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# Systematic Analysis of Challenge-Driven Improvements in Molecular Prognostic Models for Breast Cancer

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Science Translational Medicine  17 Apr 2013:
Vol. 5, Issue 181, pp. 181re1
DOI: 10.1126/scitranslmed.3006112

## Abstract

Although molecular prognostics in breast cancer are among the most successful examples of translating genomic analysis to clinical applications, optimal approaches to breast cancer clinical risk prediction remain controversial. The Sage Bionetworks–DREAM Breast Cancer Prognosis Challenge (BCC) is a crowdsourced research study for breast cancer prognostic modeling using genome-scale data. The BCC provided a community of data analysts with a common platform for data access and blinded evaluation of model accuracy in predicting breast cancer survival on the basis of gene expression data, copy number data, and clinical covariates. This approach offered the opportunity to assess whether a crowdsourced community Challenge would generate models of breast cancer prognosis commensurate with or exceeding current best-in-class approaches. The BCC comprised multiple rounds of blinded evaluations on held-out portions of data on 1981 patients, resulting in more than 1400 models submitted as open source code. Participants then retrained their models on the full data set of 1981 samples and submitted up to five models for validation in a newly generated data set of 184 breast cancer patients. Analysis of the BCC results suggests that the best-performing modeling strategy outperformed previously reported methods in blinded evaluations; model performance was consistent across several independent evaluations; and aggregating community-developed models achieved performance on par with the best-performing individual models.

## Introduction

Breast cancer is the leading female malignancy in the world (1) and is one of the first malignancies for which molecular biomarkers have exhibited promise for clinical decision making (25). Biomarkers can be used to divide the disease into predictive (the likelihood that a patient responds to a particular therapy) or prognostic (a patient’s risk for a defined clinical endpoint independent of treatment) subcategories. Such molecular subcategorization highlights the possibilities for precision medicine (6), in which biomarkers are leveraged to identify disease taxonomies that distinguish biologically relevant groupings beyond standard clinical measures and can potentially inform treatment strategies.

A decade after early achievements in the development of prognostic molecular classifiers (called signatures) of breast cancer based solely on gene expression analysis (4, 5, 7), a large number of signatures proposed as markers of clinical risk prediction either fail to surpass the performance of conventional clinical covariates or await meaningful prospective validation [for example, the MINDACT (8, 9) and TAILORx/RxPONDER (10) trials]. Although gene expression–based breast cancer prognostic tests have been successfully implemented in routine clinical use, the application of molecular data to guide clinical decision making remains controversial (11).

Slow progress to evolve useful molecular classifiers may relate to poor study design, inconsistent findings, or improper validation studies (12). Data and code that underlie a potential new disease classifier are often unavailable for diligence. In addition, the rigor and objectivity of assessing molecular models are confounded by the tendency of data generation, data analysis, and model validation to be combined within the same study. This leads to the “self-assessment trap,” in which the desire to demonstrate improved performance of a researcher’s own methodology may cause inadvertent bias in elements of study design, such as data set selection, parameter tuning, or evaluation criteria (13).

A potential response to such problems is leveraging the Internet, social media, and cloud computing technologies to make it possible for physically distributed researchers to share and analyze the same data in real time and collaboratively iterate toward improved models based on predefined objective criteria applied in blinded evaluations. The Netflix competition (14) and X-Prize (15) have successfully demonstrated that crowdsourcing novel solutions for data-rich problems and technology innovation is feasible when substantial monetary incentives are offered to engage the competitive instincts of a community of analysts and technologists. Alternatively, a game-like environment has been used to seek solutions to biological problems, such as protein (16) and RNA folding (http://eterna.cmu.edu/web), thereby appealing to a vast community of gamers. Others have demonstrated that both fee-for-service [Kaggle (17) and Innocentive (18)] and open-participation computational biology challenges (19) [CASP (20) and CAFA (21)] can be used as potential new research models for data-intensive science. Despite the wide breadth of areas covered by these competitions, a key finding is that the best models from a competition usually outperform analogous models generated using more traditional isolated research approaches (14, 17).

Building on the success of crowdsourcing efforts such as DREAM to solve important biomedical research problems, we developed and ran the Sage Bionetworks–DREAM Breast Cancer Prognosis Challenge (BCC) to determine whether predefined performance criteria, real-time feedback, transparent sharing of source code, and a blinded final validation data set could promote robust assessment and improvement of breast cancer prognostic modeling approaches. The BCC was designed to make use of the METABRIC data set, which contains nearly 2000 breast cancer samples with gene expression and copy number data and clinical information. The availability of such a large data set affords the statistical power required to assess the robustness of performance of many models evaluated in independent tests, but is subject to the trade-off of using (overall) survival time in a historical cohort as the clinical endpoint, rather than potential endpoints that could be driven by clinically actionable criteria, such as response to targeted therapies. Therefore, the aim of this Challenge was not direct clinical deployment of a full-fledged suite of complex biomarkers. Indeed, we expect this study to serve as a pilot that lays the groundwork for future breast cancer challenges designed at the outset to answer clinically actionable questions. With this in mind, the BCC resulted in the development of predictive models with better performance than standard clinicopathological measures for prediction of overall survival (OS). The performance of these models was highly consistent across multiple blinded evaluations, including a novel validation cohort generated specifically for this Challenge.

## Results

The BCC included 354 registered participants from more than 35 countries. Participants were tasked with developing computational models that predict OS of breast cancer patients based on clinical information (for example, age, tumor size, and histological grade; see Table 1), mRNA expression data, and DNA copy number data. The BCC used genomic and clinical data from a cohort of 1981 women diagnosed with breast cancer (the METABRIC data set) (22) and provided participants with authorized Web access to data from 1000 samples as a training data set, and held back the remaining samples as a test data set (see Materials and Methods for more details on the Challenge design). Participants used the data to train computational models on their own standardized virtual machine (donated to the Challenge by Google) and submitted their trained models to the Synapse computational platform (23) as an R binary object (24) and rerunnable source code, where they were immediately evaluated. The predictive value of each model was scored by calculating the concordance index (CI) of predicted death risk compared to OS in a held-out data set, and the CIs were posted on a real-time leaderboard (http://leaderboards.bcc.sagebase.org). The CI is a standard performance measure in survival analysis that quantifies the quality of ranking risk predictors with respect to survival (25). In essence, given two randomly drawn patients, the CI represents the probability that a model will correctly predict which of the two patients will experience an event before the other (for example, a CI of 0.75 for a model means that if two patients are randomly drawn, the model will order their survival correctly three of four times).

Table 1 Clinical characteristics of METABRIC and OsloVal data sets.

NA, not available.

View this table:

Throughout the 3-month orientation and training phases of the Challenge (phases 1 and 2, respectively), participants collectively submitted more than 1400 predictive models, 1400 of which successfully executed from the submitted binary object and were assigned CI scores in the test data set (Fig. 1). One unique characteristic of this Challenge with respect to previous biomedical research Challenges was that each participant’s source code was available for others to view and adapt in new models. At the end of the training phase, participants were given the opportunity to train five models each on all 1981 METABRIC samples. The overall Challenge was determined in the final phase (phase 3) by assessing up to five models per participant or team (see table S3 for a listing of those members of the Breast Cancer Challenge Consortium who submitted at least one model to phase 3 and requested to have their name and affiliation listed) against a newly generated breast cancer data set consisting of genomic and clinical data from 184 women diagnosed with breast cancer (the “OsloVal” data set) (see Materials and Methods for details on how to access the METABRIC and OsloVal data sets). The participant or team with the best-performing model in the new data set was invited to publish an article about the winning model in Science Translational Medicine, provided it exceeded the scores of preestablished benchmark models, including the first-generation 70-gene risk predictor (4, 7) and the best model developed by a group of expert researchers in a precompetition (26). This pioneering mechanism of Challenge-assisted peer review assessed performance metrics in a blinded validation test, and these results were the foremost criterion for publication in the journal.

The METABRIC and the OsloVal data sets were comparable in terms of their clinical characteristics (Table 1). Roughly three-quarters of the patients were estrogen receptor–positive (ER+), and about half were lymph node–negative (LN). The patients from the METABRIC and OsloVal cohorts received combinations of hormonal, radio-, and chemotherapy, and none were treated with more modern drugs such as trastuzumab, which specifically targets the human epidermal growth factor receptor 2/neu (HER2/neu) receptor pathway (27).

In phase 2 of the Challenge, participants trained on 1000 samples and obtained real-time feedback on model scores (CIs) in a held-out 500-sample test set (phase 1 results are displayed on the original leaderboard and are not discussed here). A reference model using a random survival forest (28) trained on the clinical covariates and genomic data was provided as a baseline. Models submitted by Challenge participants quickly exceeded the performance of this baseline model and steadily improved over time (Fig. 2A), although we note that the improvement was modest compared to the baseline model (Fig. 2A, inset).

At the end of phase 2, all models in the leaderboard were evaluated against 481 hidden validation samples from the METABRIC data set. Given that more than 1400 models were submitted, there was concern that improvement in model scores resulted from overfitting to the test set used to provide real-time feedback. However, the performance of most models in the new test set was consistent with the performance of the same models in the previous test set, with comparable score ranges (Fig. 2, B and C; Pearson correlation: 0.90). A similar outcome was observed for the five models from every team that were evaluated against the OsloVal data set, after being trained on the 1981 samples in the METABRIC data set. Again, there was little evidence of overfitting when compared to the previous METABRIC test scores (Fig. 2D; Pearson correlation: 0.59), especially for models ranked in the top quantile.

After participants trained and selected their final models, and after eliminating the ones that could not be evaluated because of run time errors, a total of 83 models were assessed and scored against the newly generated OsloVal data set. The winning team had three similar models that performed consistently better than all other scored models. The robustness of the final ranking was evaluated by sampling, without replacement, 80% of the validation set 100 times. Figure 3 shows box plots of the rankings obtained by each model across all trials ordered by the initial scores on all 184 OsloVal samples. The top three models belong to the same team and are ranked significantly better than the rest. The top model achieved a CI of 0.7561 [P = 5.1 × 10−28 compared to the fourth-ranked model, by Wilcoxon rank-sum test (29)].

The BCC top-scoring model also significantly outperformed the top model developed during a pilot precompetition phase (26) in which BCC organizers tested 60 different models based on state-of-the-art machine learning approaches and clinical feature selection strategies. The best model from the precompetition used a random survival forest trained on the clinical feature data in addition to a genomic instability index derived from the copy number data.

The top-scoring model from the precompetition would have ranked as the sixth best model in the Challenge and achieved a CI of 0.7408 (Wilcoxon paired test, P = 4.33 × 10−18 compared to the winning model from the Challenge) when trained on the full METABRIC data set and evaluated in the OsloVal data. For comparison, a research version of a 70-gene risk signature (4) was also evaluated in the OsloVal cohort and achieved a CI of 0.60. In addition, two more test models were developed that included only the clinical covariates available for the two data sets (listed in tables S1 and S2). The first model was based on boosted regression (30) and achieved a CI of 0.7001 on the validation data set, whereas the second model used random forest regression (28) and achieved a score of 0.6964. The winning Challenge model achieved a score of 0.7562, significantly higher than the two clinical-only models (Wilcoxon paired test, P = 6.1 × 10−32 for both).

Meta-analyses of predictions submitted to past DREAM Challenges have systematically demonstrated (3134) that the ensemble predictions resulting from the aggregation of the predictions of some or all the models usually perform similarly or even better than the best model. This phenomenon has been called the “wisdom of the crowds” and highlights one of the advantages of enabling research communities to work collaboratively to analyze the same data sets. The wisdom of the crowds was also at play in BCC (Fig. 4). For both cohorts, participants’ predictions were aggregated by calculating a community prediction formed by taking the average predicted rank of each patient across top n models for n = 1 … 83 (Fig. 4, A and B). Even when adding very poor predictions to the aggregate, the resulting score was robust and comparable with the top models. Robust predictors were also achieved by constructing community scores based on random subsamples of models (Fig. 4, C and D).

We evaluated model performance to determine whether there were specific clinical cases that are inherently more difficult to predict in terms of prognosis. We conducted analysis of variance (ANOVA) analyses by separating the patients from the OsloVal cohort based on the following clinical variables: age (≤50 years versus >50 years), tumor size (≤2 cm versus >2 cm), tumor grade (grades 1, 2, and 3), LN status (LN, 1 to 3 positive LNs, 4 to 9 positive LNs, >9 positive LNs), ER status (ER+ versus ER), progesterone receptor status (PR+ versus PR), HER2/neu receptor amplification status (HER2+, HER2), and OS (<5 years, 5 to 10 years, or >10 years). Of all variables considered (Fig. 5), tumor grade, LN status, OS, age, and tumor size were significantly associated with model performance (F test P values: 1.7 × 10−36, 1.1 × 10−38, 8.0 × 10−26, 9.0 × 10−7, and 5.0 × 10−5, respectively).

Breast cancer patients with high-grade tumors and large numbers of positive LNs (>9) were associated with low CI scores (Fig. 5, B and C) and explain more than 40% of CI variance (Fig. 5A), suggesting that the OS of breast cancer patients with aggressive tumors is harder to predict. By contrast, there was no significant association between model performance and ER, PR, or HER2 status (Fig. 5E).

## Discussion

The BCC was an exercise in crowdsourcing that constitutes an open distributed approach to develop predictive models with the future potential to advance precision medicine applications. By creating a large, standardized breast cancer genotypic and phenotypic resource readily accessible via Web services and a common cloud computing environment, we were able to explore whether the open sharing of predictors of disease within a Challenge environment encourages the sharing of models and ideas before publication and whether decoupling of data creation, data analysis, and evaluation would minimize analytical and self-assessment biases.

### Scientific conclusions from the Challenge

Improved model performance. The BCC results show that the best-performing model achieved significant CI improvements over currently available best-in-class methodologies, including the best model developed by a group of experts in a precompetition and a 70-gene risk signature. We note that the first-generation 70-gene risk signature used for comparison was designed as a binary risk stratifier for a specific patient subpopulation and should be viewed as a baseline and sanity check rather than a direct comparison. More significantly, the best-performing model (as well as slight variants of the same model submitted by the same BCC team) consistently outperformed all other approaches in three independent rounds of assessment and across multiple data sets. The top-scoring models used a methodology that minimized overfitting to the METABRIC training set by defining a “Metagene” feature space based on robust gene expression patterns observed in multiple external cancer data sets (35).

Robustness and generalizability of model performance. Our post hoc analysis suggests that the potential for training models overfit to the test set was not a significant confounding factor, even when participants were allowed to submit an unlimited number of models and obtain real-time scores for each submission. Comparison of the same models across multiple rounds of evaluation suggests a surprising degree of consistency between the unlimited submission phase and independent evaluations in the held-out METABRIC data and newly generated OsloVal data. This is especially remarkable because the OsloVal samples were collected from a different geographical location, by a different team, at a different time, and for a different purpose—in contrast to studies such as MAQC-II (36), in which the test sets and training sets were collected and processed by the same team and organization.

The distribution of CI scores improved with each round of evaluation, and the highest-scoring models in the OsloVal evaluation achieved higher CI scores than any of the 1400 models evaluated in either of the METABRIC phases. This apparent counterintuitive result can likely be explained by the fact that the average follow-up time for the OsloVal cohort is 4541 days, much longer than the average follow-up time of 2951 days for METABRIC. The rate of censored events is also lower in OsloVal (27.7%) than in METABRIC (55.2%); therefore, the proportion of long-term survivors scored in OsloVal is larger than that in the METABRIC validation set. Because the long-time survivors are better predicted (Fig. 5D) than the short-term survivors, the CIs associated with the METABRIC data are biased to shorter-time survivors and lower CIs.

The consistency of model scores across our three evaluations does not conclusively rule out some degree of overfitting, and, as with any scientific study, the generality of our findings will be continuously refined through sustained scrutiny in subsequent studies throughout the research community. However, the consistency of model performance across independent evaluations provides strong initial evidence that the findings of our study are likely to generalize to unseen data.

Robust performance of community models. Consistent with results of previous DREAM Challenges (3134), the current study suggests that community models, constructed by aggregating predictions across many models submitted by participants, achieve performance on par with the highest-scoring individual models, and this high performance is remarkably robust to the inclusion of many low-scoring models into the ensemble. This result suggests that crowdsourcing a biomedical prediction question as a Challenge and using the community prediction as a solution is a fast and sound strategy to attain a model with strong performance and robustness. Intuitively, such an approach leverages the best ideas from many minds working on the same problem. More specifically, different approaches that accurately model biologically relevant signals are likely to converge on similar predictions related to the true underlying biology, whereas errant predictions resulting from erroneous approaches are less likely to be consistent. Thus, ensemble models may amplify the true signals that remain consistent across multiple approaches while decreasing the effects of less-correlated errant signals.

### Contributions of BCC to the community

Challenge design. In designing BCC to incentivize crowdsourced solutions that might improve breast cancer prognostic signatures, our aims included continuous participant engagement throughout the Challenge, computational transparency, and rewarding of model generalizability. To accomplish these aims, respectively, we provided a framework that allowed users to make submissions and obtain real-time feedback on their performance, required submission of source code with each model, and provided multiple rounds of evaluation in independent data sets.

A common framework for comparing and sharing models. The model evaluation framework and CI scores calculated for models submitted during the Challenge provide a baseline set of model scores against which emerging tests, such as the PAM50 (37) risk of recurrence score and future prognostic models, may be compared.

The requirement for submission of publicly available source code provided an additional level of transparency compared to the typical Challenge design, which requires users to submit a prediction vector output from their training algorithm. We envision that sharing source code from multiple related predictive modeling Challenges will give rise to a community-developed resource of predictive models that is extended and applied across multiple projects and that might facilitate cumulative scientific progress, in which subsequent innovations build off of previous ones. It was encouraging to discover that participants learned from both well-performing and poorer-performing models. For example, the top-performing team consistently used the discussion forum to share with the other teams the prognostic ability of their Metagene features. Some of the other teams used aspects of the best-performing team’s code to improve their submissions, which in turn gave feedback to the best-performing team on the use of their methods by the other challenge participants.

Innovations from the community. Through the transparent code submission system and communication tools, such as a community discussion forum, the Challenge resulted in numerous examples of sharing and borrowing of scientific and technical insights between participants. At one point, we tested whether a cash incentive could be used to promote collaborative model improvement and offered a $500 incentive to any participant who could place atop the leaderboard by borrowing code submitted by another participant (in addition to$500 to the participant whose code was borrowed). In less than 24 hours, a participant achieved the highest-scoring model by combining the computational approaches of the previously highest-scoring model with his clinical insight of modeling LN status as a continuous, rather than binary, variable. Unanticipated innovations also emerged organically from the community, including an online game within a game (http://genegames.org/cure), in which a player and a computer avatar successively select genes as input features to a predictive model, until one model is deemed statistically significantly superior to the other in predicting survival in the held-out data sets. This game attracted 120 players within 1 week, who played more than 2000 hands.

### Limitations and extensions for future Challenges

The design of BCC included a number of simplifying assumptions intended to define a tractable prediction problem and evaluation criterion. However, it would be beneficial to account for biological and analytical complexities that were obscured by simplifications made in our experimental design.

First, we evaluated all models on the basis of a single metric, CI, which represents the most widely used statistic for evaluating survival models. However, a more complete assessment of advantages and disadvantages of each model would include additional criteria, such as model run time, trade-offs between sensitivity and specificity, or metrics more closely tied to the clinical relevance of a prognosticator.

Second, our choice to evaluate survival predictions across all samples in the cohort may obscure identification of models with advantages in particular breast cancer subtypes.

Third, by providing participants with normalized data, we tested only for modeling innovations given predefined input data, but did not assess different methods for data quality control and standardization that could contribute substantially to model improvements. A useful future Challenge design may allow participants to submit alternative methods for preprocessing raw data.

Fourth, we chose to evaluate models on the basis of OS. Other clinical endpoints, such as progression-free survival, are not currently available in the data sets used in BCC but may represent better evaluation endpoints if made available in the future. We chose OS over the other clinical endpoint available in the data set, disease-specific (DS) survival, to be consistent with decisions (38) by regulatory agencies such as the U.S. Food and Drug Administration and the European Medicines Agency. However, because of informal feedback we received from participants—that use of DS survival yielded more accurate models—we reevaluated all models and supported a separate exploratory leaderboard based on this metric. The model performance on this leaderboard suggested that using DS rather than OS as the clinical endpoint yielded improved CI scores, increased correlation with molecular features, and decreased correlation with confounding variables such as age. However, the best-performing models were consistent across both metrics.

Fifth, although prognostic models are one translational question, future Challenges that focus more directly on inferring predictive models of response to therapy may more directly affect clinical decision making. A laudable goal for future Challenges would be to directly engage the patient community and provide means to submit their own samples, help define questions, work alongside Challenge participants, and provide more direct feedback on how Challenge results can yield insights able to translate to improved patient care. In addition, providing large high-quality data sets (often prepublication) to a “crowd” of analysts is not common practice in academia and industry. However, such contributions by data generators would provide the necessary substrate for running such Challenges, with a potentially high impact on biomedical discovery.

Sixth, although statistically significant, the BCC results show that the improvement of the best-performing model is moderate with respect to the score achieved by aggregating standard clinical information. Thus, whereas molecular prognostic models derived from BCC warrant further investigation into their clinical utility, our results also suggest a new benchmark for future predictive methods derived from incorporating clinical covariates into state-of-the-art ensemble methods such as boosting or random forests. Future Challenges also may investigate the use of additional types of genomic information.

Finally, the sharing of ideas enabled by requiring submissions as rerunnable source code may ironically inhibit the diversity of innovations, effectively encouraging a monoculture as the community converges on a local optimum, modifying and extending approaches with high performance in the early stages of feedback (39). Improvements for future Challenges may include short embargo periods before sharing source code, with release possibly associated with declaring winners in stages of sub-Challenges with slightly modified data or prediction criteria. In addition, future Challenges that promote code sharing should establish well-defined criteria for assigning proper attribution (for example, in publication of winning models resulting from the Challenge) to all participants who made material intellectual contributions that were incorporated into the final winning model. More generally, it is important to develop a reward system that favors collaborative research practices that balance the currently prevalent winner-takes-all reward system.

Our results reinforce the trend from efforts such as CASP, DREAM, and others that Challenges incentivize collaboration and rapid learning, create iterative solutions in which one Challenge may feed into follow-up Challenges, motivate the generation of new data to answer clinically relevant questions, and provide the means for objective criteria to prioritize approaches likely to translate basic research into improved benefit to patients. We envision that expanding such mechanisms to facilitate broad participation from the scientific community in solving impactful biomedical problems and assessing molecular predictors of disease phenotypes could be integral to building a transparent, extensible resource at the heart of precision medicine.

## Materials and Methods

### Challenge timeline

The BCC comprised three phases spanning a period of 3 months: (i) an orientation phase, (ii) a training phase, and (iii) a validation phase.

In phase 1, participants were provided mRNA and copy number aberration (CNA) data from 997 samples from METABRIC alongside the clinical data to train predictive models of patient OS. Samples contained in this training set corresponded to those used as a training set in the original publication of the METABRIC data set (22). During this phase of the Challenge (17 July to 22 September 2012), models were evaluated in real time against a held-out data set of 500 METABRIC samples, and a leaderboard was developed to display CI scores for each model, based on predicted versus observed survival times in these 500 held-out samples.

In phase 2 (25 September to 15 October 2012), lessons learned from the previous phase were implemented: The clinical variable “LN status” was changed from binary (positive or negative) values to integer values, representing the number of affected LNs; the gene expression and CNA data were renormalized (as described below) to better correct for batch effects; and the full cohort was randomly split into a new training and a new testing set to correct for sample biases from the original split (for example, in the original split, nearly all missing clinical covariates were in the test set). During this phase, models were trained on a training set of 1000 samples and evaluated in real time against a held-out test data set of 500 samples. The resulting scores were posted on a new leaderboard and, at the end of phase 2, were evaluated against the remaining held-out 481 samples. The winner of this phase was determined on the basis of evaluation in this second test set. All official scoring was performed using the OS endpoint, although based on requests from Challenge participants, we also configured a leaderboard that allowed participants to assess their model scores against DS survival as a secondary “unofficial” evaluation.

Before submission of their final models in phase 3, participants were given the opportunity to retrain their models on the entire METABRIC data set of 1981 (for convenience, with clinical variables reduced to only those present in the OsloVal data set). Furthermore, participants were asked to select a maximum of five models per team for assessment against the OsloVal data set. All participants who submitted a model in phase 3 are listed in table S3. If a team did not choose their preferred five models, their five top-scoring models were selected by default. These models were scored on the basis of the CI predicted versus observed ranks of survival times in the OsloVal data set. The overall winner of the Challenge was determined by the top CI score in OsloVal. The significance of the top-scoring models compared to the rest was assessed on the basis of scores for multiple random subsamples without replacement of 80% of patients from the OsloVal cohort. This process was repeated 100 times, and the resulting rankings were compared with a Wilcoxon rank-sum test (29).

### Data governance

The data generators’ institutional ethics committee approved use of the clinical data within the BCC. Expression and copy number data from METABRIC were made available to Challenge participants during the duration of phases 1 to 3. Data for clinical covariates from the METABRIC cohort had been made public previously (40). Each participant who accessed the data agreed to (i) use these data only for the purposes of participating in the Challenge and (ii) not redistribute the data. Data access permissions were revoked at the completion of each BCC phase, and participants were required to reaffirm their agreement to these terms to enter each phase.

The METABRIC and OsloVal data sets have been deposited in the Synapse database (https://synapse.prod.sagebase.org/#!Synapse:syn1710250) and will be available to readers for a 6-month “validation phase” of the BCC. Those interested in accessing these data for use in independent research are directed to the following links. Expression and CNA data from OsloVal are available through an open access mechanism. All of other data are available through a controlled access mechanism.

The final Challenge scoring was based on the CIs of models, which was the same metric used in the leaderboard at earlier phases of the Challenge. The CI was predefined from the beginning of the Challenge as the performance metric to be used at final scoring. For all final submissions, every effort was made by the Sage Bionetwork team to run the submitted code in the Synapse platform. However, because of different problems in the submitted code, some of the submissions did not run successfully. Each submitted model that could be successfully run yielded a final survival prediction for the OsloVal cohort. For each of these predictions, the CI was mathematically computed in an unambiguous way by a computer program, from which the performance ranking was generated in order of descending CI. All final models can be accessed at http://leaderboards.bcc.sagebase.org/final/index.html.

### Data normalization

The Affymetrix Genome-Wide Human SNP 6.0 and Illumina HT12 Bead Chip data were normalized according to the supervised normalization of microarrays (snm) framework and Bioconductor package (41, 42). Following this framework, models were devised for each data set that expressed the raw data as functions of biological and adjustment variables. The models were built and implemented through an iterative process designed to learn the identity of important variables defined by latent factors identified via singular value decomposition. Once these variables were identified, we used the snm R package to remove the effects of the adjustment variables while controlling for the effects of the biological variables of interest.

For example, to normalize the METABRIC mRNA data, we used a model that included ER status as a biological variable and both scan date and intensity-dependent array effects as adjustment variables. The resulting normalized data consisted of the residuals from this model fit plus the estimated effects of ER status. For the relevant data sets, we list the biological and adjustment variables as follows: METABRIC mRNA: biological variable = ER status, adjustment variables = scan date and intensity-dependent array effects; METABRIC SNP: biological variable = none, adjustment variables = scan date and intensity-dependent array effects; OsloVal mRNA: biological variable = ER status, adjustment variables = Sentrix ID (43), intensity-dependent array effects; OsloVal SNP: biological variable = none, adjustment variables = scan date, intensity-dependent effects.

Summarization of probes to genes for the SNP6.0 copy number data was done as follows. First, probes were mapped to genes with information obtained from the pd.genomewidesnp.6 Bioconductor package (44). For genes measured by two probes, we defined the gene-level values as an unweighted average of the data from the two probes. For genes measured by a single probe, we defined the gene-level values as the data for the corresponding probe. For those measured by more than two probes, we devised an approach that weighted probes based on their similarity to the first eigengene as defined by taking a singular value decomposition of the probe-level data for each gene. The percent variance explained by the first eigengene was then calculated for each probe. The summarized values for each gene were then defined as the weighted mean, with the weights corresponding to the percent variance explained.

### Compute resources

A total of 2000 computational cores in the Google Cloud were provided to participants for the ~5-month duration of the Challenge, corresponding to a maximum of 7.5 million core hours if used at capacity. Specifically, each participant was provisioned an 8-core, 16-GB RAM machine, preconfigured and tested with an R computing environment and required libraries used in the Challenge. Compute resources were provisioned to each participant for dedicated use throughout the Challenge, and once capacity was reached, resources assigned to inactive users were recycled to new registrants such that all active users were provisioned compute resources. Users were also allowed to work on their own computing resources.

### OsloVal data generation

The OsloVal cohort consisted of fresh-frozen primary tumors from 184 breast cancer patients collected from 1981 to 1999 (148 from 1981 to 1989 and 36 from 1994 to 1999) at the Norwegian Radium Hospital. Tumor material collection, clinical characterization, and DNA and mRNA extraction methods are described in the Supplementary Materials and Methods.

## Supplementary Materials

www.sciencetranslationalmedicine.org/cgi/content/full/5/181/181re1/DC1

Materials and Methods

Table S1. Univariate and multivariate Cox regression statistics for the clinical covariates in the METABRIC data set.

Table S2. Univariate and multivariate Cox regression statistics for the clinical covariates in the OsloVal data set.

Table S3. The Breast Cancer Challenge Consortium: Challenge participants who submitted a model to phase 3 of the BCC.

## References and Notes

1. Acknowledgments: We thank I. R. Bergheim, E. U. Due, A. H. T. Olsen, C. Pedersen, M. La. Skrede, and P. Vu (Department of Genetics, Institute for Cancer Research) and L. I. Håseth (Department of Pathology, Oslo University Hospital, The Norwegian Radium Hospital) for their great efforts in the laboratory in collecting, cataloging, preparing, and analyzing the breast cancer tissue samples in a very short time; I. S. Jang for help on data analysis; H. Dai for assistance with the 70-gene signature evaluation; and L. Van’t Veer for helpful discussions in formulating the Challenge. The Breast Cancer Challenge Consortium members (listed in table S3) are each acknowledged for their submission of at least one computational model to phase 3 of the BCC. Funding: This work was funded by NIH/National Cancer Institute grant 5U54CA149237 and Washington Life Science Discovery Fund grant 3104672. Generation of the OsloVal data set was funded by the Avon Foundation (Foundation for the NIH grant FRIE12PD), The Norwegian Cancer Society, and The Radium Hospital Foundation. Computational resources provided to Challenge participants were donated by Google. Author contributions: A.A.M., E.B., E.H., T.C.N., H.K.M.V., O.M.R., J.G., V.N.K., S.H.F., G.S., S.A., C. Caldas, and A.-L.B.-D. conceived and designed the study. A.A.M., E.B., E.H., T.C.N., S.H.F., and G.S. drafted the manuscript. A.A.M., E.B., and E.H. performed data analysis. L.O., H.K.M.V., H.G.R., D.P., V.O.V., and A.-L.B.-D. organized clinical data and samples for the OsloVal data set. B.H.M. performed data normalization. B.S., M.R.K., M.D.F., N.A.D., B.H., and X.S. developed software infrastructure. L.M.M. organized data governance. T.P., L.Y., C. Citro, and J.H. developed Google compute resources. C. Curtis, S.A., and C. Caldas developed the METABRIC data set.
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# 5.2. Background¶

## 5.2.1. Scheduling Algorithms¶

RTEMS provides a plugin framework that allows it to support multiple scheduling algorithms. RTEMS includes multiple scheduling algorithms, and the user can select which of these they wish to use in their application at link-time. In addition, the user can implement their own scheduling algorithm and configure RTEMS to use it.

Supporting multiple scheduling algorithms gives the end user the option to select the algorithm which is most appropriate to their use case. Most real-time operating systems schedule tasks using a priority based algorithm, possibly with preemption control. The classic RTEMS scheduling algorithm which was the only algorithm available in RTEMS 4.10 and earlier, is a fixed-priority scheduling algorithm. This scheduling algorithm is suitable for uniprocessor (e.g., non-SMP) systems and is known as the Deterministic Priority Scheduler. Unless the user configures another scheduling algorithm, RTEMS will use this on uniprocessor systems.

## 5.2.2. Priority Scheduling¶

When using priority based scheduling, RTEMS allocates the processor using a priority-based, preemptive algorithm augmented to provide round-robin characteristics within individual priority groups. The goal of this algorithm is to guarantee that the task which is executing on the processor at any point in time is the one with the highest priority among all tasks in the ready state.

When a task is added to the ready chain, it is placed behind all other tasks of the same priority. This rule provides a round-robin within a priority group scheduling characteristic. This means that in a group of equal priority tasks, tasks will execute in the order they become ready or FIFO order. Even though there are ways to manipulate and adjust task priorities, the most important rule to remember is:

Note

Priority based scheduling algorithms will always select the highest priority task that is ready to run when allocating the processor to a task.

Priority scheduling is the most commonly used scheduling algorithm. It should be used by applications in which multiple tasks contend for CPU time or other resources, and there is a need to ensure certain tasks are given priority over other tasks.

There are a few common methods of accomplishing the mechanics of this algorithm. These ways involve a list or chain of tasks in the ready state.

• The least efficient method is to randomly place tasks in the ready chain forcing the scheduler to scan the entire chain to determine which task receives the processor.

• A more efficient method is to schedule the task by placing it in the proper place on the ready chain based on the designated scheduling criteria at the time it enters the ready state. Thus, when the processor is free, the first task on the ready chain is allocated the processor.

• Another mechanism is to maintain a list of FIFOs per priority. When a task is readied, it is placed on the rear of the FIFO for its priority. This method is often used with a bitmap to assist in locating which FIFOs have ready tasks on them. This data structure has $$O(1)$$ insert, extract and find highest ready run-time complexities.

• A red-black tree may be used for the ready queue with the priority as the key. This data structure has $$O(log(n))$$ insert, extract and find highest ready run-time complexities while $$n$$ is the count of tasks in the ready queue.

RTEMS currently includes multiple priority based scheduling algorithms as well as other algorithms that incorporate deadline. Each algorithm is discussed in the following sections.

## 5.2.3. Scheduling Modification Mechanisms¶

RTEMS provides four mechanisms which allow the user to alter the task scheduling decisions:

• manual round-robin selection

Each of these methods provides a powerful capability to customize sets of tasks to satisfy the unique and particular requirements encountered in custom real-time applications. Although each mechanism operates independently, there is a precedence relationship which governs the effects of scheduling modifications. The evaluation order for scheduling characteristics is always priority, preemption mode, and timeslicing. When reading the descriptions of timeslicing and manual round-robin it is important to keep in mind that preemption (if enabled) of a task by higher priority tasks will occur as required, overriding the other factors presented in the description.

### 5.2.3.1. Task Priority and Scheduling¶

The most significant task scheduling modification mechanism is the ability for the user to assign a priority level to each individual task when it is created and to alter a task’s priority at run-time, see Task Priority.

### 5.2.3.2. Preemption¶

Another way the user can alter the basic scheduling algorithm is by manipulating the preemption mode flag (RTEMS_PREEMPT_MASK) of individual tasks. If preemption is disabled for a task (RTEMS_NO_PREEMPT), then the task will not relinquish control of the processor until it terminates, blocks, or re-enables preemption. Even tasks which become ready to run and possess higher priority levels will not be allowed to execute. Note that the preemption setting has no effect on the manner in which a task is scheduled. It only applies once a task has control of the processor.

### 5.2.3.3. Timeslicing¶

Timeslicing or round-robin scheduling is an additional method which can be used to alter the basic scheduling algorithm. Like preemption, timeslicing is specified on a task by task basis using the timeslicing mode flag (RTEMS_TIMESLICE_MASK). If timeslicing is enabled for a task (RTEMS_TIMESLICE), then RTEMS will limit the amount of time the task can execute before the processor is allocated to another task. Each tick of the real-time clock reduces the currently running task’s timeslice. When the execution time equals the timeslice, RTEMS will dispatch another task of the same priority to execute. If there are no other tasks of the same priority ready to execute, then the current task is allocated an additional timeslice and continues to run. Remember that a higher priority task will preempt the task (unless preemption is disabled) as soon as it is ready to run, even if the task has not used up its entire timeslice.

### 5.2.3.4. Manual Round-Robin¶

The final mechanism for altering the RTEMS scheduling algorithm is called manual round-robin. Manual round-robin is invoked by using the rtems_task_wake_after directive with a time interval of RTEMS_YIELD_PROCESSOR. This allows a task to give up the processor and be immediately returned to the ready chain at the end of its priority group. If no other tasks of the same priority are ready to run, then the task does not lose control of the processor.

Tasks that utilize a numeric coprocessor and are created with the RTEMS_FLOATING_POINT attribute require additional operations during a context switch. These additional operations are necessary to save and restore the floating point context of RTEMS_FLOATING_POINT tasks. To avoid unnecessary save and restore operations, the state of the numeric coprocessor is only saved when a RTEMS_FLOATING_POINT task is dispatched and that task was not the last task to utilize the coprocessor.

Tasks in an RTEMS system must always be in one of the five allowable task states. These states are: executing, ready, blocked, dormant, and non-existent.

A task occupies the non-existent state before a rtems_task_create has been issued on its behalf. A task enters the non-existent state from any other state in the system when it is deleted with the rtems_task_delete directive. While a task occupies this state it does not have a TCB or a task ID assigned to it; therefore, no other tasks in the system may reference this task.

When a task is created via the rtems_task_create directive, it enters the dormant state. This state is not entered through any other means. Although the task exists in the system, it cannot actively compete for system resources. It will remain in the dormant state until it is started via the rtems_task_start directive, at which time it enters the ready state. The task is now permitted to be scheduled for the processor and to compete for other system resources.

A task occupies the blocked state whenever it is unable to be scheduled to run. A running task may block itself or be blocked by other tasks in the system. The running task blocks itself through voluntary operations that cause the task to wait. The only way a task can block a task other than itself is with the rtems_task_suspend directive. A task enters the blocked state due to any of the following conditions:

• A task issues a rtems_task_suspend directive which blocks either itself or another task in the system.

• The running task issues a rtems_barrier_wait directive.

• The running task issues a rtems_message_queue_receive directive with the wait option, and the message queue is empty.

• The running task issues a rtems_event_receive directive with the wait option, and the currently pending events do not satisfy the request.

• The running task issues a rtems_semaphore_obtain directive with the wait option and the requested semaphore is unavailable.

• The running task issues a rtems_task_wake_after directive which blocks the task for the given time interval. If the time interval specified is zero, the task yields the processor and remains in the ready state.

• The running task issues a rtems_task_wake_when directive which blocks the task until the requested date and time arrives.

• The running task issues a rtems_rate_monotonic_period directive and must wait for the specified rate monotonic period to conclude.

• The running task issues a rtems_region_get_segment directive with the wait option and there is not an available segment large enough to satisfy the task’s request.

A blocked task may also be suspended. Therefore, both the suspension and the blocking condition must be removed before the task becomes ready to run again.

A task occupies the ready state when it is able to be scheduled to run, but currently does not have control of the processor. Tasks of the same or higher priority will yield the processor by either becoming blocked, completing their timeslice, or being deleted. All tasks with the same priority will execute in FIFO order. A task enters the ready state due to any of the following conditions:

• A running task issues a rtems_task_resume directive for a task that is suspended and the task is not blocked waiting on any resource.

• A running task issues a rtems_message_queue_send, rtems_message_queue_broadcast, or a rtems_message_queue_urgent directive which posts a message to the queue on which the blocked task is waiting.

• A running task issues an rtems_event_send directive which sends an event condition to a task that is blocked waiting on that event condition.

• A running task issues a rtems_semaphore_release directive which releases the semaphore on which the blocked task is waiting.

• A timeout interval expires for a task which was blocked by a call to the rtems_task_wake_after directive.

• A timeout period expires for a task which blocked by a call to the rtems_task_wake_when directive.

• A running task issues a rtems_region_return_segment directive which releases a segment to the region on which the blocked task is waiting and a resulting segment is large enough to satisfy the task’s request.

• A rate monotonic period expires for a task which blocked by a call to the rtems_rate_monotonic_period directive.

• A timeout interval expires for a task which was blocked waiting on a message, event, semaphore, or segment with a timeout specified.

• A running task issues a directive which deletes a message queue, a semaphore, or a region on which the blocked task is waiting.

• A running task issues a rtems_task_restart directive for the blocked task.

• The running task, with its preemption mode enabled, may be made ready by issuing any of the directives that may unblock a task with a higher priority. This directive may be issued from the running task itself or from an ISR. A ready task occupies the executing state when it has control of the CPU. A task enters the executing state due to any of the following conditions:

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## Brown University: Rich Schwartz's "Countable and Uncountable Sets"

Read this document to learn about countable and uncountable sets. Focus on the several examples of uncountable subsets of R.


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## Stream: maths

If I import group_theory.coset and #print axioms it shows quotient_group._to_additive : Type This is because on line 141 there is a run_cmd that as far as I can tell adds this axiom and nothing else. I'm guessing this isn't the intended behaviour. Maybe something was unfinished.


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# Alternatives to electromagnetism for communication?

It seems, from my (weak) understanding of the standard model that the only field suitable for long range communication is the one we actually use, electromagnetism. (Gravity waves could be used as well I suppose, but gravity is too weak to be practical use.)

Questions:

1. Is this correct? It seems so, as you need a boson (to be able to produce 'intense' signals) and all the others are either unstable or strongly interacting (e.g. gluon)

2. Why is it like that? Is there some underlying fundamental reason that there are no other 'useful' bosons?

3. Does 'beyond the standard model' physics provide any insight? Any potential alternatives? (Graviphoton?)

-

## 3 Answers

Yes, this is correct. The reason for Photons being our only means of long-range communication lies in the fact that at very low energies (where we live today) the unbroken subgroup of the standard model is $$SU(3)_c \times U(1)_\mathrm{em}$$ The $SU(3)$ color part is not useful, as it is confining at low energies. The $U(1)$ electromagnetic part gives rise to a chargeless, massless boson that interacts with everyday matter, so it is perfect for that purpose.

You are correct in assuming that gravitons would in principle work as well, but their coupling to matter is just too weak for any practical applications. Same goes for Graviphotons.

Physics beyond the Standard Model always has to have the Standard Model as a low-energy limit. Therefore, even if new bosons arise, they either have to have tiny couplings (e.g. axions) or large masses (e.g. GUT gauge bosons) in order to explain why we have not seen them yet. If by now we have not had a chance to see the damn thing, how would we use it to communicate?

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it is not as correct as stated, it is a (plausible?) speculation –  Nikos M. May 29 '14 at 22:50
@NikosM. I would like to know where I went wrong, so I can correct the answer and learn more myself. Can you provide some reference? –  Neuneck May 30 '14 at 8:06
gravitons are a speculative entity (in the fashion of a boson with spin=2) of some GUT-type theories, as also of course are some super-symmetric partners of (experimentally found) elementary particles, does this answer your question? –  Nikos M. May 30 '14 at 8:11
it is not only known if gravitons exist (or exist in that format) but also not known if communication (as with photons) can be achieved even in principle, i think what i just stated is correct, if it is not so, feel (very) free to correct me –  Nikos M. May 30 '14 at 8:14
@NikosM. Agreed, gravitons are hypothetical, but they are independent of the GUT framework. Even GR predicts gravitational waves which could in principle be used for communication - as with photons you have a wave-like phenomenon. The meaningful difference would not be that they are tensor waves (this just means that a linear antenna would not work - interferometers could in principle work, though), but rather that they are non-linearly coupled. Still, not even binary pulsars can create gravit. waves strong enough to challenge the linear approximation, so no man-made technology will, either. –  Neuneck May 30 '14 at 9:21

I believe long-range communication using weak interaction ((anti)neutrinos) is also possible, at least, in principle (see my answer at Are Electromagnetic Waves The Only Means of Transmitting Information? ). I don't quite see why communication using fermions is not possible.

EDIT(05/24/2014): It looks like neutron beams can also be used for long-distance communications, at least if guiding structures are used (http://www.manep-nccr.ch/en/technological-challenges/neutrons.html )

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I discounted fermions in my question, but found this article about using neutrinos: physicsworld.com/cws/article/news/2012/mar/19/… –  DrEntropy May 24 '14 at 16:56
@DrEntropy: I guess the source you quote is based on the work quoted in my answer to the other question. –  akhmeteli May 24 '14 at 17:02

Another field used for communication is the pressure - velocity field of material media. See: http://en.wikipedia.org/wiki/Acoustic_theory. For example, long range communication through solid and liquid media are exercised by elephants and whales respectively.

-
you mean... "sound"? –  Dmitry Brant May 25 '14 at 16:22


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# Degrees of freedom (statistics)

degrees of freedomdegree of freedomEffective degrees of freedomstatistical degrees of freedomd.f.degrees of freedom in statisticsdegrees-of-freedomEffective degree of freedomeffective number of degrees of freedomResidual effective degrees of freedom
In statistics, the number of degrees of freedom is the number of values in the final calculation of a statistic that are free to vary.wikipedia
157 Related Articles

### Chi-squared distribution

chi-squaredchi-square distributionchi square distribution
The degrees of freedom are also commonly associated with the squared lengths (or "sum of squares" of the coordinates) of such vectors, and the parameters of chi-squared and other distributions that arise in associated statistical testing problems.
In probability theory and statistics, the chi-square distribution (also chi-squared or χ 2 -distribution) with k degrees of freedom is the distribution of a sum of the squares of k independent standard normal random variables.

### Statistical parameter

parametersparameterparametrization
Estimates of statistical parameters can be based upon different amounts of information or data.
For example, the family of chi-squared distributions can be indexed by the number of degrees of freedom: the number of degrees of freedom is a parameter for the distributions, and so the family is thereby parameterized.

### Student's t-distribution

Student's ''t''-distributiont-distributiont''-distribution
While Gosset did not actually use the term 'degrees of freedom', he explained the concept in the course of developing what became known as Student's t-distribution.
where \nu is the number of degrees of freedom and \Gamma is the gamma function.

### Nu (letter)

nuΝGreek letter nu
In equations, the typical symbol for degrees of freedom is ν (lowercase Greek letter nu).

### Analysis of variance

ANOVAanalysis of variance (ANOVA)corrected the means
The term is most often used in the context of linear models (linear regression, analysis of variance), where certain random vectors are constrained to lie in linear subspaces, and the number of degrees of freedom is the dimension of the subspace.
The number of degrees of freedom DF can be partitioned in a similar way: one of these components (that for error) specifies a chi-squared distribution which describes the associated sum of squares, while the same is true for "treatments" if there is no treatment effect.

### Student's t-test

t-testt''-testStudent's ''t''-test
Likewise, the one-sample t-test statistic,
degrees of freedom.

### Errors and residuals

residualserror termresidual
are residuals that may be considered estimates of the errors X i − μ.
The sum of squares of the statistical errors, divided by σ 2, has a chi-squared distribution with n degrees of freedom:

### Least squares

least-squaresmethod of least squaresleast squares method
An example which is only slightly less simple is that of least squares estimation of a and b in the model
The denominator, n − m, is the statistical degrees of freedom; see effective degrees of freedom for generalizations.

### Ordinary least squares

OLSleast squaresOrdinary least squares regression
Many non-standard regression methods, including regularized least squares (e.g., ridge regression), linear smoothers, smoothing splines, and semiparametric regression are not based on ordinary least squares projections, but rather on regularized (generalized and/or penalized) least-squares, and so degrees of freedom defined in terms of dimensionality is generally not useful for these procedures.
: The numerator, n−p, is the statistical degrees of freedom.

### Residual sum of squares

sum of squared residualssum of squares of residualsresidual sum-of-squares
In the example above, the residual sum-of-squares is

### Restricted randomization

nested datasplit plotNested factors
In some complicated settings, such as unbalanced split-plot designs, the sums-of-squares no longer have scaled chi-squared distributions.
Since there are 8 degrees of freedom for the subplot error term, this MSE can be used to test each effect that involves current.

### F-distribution

F distributionF''-distributionF'' distribution
If there is no difference between population means this ratio follows an F distribution with 2 and 3n − 3 degrees of freedom.

### Random variable

random variablesrandom variationrandom
are random variables each with expected value μ, and let
This is a chi-squared distribution with one degree of freedom.

### Welch–Satterthwaite equation

Pooled degree of freedomPooled degrees of freedomSatterthwaite approximation
For the regression effective degrees of freedom, appropriate definitions can include the trace of the hat matrix, tr(H), the trace of the quadratic form of the hat matrix, tr(H'H), the form tr(2H – H H'), or the Satterthwaite approximation, tr(H'H) 2 /tr(H'HH'H).
In statistics and uncertainty analysis, the Welch–Satterthwaite equation is used to calculate an approximation to the effective degrees of freedom of a linear combination of independent sample variances, also known as the pooled degrees of freedom, corresponding to the pooled variance.

### Tikhonov regularization

ridge regressionregularizeda squared regularizing function
Many non-standard regression methods, including regularized least squares (e.g., ridge regression), linear smoothers, smoothing splines, and semiparametric regression are not based on ordinary least squares projections, but rather on regularized (generalized and/or penalized) least-squares, and so degrees of freedom defined in terms of dimensionality is generally not useful for these procedures.
where is the residual sum of squares, and \tau is the effective number of degrees of freedom.

### Goodness of fit

goodness-of-fitfitgoodness-of-fit test
The effective degrees of freedom of the fit can be defined in various ways to implement goodness-of-fit tests, cross-validation, and other statistical inference procedures.
The chi-squared distribution has (k − c) degrees of freedom, where k is the number of non-empty cells and c is the number of estimated parameters (including location and scale parameters and shape parameters) for the distribution plus one.

### Confidence region

3-dimensional uncertainty ellipsoiderror ellipseline of variation
The more general formulation of effective degree of freedom would result in a more realistic estimate for, e.g., the error variance σ 2, which in its turn scales the unknown parameters' a posteriori standard deviation; the degree of freedom will also affect the expansion factor necessary to produce an error ellipse for a given confidence level.
Further, F is the quantile function of the F-distribution, with p and \nu = n - p degrees of freedom, \alpha is the statistical significance level, and the symbol X^\prime means the transpose of X.

### Reduced chi-squared statistic

reduced chi-squaredMean square weighted deviationChi-squared per degree of freedom
It is defined as chi-squared per degree of freedom:

### Projection matrix

hat matrixannihilator matrixobservation matrix
:where \hat{y} is the vector of fitted values at each of the original covariate values from the fitted model, y is the original vector of responses, and H is the hat matrix or, more generally, smoother matrix.
For other models such as LOESS that are still linear in the observations \mathbf{y}, the projection matrix can be used to define the effective degrees of freedom of the model.

### Semiparametric regression

regressionsemiparametric modelingsingle and multiple index models
Many non-standard regression methods, including regularized least squares (e.g., ridge regression), linear smoothers, smoothing splines, and semiparametric regression are not based on ordinary least squares projections, but rather on regularized (generalized and/or penalized) least-squares, and so degrees of freedom defined in terms of dimensionality is generally not useful for these procedures.

### Generalized least squares

feasible generalized least squaresgeneralizedgeneralized (correlated)
Many non-standard regression methods, including regularized least squares (e.g., ridge regression), linear smoothers, smoothing splines, and semiparametric regression are not based on ordinary least squares projections, but rather on regularized (generalized and/or penalized) least-squares, and so degrees of freedom defined in terms of dimensionality is generally not useful for these procedures.

### Replication (statistics)

replicationreplicatereplicates

### Sample size determination

sample sizeSampling sizessample
All the parameters in the equation are in fact the degrees of freedom of the number of their concepts, and hence, their numbers are subtracted by 1 before insertion into the equation.

### Generalized chi-squared distribution

Generalized chi-square distribution
The residual sum-of-squares \|y-Hy\|^2 has a generalized chi-squared distribution, and the theory associated with this distribution provides an alternative route to the answers provided above.
* Degrees of freedom (statistics)#Alternative

### Statistics

statisticalstatistical analysisstatistician
In statistics, the number of degrees of freedom is the number of values in the final calculation of a statistic that are free to vary.


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# If $1,000 is deposited in a certain bank account and remains  Question banks Downloads My Bookmarks Reviews Important topics Author Message TAGS: Intern Joined: 18 Aug 2010 Posts: 13 Followers: 0 Kudos [?]: 18 [3] , given: 7 If$1,000 is deposited in a certain bank account and remains [#permalink]  01 Nov 2010, 16:32
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Question Stats:

43% (02:40) correct 57% (01:43) wrong based on 373 sessions
If $1,000 is deposited in a certain bank account and remains in the account along with any accumulated interest, the dollar amount of interest, I, earned by the deposit in the first n years is given by the formula I=1,000((1+r/100)^n-1), where r percent is the annual interest rate paid by the bank. Is the annual interest rate paid by the bank greater than 8 percent? (1) The deposit earns a total of$210 in interest in the first two years
(2) (1 + r/100 )^2 > 1.15
[Reveal] Spoiler: OA

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Re: Annual interest [#permalink]  01 Nov 2010, 19:07
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If $1,000 is deposited in a certain bank account and remains in the account along with any accumulated interest, the dollar amount of interest, I, earned by the deposit in the first n years is given by the formula I=1,000((1+r/100)^n-1), where r percent is the annual interest rate paid by the bank. Is the annual interest rate paid by the bank greater than 8 percent? Given: $$I=1,000((1+\frac{r}{100})^n-1)$$. Question: is $$r>8$$. (1) The deposit earns a total of$210 in interest in the first two years --> $$I=210$$ and $$n=2$$ --> $$210=1,000((1+\frac{r}{100})^2-1)$$ --> note that we are left with only one unknown in this equation, $$r$$, and we'll be able to solve for it and say whether it's more than 8, so even withput actual solving we can say that this statement is sufficient.

(2) (1 + r/100 )^2 > 1.15 --> if $$r=8$$ then $$(1+\frac{r}{100})^2=(1+\frac{8}{100})^2=1.08^2\approx{1.16}>1.15$$ so, if $$r$$ is slightly less than 8 (for example 7.99999), $$(1+\frac{r}{100})^2$$ will still be more than 1.15. So, this statement is not sufficient to say whether $$r>8$$.

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Re: Annual interest [#permalink]  01 Nov 2010, 19:16
ah... for S2, I approached it from the other angle and had to take the square root of 1.15. I got stuck there and time was running out, so I took a guess. It's much easier to multiply 1.08 by 1.08 than to take the square root of 1.15.

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Re: Annual interest [#permalink]  02 Nov 2010, 17:53
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butterfly wrote:
It's much easier to multiply 1.08 by 1.08 than to take the square root of 1.15.

Shortcut to multiply numbers of the form (100 + a) or (100 - a)
Write $$a^2$$ on the right hand side. Add a to the original number and write it on left side. The square is ready.

e.g. $$108^2 = (100 + 8)^2$$ Write 64 on right hand side

________ 64

Add 8 to 108 to get 116 and write that on left hand side

11664 - Square of 108

e.g. $$91^2 = (100 - 9)^2$$ => ______81 => 8281
(Here, subtract 9 from 91)

Note: a could be a two digit number as well.
e.g $$112^2 = (100 + 12)^2$$ = ______44 => 12544
(Only last two digit of the square of 12 are written on the right hand side. The 1 of 144 is carried over and added to 112 + 12)

This is Vedic Math though the trick uses basic algebra.
$$(100 + a)^2 = 10000 + 200a + a^2$$
(100 + 8)^2 = 10000 + 200 x 8 + 64 = 10000 + 1600 + 64 = 11664

This is a useful trick that saves time.
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Get started with Veritas Prep GMAT On Demand for $199 Veritas Prep Reviews Manager Joined: 22 Jun 2010 Posts: 57 Followers: 1 Kudos [?]: 25 [0], given: 10 Re: Annual interest [#permalink] 03 Nov 2010, 07:59 Very neat trick Karishma! This should save me a lot of time!! THANKS Current Student Joined: 27 Jun 2012 Posts: 418 Concentration: Strategy, Finance Followers: 63 Kudos [?]: 559 [0], given: 183 Re: Annual interest [#permalink] 25 Dec 2012, 19:42 Bunuel wrote: If$1,000 is deposited in a certain bank account and remains in the account along with any accumulated interest, the dollar amount of interest, I, earned by the deposit in the first n years is given by the formula I=1,000((1+r/100)^n-1), where r percent is the annual interest rate paid by the bank. Is the annual interest rate paid by the bank greater than 8 percent?

Given: $$I=1,000((1+\frac{r}{100})^n-1)$$. Question: is $$r>8$$.

(1) The deposit earns a total of $210 in interest in the first two years --> $$I=210$$ and $$n=2$$ --> $$210=1,000((1+\frac{r}{100})^2-1)$$ --> note that we are left with only one unknown in this equation, $$r$$, and we'll be able to solve for it and say whether it's more than 8, so even withput actual solving we can say that this statement is sufficient. (2) (1 + r/100 )^2 > 1.15 --> if $$r=8$$ then $$(1+\frac{r}{100})^2=(1+\frac{8}{100})^2=1.08^2\approx{1.16}>1.15$$ so, if $$r$$ is slightly less than 8 (for example 7.99999), $$(1+\frac{r}{100})^2$$ will still be more than 1.15. So, this statement is not sufficient to say whether $$r>8$$. Answer: A. Hello Bunuel, your explanation for second DS choice suggests that, if we have only 1 variable in the equation, then we need not solve it. However, I have observed few of the GMAT problems that have similar quadratic equations (with second degree) solve to two different positive roots, hence the DS choice could not be true. I believe it would be safe to solve the equation until you know if its only going to give you "one" root. e.g. $$ax^2+bx-c=0$$, this equation will have one positive and one negative root. As rate in this case is supposed to be positive, hence only 1 root. However, if the equation resolves to $$ax^2-bx+c=0$$ then it can have two positive roots (one of which may be less than 8 and other more than 8), hence the choice may not be true. Only if both positive roots are more than 8, then the choice can be taken as true. Please advice. _________________ Thanks, Prashant Ponde Tough 700+ Level RCs: Passage1 | Passage2 | Passage3 | Passage4 | Passage5 | Passage6 | Passage7 Reading Comprehension notes: Click here VOTE: vote-best-gmat-practice-tests-excluding-gmatprep-144859.html PowerScore CR Bible - Official Guide 13 Questions Set Mapped: Click here Verbal Forum Moderator Status: Been a long time guys... Joined: 03 Feb 2011 Posts: 1420 Location: United States (NY) Concentration: Finance, Marketing GPA: 3.75 Followers: 157 Kudos [?]: 938 [1] , given: 62 Re: If$1,000 is deposited in a certain bank [#permalink]  02 Jan 2013, 20:32
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kiyo0610 wrote:
If $1,000 is deposited in a certain bank account and remains in the account along with any accumulated interest, the dollar amount of interest, I, earned by the deposit in the first n years is given by the formula I=1,000[(1+r/100)^n - 1] , where r percent is the annual interest rate paid by the bank. Is the annual interest rate paid by the bank greater than 8 percent? (1) The deposit earns a total of$210 in interest in the first two years.

(2) (1+r/100)^2 >1.15

[Reveal] Spoiler:
A

statement 1) I=$210, n=2 Putting this in the equation given in the question, we will be able to find the value of r and thereby be able to answer the question. Suffiicient. Statement 2) Using Binomial theorem, we can infer $$(1+r/100)^2 > 1.15$$ as $$(1+2r/100) > 1.15$$. On solving this relation we will get, r>7.5. But since its not given that r is an integer then r can be 7.51, 7.6,9, 11 etc. Hence insufficient. +1A _________________ Veritas Prep GMAT Instructor Joined: 16 Oct 2010 Posts: 6062 Location: Pune, India Followers: 1599 Kudos [?]: 8956 [1] , given: 195 Re: If$1,000 is deposited in a certain bank account and remains [#permalink]  14 Mar 2013, 20:16
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Responding to a pm:
Question: (1 + .08)^2 = ?

$$1.08 = \frac{108}{100}$$ (it's trickier to deal with decimal so remove it)

$$(\frac{108}{100})^2 = \frac{108^2}{10000}$$

We know how to get the square of 108
$$108^2 = 11664$$ (discussed in the post above)

So, $$(1 + .08)^2 = 11664/10000 = 1.1664$$

Or you can use (a + b)^2 = a^2 + b^2 + 2ab (the shortcut is anyway based on this formula only)

(1 + .08)^2 = 1 + .0064 + 2*1*.08 = 1.1664
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Get started with Veritas Prep GMAT On Demand for $199 Veritas Prep Reviews Verbal Forum Moderator Joined: 10 Oct 2012 Posts: 629 Followers: 66 Kudos [?]: 840 [0], given: 135 Re: If$1,000 is deposited in a certain bank [#permalink]  14 Mar 2013, 23:40
Expert's post
Marcab wrote:
kiyo0610 wrote:
If $1,000 is deposited in a certain bank account and remains in the account along with any accumulated interest, the dollar amount of interest, I, earned by the deposit in the first n years is given by the formula I=1,000[(1+r/100)^n - 1] , where r percent is the annual interest rate paid by the bank. Is the annual interest rate paid by the bank greater than 8 percent? (1) The deposit earns a total of$210 in interest in the first two years.

(2) (1+r/100)^2 >1.15

The total interest is given as I=1,000[(1+r/100)^n - 1].

From F.S 1 we have that I = 210. Thus, we have a quadratic equation and we know that it can be solved leading to a fixed value for r. Sufficient.Also, one can notice that an interest of 210$is obtained when r=10% and this is greater than 8%. Sufficient. From F.S 2, we know that n=2. And the Interest earned would be greater than 150.Thus, I=1,000[(1+r/100)^2 - 1] = $$1000[r/100*(2+\frac{r}{100})]$$. We know for r=8% we have this equal to 2.08*80 = 166.4 which is anyways greater than 150. Now, for r=7%, the expression equals 2.07*70 = 144.9. Thus, for a value between 7 and 8 , this value will change and become more than 150. Thus we wouldn't know for sure if r>8 or not. Insufficient. A. _________________ Intern Joined: 10 Apr 2012 Posts: 23 Concentration: Finance, Economics GMAT 1: 760 Q50 V44 GMAT 2: Q V GMAT 3: Q V Followers: 0 Kudos [?]: 32 [4] , given: 0 Re: If$1,000 is deposited in a certain bank account and remains [#permalink]  15 Mar 2013, 15:47
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I think the gmat is always about insight and not about arithemetic
.
Question: Is r>8%? If r can be 8 percent or greater, the statement will be insufficient.

Stmt 2) Overall interest earned over two years, is greater than 15% (If you read this far down, you know what I am talking about)
Lets say the interest was 8%, then overall compound interest earned over two years will be greater than 16% and so greater than 15%
It goes without saying that if the interest rate was greater than 8%, then the amount of interest earned over two years, will still be greater than 15%.

It took me 1:30 seconds to see this, and another 40 seconds to type this entire post because I was not satisfied with the explanations given
.No messy calculations or nail biting necessary.
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Re: If $1,000 is deposited in a certain bank account and remains [#permalink] 16 Mar 2013, 13:26 Brilliant! I was convinced the answer is D, when i realized that 7.99^2 is also a matter of concern! _________________ One Last Shot Intern Joined: 18 Jan 2013 Posts: 1 Followers: 0 Kudos [?]: 5 [0], given: 0 Re: If$1,000 is deposited in a certain bank account and remains [#permalink]  19 Mar 2013, 10:37
hi all , why not D?
From first statement, one can answer that interest rate is greater than 8%
From second second statement, one can answer that interest rate is less than 8%

So, either statement can be used to answer the question. Am I missing any thing? Please reply.
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Re: If $1,000 is deposited in a certain bank account and remains [#permalink] 19 Mar 2013, 19:39 Expert's post chandrak wrote: hi all , why not D? From first statement, one can answer that interest rate is greater than 8% From second second statement, one can answer that interest rate is less than 8% So, either statement can be used to answer the question. Am I missing any thing? Please reply. How can you say that the interest rate is less than 8% from the second statement? If r were 8%, we would have (1 + r/100 )^2 = 1.08^2 = 1.1664 Now all that statement 2 tells us is that (1 + r/100 )^2 > 1.15 We don't know whether it is less than 1.1664 or greater. Hence statement 2 alone is not sufficient. Besides, it is not possible that statement 1 tells you that r is greater than 8% and statement 2 tells you that it is less than 8%. This is a conflict. If both statements independently give you the answer, the answer you will get will be the same i.e. either both will tell that r is greater than 8% or both will tell that r is less than 8%. _________________ Karishma Veritas Prep | GMAT Instructor My Blog Get started with Veritas Prep GMAT On Demand for$199

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Re: Annual interest [#permalink]  29 Mar 2013, 07:54
Bunuel wrote:
If $1,000 is deposited in a certain bank account and remains in the account along with any accumulated interest, the dollar amount of interest, I, earned by the deposit in the first n years is given by the formula I=1,000((1+r/100)^n-1), where r percent is the annual interest rate paid by the bank. Is the annual interest rate paid by the bank greater than 8 percent? Given: $$I=1,000((1+\frac{r}{100})^n-1)$$. Question: is $$r>8$$. (1) The deposit earns a total of$210 in interest in the first two years --> $$I=210$$ and $$n=2$$ --> $$210=1,000((1+\frac{r}{100})^2-1)$$ --> note that we are left with only one unknown in this equation, $$r$$, and we'll be able to solve for it and say whether it's more than 8, so even withput actual solving we can say that this statement is sufficient.

(2) (1 + r/100 )^2 > 1.15 --> if $$r=8$$ then $$(1+\frac{r}{100})^2=(1+\frac{8}{100})^2=1.08^2\approx{1.16}>1.15$$ so, if $$r$$ is slightly less than 8 (for example 7.99999), $$(1+\frac{r}{100})^2$$ will still be more than 1.15. So, this statement is not sufficient to say whether $$r>8$$.

Bunuel,

As this is a quadratic equation , how did you concluded that we will get one value after solving this equation?
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Re: Annual interest [#permalink]  17 Jul 2013, 10:15
greatps24 wrote:
Bunuel wrote:
If $1,000 is deposited in a certain bank account and remains in the account along with any accumulated interest, the dollar amount of interest, I, earned by the deposit in the first n years is given by the formula I=1,000((1+r/100)^n-1), where r percent is the annual interest rate paid by the bank. Is the annual interest rate paid by the bank greater than 8 percent? Given: $$I=1,000((1+\frac{r}{100})^n-1)$$. Question: is $$r>8$$. (1) The deposit earns a total of$210 in interest in the first two years --> $$I=210$$ and $$n=2$$ --> $$210=1,000((1+\frac{r}{100})^2-1)$$ --> note that we are left with only one unknown in this equation, $$r$$, and we'll be able to solve for it and say whether it's more than 8, so even withput actual solving we can say that this statement is sufficient.

Bunuel,

As this is a quadratic equation , how did you concluded that we will get one value after solving this equation?

Same question for Bunuel or any of the other experts here.

My calculations:

1) $$210 = 1000 [(1+\frac{r}{100})^2-1)$$

2) $$210 = 1000 [1+\frac{2r}{100}+\frac{r^2}{10000}-1]$$

3) $$210=1000(\frac{200r+r^2}{10,000})$$

4) $$210=\frac{200r+r^2}{10}$$

5) $$2100=r(200+r)$$

How do you solve for the variable r at this point?

Any further explanation would help.

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Re: Annual interest [#permalink]  17 Jul 2013, 20:05
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Im2bz2p345 wrote:
greatps24 wrote:
Bunuel wrote:
If $1,000 is deposited in a certain bank account and remains in the account along with any accumulated interest, the dollar amount of interest, I, earned by the deposit in the first n years is given by the formula I=1,000((1+r/100)^n-1), where r percent is the annual interest rate paid by the bank. Is the annual interest rate paid by the bank greater than 8 percent? Given: $$I=1,000((1+\frac{r}{100})^n-1)$$. Question: is $$r>8$$. (1) The deposit earns a total of$210 in interest in the first two years --> $$I=210$$ and $$n=2$$ --> $$210=1,000((1+\frac{r}{100})^2-1)$$ --> note that we are left with only one unknown in this equation, $$r$$, and we'll be able to solve for it and say whether it's more than 8, so even withput actual solving we can say that this statement is sufficient.

Bunuel,

As this is a quadratic equation , how did you concluded that we will get one value after solving this equation?

Same question for Bunuel or any of the other experts here.

My calculations:

1) $$210 = 1000 [(1+\frac{r}{100})^2-1)$$

2) $$210 = 1000 [1+\frac{2r}{100}+\frac{r^2}{10000}-1]$$

3) $$210=1000(\frac{200r+r^2}{10,000})$$

4) $$210=\frac{200r+r^2}{10}$$

5) $$2100=r(200+r)$$

How do you solve for the variable r at this point?

Any further explanation would help.

~ Im2bz2p345

Solving this quadratic is a little time consuming though we will see how to do in a minute. But you don't really need to solve it to figure out that you will have only one solution.

$$2100=r(200+r)$$
$$r^2 + 200r - 2100 = 0$$

In a quadratic, $$ax^2 + bx + c = 0$$, sum of the roots = -b/a and product of the roots = c/a
Notice that the product of the roots (-2100) is negative. This means one root is positive and the other is negative. So we will have only one acceptable solution (the positive one)

Now, if you would like to solve it:
$$r^2 + 200r - 2100 = 0$$

2100 = 2*2*5*5*3*7
Now you need to split 2100 into two factors such that one is a little larger than 200 and the other is a small factor e.g. 5 or 7 or 10 etc. Once you think this way, you easily get 210 and 10

$$r^2 + 210r - 10r - 2100 = 0$$
(r + 210)(r - 10) = 0
r = -210 or 10
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Get started with Veritas Prep GMAT On Demand for $199 Veritas Prep Reviews Intern Joined: 23 Mar 2011 Posts: 30 Followers: 0 Kudos [?]: 3 [0], given: 9 Re: Annual interest [#permalink] 17 Jul 2013, 20:17 VeritasPrepKarishma wrote: Solving this quadratic is a little time consuming though we will see how to do in a minute. But you don't really need to solve it to figure out that you will have only one solution. $$2100=r(200+r)$$ $$r^2 + 200r - 2100 = 0$$ In a quadratic, $$ax^2 + bx + c = 0$$, sum of the roots = -b/a and product of the roots = c/a Notice that the product of the roots (-2100) is negative. This means one root is positive and the other is negative. So we will have only one acceptable solution (the positive one) Now, if you would like to solve it: $$r^2 + 200r - 2100 = 0$$ 2100 = 2*2*5*5*3*7 Now you need to split 2100 into two factors such that one is a little larger than 200 and the other is a small factor e.g. 5 or 7 or 10 etc. Once you think this way, you easily get 210 and 10 $$r^2 + 210r - 10r - 2100 = 0$$ (r + 210)(r - 10) = 0 r = -210 or 10 Perfect! Thank you Karishma for filling in these last steps for me. You're always a big help! As PraPon pointed out earlier in the thread, I believe it's important to bring the first statement down to the $$r^2 + 200r - 2100 = 0$$ level, otherwise there is no "guarantee" that r will only have one root. Once you realize there is will be a positive & negative solution and that you can simply "ignore" the negative solution since we checking to see if r > 8 (leaving you with truly one solution), then you can make the judgement that the first statement is sufficient. I just don't know how you can assume the solution of $$210=1,000((1+\frac{r}{100})^2-1)$$ will have only ONE positive solution (as you pointed out Karishma, if the simplified form gave you "+ 2100" instead of "- 2100," the statement would be insufficient due to having TWO positive solutions). ~ Im2bz2p345 Math Expert Joined: 02 Sep 2009 Posts: 30395 Followers: 5096 Kudos [?]: 57457 [2] , given: 8813 Re: Annual interest [#permalink] 17 Jul 2013, 22:49 2 This post received KUDOS Expert's post Im2bz2p345 wrote: greatps24 wrote: Bunuel wrote: If$1,000 is deposited in a certain bank account and remains in the account along with any accumulated interest, the dollar amount of interest, I, earned by the deposit in the first n years is given by the formula I=1,000((1+r/100)^n-1), where r percent is the annual interest rate paid by the bank. Is the annual interest rate paid by the bank greater than 8 percent?

Given: $$I=1,000((1+\frac{r}{100})^n-1)$$. Question: is $$r>8$$.

(1) The deposit earns a total of $210 in interest in the first two years --> $$I=210$$ and $$n=2$$ --> $$210=1,000((1+\frac{r}{100})^2-1)$$ --> note that we are left with only one unknown in this equation, $$r$$, and we'll be able to solve for it and say whether it's more than 8, so even withput actual solving we can say that this statement is sufficient. Bunuel, As this is a quadratic equation , how did you concluded that we will get one value after solving this equation? Same question for Bunuel or any of the other experts here. My calculations: 1) $$210 = 1000 [(1+\frac{r}{100})^2-1)$$ 2) $$210 = 1000 [1+\frac{2r}{100}+\frac{r^2}{10000}-1]$$ 3) $$210=1000(\frac{200r+r^2}{10,000})$$ 4) $$210=\frac{200r+r^2}{10}$$ 5) $$2100=r(200+r)$$ How do you solve for the variable r at this point? Any further explanation would help. ~ Im2bz2p345 Actually you don't need to solve this way: $$1,000((1+\frac{r}{100})^2-1)=210$$ $$(1+\frac{r}{100})^2-1=\frac{210}{1,000}$$ $$(1+\frac{r}{100})^2=\frac{21}{100}+1$$ $$(1+\frac{r}{100})^2=\frac{121}{100}$$ $$1+\frac{r}{100}=\frac{11}{10}$$ ($$1+\frac{r}{100}$$ cannot equal to $$-\frac{11}{10}$$ because it would men that r is negative.) $$1+\frac{r}{100}=\frac{11}{10}$$ $$\frac{r}{100}=\frac{1}{10}$$ $$r=10$$ _________________ Intern Joined: 23 Mar 2011 Posts: 30 Followers: 0 Kudos [?]: 3 [0], given: 9 Re: Annual interest [#permalink] 18 Jul 2013, 07:47 Bunuel wrote: Actually you don't need to solve this way: $$1,000((1+\frac{r}{100})^2-1)=210$$ $$(1+\frac{r}{100})^2-1=\frac{210}{1,000}$$ $$(1+\frac{r}{100})^2=\frac{21}{100}+1$$ $$(1+\frac{r}{100})^2=\frac{121}{100}$$ $$1+\frac{r}{100}=\frac{11}{10}$$ ($$1+\frac{r}{100}=\frac{11}{10}$$ cannot equal to $$-\frac{11}{10}$$ because it would men that r is negative.) $$1+\frac{r}{100}=\frac{11}{10}$$ $$\frac{r}{100}=\frac{1}{10}$$ $$r=10$$ Thank you Bunuel for this alternate calculation method! It's much easier this way you showed. Greatly appreciate your follow-up. ~ Im2bz2p345 GMAT Club Legend Joined: 09 Sep 2013 Posts: 7310 Followers: 385 Kudos [?]: 93 [0], given: 0 Re: If$1,000 is deposited in a certain bank account and remains [#permalink]  28 Jul 2014, 06:21
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# How do you use the Quotient Rule to differentiate the function f (t) = (cos (7t)) / t^5?

Jul 15, 2015

I found: $f ' \left(t\right) = - \frac{7 t \sin \left(7 t\right) + 5 \cos \left(7 t\right)}{t} ^ 6$

#### Explanation:

The Quotient Rule can be written as:
Given $f \left(t\right) = \frac{h \left(t\right)}{g \left(t\right)}$ then: $\textcolor{red}{f ' \left(t\right) = \frac{h ' \left(t\right) g \left(t\right) - h \left(t\right) g ' \left(t\right)}{g \left(t\right)} ^ 2}$

$f ' \left(t\right) = \frac{- 7 \sin \left(7 t\right) \cdot {t}^{5} - 5 {t}^{4} \cos \left(7 t\right)}{{t}^{5}} ^ 2 =$
$= \frac{- {t}^{4} \left(7 t \sin \left(7 t\right) + 5 \cos \left(7 t\right)\right)}{t} ^ 10 =$
$= - \frac{7 t \sin \left(7 t\right) + 5 \cos \left(7 t\right)}{t} ^ 6$


text:
# CPT invariance of Dirac equation

We know that Dirac equation is

$$( i \partial _\mu \gamma ^\mu - m ) \psi ~=~0.$$

How can we show that Dirac equation is invariant under CPT transformation?

-
The Dirac equation is invariant under C,P,T even separately, so of course it is invariant under CPT in this form as well. However, the true general CPT invariance shouldn't be tested on one-particle equations or classical fields such as Dirac's equation. The general argument why CPT works only makes sense on the full multiparticle quantum field theory. – Luboš Motl Jul 23 '14 at 14:57

In order to show this, figure out how the transformations $C, P, T$ act on each element in the equation individually.

Pay special attention to $C$. That's a tricky one! For a start on C you can check out this physics SE post.

-

Here is one tricky way to show that the Dirac representation is invariant under C,P,T-transformations. The free Dirac theory refers to the direct sum $\left( 0 , \frac{1}{2}\right) \oplus \left( \frac{1}{2}, 0 \right)$ of the Lorentz group irreducible spinor representations. As it can be shown, the representation $\left(\frac{n}{2}, \frac{m}{2}\right) \oplus \left(\frac{m}{2}, \frac{n}{2}\right)$ is always invariant under C, P, T transformations. So you don't need to find the explicit form of $C, P, T$-matrices for the Dirac theory if you know the spinor representation of the Lorentz group.

The short proof of the invariance of $\left( \frac{m}{2} , \frac{n}{2}\right) \oplus \left( \frac{n}{2}, \frac{m}{2} \right)$ under discrete transformations of the Lorentz group

The proof, of course, is very formal. In a few words,

spinor representation of the Lorentz group involves two Casimir operators, $$C_{1} = M_{ab}M^{ab}, \quad C_{2} = M^{\dot {a} \dot {b}}M_{\dot b \dot a}.$$ Here the generators of spinor representation of the Lorentz $M_{ab}, M_{\dot a \dot b}$ are connected with $M_{\mu \nu}$ (the vector generator of the Lorentz group) by the definite relation (here it's not important what exactly is this relation): $$\tag 1 C_{1}\left( \frac{n}{2}, \frac{m}{2}\right) = -\frac{n(n + 2)}{2}\left( \frac{n}{2}, \frac{m}{2}\right),$$ $$\tag 2 C_{2}\left( \frac{n}{2}, \frac{m}{2}\right) = -\frac{m(m + 2)}{2}\left( \frac{n}{2}, \frac{m}{2}\right).$$ Then we may introduce the general operators of $T, P, C$-inversion by their (anti)commutators with the $M_{ab}, M_{\dot {a}, \dot {b}}$ (which can be done by their clear (anti)commutators with $M_{\mu \nu}$). Finally, $$\hat{T}C_{1} = C_{2}\hat {T}, \quad \hat {P}C_{1} = C_{2}\hat {P}, \hat {T}C_{2} = C_{1}\hat {T}, \quad \hat {P}C_{2} = C_{1}\hat {P}.$$ and the similar for $C$-inversion.

So when acting on $(1)$, $(2)$ by $C, P$ or $T$ operators, we will get (for example, by $\hat {P}$) $$\tag 3 C_{1}\hat {P}\left( \frac{n}{2}, \frac{m}{2}\right) =-\frac{m(m + 2)}{2}\hat {P}\left( \frac{n}{2}, \frac{m}{2}\right),$$ $$\tag 4 C_{2}\hat {P}\left( \frac{n}{2}, \frac{m}{2}\right) =-\frac{n(n + 2)}{2}\hat {P}\left( \frac{n}{2}, \frac{m}{2}\right).$$ So we see, that $\hat {P}$ acting on $\left( \frac{n}{2}, \frac{m}{2}\right)$ changes it to $\left( \frac{m}{2}, \frac{n}{2}\right)$, so in general (except the case $m = n$) $\left( \frac{n}{2}, \frac{m}{2}\right)$ isn't invariant under $C, P, T$-transformations. But the direct sum $\left( 0 , \frac{1}{2}\right) \oplus \left( \frac{1}{2}, 0 \right)$ (in particular) is invariant, because $$\hat {P}\left(\left( \frac{n}{2}, \frac{m}{2}\right) \oplus \left( \frac{m}{2}, \frac{n}{2}\right) \right) = \left( \frac{m}{2}, \frac{n}{2}\right) \oplus \left( \frac{n}{2}, \frac{m}{2}\right)$$ (nothing have changed).

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How can it be shown the representation invariant under this representation? \left(\frac{n}{2}, \frac{m}{2}\right) \oplus \left(\frac{m}{2}, \frac{n}{2}\right) – user55944 Jul 30 '14 at 12:43
@user55944 : I have wrote the general way of proof into the answer. – Andrew McAddams Jul 30 '14 at 13:30
How can you show that the weak interaction is parity violating theory according to this representation? – user55944 Sep 13 '14 at 10:58
It's easy because it consists projectors on states $\left(\frac{1}{2} , 0 \right)$ and $\left( 0 , \frac{1}{2}\right)$ unequally. For example, W-boson interaction doesn't consist right-fermions interaction. So after making parity transformation of lagrangian we won't get initial lagrangian. – Andrew McAddams Sep 13 '14 at 11:05