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C-2: Loss Simulation Statistical Analysis in Risk Management – Two main approaches: – Maximum probable loss (or MPY) if $5 million is the maximum probable loss at the _______percent level, then the firm’s losses will be less than $_____million with probability 0.95. Same concept as “Value at risk” When to Use the Normal Distribution – Most loss distributions are not normal – From the __________ theorem, using the normal distribution will nevertheless be appropriate when – Example where it might be appropriate: Using the Normal Distribution Important property – If Losses are normally distributed with – Then Probability (Loss < ) = 0.95 Probability (Loss < ) = 0.99 Using the Normal Distribution - An Example – Worker compensation losses for Stallone Steel sample mean loss per worker = $_____ sample standard deviation per worker = $20,000 number of workers = ________ – Assume total losses are normally distributed with mean = $3 million standard deviation = – Then maximum probable loss at the 95 percent level is $3 million + = $6.3 million A Limitation of the Normal Distribution Applies only to aggregate losses, not _______losses Thus, it cannot be used to analyze decisions about per occurrence deductibles and limits Monte Carlo Simulation – Overcomes some of the shortcomings of the normal distribution approach – Overview: Make assumptions about distributions for ________ and _______ of individual losses Randomly draw from each distribution and calculate the firm’s total losses under alternative risk management strategies Redo step two many times to obtain a distribution for total losses A. Total Loss Profile 1. E(L) forecast a. single best estimate ………. b. variations from this number will occur, therefore … 2. Example for a large company.(next slide) mode, median expected = $ Pr(L) > $11,500,000 = Pr(L) > $14,000,000 = Unlimited Loss Distribution 0.35 0.3 Probability 0.25 0.2 0.15 0.1 0.05 0 9 10 11 12 13 14 15 16 Total Losses (Millions) 17 18 19 3. Uses of Total Loss Profile a. Evaluate and b. c. d. MPL (MPY) loss limits B. Monte Carlo Steps 1. Select frequency distribution 2. Select severity distribution 3. Draw from ________ distribution => N1 losses 4. Draw N1 severity values from severity distribution 5. Repeat steps____and ____ for 1000 or more iterations Ni S1 S2 … S10 … S23 … S43 … S70 Total 1 70 $ 600 $ 18,400 Iteration Number 2 1,000 23 … 43 $ 94,000 $ _____ $ 150 $ 970 $ _____ $ 2,600 $ $ 19,500 $ 1,350 $ 32,150 3,750 NA $182,000 $ 54,000 $ NA $ $ 500 NA $ Rank Order the Total Losses Iteration Percentile Total Losses 1 0.1 $ 143,000 . 100 10 1,790,000 . 500 50 2,280,000 . 700 70 ________ . 900 90 3,130,000 . 950 95 ________ . 1,000 100 3,970,000 Horizontal Layering: From One Iteration Layers for the 438th Iteration that produced 980 Severity Values 1,0005,00010000- 50,000GE Draw LT 1,000 Total 4,999 9,999 49,999 99,999 100,000 1 625 625 … 98 ________ 2,050 _________ … 251 999 4,000 _________ .. 730 789 789 … 980 999 4,000 5,000 40,000 50,000 10,001 110,000 Total 920,000 450,000 414,000 180,000 119,000 47,000 2,130,000 D. Interpretation of Results 1. Look at summary statistics: mean, sigma, percentiles 2. 3. Within Limits ,000 1 - 10 10 25 25 - 50 50 - 75 75 - 100 100 - 150 150 - 200 200 - 250 250 - 500 500 - 1,000 > 1,000 X BAR $ $ $ $ $ $ $ $ $ $ $ $ 612 326 128 65 60 26 15 23 9 1 Sigma $ $ 88 $ 92 $ 55 $ 41 $ 53 $ 32 $ 23 $ 60 $ 62 $ 8 At Limits X BAR $ $ 2,655 $ 2,981 $ 3,109 $ 3,174 $ 3,234 $ 3,260 $ 3,275 $ 3,298 $ 3,307 $ 3,307 Sigma $ $ 176 $ 239 $ 275 $ 298 $ 325 $ 340 $ 350 $ 370 $ 400 $ 404 E. Aggregates – Recap using text Simulation Example - Assumptions – Claim frequency follows a Poisson distribution Important property: Poisson distribution gives the probability of 0 claims, 1 claim, 2 claims, etc. Simulation Example - Assumptions – Claim severity follows a expected value = standard deviation = note skewness Simulation Example Assumptions Fre que ncy Distribution w ith Ex pe cte d V a lue Equa l to 30 0.25 PROBABILITY 0.2 0.15 0.1 0.05 54 48 42 36 30 24 18 12 6 0 0 Nu m b e r o f C laim s S a m ple Fre que ncy Distribution w ith Unce rta in Ex pe cte d V a lue (1000 tria ls) 0.25 PROBABILITY 0.2 0.15 0.1 0.05 54 48 42 36 30 24 18 12 6 0 0 Nu m b e r o f C laim s S a m ple Loss S e ve rity Distribution (1000 tria ls) 0.12 PROBABILITY 0.1 0.08 0.06 0.04 0.02 0 0.0075 0.6 1.2 1.8 L o s s in M illio n s 2.4 3 Simulation Example - Alternative Strategies Policy 1 2 3 $500,000 $1,000,000 none Per Occurrence Policy Limit $5,000,000 $5,000,000 none Aggregate Deductible none none $6,000,000 Aggregate Policy Limit none none $10,000,000 Premium $780,000 $415,000 Per Occurrence Deductible $165,000 Simulation Example - Results $500,000 pe r Occurre nce Re te ntion 0.2 0.18 0.16 0.16 12 13 .5 13 .5 .5 10 9 V alu e s in M illio n s $6 M illion Aggre ga te Annua l Re te ntion $1 M illion pe r Occurre nce Re te ntion 0.2 0.18 0.16 0.16 P RO BABILITY 0.2 0.18 0.14 0.12 0.1 0.08 0.06 0.14 0.12 0.1 0.08 0.06 0.04 0.04 V alu e s in M illio n s .5 10 9 5 7. 6 5 4. 3 0 .5 13 12 .5 10 9 5 7. 6 4. 3 5 1. 5 V alu e s in M illio n s 5 0 0 1. 0.02 0.02 0 P RO BABILITY 12 V alu e s in M illio n s 5 0 .5 13 12 .5 10 9 5 7. 6 5 4. 3 0 5 0.02 0 1. 0.04 0.02 7. 0.06 0.04 6 0.06 0.1 0.08 5 0.08 0.12 4. 0.1 0.14 3 0.12 5 0.14 1. P RO BABILITY 0.2 0.18 0 P RO BABILITY No Insura nce Simulation Example - Results Statistic Policy 1: Policy 2: Mean value of retained losses $______ $2,716 $2,925 $3,042 1,065 1,293 1,494 1,839 ______ 6,462 7,899 18,898 0.7% 0.1% n.a. ____% 99.9% 92.7% Standard deviation of retained losses Maximum probable retained loss at 95% level Maximum value of retained losses 4,254 11,325 Probability that losses exceed policy limits 1.1% Probability that retained losses $6 million 99.7% 5,003 12,125 Policy 3: No insurance Premium $780 $415 $165 $0 Mean total cost 3,194 3,131 3,090 3,042 5,418 6,165 6,462 Maximum probable total cost at 95% level 5,034