Actuarial Application of Monte Carlo Simulation
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This document describes using a Monte Carlo simulation to determine the optimal monthly premium price for a term life insurance policy for a husband and wife. Random variables are generated for the ages of the husband and wife at purchase and death based on statistical data. The simulation is run 1000 times with premiums from $50 to $100 monthly. Results show a $70 monthly premium, or $35 per person, has no risk of losses and ensures total premiums exceed total claims paid.
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• Calculate probabilities and cut-off values using the 68%-95%-99.7% (Empirical) Rule.
• Identify the mean, standard deviation, cut-off value, probability, and Z-score on a Normal curve.
• Use the Normal table to get percentiles (probabilities) for forward problems and to get Z-scores
in order to determine cut-offs for backward problems using both > and < in the inequalities.
• Recognize whether a story is a forward or backward Normal distribution problem, and perform
the appropriate calculations showing correct notation, the initial probability expression, and all
necessary steps.
2
Chapter 21: What is a confidence interval?
• Define statistical inference and explain when statistical inference is used.
• Explain what the confidence interval means and whether the results refer to the population or
the sample.
• Calculate the margin of error and identify the margin of error in a confidence statement.
Explain what type of error is covered in the margin of error.
• Determine whether a story is better described with a proportion or a mean.
• Use appropriate notation for proportions and means, both in the population and the sample.
• Calculate a confidence interval for a proportion and for a mean.
• Describe how increasing/decreasing the sample size or confidence level changes the margin of
error (width of the confidence interval).
• Apply cautions for using confidence inte ...
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Qnt 351 final exam august 2017 new version
QNT 351 FINAL EXAM AUGUST 2017 NEW VERSION
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QNT 351 FINAL EXAM AUGUST 2017 NEW VERSION
QNT 351 FINAL EXAM AUGUST 2017 NEW VERSION
The mean amount spent by a family of four on food is $500 per month with a standard deviation of $75. Assuming that the food costs are normally distributed, what is the probability that a family spends less than $410 per month?
• 0.1151
• 0.0362
• 0.8750
• 0.2158
As the size of the sample increases, what happens to the shape of the distribution of sample means?
• It cannot be predicted in advance.
• It is negatively skewed.
• It approaches a normal distribution.
• It is positively skewed.
What is the following table called?
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STATUse the information below to answer Questions 1 through 4..docx
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Use the information below to answer Questions 1 through 4.
Given a sample size of 36, with sample mean 670.3 and sample standard deviation 114.9, we perform the following hypothesis test.
Null Hypothesis
Alternative Hypothesis
1. What is the test statistic?
2. At a 10% significance level (90% confidence level), what is the critical value in this test? Do we reject the null hypothesis?
3. What are the border values between acceptance and rejection of this hypothesis?
4. What is the power of this test if the assumed true mean were 710 instead of 700?.
Questions 5 through 8 involve rolling of dice.
5. Given a fair, six-sided die, what is the probability of rolling the die twice and getting a “1” each time?
6. What is the probability of getting a “1” on the second roll when you get a “1” on the first roll?
7. The House managed to load the die in such a way that the faces “2” and “4” show up twice as frequently as all other faces. Meanwhile, all the other faces still show up with equal frequency. What is the probability of getting a “1” when rolling this loaded die?
8. Write the probability distribution for this loaded die, showing each outcome and its probability. Also plot a histogram to show the probability distribution.
Use the data in the table to answer Questions 9 through 11.
x
3
1
4
4
5
y
1
-2
3
5
9
9. Determine SSxx, SSxy, and SSyy.
10.
Find the equation of the regression line. What is the predicted value when
11. Is the correlation significant at 1% significance level (99% confidence level)? Why or why not?
Use the data below to answer Questions 12 through 14.
A group of students from three universities were asked to pick their favorite college sport to attend of their choice: The results, in number of students, are listed as follows:
Football
Basketball
Soccer
Maryland
60
70
20
Duke
10
75
15
UCLA
35
65
25
Supposed a student is randomly selected from the group mentioned above.
12. What is the probability that the student is from UCLA or chooses football?
13. What is the probability that the student is from Duke, given that the student chooses basketball?
14. What is the probability that the student is from Maryland and chooses soccer?
Use the information below to answer Questions 15 and 17.
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15. How many of apples have weights between 13 ounces and 15 ounces?
16. What is the probability that a randomly selected mango weighs less than 12.5 ounces?
17. A quality inspector randomly selected 100 apples from the shipment.
a. What is the probability that the 100 randomly selected apples have a mean weight less than 12.5 ounces?
b. Do you come up with the same result in Question 16? Why or why not?
18. A pharmaceutical company has developed a screening test for a rare disease that afflicted 2% of the population. Un.
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Actuarial Application of Monte Carlo Simulation
- 1. Actuarial Application of Monte Carlo Simulation A stochastic Approach to Pricing a Life Insurance Policy By: Adam Conrad
- 2. Table Of Contents • Introduction…………………………………………………………… 3 • Problem Statement…………………..............................................4 • Random Variable Generation………………………………………. – Age of Husband at Purchase……………………………………………………..6 – Age of Wife at Purchase..………………………………………………………...8 – Age of Husband at Death………………………………………………………..12 – Age of Wife at Death…………………………………………………………......14 – Premiums Collected Over Claims Paid………………………………………...16 • Monte Carlo Simulation Results.……………………………….…18 • Simulation Coding……..………………………………..................20 • Bibliography…………………………………………………………21 2
- 3. Introduction • Actuaries are mathematicians that put a price on risk. • Actuaries use statistical data to develop mathematical models. • Most actuaries work for insurance companies to set policy rates. • Assumptions made for this problem were made though research 3
- 4. Problem Statement • A husband and wife are looking to purchase the following life insurance policy: – Term: 20 Years or In the event a death – Benefit: $250,000 – Premium: Joint Monthly Payment 4
- 5. Problem Statement • If 1000 similar policies are to be sold in the next year, determine the premium price that will consistently result in total premiums being more than total claims. 5
- 6. Age of Husband at Purchase (HA) 6
- 7. Age of Husband at Purchase (HA) 7 Code Simulation Run Calculations *All future standard deviations will be calculated in the same fashion
- 8. Age of Wife at Purchase (WA) • Below is a table of the data collected on the age difference between a husband and a wife in America. 8
- 9. Age of Wife at Purchase (WA) • Assumptions: – 32.4% of wives are the same age as their husband. – 55.4% of wives are younger than their husbands. – 12.3% of wives are older than their husbands. – The average age difference is 2.3 years with the husband being older than the wife. – If the husband is older, 1 to 17 year makes up 99.7% of a normal distribution (3 standard deviations) – If the wife is older, 1 to 7 years makes up 99.7% of the a normal distribution (3 standard deviations) 9
- 10. Age of Wife at Purchase (WA) 10 Code Simulation Run
- 11. Age of Husband and Wife at Purchase 11 Simulation Run
- 12. Age of Husband at Death (HD) • Assumptions: – The average age of death of a male in America is 76 years. – Ages 65 to 100 make up 99.7% of the male deaths – The pdf is normal until it reaches the mean, and then it is exponential 12
- 13. Age of Husband at Death (HD) 13 Simulation RunCode Calculations
- 14. Age of Wife at Death (WD) • Assumptions: – The average age of death of a female in America is 81 years. – Ages 65 to 100 make up 99.7% of the female deaths – The pdf is normal until it reaches the mean, and then it is exponential 14
- 15. Age of Wife at Death (WD) 15 Simulation RunCode
- 16. Premiums Collected Over Claims Paid (POC) 16 • Lifespans of Husband and Wife after the policy purchase. Code Simulation Run
- 17. Premiums Collected Over Claims Paid (POC) 17 Events after 20 years Claim Profit Only Husband Survives $250,000 (Premium)*(Months Wife Lived) - 250000 Only Wife Survives $250,000 (Premium)*(Months Husband Lived) - 250000 Both Survive $0 (Premium)* (240 Months) The table below shows a list of possible outcomes under the policy The coding on the right determines whether or not the nth couple filed a claim before the term expired and how much premium was collected throughout the policy. It then runs the simulation 10 times so that the variance can be determined.
- 18. Monte Carlo Simulation Results 18 Selected Premium Expected Profit Standard Deviation Worst Case Profit 50.00$ 1,430,807.00$ 1,816,279.35$ (4,018,031.05)$ 60.00$ 3,267,329.01$ 1,570,115.48$ (1,443,017.43)$ 70.00$ 6,024,908.71$ 1,890,279.74$ 354,069.49$ 80.00$ 8,497,356.64$ 1,631,740.58$ 3,602,134.90$ 90.00$ 11,472,155.66$ 1,002,633.00$ 8,464,256.66$ 100.00$ 13,489,612.32$ 1,482,734.74$ 9,041,408.10$ Monte Carlo Simulation The simulation was run with premiums set at $50/month up to $100/month and the expected profits and variances were outputted. Using the variances of each run, the profit earned given the worst case scenario was calculated by subtracting 3 standard deviations from the expected value.
- 19. Monte Carlo Simulation Results 19 Selected Premium Expected Profit Standard Deviation Worst Case Profit 50.00$ 1,430,807.00$ 1,816,279.35$ (4,018,031.05)$ 60.00$ 3,267,329.01$ 1,570,115.48$ (1,443,017.43)$ 70.00$ 6,024,908.71$ 1,890,279.74$ 354,069.49$ 80.00$ 8,497,356.64$ 1,631,740.58$ 3,602,134.90$ 90.00$ 11,472,155.66$ 1,002,633.00$ 8,464,256.66$ 100.00$ 13,489,612.32$ 1,482,734.74$ 9,041,408.10$ Monte Carlo Simulation Risk was mitigated when the monthly premium was 70$/month ($35 per person) Which means that that is the minimum price that this policy can be set at so that there is no risk. The decision about how much to raise the price is dependent on the needs of the business department.
- 20. Monte Carlo Simulation (Code for Copying and Pasting into MATLAB) 20
- 21. Bibliography • http://www.cdc.gov/nchs/data/series/sr_10/sr10_260.pdf • http://www.cincinnati-oh.gov/health/environmental-health/cincinnati-neighborhood-specific- life-expectancy-data/ • http://fivethirtyeight.com/datalab/whats-the-average-age-difference-in-a-couple/ • http://www.allcountries.org/uscensus/56_married_couples_by_differences_in_ages.html • https://www.jrcinsurancegroup.com/best-age-life-insurance-for-male/ • http://www.nejm.org/doi/full/10.1056/NEJMsa1211128 • http://www3.imperial.ac.uk/statistics/msc/optionalcourses • http://www.worldlifeexpectancy.com/usa/life-expectancy-male 21