarized briefly with a focus on thyroid cancer, and we intended to help readers understand economic evaluation based on examples from previous studies. To conduct an economic evaluation for thyroid cancer, the definitions of several key elements should be discussed and clarified between researchers. The first points to consider are direct, indirect, and intangible costs [21]. Most previous studies used direct costs, but some studies also used complication costs, the value of death, and indirect costs [26-30]. Previous thyroid cancer studies used discount rates of 3% to 6%, although there were variations across countries and research periods. Costs were most commonly obtained from public expenditure databases such as Medicare or the previous literature, and the probabilities of outcomes according to interventions were derived from previous studies [26-30]. For the economic evaluation analyses, decision tree or Markov models were applied as hypothetical models. The second point to consider is the outcome. In CMA, it is common to present an excess cost per case or patient, assuming the same effect. However, most previous studies compared costs between interventions, which requires a clear distinction [31-36]. Because of the universality of the term “CEA,” regardless of the study type, numerous CUA studies have been reported as CEA studies, and the indicators were also presented as the ICER, not the ICUR [37-41]. It is possible to evaluate whether the derived ICER or ICUR is socially acceptable compared to thresholds for decision-making, as social acceptability defines the concept of willingness-to-pay. Socially acceptable marginal costs vary from country to country and over time. In studies in the United States, the cost per QALY is considered socially acceptable if it is lower than per capita gross domestic product (GDP). In 2008, the United States per capita GDP was about 48,000 dollars, and the United States threshold was therefore$50,000/QALY [42].
A highlight of recent research on thyroid cancer is the economic evaluation of surgery and active surveillance in patients with low-risk thyroid cancer [33,34,40,43]. Due to the high survival rate and good prognosis of low-risk thyroid cancer, especially papillary thyroid microcarcinoma (PTMC), the American Thyroid Association recommended that immediate surgery is not necessary for patients with low-risk tumors such as PTMC without clinically evident metastases or local invasion [44]. However, active surveillance using ultrasonography also leads to an economic burden in the long term, and patients may also live with the anxiety that cancer might develop. These points need to be considered in an economic evaluation. In this context, a prospective cohort for PTMC patients in Korea (Multicenter Prospective Cohort Study of Active Surveillance on Papillary Thyroid Microcarcinoma [MAeSTro]) was conducted to compare lobectomy and active surveillance in 2016 [45]. MAeSTro is a pragmatic cohort study with the aim of evaluating the evidence regarding whether active surveillance is helpful in reducing unnecessary surgery in patients with small, low-risk thyroid cancer. Thyroid cancer recurrence, complications, and quality of life were compared as measures of effectiveness between the two interventions. Furthermore, researchers at the Korean Thyroid Association launched a practical clinical trial (the Multicenter Randomized Controlled Study for Assessing the Usefulness of Suppressing Thyroid Stimulating Hormone Target Levels after Thyroid lobEctomy in Low to Intermediate Risk Thyroid Cancer Patients [MASTER] study) for thyroid cancer patients who underwent lobectomy for low-risk and intermediate-risk cancer to compare recurrence, quality of life, and complications (including cardiovascular disease, osteoporosis, and metabolic disease, and musculoskeletal disease) [46]. For the first study, if the effects of the two interventions are the same, the economic benefit can be evaluated through CMA. The second study aims to establish evidence about how long TSH levels should be maintained in TSH suppression therapy. The degree of levothyroxine use differs according to the maintenance of TSH levels; thus, TSH suppression therapy has both clinical benefits and risks. Therefore, in addition to CEA, an evaluation through CUA considering quality of life should be performed.

### CONCLUSIONS

In this study, we described the concept and methods of economic evaluation and reviewed previous studies focusing on thyroid cancer. Thyroid cancer has a favorable prognosis, with a high chance of long-term survival. Because of the characteristics of thyroid cancer, it is controversial whether the medical costs administered for thyroid cancer are appropriate. In particular, in countries with a national health insurance system such as Korea, it is necessary to select the most efficient and optimal method to support various medical services in a balanced way within limited financial resources. Economic evaluation is a method for making optimally cost-efficient choices in a resource-limited setting. By understanding and applying the precise concepts and methods of economic evaluation, we can produce comparative results to elucidate controversial issues in an evidence-based manner. Economic evaluation studies are expected to provide appropriate guidelines for thyroid cancer surveillance, diagnosis, and management.

### Supplementary Information

##### Supplemental Fig. S1.
Main model of economic evaluation analysis. (A) Decision tree model, (B) Markov model.
enm-2021-1164-suppl.pdf

### ACKNOWLEDGMENTS

This research was supported by a grant from the Patient-Centered Clinical Research Coordinating Center funded by the Ministry of Health & Welfare, Republic of Korea (grant number: HI19C0481, HC19C0103).
The researchers were members of the Multicenter Randomized Controlled Study for Assessing the Usefulness of Suppressing Thyroid Stimulating Hormone Target Levels after Thyroid lobEctomy in Low to Intermediate Risk Thyroid Cancer Patients (MASTER) trial conducted through the Korean Thyroid Association.

### NOTES

CONFLICTS OF INTEREST

No potential conflict of interest relevant to this article was reported.

##### Fig. 1
Main types of costs in economic evaluations.
##### Fig. 2
Basic principle of the quality-adjusted life year (QALY) in economic evaluations.
##### Table 1
Basic Concepts of Economic Evaluation
Method Description Measures Indicators
Cost-minimization analysis (CMA) Comparison of costs when the intervention group and comparison group have clearly equivalent clinical effects Only the cost is measured; the effects are not measured since these are assumed to be equal. Excess or additional cost per case
Cost-effectiveness analysis (CEA) Comparison of the relative ratio of cost and effectiveness between an intervention group and a comparison group, which are estimated as different values This method calculates the cost difference (numerator) of the two interventions per the difference in outcomes (denominator) of the two interventions as the ICER. The ICER indicates the incremental (decreased) cost per 1 unit of increased effectiveness. ICER
Cost-utility analysis (CUA) Comparison of different diseases or projects with different units of measure, considering quality of life (health status) This method calculates the cost per 1-unit increment of the effect on the health status as the ICUR. A life-year at perfect health status is considered as 1 QALY and a life-year at death is 0 QALY. ICUR
Cost-benefit analysis (CBA) Method of converting all possible alternatives into monetary values and comparing the benefits per cost (in monetary terms) of the alternatives Monetary units BCR

ICER, incremental cost-effectiveness ratio; ICUR, incremental cost-utility ratio; QALY, quality-adjusted life-year; BCR; benefit-cost ratio.

##### Table 2
Summary of Previous Studies That Conducted Economic Evaluations Focusing on Thyroid Cancer
Study Country Patients Intervention [I]
Comparison [C]
Direct cost [DC]
Indirect cost [IC]
Outcomes Analysis Results and conclusion
Lang et al. (2014) [26] Hong Kong 100,000 Hypothetical non-pregnant female patients in a cohort at age 50 years with low-risk PTC (1.5-cm cN0 PTC within one lobe) [I] TT with pCND
[C] TT alone
[DC] 20-Year cumulative cost, including procedural, complications, hospitalizations, annual routine surveillance (hospital costs published in the 2013 Government Gazette)
[IC] Not included 3% discount rate
Excess cost per case Hypothetical model based on decision-tree design TT+pCND is more expensive in the medium- and long-term in low-risk PTC patients
TT: USD 19,888.36 (20-year cumulative cost)
TT: USD 19,888.36 (20-year cumulative cost)
TT+pCND: USD 22,760.86 (20-year cumulative cost)
Incremental cost: USD 2,872.50
Sensitivity analyses: no change
Shrime et al. (2007) [27] Literature review from PubMed search 940 PTC citations published from January 1, 1966 to January 1, 2007 1 Million hypothetical low-risk PTC cases (age, metastasis, extent, and size of tumor [AMES] score <6, an AGES score <4, or adherence to the Memorial Sloan-Kettering low-risk categories) [I] HT
[C] TT
[DC] Hospital charges for inpatient operative procedure, follow-up costs, complication costs, and recurrence costs from Medicare
[IC] Costs for death from unintentional injuries: $1,130,000 based on the National Safety Council estimates Cost and ICER for CSM or RFS CSM or RFS under the same cost assumption CSM and RFS were derived from a literature review (31 studies were finally selected) Monte Carlo microsimulation using the fixed probability estimates of complications and recurrence derived from the 31 eligible studies Sensitivity analyses using a 6% annual discount and cost-to-cost ratio estimated from the 31 eligible studies Threshold analyses: comparison of net monetary benefits obtained from TT and HT using a variable range of WTP values TT dominates HT as initial treatment for low-risk PTC 20-Year ICER for overall survival: [I] HT,$15,184; [C] TT, $13,979 20-Year ICER for RFS: [I] HT,$15,464; [C] TT, $20,005 TT is consistently cost-effective in Monte Carlo simulations Sensitivity analyses: HT was sensitive to recurrence rates and follow-up costs Threshold analyses: robust when compared with WTP Guo et al. (2018) [28] China 256 PTC patients who underwent TT and BND [I] TT with simultaneous BND [C] TT with two-stage BND [DC] Hospital charges for surgery and related charges during hospital stays were obtained [IC] Accommodation, meals, and transportation fees Cost for DFS during a mean follow-up of 5 years Cost-effectiveness analysis was performed using a retrospective cohort. Sensitivity analyses: not done Threshold analyses: not done Discount rate was not applied Simultaneous BND was more cost-effective for the management of PTC patients than two-stage BND Two-stage BND was recommended in cases of internal jugular vein invasion 5-Year cost of surgery: simultaneous BND,$4,145.32; two-stage BND, $7,352.53 Cost per DFS: simultaneous BND,$4,423.34; two-stage BND, $8,158.25 Walgama et al. (2020) [29] USA Hypothetical T2N0M0 PTC patients with undiscovered ipsilateral vocal fold paralysis. [I] Preoperative laryngoscopy prior to TT [C] No preoperative laryngoscopy prior to TT [DC] Medicare reimbursement data in Medicare from 2019 and the recent medical literature [IC] Not included 3% discount rate Cost-effectiveness measured by ICER (ΔCost/ΔQALY) Ipsilateral and contralateral vocal fold paralysis, as well as hypoparathyroidism over 30 years Decision tree model with literature-based probabilities Sensitivity analyses: varying the rate of VCP and selected variables. Monte Carlo simulation was performed to determine the overall uncertainty of the model. Threshold analyses: threshold of$100,000/QALY
Fiberoptic laryngoscopy is cost-effective prior to TT in asymptomatic, low-risk thyroid cancer patients
Preoperative laryngoscopy has an ICER of $45,193/QALY compared to no laryngoscopy Sensitivity analyses: laryngoscopy is cost-effective in 90.9% of cases Threshold analyses: the intervention is cost-effective if the incidence of vocal fold paralysis is at least 0.57%, or when the permissible rate of HT in cases of incidental paralysis is at least 41% Kuo et al. (2018) [30] USA A euthyroid 40-year-old patient with an incidentally discovered thyroid nodule measuring 1 cm. The reference case did not have any clinical or radiologic evidence of cervical nodal metastases, neck surgery, radiation, or family history of thyroid cancer. [I] Ultrasound surveillance [C] Immediate biopsy [DC] Medicare reimbursement data in 2016, hospitalization cost data from the Nationwide Inpatient Sample, and a literature review. A 3% annual discount rate was applied to all future costs. [IC] Not included 40-Year quality-adjusted life expectancy measured by ICER (ΔCost/ΔQALY) Effectiveness was measured in QALY. Each transition-state in the model was assigned a quality-adjustment factor based on literature and previous work. Event outcomes were natural history of thyroid nodules and thyroid cancer, test performance of thyroid FNA, and surgical outcomes of thyroidectomy and neck dissection. Markov transition-state model with probabilities obtained from the literature and expert opinion Sensitivity analyses: one-way and two-way sensitivity analyses and a Monte Carlo simulation were performed Threshold analyses: threshold of$100,000/QALYs
US surveillance is more cost-effective than immediate biopsy for 1.0-cm thyroid nodules with an intermediate-suspicion sonographic pattern
US surveillance was 1,829 less costly and 0.016 QALYs more effective than immediate biopsy Sensitivity analyses: immediate biopsy became cost-effective when the probability of malignancy increased from 15% to 84% or the cost of US increased from129 to \$793.
Threshold analyses: robust when compared with WTP

PTC, papillary thyroid cancer; TT, total thyroidectomy; pCND, prophylactic central neck dissection; USD, US dollar; AGES, Age Grade Extent Size; HT, hemi-thyroidectomy; ICER, incremental cost-effectiveness ratio; CSM, cause-specific mortality; RFS, recurrence-free survival; WTP, willingness to pay; BND, bilateral neck dissection; DFS, disease-free survival; QALY, quality-adjusted life-year; VCP, vocal cord paralysis; FNA, fine needle aspiration; US, United States.

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33. Lin JF, Jonker PK, Cunich M, Sidhu SB, Delbridge LW, Glover AR, et al. Surgery alone for papillary thyroid microcarcinoma is less costly and more effective than long term active surveillance. Surgery 2020;167:110-6.
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34. Oda H, Miyauchi A, Ito Y, Sasai H, Masuoka H, Yabuta T, et al. Comparison of the costs of active surveillance and immediate surgery in the management of low-risk papillary microcarcinoma of the thyroid. Endocr J 2017;64:59-64.
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35. Boltz MM, Hollenbeak CS, Schaefer E, Goldenberg D, Saunders BD. Attributable costs of differentiated thyroid cancer in the elderly Medicare population. Surgery 2013;154:1363-70.
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36. Lubitz CC, Kong CY, McMahon PM, Daniels GH, Chen Y, Economopoulos KP, et al. Annual financial impact of well-differentiated thyroid cancer care in the United States. Cancer 2014;120:1345-52.
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37. Lang BH, Wong CK. A cost-effectiveness comparison between early surgery and non-surgical approach for incidental papillary thyroid microcarcinoma. Eur J Endocrinol 2015;173:367-75.
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38. Lee L, How J, Tabah RJ, Mitmaker EJ. Cost-effectiveness of molecular testing for thyroid nodules with atypia of undetermined significance cytology. J Clin Endocrinol Metab 2014;99:2674-82.
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39. Wilson L, Huang W, Chen L, Ting J, Cao V. Cost effectiveness of lenvatinib, sorafenib and placebo in treatment of radioiodine-refractory differentiated thyroid cancer. Thyroid 2017;27:1043-52.
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40. Venkatesh S, Pasternak JD, Beninato T, Drake FT, Kluijfhout WP, Liu C, et al. Cost-effectiveness of active surveillance versus hemithyroidectomy for micropapillary thyroid cancer. Surgery 2017;161:116-26.
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41. Zanocco K, Heller M, Elaraj D, Sturgeon C. Cost effectiveness of intraoperative pathology examination during diagnostic hemithyroidectomy for unilateral follicular thyroid neoplasms. J Am Coll Surg 2013;217:702-10.
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42. Cleemput I, Neyt M, Thiry N, De Laet C, Leys M. Threshold values for cost-effectiveness in health care. Belgian Health Care Knowledge Centre. HTA reports. Brussels: KCE; 2008. p. 88.
43. Lohia S, Hanson M, Tuttle RM, Morris LG. Active surveillance for patients with very low-risk thyroid cancer. Laryngoscope Investig Otolaryngol 2020;5:175-82.
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44. Haugen BR, Alexander EK, Bible KC, Doherty GM, Mandel SJ, Nikiforov YE, et al. 2015 American Thyroid Association management guidelines for adult patients with thyroid nodules and differentiated thyroid cancer: the American Thyroid Association Guidelines Task Force on thyroid nodules and differentiated thyroid cancer. Thyroid 2016;26:1-133.
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46. Lee EK, Kang YE, Park YJ, Koo BS, Chung KW, Ku EJ, et al. A multicenter, randomized, controlled trial for assessing the usefulness of suppressing thyroid stimulating hormone target levels after thyroid lobectomy in low to intermediate risk thyroid cancer patients (MASTER): a study protocol. Endocrinol Metab (Seoul) 2021;36:574-81.
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text:
# A Lot Of Rooks!

Dimitri places the maximum possible number of Rooks on a $$8\times 8$$ chessboard in such a way that there's no two Rooks that attack each other. In how many ways can he do that?

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#### Phase transition in noisy high-dimensional random geometric graphs

##### Suqi Liu, Miklos Z. Racz

We study the problem of detecting latent geometric structure in random graphs. To this end, we consider the soft high-dimensional random geometric graph $\mathcal{G}(n,p,d,q)$, where each of the $n$ vertices corresponds to an independent random point distributed uniformly on the sphere $\mathbb{S}^{d-1}$, and the probability that two vertices are connected by an edge is a decreasing function of the Euclidean distance between the points. The probability of connection is parametrized by $q \in [0,1]$, with smaller $q$ corresponding to weaker dependence on the geometry; this can also be interpreted as the level of noise in the geometric graph. In particular, the model smoothly interpolates between the spherical hard random geometric graph $\mathcal{G}(n,p,d)$ (corresponding to $q = 1$) and the Erd\H{o}s-R\'enyi model $\mathcal{G}(n,p)$ (corresponding to $q = 0$). We focus on the dense regime (i.e., $p$ is a constant). We show that if $nq \to 0$ or $d \gg n^{3} q^{2}$, then geometry is lost: $\mathcal{G}(n,p,d,q)$ is asymptotically indistinguishable from $\mathcal{G}(n,p)$. On the other hand, if $d \ll n^{3} q^{6}$, then the signed triangle statistic provides an asymptotically powerful test for detecting geometry. These results generalize those of Bubeck, Ding, Eldan, and R\'acz (2016) for $\mathcal{G}(n,p,d)$, and give quantitative bounds on how the noise level affects the dimension threshold for losing geometry. We also prove analogous results under a related but different distributional assumption, and we further explore generalizations of signed triangles in order to understand the intermediate regime left open by our results.

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Show Summary Details
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Journal für die reine und angewandte Mathematik

Managing Editor: Weissauer, Rainer

Ed. by Colding, Tobias / Huybrechts, Daniel / Hwang, Jun-Muk / Williamson, Geordie

IMPACT FACTOR 2018: 1.859

CiteScore 2018: 1.14

SCImago Journal Rank (SJR) 2018: 2.554
Source Normalized Impact per Paper (SNIP) 2018: 1.411

Mathematical Citation Quotient (MCQ) 2017: 1.49

Online
ISSN
1435-5345
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Convergence to equilibrium of gradient flows defined on planar curves

Matteo Novaga
/ Shinya Okabe
Published Online: 2015-06-05 | DOI: https://doi.org/10.1515/crelle-2015-0001

Abstract

We consider the evolution of open planar curves by the steepest descent flow of a geometric functional, with different boundary conditions. We prove that, if any set of stationary solutions with fixed energy is finite, then a solution of the flow converges to a stationary solution as time goes to infinity. We also present a few applications of this result.

References

• [1]

J. W. Barrett, H. Garcke and R. Nürnberg, Parametric approximation of isotropic and anisotropic elastic flow for closed and open curves, Numer. Math. 120 (2012), 489–542.

• [2]

G. Bellettini, C. Mantegazza and M. Novaga, Singular perturbations of mean curvature flow, J. Differential Geom. 75 (2007), 403–431.

• [3]

H. Brezis, Opérateurs maximaux monotones et semi-groupes de contractions dans les espaces de Hilbert, North-Holland, Amsterdam 1973. Google Scholar

• [4]

A. Dall’Acqua and P. Pozzi, Willmore–Helfrich ${L}^{2}$-flow of curves with natural boundary conditions, Comm. Anal. Geom. 22 (2014), no. 4, 617–669. Google Scholar

• [5]

K. Deckelnick and H.-C. Grunau, Boundary value problem for the one-dimensional Willmore equation, Calc. Var. Partial Differential Equations 30 (2007), 293–314.

• [6]

G. Dziuk, E. Kuwert and R. Schätzle, Evolution of elastic curves in ${ℝ}^{n}$: Existence and computation, SIAM J. Math. Anal. 33 (2002), 1228–1245. Google Scholar

• [7]

Y. Giga and R. V. Kohn, Characterizing blowup using similarity variables, Indiana Univ. Math. J. 36 (1987), no. 1, 1–40.

• [8]

N. Koiso, On the motion of a curve towards elastica, Acta de la table ronde de géométrie différentielle (Luminy 1992), Sémin. Congr. 1, Société Mathématique de France, Paris (1996), 403–436. Google Scholar

• [9]

C.-C. Lin, ${L}^{2}$-flow of elastic curves with clamped boundary conditions, J. Differential Equations 252 (2012), 6414–6428.

• [10]

A. Linnér, Some properties of the curve straightening flow in the plane, Trans. Amer. Math. Soc. 314 (1989), no. 2, 605–618.

• [11]

M. Novaga and S. Okabe, Curve shortening-straightening flow for non-closed planar curves with infinite length, J. Differential Equations 256 (2014), no. 3, 1093–1132.

• [12]

S. Okabe, The motion of elastic planar closed curves under the area-preserving condition, Indiana Univ. Math. J. 56 (2007), no. 4, 1871–1912.

• [13]

S. Okabe, The dynamics of elastic closed curves under uniform high pressure, Calc. Var. Partial Differential Equations 33 (2008), no. 4, 493–521.

• [14]

A. Polden, Curves and surfaces of least total curvature and fourth-order flows, Ph.D. thesis, University of Tübingen, 1996. Google Scholar

• [15]

L. Simon, Asymptotics for a class of nonlinear evolution equations, with applications to geometric problems, Ann. of Math. (2) 118 (1983), no. 3, 525–571.

• [16]

Y. Wen, ${L}^{2}$-flow of curve straightening in the plane, Duke Math. J. 70 (1993), no. 3, 683–698. Google Scholar

• [17]

Y. Wen, Curve straightening flow deforms closed plane curves with nonzero rotation number to circles, J. Differential Equations 120 (1995), no. 1, 89–107.

Revised: 2014-12-26

Published Online: 2015-06-05

Published in Print: 2017-12-01

The second author was partially supported by the JSPS Strategic Young Researcher Overseas Visits Program for Accelerating Brain Circulation and by Grant-in-Aid for Young Scientists (B) (No. 24740097).

Citation Information: Journal für die reine und angewandte Mathematik (Crelles Journal), Volume 2017, Issue 733, Pages 87–119, ISSN (Online) 1435-5345, ISSN (Print) 0075-4102,

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© 2017 Walter de Gruyter GmbH, Berlin/Boston.


text:
# What is the arclength of f(x)=1/sqrt((x-1)(2x+2)) on x in [6,7]?

Jun 9, 2018

$1.000153728$

#### Explanation:

Wrting your $f \left(x\right)$ in the form

$f \left(x\right) = \frac{1}{\sqrt{2}} \cdot \frac{1}{\sqrt{{x}^{2} - 1}}$

then $f ' \left(x\right)$ is given by

$f ' \left(x\right) = - \frac{1}{2} \cdot \sqrt{2} \frac{x}{{x}^{2} - 1} ^ \left(\frac{3}{2}\right)$
then our integralo is given by

$\frac{1}{2} \cdot {\int}_{6}^{7} \sqrt{4 + 2 \cdot {x}^{2} / {\left({x}^{2} - 1\right)}^{3}} \mathrm{dx}$
A numerical method gives
$\approx 1.000153728$


text:
# Synopsis: W Marks the Spot

Two experimental collaborations at Fermilab report a new measurement of the W boson mass that leads to a better prediction of the mass of the Higgs boson.

Long before experimentalists at the Large Hadron Collider reported hints of the Higgs boson in particle collisions (see 13 March 2012 Viewpoint), physicists knew roughly what the Higgs mass had to be from measurements of the W boson. That’s because according to the standard model, the W boson, one of the particles that mediates the weak interaction, can emit a virtual Higgs boson and reabsorb it, which alters the W boson’s mass. The mass of the W boson also shifts due to a virtual process containing a top and bottom quark. So with a precise measurement of the W mass, and a good measurement of the top quark mass, it is possible to predict the mass of the Higgs boson.

Now the CDF and D0 Collaborations at Fermilab are each reporting in Physical Review Letters their new measurements of the W mass using datasets containing a total of about 2 million W decays to an electron or muon and a neutrino. By analyzing the kinematics from this large sample, the two experiments achieve a combined precision of about $0.02%$.

These new values narrow the allowed range in top-W mass space. The band of top-W masses corresponding to the 115–127 GeV range of Higgs masses, allowed by direct searches, goes right through the allowed region determined by CDF and D0. If the LHC does find the Higgs boson in the 115–127 GeV mass window, it will be yet another success for the predictions of the standard model. – Robert Garisto

More Features »

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## Subject Areas

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Confirming previous controversial results, the MiniBooNE experiment detects a signal that is incompatible with neutrino oscillations involving just the three known flavors of neutrinos. Read More »

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### Viewpoint: Dissecting the Mass of the Proton

A calculation determines four distinct contributions to the proton mass, more than 90% of which arises entirely from the dynamics of quarks and gluons. Read More »


text:
# What is the effect of using survey sample weights for a sub-sample?

If a sub-sample of the survey sample, selected based on certain demographic characteristics of the data (e.g. age, race etc.), is used, which means the sub-sample might not be representative of the population anymore, is it better to not use original sampling survey weights provided or is it still better to use the survey weights calibrated for the original survey sample data?

P.S. If I choose to generate my own weights based on a sub-sample data, what is the best methodology for this, and could someone please point me to the references that might be helpful?

If the option of generating own weights isn't available, what would be the second best option?

new_subsample <- subset( full_sample , some_demographic_group == TRUE )


text:
2-dimensional Lebesgue-Stieltjes measure (Bivariate Density to Distribution Function)

Consider a probability space $(\Omega,\mathcal{F},\mathbb{P})$ along with a random variable $X:\Omega\rightarrow\mathbb{R}$ such that $(\mathbb{R},\mathcal{B},\mu)$ is an induced probability space where $\mathcal{B}$ is to denote the Borel $\sigma$-algebra and $\mu$ is a pushforward measure such that $\mu=\mathbb{P}\circ X^{-1}$. Hence, $\mathbb{P}(X\in B)=\mu(B)$ for all $B\in\mathcal{B}$.

It then follows that $\mu$ - as a Lebesgue-Stieltjes measure $\mu(a,b]=F(b)-F(a)$ - is linked to a probability distribution function $F$ such that $F$ is monotonically increasing, right-continuous, and bounded to take values in $\,[0,1]$ with $F(-\infty)=0$ and $F(\infty)=1$.

I do not really feel comfortable with this result, but I was told that

$$\text{d}\mu=\mu(x,x+\text{d}x]=F(x+\text{d}x)-F(x)=\text{d}F.$$

Therefore, one can define the Lebesgue-Stieltjes integral of an arbitrary function $g$ as

$$\int_Bg\,\text{d}F=\int_{B}g\,\text{d}\mu\qquad\forall\quad B\in\mathcal{B}$$

where the righthandside is just the conventional Lebesgue integral w.r.t $\mu$.

Assuming $\mu\ll\lambda$ where $\lambda$ is the 1-dim Lebesgue measure, $F$ is differentiable such that $\text{d}F=F^\prime\,\text{d}x$. It thus follows that $$\int_Bg\,F^\prime\,\text{d}x=\int_{B}g\,\text{d}\mu\qquad\forall\quad B\in\mathcal{B}.$$ Moreover, since $\mu\ll\lambda$ the Radon-Nikodym theorem dictates that there exists a density function $f$ such that $$\int_{B}g\,\text{d}\mu=\int_{B}g\,f\,\text{d}\lambda$$

And finally, since $\lambda$ is a Lebesgue-Stieltjes measure with $F$ such that $x\mapsto x$ such that $\text{d}\lambda=\text{d}x$

$$\int_{B}g\,F^\prime\,\text{d}x=\int_{B}g\,f\,\text{d}\lambda=\int_Bg\,f\,\text{d}x$$ Since this statement holds true for an arbitrary function $g$ as well as for every set $B\in\mathcal{B}$ it would seem to follow that $F^\prime=f$ such that the pdf of a continuous random variable $X$ is simply the derivative of the corresponding cdf.

I would like to be able to transfer this line of reasoning to a bivariate case such that where a vector-valued measurable function $\boldsymbol{X}:\Omega\rightarrow\mathbb{R}^2$ gives a random variable that allows for the construction of an induced probability space $(\mathbb{R}^2,\mathcal{B}_2,\mu)$. At this point $\mathcal{B}_2=\mathcal{B}\otimes\mathcal{B}$ is the product $\sigma$-algebra of $\mathcal{B}$ and $\mu$ is a pushforward measure $\mu=\mathbb{P}\circ \boldsymbol{X}^{-1}$.

Starting from

$\mu(a,b]\times(c,d]=F(b,d)-F(a,d)-F(b,c)+F(a,c)$

(which is just my starting point since $\mathbb{P}(a<X\leq b,c<Y\leq d)=F(b,d)-F(a,d)-F(b,c)+F(a,c)$
and not because I would have read anywhere that the 2-dim Lebegue-Stieltjes measure is defined this way)

I would like to see why $$\text{d}\mu=\text{d}F$$

such that I can arrive at a statement to the effect that

$$\int_{B}g\,\dfrac{\partial^2 F}{\partial x\partial y}\,\text{d}x\,\text{dy}=\int_Bg\,f\,\text{d}x\,\text{d}y.$$

Thank you so much for taking the time to read my question. I would appreciate any pointers you might have for me - especially on why $\text{d}\mu=\text{d}F$ should be true in the bivariate case.

Best regards,
Jon

It seems to me that you are going in circles.

Let $\mu$ denote some probability measure on $(\mathbb R^2,\mathcal B_2)$.

Then $\mu$ induces a function $F:\mathbb R^2\to\mathbb R$ that is prescribed by: $$(b,d)\mapsto P((-\infty,b]\times(-\infty,d])$$

This function on its turn induces a (unique) measure $\mu_F$ that is characterized by: $$\mu_F((a,b]\times(c,d])=F(b,d)-F(a,d)-F(b,c)+F(a,c)$$ Observe that we could also write this as:$$\mu_F((a,b]\times(c,d])=\mu((a,b]\times(c,d])$$

So actually now we have two measures $\mu$ and $\mu_F$ that coincide on the collection: $$\mathcal A=\{(a,b]\times(c,d]\mid -\infty<a<b<\infty, -\infty<c<d<\infty\}$$

This collection is closed under intersection so $\mu$ and $\mu_F$ will also coincide on the sigma-algebra generated by $\mathcal A$ which is $\mathcal B_2$.

In a way this closes the circle: $\mu=\mu_F$.

The expression $\text{d}\mu=\text{d}F$ is just a notation for $\mu=\mu_F$.

• Thank you so very much for your answer. So $\text{d}\mu=\text{d}F$ is essentially notational justification for writing $$\int_{B}\,g\,\text{d}\mu=\int_B\,g\,\text{d}\mu_F=\int_B\,g\,\text{d}F?$$ I am afraid, however, since the lhs is evaluated as a Lebesgue-integral and the rhs is a Stieltjes-integral and thus an integral w.r.t. a function $F:\mathbb{R}^2\rightarrow[0,1]$ rather than a measure $\mu:\mathcal{B}\rightarrow[0,1]$, I have to admit, that I still have a hard time seeing why the two need to be equal just as, by definition, $F(b,d)=\mu(-\infty,b]\times(-\infty,d]$. – J.Beck Jul 24 '18 at 14:30
• According to my view every integral is based on a measure. We can write $\int gdF$ but actually that is only a notation for $\int gd\mu_F$ where $\mu_F$ is a unique measure generated by $F$. The notation $d\mu=dF$ tells us that we are dealing with the same measure. Also have a look here. – drhab Jul 24 '18 at 14:55
• If I may be a bit tenacious. Presuming $\mu_F$ is absolutely continuous w.r.t. to $\lambda$, then $F$ is a real-valued continuous function and the Lebesgue-Stieltjes integral specializes to a Rieman-Stieltjes integral. Hence, we can write $\text{d}F=F^\prime\,\text{d}x$ and end up with $$\int\,g\,\text{d}\mu=\int\,g\,\text{d}\mu_F=\int\,g\,\text{d}F=\int\,g\,F^\prime\,\text{d}x$$ which is a simple Rieman integral. I fail to see, how it can just be a question of notation to go from $\int g\,\text{d}\mu$ to $\int g\,F^\prime\,\text{d}x$ – J.Beck Jul 24 '18 at 15:05
• For the last equality $\mu_F$ must be absolutely continuous wrt Lebesguemeasure $\lambda$. Then a density $f$ exists with $\int gdF=\int gfd\lambda$ or in other notation $=\int gfdx$. For this the derivative of $F$ does not have to exist. If it does then $g=F'$ can be chosen as density (there are more than one candidates). – drhab Jul 24 '18 at 15:13
• What I basically want to say is that I feel that there must be a proof to show that this is true iff $F(b,d)=\mu(-\infty,b]\times(-\infty,d].$ – J.Beck Jul 24 '18 at 15:13

Thank to the helpful input of drhab, I believe I have pinpointed my problem:

For the sake of argument I confine my attention to the one-dim case. Let g be a continuous function and $\mu\ll\lambda$ (and therefore also $\mu_F\ll\lambda$). It follows that $F$ is continuous and differentiable.

By the Radon-Nikodym theorem there exists (more than one) density function $f$ such that $\text{d}\mu=f\,\text{d}\lambda$ and $\text{d}\mu_F=f\,\text{d}\lambda$. Hence one can write $$\int_Bg\,\text{d}\mu=\int_Bg\,f\,\text{d}\lambda.$$

Moreover, since $F$ is a differentiable function, it is also true that - in a usual calculus sense - $\text{d}F=F^\prime\,\text{d}x$. Just substituting that in yields $$\int_Bg\,\text{d}F=\int_Bg\,F^\prime\,\text{d}x.$$ On their own, I am okay with both of these statements. I am struggling with the fact, that the notational convention above says that they are equal such that since $$\int_Bg\,\text{d}\mu=\int_Bg\,\text{d}F\quad\text{ it follows that }\quad\int_Bg\,f\,\text{d}\lambda=\int_Bg\,F^\prime\,\text{d}x.$$

I suppose it would already help, if I could see a rigorous proof that

$\,\,\,$a) $\quad\text{d}\lambda=\text{d}x$
(I suppose this says that Rieman and Lebesgue integral coincide provided the Rieman integral exists)

$\,\,\,$b) $\quad F^\prime$ is, in fact, a Radon-Nikodym derivative of $\mu_F$ with respect to $\lambda$.

I am afraid, I could show b) only by relying on the notational convention that $\text{d}\mu_F=\text{d}F$ and $\text{d}\lambda=\text{d}x$ (where $\text{d}x$ is essentially $\text{d}F$ of the distribution function prescribed by $x\mapsto x$, i.e. the distribution function corresponding to the Lebesgue measure). Then one could that argue that $$\dfrac{\text{d}\mu_F}{\text{d}\lambda}=f\quad\text{and}\quad\dfrac{\text{d}F}{\text{d}x}=F^\prime$$ But since $\text{d}\mu_F=\text{d}F$ and $\text{d}\lambda=\text{d}x$ $$\dfrac{\text{d}\mu_F}{\text{d}\lambda}=\dfrac{\text{d}F}{\text{d}x}\quad\text{such that}\quad f=F^\prime$$ However, this does not answer the question why the derivative of a distribution function is a Radon-Nikodym derivative of the measure induced by said distribution function with respect to Lebesgue measure.

Once again, thank you so much.

Best regards,
Jon

PS. Having read the wikipedia article on "Absolute Continuity", the third paragraph in the section on the "Relation between the two notions of absolute continuity" asserts that
If the absolute continuity holds then the Radon–Nikodym derivative of μ is equal almost everywhere to the derivative of F .

This is precisely what I need to see proved in order to resolve my struggle. Wikipedia cites Royden 1988, Problem 12.17(b) on page 303.


text:
Question

# What is free capacity in the Avanti - Vidisha pipeline?

A
300
B
200
C
100

Solution

## The correct option is B 0Following is the requirements at different cities: Total requirement Jyotishamati + Panchal  = 700 + 400 = 1100.  Out of this, 300 comes through Vaishali – Jyotishamati pipeline.  So remaining 800 shall come through Vidisha – Jyotishmati pipeline. Now, the requirement at Vidisha = 200.  Hence Avanti Vidisha pipeline shall carry 200 (Requirement at Vidisha) + 800 (to be sent further to Jyotishmati) = 1000 So, this is the final situation. Now all the questions can be answered. Each pipeline can carry maximum of 1000. Since Avanti- Vidisha pipeline carries 1000, free capacity in Avanti – Vidisha is zero.  Hence option (d) is the answer.

Suggest corrections


text:
# KTREE- Editorial

Author: Nishit Sharma

Tester: Aman Singhal

Editorialist: Nishit Sharma

Easy-Medium.

### PREREQUISITES:

DFS,Bitwise operations

### PROBLEM:

Given an integer K and a tree with N nodes rooted at 1, where the ith node has the value A[i] . Choose any leaf node and for every node i on the simple path from root to the leaf node choose a value D such that 2^{A[i]-1}\leq D \lt 2^{A[i]}. Determine whether you can obtain the value K after performing BITWISE XOR or BITWISE AND of the chosen integers.

### Quick Explanation:

Do a DFS on the tree and maintain a map to store the count of all values that occur on a path. Let Big and Small be the largest and the smallest values on the path, M be the most significant bit of K.
The answer will be “YES” if on reaching any leaf node any one of the following conditions is true.
1.Big-1 \gt M and No of times Big occurs on the path is even.
2.Big-1=M and No of times Big occurs on the path is odd.
3.Small-1\gt M and Not all nodes on the path have value Small.
4.Small-1=M

### OBSERVATION 1:

Tap to view

The answer only depends on the most significant bit of K and the largest and smallest values on the path from root to leaf.

### OBSERVATION 2:

Tap to view

We have to check all leaf nodes to get the correct answer. Mainting a map or a Dictionary while doing the DFS is an optimal way to keep track of values on any path.

### Hint 1 :

Tap to view

Try to solve this question for the most significant bit of K, try if you can get the most significant bit in your answer to be the same as the most significant bit in K. Assure yourself that the most significant bit in your anwer will depend on the largest value on the path from root to leaf for the XOR operation and the smallest value on the path for the AND operation.

### EXPLANATION:

Tap to view

Now let us see why the answer only depends on the most significant bit of K.
Consider any arbitary node with value X. For this node we are allowed to choose any value between 2^{X-1} and 2^{X}-1 (inclusive). Using basic maths we can prove that we can set or unset any bit from 0thbit to (X-2)th bit in any way we want i.e. only the (X-1)th bit is fixed in all numbers from 2^{X-1} and 2^{X}-1.Hence,solving the problem for only the most significant bit is enough because if we can set the most significant bit then we can set all the other bits according to our need.

Now we will try to solve the problem using only the XOR operation, we shall deal with the AND operation separately.
Using the above stated fact we can see that using the largest value( Big )on the path we can control all bits except for the (Big-1)th bit by doing the XOR operation.
However, the largest value Big may occur multiple times on the path, now two more cases arise here:

1. Big occurs odd number of times:
In this case the (Big-1)th bit wil be set in our answer, hence if Big-1 is equal to the most significant bit of K, then the answer is “YES”.
2.Big occurs even number of times:
In this case the (Big-1)th bit will not be set in our answer but we can still control all bits from 0 to Big-2, hence if Big-1 is greater than the most significant bit of K, then the answer is “YES”.

Now coming to the AND operation, in this operation the the most significant bit will be goverened by the smallest(Small) value on the path from root to leaf.
An important edge case here is that we can control the (Small-1)th bit in our answer if there is atleast one node on the path will value greater than Small.In other words the (Small-1)th will be compulsorily set in our answer if and only if all values on the path are equal to Small. So we again have two cases here:

1. There is atleast one node with value greater than Small: Then the answer is “YES” if Small-1 is greater than or equal to most significant bit of K.
2. All nodes on the path have value Small: The answer is “YES” if and only if Small-1 is equal to the most significant bit of K.

Coming to the implementation, we will visit all leaf nodes and then check if any of the above mentioned conditions is true. This can be done using DFS while maintaining a ordered map/dicionary to keep track of the smallest and the largest values on the path. See the setters solution to get a better idea of the implementation.

#### Bonus:

Try to print the path from root to leaf and the corresponding values chosen for each node.

#### CODE:

Setter's Solution(C++)
#include<bits/stdc++.h>
#define ll long long int
#define fab(a,b,i) for(int i=a;i<b;i++)
#define pb push_back
#define endl "\n"
#define f first
#define se second
#define MOD 1000000007
#define quick ios_base::sync_with_stdio(false);cin.tie(NULL);cout.tie(NULL)
using namespace std;

const int N=5e5+3;
map<ll,int> m;
vector<int> a;
multiset<int> s;
vector<int> v[N];
ll k;
int n;
int msb;

void init()
{
fab(0,n+1,i)
v[i].clear();
a.clear();
m.clear();

}
void findmsb()
{
fab(0,64,i)
{
if(k&(1LL<<i))
msb=i;
}

}

bool dfs( int src, int par)
{
m[a[src-1]]++;
bool ok=0;
for(auto i: v[src])
{
if(i^par)
{
ok|=dfs(i,src);
}
}
if(v[src].size()==1 and v[src][0]==par)
{
//this is a leaf node

assert(m.size()>0);
auto big=m.rbegin();
auto small=m.begin();
if(big->first > msb and (big->second)%2==0 )
ok|=1;
if(big->first==msb and (big->second)%2==1 )
ok|=1;
if(small->first > msb and m.size()!=1)
ok|=1;
if(small->first==msb)
ok|=1;

}
m[a[src-1]]--;
if(m.find(a[src-1])!=m.end() and m[a[src-1]]==0)
m.erase(a[src-1]);
return ok;

}

int main()
{
quick;

int t;
cin>>t;

while(t--)
{
cin>>n>>k;

init();

a= vector<int> (n);
fab(0,n-1,i)
{
int x,y;
cin>>x>>y;

assert(x>0 and x<=n and y>0 and y<=n);
v[x].pb(y);
v[y].pb(x);

}

fab(0,n,i)
{   cin>>a[i];
a[i]--;

assert(a[i]<64 );
assert(a[i]>=0);
}

findmsb();

bool ok=dfs(1,-1);
cout<<(ok?"YES":"NO")<<endl;

}

cerr << "time taken : " << (float)clock() / CLOCKS_PER_SEC << " secs" << endl;
return 0;
}

Tester's Solution(Python)
import sys,collections
def main():
T=int(input())
for _ in range(T):
N,K=map(int,input().split())
Tree={} #Dic for storing tree
for j in range(N):
Tree[j]=[]

for i in range(N-1):
u,v=map(int,input().split())
Tree[u-1].append(v-1)
Tree[v-1].append(u-1)

A=list(map(int,input().split()))

vis=[0 for i in range(N)]   #to mark visited vertices 0 for visited and 1 for not visited
maxval=[[0,0] for i in range(N)]   #Nx2 list where each i stores max value till now and its count
minval=[0 for i in range(N)]   #Nx2 list where each i stores min value till now
lfnode=[]   #list to store leaf nodes

#Appending node 1
vis[0]=1
Q=collections.deque([0])
maxval[0][0],maxval[0][1]=A[0],1
minval[0]=A[0]

while(len(Q)!=0):
a=Q.pop()
mv1=maxval[a][0]
mv2=minval[a]

flag=0 #to check leaf node

for i in Tree[a]:
if (vis[i]==0):
vis[i]=1
flag=1          #not a leaf node
v=A[i]
Q.append(i)

#Comparing maximum value of parent node
if (mv1<v):
maxval[i][0],maxval[i][1]=v,1

elif(mv1==v):
maxval[i][0],maxval[i][1]=mv1,maxval[a][1]+1

else:
maxval[i][0],maxval[i][1]=maxval[a][0],maxval[a][1]

#Comparing minimum value of parent node
if (mv2>v):
minval[i]=v
elif(v==mv2):
minval[i]=mv2
else:
minval[i]=minval[a]

if (flag==0):
lfnode.append(a)

flag=0  #For answer if 0 then NO else YES

K1=len(bin(K))-2 #length of K

#print(lfnode,val)
for i in lfnode:
v1,v2=maxval[i][0],maxval[i][1]

if (v1>K1 and v2%2==0):
flag=1
elif(v1==K1 and v2%2==1):
flag=1

v11=minval[i]
if (v11>K1 and v11!=v1):
flag=1
elif(v11==K1):
flag=1

if(flag==1):
break

if (flag==1):
print("YES")
else:
print("NO")

main()

If anything is unclear please let me know in the comments, it will help me improve.

6 Likes

no need to use map/dict. just some small arithmetic can be done.
#include <bits/stdc++.h>
using namespace std;
#define ll long long
bool dfs(vector tree[],ll A[],ll M,ll v=0,ll p=-1,ll Big=0,ll Small=65,ll countBig=0){
if(Big==A[v])
countBig++;
else if(Big<A[v]){
Big=A[v];
countBig=1;
}
Small=min(Small,A[v]);
if(tree[v].size()==1 && tree[v][0]==p){
if(Small-1==M || (Small-1>M && Big!=Small))
return true;
else if((Big-1==M && countBig%2==1) || (Big-1>M && countBig%2==0))
return true;
else return false;
}
for(ll x:tree[v])
if(x!=p && dfs(tree,A,M,x,v,Big,Small,countBig))
return true;
return false;
}
ll findmsb(ll K){
for(ll i=63;i>=0;i–)
if(K&(1LL<<i))
return i;
}
void solve(){
ll N,K;
cin >> N >> K;
vector tree[N];
for(ll i=0;i<N-1;i++){
ll u,v;
cin >> u >> v;
–u,–v;
tree[u].push_back(v);
tree[v].push_back(u);
}
ll A[N];
for(ll i=0;i<N;i++)
cin >> A[i];
ll M=findmsb(K);
dfs(tree,A,M) ? cout << “YES” : cout << “NO”;
cout << endl;
}
int main(){
ios_base::sync_with_stdio(false); cin.tie(NULL); cout.tie(NULL);
int T;
cin >> T;
while(T–)
solve();
}


text:
Chapter 186

Methylxanthines (including theophylline, theobromine, caffeine, and nicotine) are plant-derived alkaloids with ubiquitous use in beverages (caffeine in coffee and soda), foods (theobromine in chocolate), tobacco products (nicotine), and medications (theophylline and caffeine). They have shared pharmacologic properties and very similar pharmacodynamic effects. Newer methylxanthine derivatives include pentoxifylline (an anti-inflammatory) and doxofylline.1

### Epidemiology

Theophylline (1,3-dimethylxanthine) and its water-soluble salt, aminophylline, were used extensively in the past for the treatment of asthma and chronic obstructive pulmonary disease. However, theophylline’s use has declined due to its narrow therapeutic window and the development of safer agents. The number of theophylline toxicity cases reported to the American Association of Poison Control Centers decreased from 2609 exposures with 20 deaths in 1997 to only 357 exposures with 2 deaths in 2008.2,3 Theophylline is still used in patients with debilitating bronchospastic disease, particularly outside the U.S., and has been studied for the treatment of other diseases, including acute mountain sickness and contrast-induced nephropathy.4–6

Theobromine (3,7-dimethylxanthine) is found in the seeds of Theobroma cacao, from which chocolate and cocoa are derived; Commelia thea, from which teas are steeped; and is an ingredient in numerous “energy drinks.” There are very few cases of human toxicity, but it has been associated with atrial fibrillation.7

Caffeine (1,3,7-trimethylxanthine), also a methylxanthine and structural analog of adenosine, is the most commonly used psychoactive drug in the world and the only one that can be legally purchased by children. It is found at varying amounts in beverages and “energy-enhanced” foods, such as candy bars, potato chips, and oatmeal (Table 186-1).8 Other uses for caffeine include apnea of prematurity, analgesic adjuncts, appetite suppression for weight loss, sleep prevention, and diuresis. In 2008, the American Association of Poison Control Centers received reports of 4852 caffeine exposures with one death.3 Many “energy drinks” also use guarana, which contains relatively high concentrations of caffeine and other methylxanthines.9

Table 186-1 Caffeine Content of Various Products

### Pharmacology

Theophylline is usually taken orally, although its absorption may be affected by food. It is available via elixir, or extended-release and controlled-release tablets. Despite being designed for once-daily dosing, controlled-release tablets can result in erratic or rapid absorption. Theophylline can also be administered IV as aminophylline. Caffeine is most commonly consumed PO; however, it can be administered PR or parenterally (IV, IM, or SC).

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text:
# Centrifugal Force¶

This boundary conditions is used to apply the forces that act on a structure when it rotates around a fixed axis. The axis itself has to be fixed, but the rotational speed is allowed to vary with time (or frequency for a harmonic analysis). The user has to specify a volume on which the force acts as well as the rotational speed (in radians per second) and the axis of rotation.

## Rotational speed¶

The rotational speed can be either constant, or a time-dependent (or frequency-dependent respectively). The variable speed has to be given by a function or a table.

## Rotational axis¶

The rotational axis is defined by its direction and a point on the axis.

### Point on axis¶

The user has to define an arbitrary point on the axis of rotation in the global coordinate system of the mesh.

### Axis direction¶

The direction of the rotational axis is defined by three directional values in the global coordinate system. This direction vector doesn’t need to have unit length.

### Example¶

Deformation of a turbo charger impeller at 100 000 rpm


text:
# sklearn.cluster.kmeans_plusplus¶

sklearn.cluster.kmeans_plusplus(X, n_clusters, *, x_squared_norms=None, random_state=None, n_local_trials=None)[source]

Init n_clusters seeds according to k-means++

New in version 0.24.

Parameters
X{array-like, sparse matrix} of shape (n_samples, n_features)

The data to pick seeds from.

n_clustersint

The number of centroids to initialize

x_squared_normsarray-like of shape (n_samples,), default=None

Squared Euclidean norm of each data point.

random_stateint or RandomState instance, default=None

Determines random number generation for centroid initialization. Pass an int for reproducible output across multiple function calls. See Glossary.

n_local_trialsint, default=None

The number of seeding trials for each center (except the first), of which the one reducing inertia the most is greedily chosen. Set to None to make the number of trials depend logarithmically on the number of seeds (2+log(k)).

Returns
centersndarray of shape (n_clusters, n_features)

The inital centers for k-means.

indicesndarray of shape (n_clusters,)

The index location of the chosen centers in the data array X. For a given index and center, X[index] = center.

Notes

Selects initial cluster centers for k-mean clustering in a smart way to speed up convergence. see: Arthur, D. and Vassilvitskii, S. “k-means++: the advantages of careful seeding”. ACM-SIAM symposium on Discrete algorithms. 2007

Examples

>>> from sklearn.cluster import kmeans_plusplus
>>> import numpy as np
>>> X = np.array([[1, 2], [1, 4], [1, 0],
...               [10, 2], [10, 4], [10, 0]])
>>> centers, indices = kmeans_plusplus(X, n_clusters=2, random_state=0)
>>> centers
array([[10,  4],
[ 1,  0]])
>>> indices
array([4, 2])


text:
# How to connect the nine dots with four straight lines?

1,735 views

Nine dots are placed in three rows of each three dots, as shown in the picture. These nine dots must be connected by four straight, connected lines (i.e. without 'lifting up the pen' in between).

posted Feb 19, 2014

Similar Puzzles

Using only 3 straight lines and without lifting your pencil, can you connect all the dots?

Can you solve the 12 dot puzzle? The challenge is to connect every dot with five straight lines. However, each line must connect to the last (think about a pen you cannot pick up from the paper). The twist to this is that the last line must end at the beginning of the first line.

+1 vote

Using points, straight lines, and possibly the curve of a parabola, construct a piecewise function whose graph looks like a smiling face.


text:
# Math Help - venn diagram

1. ## venn diagram

Is the venn diagram for $(A - (A \cap B)) \cup (B - (A \cap B))$, the same as $(A-B) \cup (B-A)$

where as $A \cap B \neq \varnothing$

2. Originally Posted by Discrete
Is the venn diagram for $(A - (A \cap B)) \cup (B - (A \cap B))$, the same as $(A-B) \cup (B-A)$

where as $A \cap B \neq \varnothing$
yes, both will look like this: (the shaded area is the desired set)

3. Originally Posted by Discrete
Is the venn diagram for $(A - (A \cap B)) \cup (B - (A \cap B))$, the same as $(A-B) \cup (B-A)$
where as $A \cap B \neq \varnothing$
Yes and we do not even need $A \cap B \neq \varnothing$
This is known as the symmetric difference.

4. Originally Posted by Discrete
Is the venn diagram for $(A - (A \cap B)) \cup (B - (A \cap B))$, the same as $(A-B) \cup (B-A)$

where as $A \cap B \neq \varnothing$
The reason is :
$(A-(A\cap B))\cup(B - (A \cap B))$ = $((A-A)\cup(A-B))\cup((B-A)\cup(B-B))$ = $\varnothing\cup (A-B)\cup (B-A)\cup\varnothing = (A-B) \cup (B-A)$

5. Jhevon, what software did you use to draw that Venn diagram?

6. Originally Posted by rualin
Jhevon, what software did you use to draw that Venn diagram?
regular, good-old-fashioned MS-Paint. I use that for pretty much all my diagrams besides graphs.

7. Hello, Discrete!

Here's another proof . . .

Prove: . $[A - (A \cap B)] \cup [B - (A \cap B)] \;=\;(A-B) \cup (B-A)$

We need (among other definitions and properties):

. . [1] . $P - Q \;=\;P \cap\, \overline{Q}$ . (Def. of set subtraction)
. . [2] . $P \cap \overline P \:=\:f$
. . [3] . $P \cup f \:=\:P$

We have: . $[A - (A \cap B)] \cup [B - (A \cap B)]$

. . . . . $= \;\left[A \cap (\overline{A \cap B})\right] \cup \left[B \cap \left(\overline{A \cap B}\right)\right]$ . . . . . . . . . . . . . . . [1]

. . . . . $= \;\left[A \cap \left(\overline A \cup \overline B\right)\right] \cup \left[B \cap \left(\overline A \cup \overline B\right)\right]$ . . . . . . . . . . . DeMorgan's Law

. . . . . $= \;\left[\left(A \cap \overline A\right) \cup \left(A \cap \overline B\right)\right] \cup \left[\left(B \cap \overline A\right) \cup \left(B \cap \overline B\right)\right]$ . . Distributive Property

. . . . . $= \;\left[f \cup \left(A \cap \overline B\right)\right] \cup \left[\left(B \cap \overline A\right) \cup f\right]$ . . . . . . . . . . . . . . . [2]

. . . . . $= \;\left(A \cap \overline B\right) \cup \left(B \cap \overline A\right)$ . . . . . . . . . . . . . . . . . . . . . . [3]

. . . . . $= \;(A - B) \cup (B - A)$ . . . . . . . . . . . . . . . . . . . . . . [1]


text:
Recurrence relation for trinomial Apery numbers

It is well known (Beukers 1987) that the Apery numbers $$A_n\equiv A_n^{(2)}=\sum\limits_{k=0}^n\binom{n}{k}^2\binom{n+k}{k}^2$$ satisfy the fancy recurrence relation $$n^3A_n=(34n^3-51n^2+27n-5)A_{n-1}-(n-1)^3A_{n-2},\;\;\;\;\;\;\;\;\;\;\;\;\;(1)$$ with $A_0=1$ and $A_1=5$.

Introducing trinomial coefficients $$\binom{n}{k,\;l}=\frac{n!}{k!\;l!\;(n-k-l)!},$$ let's generalize the Apery numbers as follows $$A_n^{(3)}=\sum\limits_{k=0}^n\sum\limits_{l=0}^{n-k}\binom{n}{k,\;l}^2\binom{n+k+l}{k,\; l}^2.$$ Do these numbers satisfy some analog of the recurrence relation (1)?

The first $A_n^{(3)}$ numbers are: $$\begin{array}{l} 9\\ 721\\ 82089\\ 12230001\\ 2120202009\\ 406989480241\\ 84181340789289\\ 18415254766978801\\ 4208936841232398009\\ \end{array}$$ Note that $$\hspace{50mm}A_n^{(3)}\equiv 0 \;(\mathrm{mod}\; 9),\hspace{50mm} (2)$$ if $n=1,3,4,5,7,9$. This sequence belongs to the so called vile numbers http://oeis.org/A003159 Is the congruence (2) true for any vile number $n$? Note also that $A_5^{(3)}\equiv A_1^{(3)}\;(\mathrm{mod}\; 3^3)$, $A_8^{(3)}\equiv A_2^{(3)}\;(\mathrm{mod}\; 3^3)$ and $A_9^{(3)}\equiv A_1^{(3)}\;(\mathrm{mod}\; 5^3)$. What about Beukers-like congruence $$\hspace{50mm}A_{np-1}^{(3)}\equiv A_{n-1}^{(3)} \;(\mathrm{mod}\; p^3),\hspace{50mm} (3)$$ for any prime $p$ and positive integer $n$, is it true?

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You can express the trinomials as a product of binomials. For sums of products of binomials, there are computer packages for finding recurrence relations due to Zeilberger and others. See the book "A=B":math.upenn.edu/~wilf/AeqB.html –  Ian Agol Jul 28 '13 at 22:51
In your formula for $A^{(3)}_n$ you use trinomials with $l+k>n$. They are not covered by your definition. –  Uwe Stroinski Jul 29 '13 at 15:50
Thanks! I have corrected the typo. –  Zurab Silagadze Jul 29 '13 at 16:50
Congruence (2) is valid for $A^{(3)}_{10}$. –  Uwe Stroinski Jul 30 '13 at 12:26
For $n\leq 175$ we have that $A^{(3)}_n$ is not divisible by $9$ for $n\in\{2,6,8,18,20,24,26,54,56,60,62,72,74,78,80,162,164,168,170\}$. Note the big gap between $80$ and $162$. –  Uwe Stroinski Jul 30 '13 at 14:49


text:
# Automatic selection of nuber or factors in PCA based on the Cattells scree plot in R

I would like to automatically select the number of factors after factor analysis (PCA). I mean the graphical selection method according to the Cattel criterion, which determines where the steepness of the graph ends:

Therefore, I want to check where the first steepness of the graph ends, withouth looking on plot. For any set of data (df) I have at my disposal percentages of explained variance, raw values square roots of the eigenvalues, and so on:

# df is my data set
pca_results <- prcomp(df)
# here i can get plot values
plot_values <- pca_results$$sdev # ... and change plot values to percents of variance explained variance_percents <- (pca_results$$sdev^2 / sum(pca_results\$sdev^2)) * 100
# i can take values higher than 1 (Kaiser's simple criterion)
pca_number_of_factors <- length(plot_values[plot_values > 1])
# get some variance percents
first_f_var_perc <- variance_percents[1]
second_f_var_perc <- variance_percents[2]
third_f_var_perc <- variance_percents[3]
# or mark steppness level
PCA_steepness = 1 - second_f_var_perc / first_f_var_perc

I'm looking for a simple mathematical way to determine the first steepness collapse of a graph. Something like "number of factors" selected automatically in this way. Could someone help me in this regard?

Of course, I am aware of the disadvantages of this approach (for example, a few slopes of the graph), but I would just like to find out how it can be solved geometrically-numerically, because frankly, I don't have much in mind.

OK, to be honest i don't know if i'm not missing something from simple trigonomethry, but here's my solution.

From the standardized set of plot_values obtained in the previous code, you can create a simple plot and look at it:

plot(plot_values)
plot(plot_values, type="l")

What visually makes a decrease in the steepness of the graph? That a line is at an angle of less than 45 degrees to the horizontal axis. Here we got first line with more than 45 degrees.

Given that the graph/plot points are standardized, the distances between the individual objects are the same (they are 1, that is line a) then the steepness ends where the direct value relative to the next point (line c) is simply less than the y-axis 1. Then the line b will be directed at an angle less than 45 degrees and its steepness will end. So we can find it like this:

# make new variable
plot_values2 = plot_values
# let's leave the heights of the triangles the same,
# that is, let's subtract the next value from the previous one
for (i in 1:(length(plot_values2)-1))
{ plot_values2[i] = plot_values2[i] - plot_values2[i+1] }
# with last value = 0 not to distort our results
plot_values2[length(plot_values2)] = 0
# let's get first result lower than 1, and that's it
number_of_PCA_factors = (match(TRUE, (plot_values2 < 1))) - 1
# and for any problems:
if (number_of_PCA_factors < 1) {number_of_PCA_factors = NA}

If there is something wrong with this set of code, please let me know or correct it.


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# Permutations and Combinations Help!

• Nov 25th 2009, 12:26 PM
TheCB4
Permutations and Combinations Help!
Hello, what are some ways/tips that i can use to help me distinguish between permutation and combination word problems?
• Nov 25th 2009, 01:29 PM
Plato
Quote:

Originally Posted by TheCB4
Help me distinguish between permutation and combination word problems?

If it involves order it is a permutation.
If only content matters then it is a combination.

How many ways can you make a four-letter string using the letters in $S~O~U~T~H~E~R~N$?
That is a permutation.

How many ways can you select four letters from $S~O~U~T~H~E~R~N$?
That is a combination.
• Nov 25th 2009, 06:38 PM
Soroban
Hello, TheCB4!

Quote:

What are some ways/tips that i can use to help me distinguish
between permutation and combination word problems?

In a permutation, the order of the objects is important.

Suppose there are 10 books on the table
. . and you may select 4 of them.

If you choose the 4 books and place them in a row on a shelf,
. . it is a permutation.

There are: . $_{10}P_4 \:=\:\frac{10!}{6!} \:=\:10\cdot9\cdot8\cdot7 \:=\:5040$ possible arrangments.

If you choose the 4 books and toss them into a shopping bag,
. . it is a combination.

There are: . $_{10}C_4 \:=\:\frac{10!}{4!\,6!} \:=\:210$ possible selections.


text:
Two airplanes leave an airport at the same time, and the angle between their flight paths is 40*.An hour later, one plane has traveled 300 miles, while the other has travelled 200 miles,each in a straight line. How far apart are the planes at this time?

2. Originally Posted by cinbad
Two airplanes leave an airport at the same time, and the angle between their flight paths is 40*.An hour later, one plane has traveled 300 miles, while the other has travelled 200 miles,each in a straight line. How far apart are the planes at this time?
This is an allication of the cosine rule. You have a triangle with two
sides one 300 and the other 200 miles, and the angle between them
is 40 degrees.

$
c^2=a^2+b^2-2ab \cos (\theta)
$

Substitute $a=200$, $b=300$ and $\theta=40^{\circ}$ into the above and that's it

RonL

3. Thank you very much Captain