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# Camarines Sur Chi Square Of Independence Manual Solving

## dumath AP Stats Spring 2019 Notes

### Contingency Tables Free Statistics Book

dumath AP Stats Spring 2019 Notes. According to the Chi-Square test of independence if the calculated value is greater than the table value, reject the null hypothesis. StatDistributions.com вЂ” Chi-square distribution calculator, Graduate statistics students and business managers will find this e-manual to be, by far, the easiest and fastest way to master graduate-level statistics and to apply advanced statistics in Excel to solve difficult, real-world problems, homework assignments, and exam questions..

### SPSS Chi-Square Independence Test Tutorial

S.5 Power Analysis STAT ONLINE. For this example, the degrees of freedom is (2-1)(4-1) = 3. The Chi Square calculator can be used to determine that the probability value for a Chi Square of 16.55 with three degrees of freedom is equal to 0.0009. Therefore, the null hypothesis of no relationship between diet and outcome can be rejected., Welcome to ChartitNow. Your one stop, online shop for all of your manufacturing charts, templates and statistical analysis needs for your day to day operations. Apply these easy to use professional templates to monitor and improve your manufacturing operations..

Graduate statistics students and business managers will find this e-manual to be, by far, the easiest and fastest way to master graduate-level statistics and to apply advanced statistics in Excel to solve difficult, real-world problems, homework assignments, and exam questions. AP Stats: Spring 2019 Class Notes . Hello all you Manual Statisticians! For the New Year, I plan to upload notes by chapter. If you do not find what you are looking for, please let me know and I'll try to include it.

The homogeneity chi-square test statistics is computed exactly the same as the test for independence using contingency table as when determining the independence of charateristics chi-square statistics. The only difference between the test for independence вЂ¦ Calculate the chi square p value Excel: Steps. Step 1: Calculate your expected value. The expected value in chi-square is found by dividing your counts (the number of responses or data items) by the number of categories. There are twelve categories (zodiac signs) in the question, so:

Some having intermittent demand. I have applied the Goodness of fit test by dividing into bins and then carrying out a Chi Square test. My series has a lot of zero values corresponding to each SKU and intermittent behavior is quite evident but my Goodness of fit test shows it as Normal distribution. Precision Consulting-- Offers dissertation help, editing, tutoring, and coaching services on a variety of statistical methods including ANOVA, Multiple Linear Regression, Structural Equation Modeling, Confirmatory Factor Analysis, and Hierarchical Linear Modeling.If you're stuck on your proposal, methodology, or statistical phase of your dissertation, you might want to contact them.

Some having intermittent demand. I have applied the Goodness of fit test by dividing into bins and then carrying out a Chi Square test. My series has a lot of zero values corresponding to each SKU and intermittent behavior is quite evident but my Goodness of fit test shows it as Normal distribution. 10/17/2014В В· Objective To examine autonomy-promoting parenting and independent problem-solving in children with food allergy. Methods 66 children with food allergy, aged 3вЂ“6 years, and 67 age-matched healthy peers and their mothers were videotaped while completing easy and difficult puzzles.Coders recorded time to puzzle completion, childrenвЂ™s direct and indirect requests for help, and maternal help

Note that the chi-square test is more commonly used in a very different situation -- to analyze a contingency table. This is appropriate when you wish to compare two or more groups, and the outcome variable is categorical. For example, compare number of patients with postoperative infections after two kinds of operations. If you need to analyze P Value Calculator Use this calculator to compute a P value from a Z, t, F, r, or chi-square value that you obtain from a program or publication.

A chi-square test for independence compares two variables in a contingency table to see if they are related. In a more general sense, it tests to see whether distributions of categorical variables differ from each another. A very small chi square test statistic; means that your observed data fits your expected data extremely well. In other According to the Chi-Square test of independence if the calculated value is greater than the table value, reject the null hypothesis. StatDistributions.com вЂ” Chi-square distribution calculator

Chi-Square - Test of Independence. This use of Chi-square is somewhat different. In this case we have two (or more) variables, both of which are categorical, and we want to determine if they are independent or related. In these problems we first lay out our actual вЂ¦ Only proportions and denominators available. But how do you do a Chi-square test when you only have proportions and denominators available? For example, you know from the literature that 33.0% of 276 people using medication A got the disease, while 34.4% of вЂ¦

Precision Consulting-- Offers dissertation help, editing, tutoring, and coaching services on a variety of statistical methods including ANOVA, Multiple Linear Regression, Structural Equation Modeling, Confirmatory Factor Analysis, and Hierarchical Linear Modeling.If you're stuck on your proposal, methodology, or statistical phase of your dissertation, you might want to contact them. Graduate statistics students and business managers will find this e-manual to be, by far, the easiest and fastest way to master graduate-level statistics and to apply advanced statistics in Excel to solve difficult, real-world problems, homework assignments, and exam questions.

Welcome to ChartitNow. Your one stop, online shop for all of your manufacturing charts, templates and statistical analysis needs for your day to day operations. Apply these easy to use professional templates to monitor and improve your manufacturing operations. Welcome to ChartitNow. Your one stop, online shop for all of your manufacturing charts, templates and statistical analysis needs for your day to day operations. Apply these easy to use professional templates to monitor and improve your manufacturing operations.

If the chi-square difference statistic is not significant between the original and constrained models, then we conclude that the model has measurement invariance across groups. 54 Perform A Chi-square Test Of Independence Using A 5% Level Of Significance. This problem has been solved! See the answer. I need help solving this problem please. Show transcribed image text. Expert Answer . 1)here null hypothesis:gender in school and gender are independent alternate hypothesis:gender in school and gender are dependent 2)

According to the Chi-Square test of independence if the calculated value is greater than the table value, reject the null hypothesis. StatDistributions.com вЂ” Chi-square distribution calculator The calculator below should be self-explanatory, but just in case it's not: your chi-square score goes in the chi-square score box, you stick your degrees of freedom in the DF box (df = (N Columns-1)*(N Rows-1) for chi-square test for independence), select your significance level, then press the button.

After a significant result from the chi-square test of independence, you can perform one of several follow-up tests to pinpoint the cause of the significant result. Further information about this topic can be found by clicking on the following links: Post-hoc testing after chi-square independence testing; Real Statistic support for post-hoc testing P Value Calculator Use this calculator to compute a P value from a Z, t, F, r, or chi-square value that you obtain from a program or publication.

The calculator below should be self-explanatory, but just in case it's not: your chi-square score goes in the chi-square score box, you stick your degrees of freedom in the DF box (df = (N Columns-1)*(N Rows-1) for chi-square test for independence), select your significance level, then press the button. Welcome to ChartitNow. Your one stop, online shop for all of your manufacturing charts, templates and statistical analysis needs for your day to day operations. Apply these easy to use professional templates to monitor and improve your manufacturing operations.

P Value Calculator Use this calculator to compute a P value from a Z, t, F, r, or chi-square value that you obtain from a program or publication. The following Excel-generated graph shows the Chi-SquareвЂ™s PDF (Probability Density Function) for 20 degrees of freedom as the X value (the Chi-Square Statistic вЂ“ О§ 2) goes from 0 to 25. Note that the graph of the Chi-SquareвЂ™s PDF resembles a wave rolling from left to right across the grid as the degrees of freedom are increased.

For this example, the degrees of freedom is (2-1)(4-1) = 3. The Chi Square calculator can be used to determine that the probability value for a Chi Square of 16.55 with three degrees of freedom is equal to 0.0009. Therefore, the null hypothesis of no relationship between diet and outcome can be rejected. Degrees of freedom for Chi-squared test. Ask Question Asked 8 years, 1 are you trying to test. The correct degrees of freedom depends on that. Most answers assume that you are interested in independence The degrees of freedom for chi square test in contingency table is determined by the number of 'expected observations' estimated

Perform A Chi-square Test Of Independence Using A 5% Level Of Significance. This problem has been solved! See the answer. I need help solving this problem please. Show transcribed image text. Expert Answer . 1)here null hypothesis:gender in school and gender are independent alternate hypothesis:gender in school and gender are dependent 2) Welcome to ChartitNow. Your one stop, online shop for all of your manufacturing charts, templates and statistical analysis needs for your day to day operations. Apply these easy to use professional templates to monitor and improve your manufacturing operations.

Perform A Chi-square Test Of Independence Using A 5% Level Of Significance. This problem has been solved! See the answer. I need help solving this problem please. Show transcribed image text. Expert Answer . 1)here null hypothesis:gender in school and gender are independent alternate hypothesis:gender in school and gender are dependent 2) After a significant result from the chi-square test of independence, you can perform one of several follow-up tests to pinpoint the cause of the significant result. Further information about this topic can be found by clicking on the following links: Post-hoc testing after chi-square independence testing; Real Statistic support for post-hoc testing

AP Stats: Spring 2019 Class Notes . Hello all you Manual Statisticians! For the New Year, I plan to upload notes by chapter. If you do not find what you are looking for, please let me know and I'll try to include it. 5/9/2017В В· How to Calculate a Chi Square: Formula & Example. It's a common mistake for students to take the square root, as you would if you were solving for chi. However, the value that we are looking

The homogeneity chi-square test statistics is computed exactly the same as the test for independence using contingency table as when determining the independence of charateristics chi-square statistics. The only difference between the test for independence вЂ¦ Graduate statistics students and business managers will find this e-manual to be, by far, the easiest and fastest way to master graduate-level statistics and to apply advanced statistics in Excel to solve difficult, real-world problems, homework assignments, and exam questions.

For this example, the degrees of freedom is (2-1)(4-1) = 3. The Chi Square calculator can be used to determine that the probability value for a Chi Square of 16.55 with three degrees of freedom is equal to 0.0009. Therefore, the null hypothesis of no relationship between diet and outcome can be rejected. Welcome to ChartitNow. Your one stop, online shop for all of your manufacturing charts, templates and statistical analysis needs for your day to day operations. Apply these easy to use professional templates to monitor and improve your manufacturing operations.

If the chi-square difference statistic is not significant between the original and constrained models, then we conclude that the model has measurement invariance across groups. 54 Using Your TI-83/84 Calculator for Hypothesis Testing: The Chi-Square Goodness-of-Fit Test Dr. Laura Schultz If births were uniformly distributed across the week, we would expect that about 1/7 of all births occur during each day of the week. How closely do the observed number of вЂ¦

### GraphPad QuickCalcs chi square calculator

The Friedman Test Welcome to oak.ucc.nau.edu. The calculator below should be self-explanatory, but just in case it's not: your chi-square score goes in the chi-square score box, you stick your degrees of freedom in the DF box (df = (N Columns-1)*(N Rows-1) for chi-square test for independence), select your significance level, then press the button., A chi-square test for independence compares two variables in a contingency table to see if they are related. In a more general sense, it tests to see whether distributions of categorical variables differ from each another. A very small chi square test statistic; means that your observed data fits your expected data extremely well. In other.

How to do a Chi-square test when you only have proportions. 10/17/2014В В· Objective To examine autonomy-promoting parenting and independent problem-solving in children with food allergy. Methods 66 children with food allergy, aged 3вЂ“6 years, and 67 age-matched healthy peers and their mothers were videotaped while completing easy and difficult puzzles.Coders recorded time to puzzle completion, childrenвЂ™s direct and indirect requests for help, and maternal help, The following Excel-generated graph shows the Chi-SquareвЂ™s PDF (Probability Density Function) for 20 degrees of freedom as the X value (the Chi-Square Statistic вЂ“ О§ 2) goes from 0 to 25. Note that the graph of the Chi-SquareвЂ™s PDF resembles a wave rolling from left to right across the grid as the degrees of freedom are increased..

### GraphPad QuickCalcs chi square calculator

dumath AP Stats Spring 2019 Notes. Precision Consulting-- Offers dissertation help, editing, tutoring, and coaching services on a variety of statistical methods including ANOVA, Multiple Linear Regression, Structural Equation Modeling, Confirmatory Factor Analysis, and Hierarchical Linear Modeling.If you're stuck on your proposal, methodology, or statistical phase of your dissertation, you might want to contact them. Note that the chi-square test is more commonly used in a very different situation -- to analyze a contingency table. This is appropriate when you wish to compare two or more groups, and the outcome variable is categorical. For example, compare number of patients with postoperative infections after two kinds of operations. If you need to analyze.

• Comprehensive Guide to Chi-SquareTests For Independence
• Comprehensive Guide to Chi-SquareTests For Independence

• If the chi-square difference statistic is not significant between the original and constrained models, then we conclude that the model has measurement invariance across groups. 54 10/13/2010В В· An explanation of how to compute the chi-squared statistic for independent measures of nominal data. For an explanation of significance testing in general, s...

Only proportions and denominators available. But how do you do a Chi-square test when you only have proportions and denominators available? For example, you know from the literature that 33.0% of 276 people using medication A got the disease, while 34.4% of вЂ¦ For this example, the degrees of freedom is (2-1)(4-1) = 3. The Chi Square calculator can be used to determine that the probability value for a Chi Square of 16.55 with three degrees of freedom is equal to 0.0009. Therefore, the null hypothesis of no relationship between diet and outcome can be rejected.

10/13/2010В В· An explanation of how to compute the chi-squared statistic for independent measures of nominal data. For an explanation of significance testing in general, s... Dissertation Topic Selection: Help you swiftly select a topic that is researchable, fits within your discipline, and gets approved. There are several approaches to topic development. One is looking for archival data, backing-out research questions, and developing a methodology from there.

Perform A Chi-square Test Of Independence Using A 5% Level Of Significance. This problem has been solved! See the answer. I need help solving this problem please. Show transcribed image text. Expert Answer . 1)here null hypothesis:gender in school and gender are independent alternate hypothesis:gender in school and gender are dependent 2) If the independence assumption is violated, the test is likely to yield inaccurate results. It should be noted that the analysis permits dependency among scores within a set. Assumption 2: The Chi-Square values for the Friedman test yield relatively accurate results to the extent that the sample size is large.

10/17/2014В В· Objective To examine autonomy-promoting parenting and independent problem-solving in children with food allergy. Methods 66 children with food allergy, aged 3вЂ“6 years, and 67 age-matched healthy peers and their mothers were videotaped while completing easy and difficult puzzles.Coders recorded time to puzzle completion, childrenвЂ™s direct and indirect requests for help, and maternal help The calculator below should be self-explanatory, but just in case it's not: your chi-square score goes in the chi-square score box, you stick your degrees of freedom in the DF box (df = (N Columns-1)*(N Rows-1) for chi-square test for independence), select your significance level, then press the button.

Perform A Chi-square Test Of Independence Using A 5% Level Of Significance. This problem has been solved! See the answer. I need help solving this problem please. Show transcribed image text. Expert Answer . 1)here null hypothesis:gender in school and gender are independent alternate hypothesis:gender in school and gender are dependent 2) P Value Calculator Use this calculator to compute a P value from a Z, t, F, r, or chi-square value that you obtain from a program or publication.

According to the Chi-Square test of independence if the calculated value is greater than the table value, reject the null hypothesis. StatDistributions.com вЂ” Chi-square distribution calculator P Value Calculator Use this calculator to compute a P value from a Z, t, F, r, or chi-square value that you obtain from a program or publication.

Using Your TI-83/84 Calculator for Hypothesis Testing: The Chi-Square Goodness-of-Fit Test Dr. Laura Schultz If births were uniformly distributed across the week, we would expect that about 1/7 of all births occur during each day of the week. How closely do the observed number of вЂ¦ Calculate the chi square p value Excel: Steps. Step 1: Calculate your expected value. The expected value in chi-square is found by dividing your counts (the number of responses or data items) by the number of categories. There are twelve categories (zodiac signs) in the question, so:

P Value Calculator Use this calculator to compute a P value from a Z, t, F, r, or chi-square value that you obtain from a program or publication. P Value Calculator Use this calculator to compute a P value from a Z, t, F, r, or chi-square value that you obtain from a program or publication.

10/13/2010В В· An explanation of how to compute the chi-squared statistic for independent measures of nominal data. For an explanation of significance testing in general, s... Using Your TI-83/84 Calculator for Hypothesis Testing: The Chi-Square Goodness-of-Fit Test Dr. Laura Schultz If births were uniformly distributed across the week, we would expect that about 1/7 of all births occur during each day of the week. How closely do the observed number of вЂ¦

Note that the chi-square test is more commonly used in a very different situation -- to analyze a contingency table. This is appropriate when you wish to compare two or more groups, and the outcome variable is categorical. For example, compare number of patients with postoperative infections after two kinds of operations. If you need to analyze If the chi-square difference statistic is not significant between the original and constrained models, then we conclude that the model has measurement invariance across groups. 54

## Comprehensive Guide to Chi-SquareTests For Independence

Contingency Tables Free Statistics Book. Dissertation Topic Selection: Help you swiftly select a topic that is researchable, fits within your discipline, and gets approved. There are several approaches to topic development. One is looking for archival data, backing-out research questions, and developing a methodology from there., Using Your TI-83/84 Calculator for Hypothesis Testing: The Chi-Square Goodness-of-Fit Test Dr. Laura Schultz If births were uniformly distributed across the week, we would expect that about 1/7 of all births occur during each day of the week. How closely do the observed number of вЂ¦.

### S.5 Power Analysis STAT ONLINE

How to do a Chi-square test when you only have proportions. 10/13/2010В В· An explanation of how to compute the chi-squared statistic for independent measures of nominal data. For an explanation of significance testing in general, s..., For this example, the degrees of freedom is (2-1)(4-1) = 3. The Chi Square calculator can be used to determine that the probability value for a Chi Square of 16.55 with three degrees of freedom is equal to 0.0009. Therefore, the null hypothesis of no relationship between diet and outcome can be rejected..

Graduate statistics students and business managers will find this e-manual to be, by far, the easiest and fastest way to master graduate-level statistics and to apply advanced statistics in Excel to solve difficult, real-world problems, homework assignments, and exam questions. Perform A Chi-square Test Of Independence Using A 5% Level Of Significance. This problem has been solved! See the answer. I need help solving this problem please. Show transcribed image text. Expert Answer . 1)here null hypothesis:gender in school and gender are independent alternate hypothesis:gender in school and gender are dependent 2)

Welcome to ChartitNow. Your one stop, online shop for all of your manufacturing charts, templates and statistical analysis needs for your day to day operations. Apply these easy to use professional templates to monitor and improve your manufacturing operations. Calculate the chi square p value Excel: Steps. Step 1: Calculate your expected value. The expected value in chi-square is found by dividing your counts (the number of responses or data items) by the number of categories. There are twelve categories (zodiac signs) in the question, so:

Degrees of freedom for Chi-squared test. Ask Question Asked 8 years, 1 are you trying to test. The correct degrees of freedom depends on that. Most answers assume that you are interested in independence The degrees of freedom for chi square test in contingency table is determined by the number of 'expected observations' estimated According to the Chi-Square test of independence if the calculated value is greater than the table value, reject the null hypothesis. StatDistributions.com вЂ” Chi-square distribution calculator

P Value Calculator Use this calculator to compute a P value from a Z, t, F, r, or chi-square value that you obtain from a program or publication. Perform A Chi-square Test Of Independence Using A 5% Level Of Significance. This problem has been solved! See the answer. I need help solving this problem please. Show transcribed image text. Expert Answer . 1)here null hypothesis:gender in school and gender are independent alternate hypothesis:gender in school and gender are dependent 2)

AP Stats: Spring 2019 Class Notes . Hello all you Manual Statisticians! For the New Year, I plan to upload notes by chapter. If you do not find what you are looking for, please let me know and I'll try to include it. 10/17/2014В В· Objective To examine autonomy-promoting parenting and independent problem-solving in children with food allergy. Methods 66 children with food allergy, aged 3вЂ“6 years, and 67 age-matched healthy peers and their mothers were videotaped while completing easy and difficult puzzles.Coders recorded time to puzzle completion, childrenвЂ™s direct and indirect requests for help, and maternal help

Degrees of freedom for Chi-squared test. Ask Question Asked 8 years, 1 are you trying to test. The correct degrees of freedom depends on that. Most answers assume that you are interested in independence The degrees of freedom for chi square test in contingency table is determined by the number of 'expected observations' estimated Only proportions and denominators available. But how do you do a Chi-square test when you only have proportions and denominators available? For example, you know from the literature that 33.0% of 276 people using medication A got the disease, while 34.4% of вЂ¦

For this example, the degrees of freedom is (2-1)(4-1) = 3. The Chi Square calculator can be used to determine that the probability value for a Chi Square of 16.55 with three degrees of freedom is equal to 0.0009. Therefore, the null hypothesis of no relationship between diet and outcome can be rejected. For this example, the degrees of freedom is (2-1)(4-1) = 3. The Chi Square calculator can be used to determine that the probability value for a Chi Square of 16.55 with three degrees of freedom is equal to 0.0009. Therefore, the null hypothesis of no relationship between diet and outcome can be rejected.

Consider a research experiment where the p-value computed from the data was 0.12.As a result, one would fail to reject the null hypothesis because this p-value is larger than $$\alpha$$ = 0.05.However, there still exist two possible cases for which we failed to reject the null hypothesis: Note that the chi-square test is more commonly used in a very different situation -- to analyze a contingency table. This is appropriate when you wish to compare two or more groups, and the outcome variable is categorical. For example, compare number of patients with postoperative infections after two kinds of operations. If you need to analyze

If the independence assumption is violated, the test is likely to yield inaccurate results. It should be noted that the analysis permits dependency among scores within a set. Assumption 2: The Chi-Square values for the Friedman test yield relatively accurate results to the extent that the sample size is large. The calculator below should be self-explanatory, but just in case it's not: your chi-square score goes in the chi-square score box, you stick your degrees of freedom in the DF box (df = (N Columns-1)*(N Rows-1) for chi-square test for independence), select your significance level, then press the button.

Using Your TI-83/84 Calculator for Hypothesis Testing: The Chi-Square Goodness-of-Fit Test Dr. Laura Schultz If births were uniformly distributed across the week, we would expect that about 1/7 of all births occur during each day of the week. How closely do the observed number of вЂ¦ According to the Chi-Square test of independence if the calculated value is greater than the table value, reject the null hypothesis. StatDistributions.com вЂ” Chi-square distribution calculator

AP Stats: Spring 2019 Class Notes . Hello all you Manual Statisticians! For the New Year, I plan to upload notes by chapter. If you do not find what you are looking for, please let me know and I'll try to include it. Dissertation Topic Selection: Help you swiftly select a topic that is researchable, fits within your discipline, and gets approved. There are several approaches to topic development. One is looking for archival data, backing-out research questions, and developing a methodology from there.

4/11/1998В В· Calculating and Interpreting Expected and Chi-Square Tables Date: 04/10/98 at 22:31:30 From: Lola Kelly Subject: Stats- chi squared How do you get the frequency expected when doing chi squared? I have a 3-by-3 table. Help. Please. 10/13/2010В В· An explanation of how to compute the chi-squared statistic for independent measures of nominal data. For an explanation of significance testing in general, s...

Solving for Parameter Values that Satisfy the Least-Square Criterion. Chi-Square and Tests of Contingency tables. Review of Contingency Tables. Finding the Exact Significance Level for A Chi-Square Statistic. Interpreting a Significant Chi-Square. Summary. Testing a Single Correlation Coefficient. The Hypothesis and Nature of the Effects. Consider a research experiment where the p-value computed from the data was 0.12.As a result, one would fail to reject the null hypothesis because this p-value is larger than $$\alpha$$ = 0.05.However, there still exist two possible cases for which we failed to reject the null hypothesis:

Some having intermittent demand. I have applied the Goodness of fit test by dividing into bins and then carrying out a Chi Square test. My series has a lot of zero values corresponding to each SKU and intermittent behavior is quite evident but my Goodness of fit test shows it as Normal distribution. 10/13/2010В В· An explanation of how to compute the chi-squared statistic for independent measures of nominal data. For an explanation of significance testing in general, s...

Perform A Chi-square Test Of Independence Using A 5% Level Of Significance. This problem has been solved! See the answer. I need help solving this problem please. Show transcribed image text. Expert Answer . 1)here null hypothesis:gender in school and gender are independent alternate hypothesis:gender in school and gender are dependent 2) The calculator below should be self-explanatory, but just in case it's not: your chi-square score goes in the chi-square score box, you stick your degrees of freedom in the DF box (df = (N Columns-1)*(N Rows-1) for chi-square test for independence), select your significance level, then press the button.

Calculate the chi square p value Excel: Steps. Step 1: Calculate your expected value. The expected value in chi-square is found by dividing your counts (the number of responses or data items) by the number of categories. There are twelve categories (zodiac signs) in the question, so: The calculator below should be self-explanatory, but just in case it's not: your chi-square score goes in the chi-square score box, you stick your degrees of freedom in the DF box (df = (N Columns-1)*(N Rows-1) for chi-square test for independence), select your significance level, then press the button.

Perform A Chi-square Test Of Independence Using A 5% Level Of Significance. This problem has been solved! See the answer. I need help solving this problem please. Show transcribed image text. Expert Answer . 1)here null hypothesis:gender in school and gender are independent alternate hypothesis:gender in school and gender are dependent 2) 10/13/2010В В· An explanation of how to compute the chi-squared statistic for independent measures of nominal data. For an explanation of significance testing in general, s...

Welcome to ChartitNow. Your one stop, online shop for all of your manufacturing charts, templates and statistical analysis needs for your day to day operations. Apply these easy to use professional templates to monitor and improve your manufacturing operations. Consider a research experiment where the p-value computed from the data was 0.12.As a result, one would fail to reject the null hypothesis because this p-value is larger than $$\alpha$$ = 0.05.However, there still exist two possible cases for which we failed to reject the null hypothesis:

Calculate the chi square p value Excel: Steps. Step 1: Calculate your expected value. The expected value in chi-square is found by dividing your counts (the number of responses or data items) by the number of categories. There are twelve categories (zodiac signs) in the question, so: Using Your TI-83/84 Calculator for Hypothesis Testing: The Chi-Square Goodness-of-Fit Test Dr. Laura Schultz If births were uniformly distributed across the week, we would expect that about 1/7 of all births occur during each day of the week. How closely do the observed number of вЂ¦

Only proportions and denominators available. But how do you do a Chi-square test when you only have proportions and denominators available? For example, you know from the literature that 33.0% of 276 people using medication A got the disease, while 34.4% of вЂ¦ Graduate statistics students and business managers will find this e-manual to be, by far, the easiest and fastest way to master graduate-level statistics and to apply advanced statistics in Excel to solve difficult, real-world problems, homework assignments, and exam questions.

If the chi-square difference statistic is not significant between the original and constrained models, then we conclude that the model has measurement invariance across groups. 54 Chi-Square - Test of Independence. This use of Chi-square is somewhat different. In this case we have two (or more) variables, both of which are categorical, and we want to determine if they are independent or related. In these problems we first lay out our actual вЂ¦

### Comprehensive Guide to Chi-SquareTests For Independence

You WILL Master Advanced Statistics in Excel. Some having intermittent demand. I have applied the Goodness of fit test by dividing into bins and then carrying out a Chi Square test. My series has a lot of zero values corresponding to each SKU and intermittent behavior is quite evident but my Goodness of fit test shows it as Normal distribution., A chi-square test for independence compares two variables in a contingency table to see if they are related. In a more general sense, it tests to see whether distributions of categorical variables differ from each another. A very small chi square test statistic; means that your observed data fits your expected data extremely well. In other.

S.5 Power Analysis STAT ONLINE. After a significant result from the chi-square test of independence, you can perform one of several follow-up tests to pinpoint the cause of the significant result. Further information about this topic can be found by clicking on the following links: Post-hoc testing after chi-square independence testing; Real Statistic support for post-hoc testing, The calculator below should be self-explanatory, but just in case it's not: your chi-square score goes in the chi-square score box, you stick your degrees of freedom in the DF box (df = (N Columns-1)*(N Rows-1) for chi-square test for independence), select your significance level, then press the button..

### dumath AP Stats Spring 2019 Notes

dumath AP Stats Spring 2019 Notes. Only proportions and denominators available. But how do you do a Chi-square test when you only have proportions and denominators available? For example, you know from the literature that 33.0% of 276 people using medication A got the disease, while 34.4% of вЂ¦ Some having intermittent demand. I have applied the Goodness of fit test by dividing into bins and then carrying out a Chi Square test. My series has a lot of zero values corresponding to each SKU and intermittent behavior is quite evident but my Goodness of fit test shows it as Normal distribution..

Note that the chi-square test is more commonly used in a very different situation -- to analyze a contingency table. This is appropriate when you wish to compare two or more groups, and the outcome variable is categorical. For example, compare number of patients with postoperative infections after two kinds of operations. If you need to analyze 10/13/2010В В· An explanation of how to compute the chi-squared statistic for independent measures of nominal data. For an explanation of significance testing in general, s...

The calculator below should be self-explanatory, but just in case it's not: your chi-square score goes in the chi-square score box, you stick your degrees of freedom in the DF box (df = (N Columns-1)*(N Rows-1) for chi-square test for independence), select your significance level, then press the button. SPSS Step-by-Step 7 SPSS Tutorial and Help 10. Click Show Me. SPSS now opens the tutorial to the chi-square topic in the form of an Internet page. 11. Click Next. In addition to an example of how to use a chi-square test, the win-

Solving for Parameter Values that Satisfy the Least-Square Criterion. Chi-Square and Tests of Contingency tables. Review of Contingency Tables. Finding the Exact Significance Level for A Chi-Square Statistic. Interpreting a Significant Chi-Square. Summary. Testing a Single Correlation Coefficient. The Hypothesis and Nature of the Effects. Welcome to ChartitNow. Your one stop, online shop for all of your manufacturing charts, templates and statistical analysis needs for your day to day operations. Apply these easy to use professional templates to monitor and improve your manufacturing operations.

Degrees of freedom for Chi-squared test. Ask Question Asked 8 years, 1 are you trying to test. The correct degrees of freedom depends on that. Most answers assume that you are interested in independence The degrees of freedom for chi square test in contingency table is determined by the number of 'expected observations' estimated 4/11/1998В В· Calculating and Interpreting Expected and Chi-Square Tables Date: 04/10/98 at 22:31:30 From: Lola Kelly Subject: Stats- chi squared How do you get the frequency expected when doing chi squared? I have a 3-by-3 table. Help. Please.

If the chi-square difference statistic is not significant between the original and constrained models, then we conclude that the model has measurement invariance across groups. 54 The homogeneity chi-square test statistics is computed exactly the same as the test for independence using contingency table as when determining the independence of charateristics chi-square statistics. The only difference between the test for independence вЂ¦

Using Your TI-83/84 Calculator for Hypothesis Testing: The Chi-Square Goodness-of-Fit Test Dr. Laura Schultz If births were uniformly distributed across the week, we would expect that about 1/7 of all births occur during each day of the week. How closely do the observed number of вЂ¦ AP Stats: Spring 2019 Class Notes . Hello all you Manual Statisticians! For the New Year, I plan to upload notes by chapter. If you do not find what you are looking for, please let me know and I'll try to include it.

SPSS Step-by-Step 7 SPSS Tutorial and Help 10. Click Show Me. SPSS now opens the tutorial to the chi-square topic in the form of an Internet page. 11. Click Next. In addition to an example of how to use a chi-square test, the win- 4/11/1998В В· Calculating and Interpreting Expected and Chi-Square Tables Date: 04/10/98 at 22:31:30 From: Lola Kelly Subject: Stats- chi squared How do you get the frequency expected when doing chi squared? I have a 3-by-3 table. Help. Please.

AP Stats: Spring 2019 Class Notes . Hello all you Manual Statisticians! For the New Year, I plan to upload notes by chapter. If you do not find what you are looking for, please let me know and I'll try to include it. Only proportions and denominators available. But how do you do a Chi-square test when you only have proportions and denominators available? For example, you know from the literature that 33.0% of 276 people using medication A got the disease, while 34.4% of вЂ¦

Consider a research experiment where the p-value computed from the data was 0.12.As a result, one would fail to reject the null hypothesis because this p-value is larger than $$\alpha$$ = 0.05.However, there still exist two possible cases for which we failed to reject the null hypothesis: Consider a research experiment where the p-value computed from the data was 0.12.As a result, one would fail to reject the null hypothesis because this p-value is larger than $$\alpha$$ = 0.05.However, there still exist two possible cases for which we failed to reject the null hypothesis:

Chi-Square - Test of Independence. This use of Chi-square is somewhat different. In this case we have two (or more) variables, both of which are categorical, and we want to determine if they are independent or related. In these problems we first lay out our actual вЂ¦ SPSS Step-by-Step 7 SPSS Tutorial and Help 10. Click Show Me. SPSS now opens the tutorial to the chi-square topic in the form of an Internet page. 11. Click Next. In addition to an example of how to use a chi-square test, the win-

P Value Calculator Use this calculator to compute a P value from a Z, t, F, r, or chi-square value that you obtain from a program or publication. 4/11/1998В В· Calculating and Interpreting Expected and Chi-Square Tables Date: 04/10/98 at 22:31:30 From: Lola Kelly Subject: Stats- chi squared How do you get the frequency expected when doing chi squared? I have a 3-by-3 table. Help. Please.

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text:
Home » Uncategorized » GATE Electromagnetic question and solution

# GATE Electromagnetic question and solution

In this post, we will see how to approach an electromagnetics question which had appeared in GATE 1995 for 2 marks.

Soln: First, let’s understand the question and draw a rough figure based on the question

As you see, the current direction is towards Z axis. Now the first question is to find out the direction of Magnetic field.

Can the Magnetic field exist in Z direction – No because Magnetic field is perpendicular to the direction of current.

Can the Magnetic field exist in Y direction? Let’s analyse that. Just take two current lines flowing from a plane Y=0, in this case, current is coming out of OUR computer screen, so according to right hand rule, the magnetic field will be in the anti-clockwise direction.

As you can see the magnetic field at the point due to the 2 current lines is only towards -x direction, thus magnetic field due to infinite current lines contributes only in -x direction and all the y components will be cancelled out.

Now In order to calculate the magnetic field due to the uniform current flowing in the Y=0 plane, we need to consider the small portion of the surface and draw appropriate ‘Amperian loop’ enclosing the current and apply Ampere’s law. In this case the amperian loop will be a rectangular box which is parallel to XY plane and extending an equal distance above and below the surface as shown in the above figure(1).

Applying Ampere’s law

$\oint_{C} \vec B. \vec dl$ = 2Bl=$\mu_o I_{enc}$=$\mu_o Kl$ where K= Current/metre

So in this case

B=$-\mu_o/2 K$
H= -K/2 $a^x$ where K=30 mA/m

So the correct Answer is H= -15 $a^x$ mA/m

format: plain


text:
853 views

edited
• The total spin resulting from a d7 configuration is:
• The orbital with zero orbital angular momentum is:
• Which of the following is electronic configuration of Cu2+(Z=29)?

+1 vote
by (5.0k points)
selected

a) d7 means the configuration. In other words, there will be 3 unpaired electrons of a 3/2 spin.

b) The correct answer is s- orbital

• When l = 0, it is evident from that the angular momentum of the electron is zero.
• The atomic orbitals which describe these states of zero angular momentum are called s orbitals.
• The s orbitals are distinguished from one another by stating the value of n, the principal quantum number.

c) The correct option is c

After the 4s is full we put the remaining six electrons in the 3d orbital and end with 3d9. Therefore the expected electron configuration for Copper will be 1s2 2s2 2p2 3s2 3p2 4s1 3d10.

[Ar] 4s1 3d10

by (57 points)
+1


text:
7.3k views

Consider the following statements about the context free grammar

$$G = \left \{ S \rightarrow SS, S \rightarrow ab, S \rightarrow ba, S \rightarrow \epsilon \right \}$$

1. $G$ is ambiguous
2. $G$ produces all strings with equal number of $a$’s and $b$’s
3. $G$ can be accepted by a deterministic PDA.

Which combination below expresses all the true statements about $G$?

1. I only
2. I and III only
3. II and III only
4. I, II and III

edited | 7.3k views
–3
option D correct
+4
yes. You can have a DPDA which puts a A when it reads a 'a' and deletes it when the next 'b' is read. similarly, it can put a 'b" on the stack when a b is read and delete it when the next 'a' is read. In the language generated ab occur together as well as ba. (ii) G is ambiguous. ababab can be generated in 2 ways. You do not talk about a grammar being inherently ambiguous, only a language as inherently ambiguous. Inherently ambiguous CFL can be accepted by NPDA but not DPDA

Prof. Kamala Krithivasan.
0
can we say that all DCFL are unambiguous and NDCFL are inherently ambiguous?
0

@codingo1234 You can't say

All DCFL's are unambiguous

because "Ambiguity" is a property of a grammar, not a language.

What you can say though, is

All DCFG's are unambiguous

because DCFG $\rightarrow \exists$ a DPDA $\rightarrow \exists$ an LR(1) grammar and LR(k) grammars are necessarily unambiguous.

But here's the catch: "Inherent Ambiguity", as opposed to plain "Ambiguity", is a property of a language, not a grammar. From Sipser:

As for the question

Are all NDCFL's inherently ambiguous?

The answer is NO. A NDCFL may or may not be inherently ambiguous. That's because a language being NDCF does not mean that it can be generated only by ambiguous grammars. For Example, the language $\{ ww^R \vert w \in (a+b)^* \}$ is non-deterministic (no deterministic PDA exists that can accept it), but it can be generated by the unambiguous grammar $S \rightarrow aSa \vert bSb \vert \epsilon$

0

@ thanks that cleared many things

0

This is the original question and nowhere $3^{rd}$ production seems to be $\epsilon$. It seems to be a terminal of length 1. Then how grammar can be ambiguous

0
^ is meant to represent lambda, which is another way of writing the null string.

1. True. $G$ is ambiguous. E.g. the string $ab$ has multiple derivation trees like $S \rightarrow SS \rightarrow abS \rightarrow ab$, and $S \rightarrow ab$.

2. False. $G$ does not produce all strings with equal no. of $a$s and $b$s. ($aabb$ cannot be generated).

3. True. The given grammar $G$ generates the language $(ab+ba)^*$, which is Regular and therefore also DCFL. So, a D-PDA can be designed for $G$.

Hence, the answer is option B.

by Boss (31k points)
edited by
0
@Arjun sir,
I request you to please this question which i asked today at stackexchange
http://cs.stackexchange.com/questions/68455/dcfl-with-prefix-property-have-lr0-grammar

This is about LR and DCFl.
0
I think B is also true

it says equal no. of a's and b's and it generates equal no. of a's and b's

correct me if I am wrong
+1
@ARSHAD no it will not produce all strings having equal no of a's and b's such as aabb.
0
According to regular expression DPDA is right. But DPDA is not allowed for ambiguity it's NPDA which handle such issues  while here its working fine. How can it happened ? I am confused regarding this issue.
0

Arjun Sir, Sorry but I still  have a doubt:

I understand that an unambiguous grammar can either generate a DCFL or a NDCFL.

But, you also wrote the following:

A grammer can be

• ambiguous -- yes, and the language can be finite, regular, deterministic context-free or non-deterministic context-free

I am not able to understand the above line. How can an ambiguous grammar be finite, regular, deterministic context-free.

Also, could you please tell what is an ambiguous language? I understand what an ambiguous grammar can be. (The grammar which can one more than parse tree), but I do not know what is an ambiguous language.

And also, for the above question, is there an unambiguous grammar available cause it is a regular language?

Thank you

0
Can someone please give example for Ambigous grammar generating NCFL..
0

It is simple.

$S \rightarrow A\; | \; a\; | \;\epsilon$

$A\rightarrow \; a$

It is ambiguous grammar because for string 'a' , There are 2 possible parse trees :- $S\rightarrow A\rightarrow a$ and  $S\rightarrow a$ and language is finite. So it is regular, DCFL and  NDCFL. After simplification we can write its equivalent unambiguous grammar but originally it is ambiguous. Word 'ambiguous' is associated with grammar and 'inherently-ambiguous' is associated with languages for which no unambiguous grammar is possible.

The equivalent unambiguous grammar for the given question can be :-

$S\rightarrow A\; | B \; | \; \epsilon$

$A\rightarrow abA\; | \; abB\;|\; ab$

$B\rightarrow baB\; | \; baA\;|\; ba$

0

@ankitgupta.1729

We can say DCFL is NOT inherently ambiguous as there always exits LR(1) grammar for every DCFL which is unambigous..

But this may not be true for NCFL..it may not have atleast one unambigous G for it hence it may /may not be inherently ambigous language

am I right ?

Can you please given one grammar which is ambigous and generating NCFL..(not relevant to ques)

0

@jatin , yes. right. I have already given one example. You can check Praveen sir's answer for option (C) where language is inherently ambiguous NCFL  here and  here.

0
by Loyal (8.7k points)
+1
The question said that using grammar G's production and then generating string contain equal number of a's and b's not over tha alphabet {a,b}

So in this sense it may be true all three options....

please correct me if I am wrong.....

1
2


text:
# bitbanging i2c problem with ACK from slave

I am trying to implement bit-banging i2c to communicate between an atmega128A's GPIO and an SHT21 (I2C bus was used for some other devices). The first task is to send a write sequence to the SHT21. Using an oscilloscope, I can see that the sequence sending out from the atmega has the correct start signal, correct order of bits, correct stop signal, signal levels looks correct. The serial analyzer from the scope reads out correct I2C message: S80W~A (SHT21 address is 0x80, writing so last bit is 0). Yet there is not any ACK response from the SHT21. The SDA and SCL are both pulled up to 3.3V via 6.7K resistors.

I really need help finding out what is wrong.

code sending the first byte repeatly if no ACK receive:

 i2c_start();
while(!i2c_send(0x80)){
i2c_stop();
i2c_start();
}

code of i2c protocol:

void i2c_start ()
{
// Pull data high and sck low to ensure transition of start sequence
DATA_HIGH ();

SCK_HIGH ();
DATA_LOW ();
SCK_LOW ();
}
void i2c_stop ()
{
DATA_LOW ();
SetBit (DDRD, 7);
//pull data high while clck is high to end transmission.
SCK_HIGH ();
DATA_HIGH ();
}

uchar i2c_send (uchar command)
{
uchar i;

for (i = 0; i < 8; i++)     // Send command
{

if ((command & 0x80) == 0x80)
DATA_HIGH ();
else
DATA_LOW ();
SCK_LOW ();
SCK_HIGH ();
SCK_LOW ();

command = command << 1;
}

Delay(5);
ClrBit (DDRD, 7);       // Change Data Direction
//SetBit (PORTD, 7);      // Turn on internal pull-up

//SCK_LOW ();
SCK_HIGH ();

if (!CheckBit (PIND, 7))     // Read ACK
{
SCK_LOW ();
SetBit (DDRD, 7);
return 1;
}
else
{
SCK_LOW ();
SetBit (DDRD, 7);
return 0;               // NO ACK
}

}

• Does the device respond if you put it on the proper I2C bus? Mar 14, 2014 at 23:00
• I assume you are asking about the SHT21, yes it responses correctly on a proper I2C bus Mar 14, 2014 at 23:07
• @user3421560 Can you edit your question and add a few more things: snippet of your bit-banging code, oscilloscope screenshot of the first byte on your I2C (both SDA and SCL, if possible) ? Mar 14, 2014 at 23:16
• You asked the same question on Stackoveflow and I'm repeating my comment: are you actually putting the GPIOs in open-collector/drain mode? In other words, when you call SCK_HI or DATA_HI, does your GPIO just release the line (correct) or force it high (bad)? Mar 15, 2014 at 0:41
• @DoxyLover Thank you so much for your comment. I put it here since someone left a comment saying it would be better luck to try over here. About the problem. In the code I set SCK and SDA high by calling setbit(port, pinnumber). It does seem like it is forcing the line high. I will try to fix it asap Mar 15, 2014 at 1:53


text:
# Creating a plot without an x-axis and automatic ticks for the y-axis

I am trying to plot a function with specific options for the appearance of the plot. I want the X axis to not show up, and the Y axis to be labeled normally (i.e. numbers outside the plot).. Yet, I don't get any numbers in the Y axis' ticks:

The code I am using is the following (the function to be plotted is different, but still returns this behavior):

Plot[5 t + 3, {t, 0, 10},
PlotRange -> All,
AxesOrigin -> {10, 0},
LabelStyle -> {FontFamily -> "Arial", Bold, 40},
Axes -> {False, True},
PlotStyle -> {{Thickness[0.007], Darker[Blue, 0.5]}},
AxesStyle -> AbsoluteThickness[3], ImageSize -> {640, 480}
]

-
That doesn't help me. Either I get blank space within the plot when I change AxesOrigin, and resizing does no change to the plot :S –  Sosi May 31 '12 at 14:23

The problem with your plot is that you are using the y-axis on the left, but positioning it on the right by shifting the origin. However, its behaviour is still that of a left-axis (i.e., ticks are to the left of the line). What you need is a right-axis, where the ticks are to the right (with the origin remaining where it should be).

You can do this by using Frame and setting only the right frame to be visible. Note that all the Axes* options will now be named Frame*.

Plot[5 t + 3, {t, 0, 10},
LabelStyle -> {FontFamily -> "Arial", Bold, 40},
Frame -> {{False, True}, {False, False}},
FrameTicks -> {{None, All}, {None, None}},
PlotStyle -> {{Thickness[0.007], Darker[Blue, 0.5]}},
FrameStyle -> AbsoluteThickness[3], ImageSize -> {640, 480},
PlotRange -> All
]

-
Oh, I see!! Thanks, I will check it out now! Looks great! Edit:Indeed, it works great! Thanks again! –  Sosi May 31 '12 at 14:31
AxesOrigin is unnecessary. Only nitpicking because you beat me by 1 sec :) –  István Zachar May 31 '12 at 14:33

You don't need to switch from axes to frame. And you may not be able to use a frame if the axis happens to be inside the plot instead of at the border.

So the way to get what you want with the minimum changes to your original code is this:

Plot[5 t + 3, {t, 0, 10},
PlotRange -> All,
AxesOrigin -> {10, 0},
LabelStyle -> {FontFamily -> "Arial", Bold, 40},
PlotStyle -> {{Thickness[0.007], Darker[Blue, 0.5]}},
AxesStyle -> {Opacity[0], AbsoluteThickness[3]},
ImageSize -> {640, 480}]

All I did is to remove Axes -> {False, True}, and then modified the AxesStyle to make the x axis invisible.

To see how this differs from the solution using Frame, I'll move the axis into the middle:

Plot[5 t + 3, {t, 0, 10},
PlotRange -> All,
AxesOrigin -> {5, 0},
LabelStyle -> {FontFamily -> "Arial", Bold, 40},
PlotStyle -> {{Thickness[0.007], Darker[Blue, 0.5]}},
AxesStyle -> {Opacity[0], AbsoluteThickness[3]},
ImageSize -> {640, 480}]

-
"... Y axis to be labeled normally (i.e. numbers outside the plot)..." That's why I used a frame. Yours still puts the numbers inside the plot. Of course, you could possibly control it with a tick function, but then it isn't minimal anymore :) –  rm -rf May 31 '12 at 19:39
@R.M Well, if you're going to take things literally: I could just change mine to AxesOrigin -> {0, 0}... –  Jens May 31 '12 at 20:02
Nah, I interpreted their question as wanting to place the axis to the right, but since they didn't know how to, they simply moved it as far right as possible. I do agree that yours is useful, since you can't have frames in arbitrary positions –  rm -rf May 31 '12 at 20:18
Indeed I wanted to have the axis on the right-hand side. Yet, thanks for your help Jens, I learnt something now! :) –  Sosi Jun 1 '12 at 9:54


text:
Message-formatting code with pluralization, revised to be more functional

I'm a C++ dev, and I've recently started working my way through Clean Code*. Whenever I encounter an example I think I could improve, I try to re-implement it in C++.

On pp. 28-29 there is an example with a message building class, which looks like this when implemented in C++ (includes and namespaces omitted):
.h

class GuessStatisticsMessage
{
public:
GuessStatisticsMessage() = default;
~GuessStatisticsMessage() = default;

std::string make(char candidate, int count);

private:
void createPluralDependentMessageParts(int count);
void createThereAreManyLetters(int count);
void createThereIsOneLetter();
void createThereAreNoLetters();

std::string number;
std::string verb;
std::string pluralModifier;
};

.cpp

std::string GuessStatisticsMessage::make(char candidate, int count)
{
createPluralDependentMessageParts(count);
return fmt::format("There {} {} {}{}", verb, number, candidate, pluralModifier);
}

void GuessStatisticsMessage::createPluralDependentMessageParts(int count)
{
if (count == 0) {
createThereAreNoLetters();
} else if (count == 1) {
createThereIsOneLetter();
} else {
createThereAreManyLetters(count);
}
}

void GuessStatisticsMessage::createThereAreManyLetters(int count)
{
verb = "are";
number = std::to_string(count);
pluralModifier = "s";
}

void GuessStatisticsMessage::createThereIsOneLetter()
{
verb = "is";
number = "one";
pluralModifier = "";
}

void GuessStatisticsMessage::createThereAreNoLetters()
{
verb = "are";
number = "no";
pluralModifier = "s";
}

In an attempt to make this code more functional I ended up with the following:
.h

class GuessStatisticsMessageAlt
{
public:
GuessStatisticsMessageAlt() = default;
~GuessStatisticsMessageAlt() = default;

std::string make(char candidate, int count);

private:
struct MessageComponents
{
std::string verb;
std::string number;
std::string pluralModifier;
};

MessageComponents createPluralDependentMessageParts(int count);
MessageComponents createThereAreManyLetters(int count);
MessageComponents createThereIsOneLetter();
MessageComponents createThereAreNoLetters();
};

.cpp

std::string GuessStatisticsMessageAlt::make(char candidate, int count)
{
auto messageComponents = createPluralDependentMessageParts(count);
return fmt::format("There {} {} {}{}", messageComponents.verb, messageComponents.number,
candidate, messageComponents.pluralModifier);
}

GuessStatisticsMessageAlt::MessageComponents GuessStatisticsMessageAlt::createPluralDependentMessageParts(int count)
{
if (count == 0) {
return createThereAreNoLetters();
} else if (count == 1) {
return createThereIsOneLetter();
} else {
return createThereAreManyLetters(count);
}
}

GuessStatisticsMessageAlt::MessageComponents GuessStatisticsMessageAlt::createThereAreManyLetters(int count)
{
return { "are", std::to_string(count), "s" };
}

GuessStatisticsMessageAlt::MessageComponents GuessStatisticsMessageAlt::createThereIsOneLetter()
{
return { "is", "one", "" };
}

GuessStatisticsMessageAlt::MessageComponents GuessStatisticsMessageAlt::createThereAreNoLetters()
{
return { "are", "no", "s" };
}

Since at this point there was no real need for a class, I continued, reaching the final result:
.h

std::string makeGuessStatisticsMessage(char candidate, int count);

.cpp

namespace {

struct MessageComponents
{
std::string verb;
std::string number;
std::string pluralModifier;
};

MessageComponents createThereAreManyLetters(int count)
{
return { "are", std::to_string(count), "s" };
}

MessageComponents createThereIsOneLetter()
{
return { "is", "one", "" };
}

MessageComponents createThereAreNoLetters()
{
return { "are", "no", "s" };
}

MessageComponents createPluralDependentMessageParts(int count)
{
if (count == 0) {
return createThereAreNoLetters();
}
else if (count == 1) {
return createThereIsOneLetter();
}
else {
return createThereAreManyLetters(count);
}
}

}

std::string makeGuessStatisticsMessage(char candidate, int count)
{
auto messageComponents = createPluralDependentMessageParts(count);
return fmt::format("There {} {} {}{}", messageComponents.verb, messageComponents.number,
candidate, messageComponents.pluralModifier);
}

The point is, I grapple a bit with when to apply different programming paradigms and find it difficult to decide if my re-implementation is actually an improvement in this case. In particular, hiding so much stuff in the anonymous namespace is something that bothers me somewhat.

So the primary question is - is this really an improvement? And the secondary - how can I learn to judge this better?

I hope it isn't too opinion-based, as this is a real struggle for me.

* Clean Code: A Handbook of Agile Software Craftsmanship, Robert C. Martin

• Getting rid of the class is definitely a step in the right direction.

• Assuming count can never be negative, we should use an unsigned integer type for it.

• I don't think the MessageComponents class is a good idea. It makes the design very inflexible. If we ever want to change the message, we'll have to change every single function there.

• We're programming a very specific output message, and the code doesn't seem to be intended for re-use, so I don't think we need the named functions either.

I'd suggest something more like:

std::string makeStatisticsMessage(char letter, unsigned int count)
{
auto const verb = (count == 1) ? "is" : "are";
auto const number = (count == 0) ? "no" : (count == 1) ? "one" : std::to_string(count);
auto const plural = (count == 1) ? "" : "s";
return fmt::format("There {} {} '{}'{}.", verb, number, letter, plural);
}

Although we now check count multiple times, I'd argue that it's much easier to follow the logic for each component separately.

It also means that we can abstract or change the behavior for each component later (e.g. if we decide we want to print all the numbers as words instead of digits).

• One reason to use a class rather than a simple function could be a redesign to allow multiple languages for the output. In this case I would define a interface with a virtual function makeStatisticsMessage() and several inherited classes for the specific languages which should be supported. – πάντα ῥεῖ Jul 21 '19 at 12:15
• Also to enhance readability of your conditional logic, I'd recommend to avoid nested ternary operations in favor of if / else statements. The result would be the same and it's much better to follow for anyone (beginner or expert). – πάντα ῥεῖ Jul 21 '19 at 12:22
• I've accepted this answer since the other one builds on it. Both of them were helpful, though. – SG_90 Jul 23 '19 at 18:19

While I am supporting @user673679's approach a lot in terms that putting everything into a single function, I still have concerns:

1. Using an interface might be better if you want to refactor the functionality to support multiple languages:

struct IGuessStatisticsMessage {
virtual std::string makeStatisticsMessage(char letter, unsigned int count) = 0;
virtual ~IGuessStatisticsMessage() {}
};

This would allow to provide implementations for different languages easier:

struct GuessStatisticsMessage_EN() : public IGuessStatisticsMessage {
std::string makeStatisticsMessage(char letter, unsigned int count) {
auto const verb = (count == 1) ? "is" : "are";
auto const number = (count == 0) ? "no" : (count == 1) ? "one" :
std::to_string(count);
auto const plural = (count == 1) ? "" : "s";
return fmt::format("There {} {} '{}'{}.", verb, number, letter, plural);
}
};

struct GuessStatisticsMessage_DE() : public IGuessStatisticsMessage {
std::string makeStatisticsMessage(char letter, unsigned int count) {
auto const number = (count == 0) ? "kein" : (count == 1) ? "ein" :
std::to_string(count);
auto const plural = (count == 1) ? "" : "s";
return fmt::format("Es gibt {} '{}'{}.", number, letter, plural);
}
};

2. Also I am not a friend of nested ternary expressions, they generally tend to be hard to read. I'd rather use some code like this:

std::string makeStatisticsMessage(char letter, unsigned int count) {
auto const verb = (count == 1) ? "is" : "are";
std::string number;
if(count == 0) {
number = "no";
else if (count == 1) {
number = "one";
}
else {
number = std::to_string(count);
}
auto const plural = (count == 1) ? "" : "s";
return fmt::format("There {} {} '{}'{}.", verb, number, letter, plural);
}

• Thanks for the contribution. Would it make sense to go a step further by getting rid of the verb/number/plural vars altogether and returning a complete string depending on the count (having multiple return statements), or would you insist on keeping the vars? – SG_90 Jul 23 '19 at 18:16
• @SG_90 The (local) variables are a minor point. Keeping a replacable interface is a major. – πάντα ῥεῖ Jul 23 '19 at 18:21


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UV radiation having a 300-nm wavelength falls on uranium metal, ejecting 0.500-eV electrons. What is the binding energy of electrons to uranium metal?
Question by OpenStax is licensed under CC BY 4.0.
$3.64 \textrm{ eV}$
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This is College Physics Answers with Shaun Dychko. Photons with a wavelength of 300 nanometers incident on Uranium eject electrons that have kinetic energy of 0.500 electron volts. And we assume this is the maximum kinetic energy they can have because that's what this formula is all about—the maximum kinetic energy. This is [29.5]. So maximum kinetic energy of the electrons ejected are Planck's constant times the frequency of the photon minus the binding energy. Now we need to replace frequency, with something in terms of lambda because we are given the wavelength. So the wave equation says that the speed of the wave is frequency times lambda and then we divide both sides by lambda and we solve for frequency is c over λ. So f gets replaced with c over λ here. Then we are gonna add binding energy to both sides. And then also subtract kinetic energy from both sides. Then we have binding energy then is hc over λ minus kinetic energy of the electrons. So the binding energy of Uranium is Planck's constant times the speed of light divided by this wavelength—300 times 10 to the minus 9 meters— minus 0.5 electron volts giving 3.64 electron volts. And if you prefer less button pushing in your calculations, you could replace this numerator with 1240 electron volt nanometers which is equation [29.14] and then divide by 300 nanometers and you would get the same answer.


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Kickstart 2020 with new opportunities! - hours only!Up to 80% off on all courses and bundles.-Close
Introduction
FIRST_VALUE, LAST_VALUE, NTH_VALUE
Summary
25. Summary

## Instruction

Excellent!! That's all we wanted to teach you today. Let's review what we've learned:

• LEAD(x) and LAG(x) give you the next/previous value in the column x, respectively.
• LEAD(x,y) and LAG(x,y) give you the value in the column x of the row which is y rows after/before the current row, respectively.
• FIRST_VALUE(x) and LAST_VALUE(x) give you the first and last value in the column x, respectively.
• NTH_VALUE(x,n) gives you the value in the column x of the n-th row.
• LAST_VALUE and NTH_VALUE usually require the window frame to be set to ROWS BETWEEN UNBOUNDED PRECEDING AND UNBOUNDED FOLLOWING.

Now, are you ready for a short quiz?

Click to start.


text:
Agda-2.6.1: A dependently typed functional programming language and proof assistant

Agda.TypeChecking.LevelConstraints

Synopsis

# Documentation

Arguments

 :: Constraint Constraint c to simplify. -> [Constraint] Other constraints, enable simplification. -> Maybe [Constraint] Just: list of constraints equal to the original c. Nothing: no simplification possible.

simplifyLevelConstraint c cs turns an c into an equality constraint if it is an inequality constraint and the reverse inequality is contained in cs.

The constraints don't necessarily have to live in the same context, but they do need to be universally quanitfied over the context. This function takes care of renaming variables when checking for matches.


text:
Extremes of L^{p}-norm of Vector-valued Gaussian processes with Trend

# Extremes of Lp-norm of Vector-valued Gaussian processes with Trend

Long Bai Long Bai, Department of Actuarial Science, University of Lausanne
UNIL-Dorigny, 1015 Lausanne, Switzerland

Abstract: Let be a Gaussian vector process and be a continuous function. The asymptotics of distribution of , the norm for Gaussian finite-dimensional vector, have been investigated in numerous literatures. In this contribution we are concerned with the exact tail asymptotics of with trend over . Both scenarios that is locally stationary and non-stationary are considered. Important examples include and chi-square processes with trend, i.e., . These results are of interest in applications in engineering, insurance and statistics, etc.

Keywords: Tail asymptotics; -norm; vector-valued Gaussian process; fractional Brownian motion; Pickands constant; Piterbarg constant.

AMS Classification: Primary 60G15; secondary 60G70

## 1. Introduction

In engineering sciences, extreme values of non-linear functions of multivariate Gaussian processes are of interest in dealing with the safety of structures, see [EXCRA1980] and the references therein. Probabilistic structural analysis to answer the question is: what is the probability that a certain mechanical (or other) structure will survive when it is subject to a random load. The load is then usually defined by some -dimensional vector process , and one seeks the probability that exceeds some more or less well-defined safe region, which is specific for the structure as

 (1) P{Y(t)∉Su(t),for some t∈[0,T]},

where the time-dependent safety region is defined by

 Su(t)={(x1,⋯,xn)∈Rn:∥x∥p≤h(t,u)}

with some continuous function and , the norm, i.e.,

 ||x||p={(∑ni=1|xi|p)1/p, p∈[1,∞),max(|x1|,…,|xn|), p=∞,

in the space .
Assume that where are independent copies of a centered Gaussian process which has continuous trajectories, variance function and correlation function and

 (2) d=(d1,…,dn),  1=d1=⋯=dm>dm+1≥dm+2≥⋯≥dn>0,1≤m≤n.

In the framework of (1), set , then we can rewrite (1) as

 P{∃t∈[0,T]Z(t)>h(t,u)}

where

 (3) Z(t):=Zp(t):=∥X(t)∗d∥p,

and hereafter, we call the norm process.
When , for a positive constant , as in the convention is called the chi process when and the chi-square process when .
Further, as the Gaussian processes, we can introduce the stationary, locally-stationary, and non-stationary norm processes according to the stationary, locally-stationary, and non-stationary properties of , respectively.
The investigate of

 P{∃t∈[0,T]Z2(t)>u}=P{supt∈[0,T]Z2(t)>u},as u→∞

is initiated by the studies of high excursions of envelope of a Gaussian process, see e.g., [BN1969] and generalized in [Lindgren1980a, Lindgren1980b, Lindgren1989]. When is stationary with and

 r(s,t)=1−a|t−s|α+o(|t−s|α),  |t−s|→0, α∈(0,2],

[Albin1990, Albin1992] develop the Berman’s approach in [Berman82] to obtain an asymptotic behavior of large deviation probabilities of the stationary chi-square processes.
Further, if there exists unique satisfies and

where and are positive constants related to , the tail asymptotic behavior of the non-stationary and are investigated in [Pitchi1994] and [FatalovA1993], respectively, under the application of the so-called ”double-sum method” in [Pit96].

Some recent contributions are focused on more general scenarios of chi process and chi-square process with , i.e.,

 P{∃t∈[0,T]Zc2(t)>h(t,u)}=P{supt∈[0,T](Zc2(t)+g(t))>u}, c=1,2,

where the continuous function is generally considered as a trend or a drift.
When are non-stationary Gaussian processes, , the non-stationary chi processes with trend, and , the non-stationary chi-square processes with trend, are studied in [EnkelejdJi2014Chi] and [PL2015], respectively.
When are locally-stationary Gaussian processes, [LJ2017] obtains the extreme of the supremum of with trend, see, e.g., [Berman92, Hus90] for more details about locally stationary Gaussian processes.

Considering both the locally stationary and non-stationary norm processes, the contribution of this paper concerns an exact asymptotic behavior of large deviation probabilities for with , constant and a continuous function, which contains the aforementioned results.
Organisation of the rest of the paper: In Section 2, the notation and some preliminaries are given. Our main results are displayed in Section 3. Following in Section 4 are two applications related to insurance and statistics. Finally, we present the proofs in Section 5 and several lemmas in Section 6.

## 2. Notation and preliminaries

First we introduce some notation, starting with the well-known Pickands constant defined by

where are constants and is a standard fractional Brownian motion (fBm) with Hurst index Further, define for non-negative continuous function.

 Pfα,a[−S1,S2]=E{supt∈[−S1,S2]e√2aBα(t)−a|t|α−f(t)},

and

 Pfα,a[0,∞)=limS2→∞Pfα,a[0,S2],  Pfα,a(−∞,∞)=limS1,S2→∞Pfα,a[−S1,S2].

The exact values of are known for and , namely,

 Pbt1,a[0,∞)=1+abandPbt22,a[0,∞)=12(1+√1+ab).

See [PicandsA, Pit72, debicki2002ruin, DI2005, DE2014, DiekerY, DEJ14, Pit20, Tabis, DM, SBK, GeneralPit16] for various properties of and .

Through this paper means asymptotic equivalence when the argument tends to or . We notice that denotes the tail distribution function of an random variable and .
For the norm process in (3) and a continuous function , we shall investigate the asymptotics of

 (4) P{supt∈[0,T](Zc(t)+g(t))>u},u→∞,

with a constant. As in [FatalovA1993, Pitchi1994], for , using the duality property of norm we find

 P{supt∈[0,T]Zc(t)>u}=P{supt∈[0,T]Z(t)>u1/c}=P{sup(t,v)∈[0,T]×SqY(t,v)>u1/c},

where is a centered Gaussian field defined on cylinder with

 (5) Sq={v∈Rn:||v||q=1},

where if , if and if .

###### Lemma 2.1.

On , attains its maximum at:
(i) for at points where ( stands at the i-th position), ( stands at the i-th position), ;
(ii) for at points on , ;
(iii) for at points , where

 z=(z1,…,zn), zi=±(di/d)2/(q−2),d=[n∑i=1d2p/(2−p)i](2−p)/2p,

( we take all possible combinations of signs ”+” and ”-” ), where .

The proof can be easily carried out by method of Lagrangian multipliers or referring to [FatalovA1993] [Lemma 3.1].
Next by [randomChaos2015], we have the following lemma.

###### Lemma 2.2.

For the norm process in (3), if for some , then we have that as

 P{Zc(t0)>u}∼Ψ(u1/cd)⎧⎪ ⎪ ⎪ ⎪ ⎪⎨⎪ ⎪ ⎪ ⎪ ⎪⎩2n(2−p)(1−n)/2, if  p∈[1,2),√2π2(2−m)2um−1cΓ(m/2)∏ni=m+1(1−d2i)−12, if  p=2,2m, if  p∈(2,∞],

with the convention and the same as in Lemma 2.1.

## 3. Extremes of Lp norm processes with trend

In this section, recall that in (3) is the norm process and ’s are independent copies of with continuous trajectories, variance functions and correlation functions .

### 3.1. Extremes of non-stationary Lp norm processes with trend

As in [Threshold2016], if is non-stationary, we introduce the following assumptions:

• attains its maximum on at the unique point and

 σ(t)=1−b|t−t0|β+o(|t−t0|β),    t→t0

for some positive constants .

• for some constants and

Further, we introduce a bounded measurable trend function which satisfies

• for some constants .

###### Theorem 3.1.

If assumptions (i)-(iii) are satisfied, then for in (2) and in Lemma 2.1, we have as

 P{supt∈[0,T](Zc(t)+g(t))>u}∼P{Zc(t0)>u}⎧⎪ ⎪ ⎪ ⎪⎨⎪ ⎪ ⎪ ⎪⎩u2α∗−2β∗a1/αd−2/αHα∫∞Qe−f(t)dt, if  α∗<β∗,Pf(t)α,ad−2[Q,∞), if  α∗=β∗,1, if  α∗>β∗,

where , , , and if , if .

###### Remarks 3.2.

i) In Theorem 3.1, if we assume that , we get the extremes of centered non-stationary norm processes i.e.,

 P{supt∈[0,T]Zc(t)>u}, u→∞.

ii) Following the similar arguments as in the proof of Theorem 3.1, the result in Theorem 3.1 still holds for if and if .

### 3.2. Extremes of locally stationary Lp norm processes with trend

If is locally stationary, as in [Threshold2016], we shall suppose that:

• where are positive continuous function on .

Before giving the scenarios with trend, we consider the extremes of the centered locally stationary norm processes.

###### Theorem 3.3.

Assume that , i.e., unit variance and covariance function satisfies assumptions (iv) and (v). Then we have for

 P{supt∈[0,T]Zc(t)>u}∼∫T0(a(t))1αdtd−2αHαu2αcP{Zc(0)>u}, u→∞,

where is the same as in Lemma 2.1.

###### Theorem 3.4.

Assume that , i.e., unit variance and correlation function satisfies assumptions (iv) and (v). Assume that is a continuous function which attains its maximum at a unique point satisfying assumption (iii) for some constants . Further, set and and is the same as in Lemma 2.1.
If , then we have as

where and if , if .
If , then we have

 P{supt∈[0,T](Zc(t)+g(t))>u}∼∫T0(a(t))1αeg(t)2d2dtd−2αHαu2α∗P{Zc(0)>u}, u→∞.

If , then we have

###### Remark 3.5.

By the proof, we notice that for the case in Theorem 3.4, the result always holds for any continuous function . When , the result holds for any bounded function .

###### Example 3.6.

For in (3) with the independent fractional Brownian motions, we have as

 P{supt∈[0,1]Z(t)−√1−t>u}=P{Z(1)>u}⎧⎪ ⎪ ⎪ ⎪⎨⎪ ⎪ ⎪ ⎪⎩u2α−2(12d2)1/αHα∫∞0e−f(t)dt, if  α<1,Pf(t)α,d−2/2[0,∞), if  α=1,1, if  α>1,

where and is the same as in Lemma 2.1

Following example is a special case of Theorem 3.4, which is corresponded with [LJ2017] [Theorem 2.1].

###### Example 3.7.

In Theorem 3.4, assume that , and is a continuous function, then we have

 P{supt∈[0,T](Z22(t)+g(t))>u}∼∫T0(a(t))1αeg(t)2dtHα21−m/2∏ni=m+1(1−d2i)−12Γ(m/2)um−12+1αe−u2, u→∞.

## 4. Applications

### 4.1. Ruin probability of a risk model

In theoretical insurance modelling a surplus process can be defined by

 U(t)=u+wt−X(t),   t≥0,

see [MR1458613], where is the initial reserve, is the rate of premium and the stochastic process denotes the aggregate claims process. See [rolski2009stochastic, DHJ13a, HX2007, ParisianBrownianfinite2017, Threshold2016, ParisianInfinite2018] for more studies on related risk models. Here we investigate

 X(t)=n∑i=1∣∣diBiα(t)∣∣2,t≥0,

where is the same as in (2) and are independent fractional Brownian motions. can be considered as the sum of independent claims or payments until time . The corresponding ruin probability over a finite-time horizon is defined as

 P{inft∈[0,1]U(t)<0}.

We present next approximation of this ruin probability.

###### Proposition 4.1.

We have as

 P{inft∈[0,1]U(t)<0}∼um/2−1e−u+w221−m/2Γ(m/2)n∏i=m+1(1−d2i)−1/2⎧⎪⎨⎪⎩u1/α−121−1/αHα, if  α<1,2, if  α=1,1, if  α>1.

Besides in risk modelling, the norm processes, especially the chi-square processes, are also widely utilized in hypothesis testing, see [HTDa1987, SCSP2014] and the reference there. Next we give an example.

### 4.2. The Ornstein-Uhlenbeck chi-square process in Quantitative Trait Locus detection

A Quantitative Trait Locus (QTL) denotes a gene with quantitative effect on a trait. The method used by most of geneticists in order to detect a QTL on a chromosome, is the Interval Mapping proposed by [LanderBO1989]. Using the Haldane distance and modelling in [Haldane1919], each chromosome is represented by a segment . The distance on is called the genetic distance. At each location , using the ”genome information” brought by genetic markers, a likelihood ratio test (LRT) is performed, testing the presence of a QTL at this position. [ADR2014] prove that when the number of genetic markers and the number of progenies tends to infinity, the limiting process of the LRT process is an Ornstein-Uhlenbeck chi-square process under the null hypothesis of the absence of QTL on the interval . In order to take decision about the presence of a QTL on , we need to calculate the critical value for the supremum of an Ornstein-Uhlenbeck chi-square process, i.e.,

 supt∈[0,T]S(t),

where the Ornstein-Uhlenbeck chi-square process is

 S(t)=n∑i=1Vi(t)2

and are independent identically stationary Gaussian processes with covariance function given by

 Cov(Vi(s),Vi(t))=e−2|t−s|.
###### Proposition 4.2.

We have as

 P{supt∈[0,T]S(t)>u}∼22−n/2Γ(n/2)Tun/2e−u/2.

## 5. Proofs

During the following proofs, are some positive constants which can be different from line by line and for interval we denote

 Lu(Δ1):=P{supt∈Δ1(Zc(t)+g(t))>u},  Lu(Δ1,Δ2):=P{supt∈Δ1(Zc(t)+g(t))>u,supt∈Δ2(Zc(t)+g(t))>u},

and

 Ku(Δ1):=P{supt∈Δ1Zc(t)>u},  Ku(Δ1,Δ2):=P{supt∈Δ1Zc(t)>u,supt∈Δ2Zc(t)>u}.

Proof of Theorem 3.1: We first present the proof for the case .
Set , with and for large enough

 Y(t,v)=n∑i=1diviXi(t),(t,v)∈R×Sq

with the same as in (5) which is a centered Gaussian field.
We have for some small and large enough

 (6) Lu([0,δ(u)])≤Lu([0,T])≤Lu([0,δ(u)])+Lu([δ(u),θ])+Lu([θ,T]).

We first give the upper bounds of and .
Set and . Then by Borell inequality as in [AdlerTaylor] and Lemma 2.2 for large

 (7) Lu([θ,T]) ≤ P{supt∈[θ,T]Z(t)>(u−gm)1/c} ≤ P{sup(t,v)∈[θ,T]×SqY(t,v)>(u−gm)1/c} ≤ exp⎛⎝−((u−gm)1/c−Q1)22V∗Y⎞⎠ = o(P{Zc(0)>u}), u→∞,

where and

 V∗Y:=sup(t,v)∈[θ,T]×Sq% Var(Y(t,v))≤(supt∈[θ,T]σ2(t))d2=σ2θd2

By assumptions (i) and (iii), we know that for some

 (8) u−g(t)σc(t)≥(u+w(1−ε1)|t|γ)(1+(1−ε1)bc|t|β)≥u(1+w(1−ε1)u|t|γ+(1−ε1)bc|t|β), (9) u−g(t)σc(t)≤(u+w(1+ε1)|t|γ)(1+(1+ε1)bc|t|β)≤u(1+w(1+ε1)u|t|γ+(1+ε1)bc|t|β)

hold for when small enough, then

 inft∈[δ(u),θ](u−g(t))2/cσ2(t) ≥inft∈[δ(u),θ]u2/c(1+w(1−ε1)u|t|γ+(1−ε1)bc|t|β)2/c (10) ≥u2/c+Q2(lnu)(ρβ)∨(ργ).

Denote with . By assumption (ii), we have that

 E⎧⎨⎩((n∑i=1divi¯¯¯¯¯Xi(t))−(n∑i=1div′i¯¯¯¯¯Xi(s)))2⎫⎬⎭ ≤2E⎧⎨⎩((n∑i=1divi¯¯¯¯¯Xi(t))−(n∑i=1divi¯¯¯¯¯Xi(s)))2⎫⎬⎭ +2E⎧⎨⎩((n∑i=1divi¯¯¯¯¯Xi(s))−(n∑i=1div′i¯¯¯¯¯Xi(s)))2⎫⎬⎭ ≤4E{n∑i=1(¯¯¯¯¯Xi(t)−¯¯¯¯¯Xi(s))2}+4E{n∑i=1(vi−v′i)2(¯¯¯¯¯Xi(s))2} ≤Q3|s−t|α+Q4n∑i=1|vi−v′i|2 ≤Q5(|s−t|α+n∑i=1|vi−v′i|α)

holds for and . Thus it follows from [Pit96] [Theorem 8.1], (5) and Lemma 2.2 that

 Lu([δ(u),θ]) ≤P{sup(t,v)∈[δ(u),θ]×Sqn∑i=1divi¯¯¯¯¯Xi(t)>infs∈[δ(u),θ](u−g(s))1/cσ(s)} ≤Q6u2(n+1)αΨ(infs∈[δ(u),θ](u−g(s))1/cdσ(s)) ≤Q6d√2πu2(n+1)α−2cexp(−12d(u2/c+Q2(lnu)(ρβ)∨(ργ))) (11) =o(P{Zc(0)>u}), u→∞.

Thus by (7), (5) and the fact that for positive, we have

 (12) Lu([δ(u),θ])=o(Lu([0,δ(u)])),  Lu([θ,T])=o(Lu([0,δ(u)])), u→∞,

which combined with (6) imply

 (13) Lu([0,T])∼Lu([0,δ(u)]), u→∞.

Now we focus on the asymptotic of , as .
Denote for any and some

 Ik(u)=[ku−2/α∗λ,(k+1)u−2/α∗λ],  k∈N,  N(u)=⌊(lnu)2qβ∗u2α∗−