SOA EXAM STAM Detailed Solutions

SOA 211

"Mixtures and Splices gx= new density func gx = c bw 0<x<3 gx= k* exp pdf theta=4 x>3 MAIN IDEA W SPLICING PROBLEMS: Since the function is continuous, both parts of the function must have the same value at x=3, and the function must also integrate to 1. at x=3, c=c k= ke^-3/4 c=ke^-3/4 ∫0 to 3 cdx+∫3 to infinity ke−x/4dx 3c+4ke^-.75 using the fact we defined before, c=ke^-.75 3c+4c=1 7c=1 c=1/7 F3=3c=3/7 MY MISTAKE: not sure why gx for 0<x<3 is c and not 1/3. CA ANSWER: "the first bullet is trying to tell us that the PDF is a CONSTANT for 0<x<3. You can just start with c. the idea is that c'= c*1/3 and k' = k/4 for the exponential portion. ""The PDF is uniform over [0, 3]" means it is constant over [0, 3]. It does not necessarily mean it has a PDF of a uniform distribution with range [0, 3], which is 1/3." AHHHHH OKAY. UNIFORM BETWEEN. NOT NECESSARILY A UNIFORM DISTRIBUTION PDF. WOW. KINDA RUDE. WHAT AN AWFUL ASSUMPTION.

SOA STAM 85 Computer maintenance costs for a department are modeled as follows: The distribution of the number of maintenance calls each machine will need in a year is Poisson with mean 3. The cost for a maintenance call has mean 80 and standard deviation 200. The number of maintenance calls and the costs of the maintenance calls are all mutually independent. The department must buy a maintenance contract to cover repairs if there is at least 10% probability that aggregate maintenance costs in a given year will exceed 120% of the expected costs. Using the normal approximation for the distribution of the aggregate maintenance costs, calculate the minimum number of computers needed to avoid purchasing a maintenance contract. A

*""Aggregate Loss Models I got this one wrong because i didnt consider the number of calls PER COMPUTER. we know this because "number of maintenance calls each machine will need" in other words, m*E(N) gives us the expected number of calls for each computer. ES = m*EN*EX ES=240m VS= 139200m 10% prob losses exceed 1.2ES is 288m P(Z>288m-240m/rad(139200m)) = 0.10 1.282 < .1287m^1/2 m > 99.3 = 100 normal approximation is appropriate for large sample sizes and x is unknown dist. CA says: *"If Pr(S>1.2E[S])≥0.1, the manufacturer needs to buy a maintenance contract. The manufacturer wants to avoid that (said in the last sentence). So, we need to find the minimum value of mm such that Pr(S>1.2E[S])<0.1*" "*Var[∑i=1Si]=∑i=1,mVar[Si]. If Si's are i.i.d., then Var[Si]=Var[S] for all i and thus ∑i=1,mVar[Si]=mVar[S]]. In this question, we need to find the variance of the sum of the total costs of m computers, which is the first case.*" this is an important concept to not miss! *my suggestion to myself: take the extra couple seconds to expand the formula for scenarios: vs1+vs2+vs3+....+vsm =m * vsi whereas if it was V(ms), would be m^2VS im sure theres a proof somewhre for the second one lol*

Understanding limited expected value and limited expected severity ex) C_TC0280 The empirical limited expected value function for this sample evaluated at 300 is 281.90. Calculate the empirical limited expected value function for this sample evaluated at 1,000.

According to CA, "the empirical limited expected value at d is the sum of claims below d plus d for every claim above 300, all divided by the number of claims summed in the numerator." in table, we see that x and y are in intervals that are less than 300. the other claims are 300 or higher. we cap those at 300 bc we know they are AT LEAST 300 divide by 100 bc 100 total claims observed we need to solve for x+y to plug in and find e x ^ 1000

SOA STAM 53 Calculate the variance of the aggregate loss.

Aggregate Loss Models The idea here is to look at all possible claim combinations and creating new probabilities of these actual combinations occurring. the new probabilities should sum to 1. According to Coaching Actuaries, the probability for 250 includes (2 ncr 1) because there are two permutations of 250 and 50 (250, 50 or 50,250). possibilities are 0,25,100,150,250,400 ES = 105 VS = 8100

SOA STAM 167

Aggregate Loss Models you could have gotten this problem right if you took the time to define definitions correctly. N- number of boats that need REPAIR so, N is not the given number of boats EN = number of boats * prob that boat needs repair also, looking at the table, we have conditional data EN|boat type ex E(N|Power boat) = 100*0.3 = 30 Once we find the individual components, we can just sumproduct to get ES=189000 VS = 401.19(10)^6 y= 209030 its important to recognize the difference between weights and probabilities. we were given probs of repairs. these are not the weights for each expected value. this is why my definitions were off. also, this is a portfolio of boats. out of the portfolio, we have 100 p, 300 s, 50 l. S is the sum of these boats all together. VS = sum100 VP + sum300 VS + sum50 VL 100[ 0.3*10k+ 300^2(0.3)(1-0.3)] + ... idk how. we cant use the variances they gave us. VS = v(sum si) = sum V[si] =Sum( Es^2 - (ES)^2) =Sum(Es^2) - sum? ES^2 - sumE(Si)^2 var = 2nd moment - sum each types mean squared instead of the total mean squared idea: in many problems, we are asked about the repair cost of one boat. the distribution of the repair cost of a randomly picked boat is a mixture. we are asked about total repair costs of ALL boats. N is binomial bc either needs repair or does not. m=100 prob repair q=0.3 VN = mq(1-q) =21 "the real trick is knowing what the second moment of n is" E[N^2] does not equal 100^2 * 0.3. That implies that there are either 0 or 100 repairs, which is not true. for each boat, number of repairs is bernoulli by context, there is either one repair or zero repairs. OKAY: if asked about randomly selected boat, we use LOTE or LOTV. since the boat could be three different kinds, its a mixture. we are asked for ALL BOATS, so this is a sum. OKAY OKAY: WE ALWAYS USE COMPOUND VARIANCE FORMULA TO CALCULATE THE VARIANCE OF AGG LOSSES MIXTURE IS THE ONE WHERE WE HAVE TO DO THE BASIC EX^2 - EX^2 TO FIND VARIANCE CA responses pt 2: 1. the total repair costs comes from the sum of the repair costs. VS = V(p+s+l) 2. Repair costs are independent of eachother VS = vp + vs + vl 3. If we were looking at a randomly selected boat, could take on three different kinds. we would have to account for weights and likelihood of boat being selected. this would be a mixture.

SOA STAM 24 The probability that an insured will have exactly one claim is θθ. The prior distribution of θθ has probability density function: π(θ)=32θ√,0<θ<1π(θ)=32θ,0<θ<1 A randomly chosen insured is observed to have exactly one claim. Calculate the posterior probability that θθ is greater than 0.60.

Bayesian Credibility

SOA STAM 272 The number of claims made by an individual in any given year has a binomial distribution with parameters m=4m=4 and qq. The prior distribution of qq has probability density function π(q)=6q(1−q),0<q<1π(q)=6q(1−q),0<q<1 Two claims are made in a given year. Calculate the mode of the posterior distribution of qq.

Bayesian Credibility

SOA STAM 48

Buhlmann Credibility MUST REMEMBER: when calculating the individual conditional expected values, we expect that the probabilities used should sum to 1. if they dont, we would scale the probabilities to make them sum to 1! thats the main reason why we divide by the sum of probabilities. ex) 0.4/0.6 = 2/3 0.1/0.6 = 1/6 1/6*2 + 2/3 = 1. NOW THEY CAN BE USED REGULARLY!! IF PROBABILITIES FROM START DO NOT SUM TO 1, EVEN FOR THE INDIVIDUAL EXPECTED VALUES, MUST SCALE RIGHT AWAY. "Because (0.4 + 0.1 + 0.1) + (0.1 + 0.2 + 0.1) = 1, all we need to do is calculate the expected value as if these 6 probabilities were part of one random variable, i.e. mu = 0(0.4) + 1(0.1) + 2(0.1) + 0(0.1) + 1(0.2) + 2(0.1) = 0.7." this makes sense YAY i wanted to approach this as bayesian, but it is indeed buhlmann. u = E(Ex|theta)) E(x|th=0)*p(th=0)+Ex|th=1)*p(th=1) ^this is your law of total expectation. this should be intuitive. p(th=0) = 0.4+0.1+0.1 (i guess think law of total prob?) p(th=1) = 0.1+0.2+0.1 u=0.7 v= Vx|theta=0 (E(x^2|th=0)-e(x|th=0)^2) * p(th=0) v= 0.55 a can be done w formulas or bernoulli a=0.06 k=55/6 z=60/115 pc = z(10/10)+(1-z)(0.7) = .8565 "0.4,0.1,0.1 dont add to 1. we divide by the sum to scale each probability proportionally so the total adds to 1. do this for the other probabilities. only do this if we use this approach. bayesian actually works, but so tedious

SOA STAM 67

Buhlmann Credibility Must understand collective risk model vs. inidivudal risk model. Collective Risk Model: all existing x1-xn are iid rv n and any existing xi are independent in this case, dist of x is called the compound distribution Individual Risk Model: simplify the frequency, generalize the severity instead of rv, the frequency is now a constant, n, that is used to represent the number of policyholders. xi is agg loss amount of ith policyholder S= sumxi x1-xn are independent BUT MAY NOT BE IDENTICAL RV. they may have different distributions. Anyways, we are interested in the aggregate loss X, which has frequency and severity components. E(X|r) = E(n|r)*E(y) = 100r 500*0.2r = 100r v=V(x|r) remember, x is technically agg S VS = envx+vnex^2 E(n|r)vy+vn|r*ey^2 210,000r Why do we do this? Because x is defined as aggregate. I was trying to solve this by doing e(v(n|r)) this is wrong because..... i think bc its aggregate and not conditional....

SOA STAM 70 You are given the following information on claim frequency of automobile accidents for individual drivers: You are also given: Each driver's claims experience is independent of every other driver's. There are an equal number of business and pleasure drivers. Calculate the Bühlmann credibility factor for a single driver.

Buhlmann Credibility The key to solving this problem correctly is to think about joint probabilities!! In the table, we have it split 50/50 by business use and pleasure use. across the rows, we have it split by rural and urban. this is NOT 50/50. We are given the individual and total expected values and claim variance by business use and pleasure use. This is not unconditional because it is split up by rural and urban within. Step 1 is to solve for unconditional probabilities of rural and urban We can use either business or pleasure expected value to solve. Since there are only two choices, assign p=rural and 1-p = urban. Set up weighted average formula p(1.5)+(1-p)2.5 = 2.3 1.5p+2.5-2.5p = 2.3 -p= -0.2 p=0.2 = rural 1-p=0.8 = urban Step 2 - Use these class probabilities to solve for unconditional joint probabilities P(business and rural) = P(business | rural) * p(rural) = (0.2)(0.5) = 0.1 p( b and u) = 0.4 p(p and r) = 0.1 p(p and u) = 0.4 Remember that prob business and prob pleasure are 50/50. Step 3- Use these probabilities to calculate mean and variance E(claims) = 2.05 v(claims) = 0.93 In the table, we see there are 4 possible expected claim choices, so we must have 4 probabilities to multiply by. Step 4- Solve for k k=4.18 Step 5- Solve for z n=1 bc we have 1 single driver Z=1/1+4.18 = .193

SOA STAM 140 You are given the following random sample of 30 auto claims: You test the hypothesis that auto claims follow a continuous distribution F(x)F(x) with the following percentiles: You group the data using the largest number of groups such that the expected number of claims in each group is at least 5. Calculate the chi-square goodness-of-fit statistic.

Chi-Square Goodness-of-Fit The main issue with this is understanding how to group. The best place to start is by using the given Fx values and checking what the current expected number of claims is. KEY FORMULA: Enj = n*pi When you use the current Fx, you will notice that some groups have less than 5. ALSO WORTH NOTING: We expect the groups to be contiguous, meaning they are grouped gradually in increasing order? (Dont group first interval and last interval together to make them reach 5. That wouldnt make sense.) Unless stated, always assume contiguous for the exam! When regrouping the first 2 groups together and the last 3 groups together, we are left with 4 total groups. Next step is to recount the number of observed in the new intervals. After, add the grouped expected counts to calculate the new expected counts. Now, we have everything we need to plug into the chi sq formula. Total should be 6.65, the test statistic!

Maximum Likelihood Estimators (CA Summary)

Complete data describes data where we

SOA STAM 311

Homeowners Insurance Basic formulas. Recognize that the original home value is unnecessary info.

SOA STAM 44 Losses follow an exponential distribution with mean θθ. A random sample of 20 losses is distributed as follows: Calculate the maximum likelihood estimate of θθ.

Maximum Likelihood Estimators

SOA STAM 79 Losses come from a mixture of an exponential distribution with mean 100 and with probability pp and an exponential distribution with mean 10,000 with probability 1−p1−p. Losses of 100 and 2,000 are observed. Determine the likelihood function of pp.

Maximum Likelihood Estimators This is somewhat conceptual. According to Coaching Actuaries, "the probability density function of a mixture is the weighted average of the individual density functions. Therefore, likelihood function L(100,200) = f(100)*f(200)" "The official formula to determine the likelihood function is: L(θ)=∏f(x)" My mistake is that I added the data of 100 and 2000, when the likelihood should always be multiplied!

SOA STAM 307 Five models are fitted to a sample of n=260n=260 observations with the following results: Determine the model favored by the Akaike Information Criterion. A

Score-Based Approaches AIC criteria is l-r l is log-likelihood r is # estimated parameters Select the model with the highest AIC value

SOA STAM 60 Calculate E[X5∣S]E[X5∣S] using Bayesian analysis.

Bayesian Credibility Main idea: use the experience S to update our probabilities of heads. Goal: Expected value of 5th toss, given the experience. Step 1- Use the experience to calculate the likelihood of the goal, given the data. P(data | coin) ex) P(data|C1-4) = 0.5^4= 0.0625. The probability of S occurring will vary depending on which coin was selected and tossed. The probability of heads is consistent between tosses of the same coin. Step 2- Identify the prior probabilities Since our data depends on which coin was selected, our prior distribution is the probability of selecting the coin. ex) P(c1-4) = 4/6 Step 3) Multiply prior*likelihood and sum to get Probability of S (Law of Total Probability) P(S) = P(S and c1-4)+P(s and c5)+p(s and c6) Note that we are only given model dist S|coin and coin, so we use bayes theorem to rearrange and solve for the joint probability for the law of total probability formula. P(S)= 0.06120 Step 4 - Calculate the posterior probabilities the idea here is to use S to update our probability of coin. We use the experience to update prior to posterior! P(coin 1-4 | S) = [(PS|c1-4)*P(c1-4)]/P(s) keep in mind that posterior probabilities are used in weighted average and should sum to 1 Step 5 - Get the means to use to calculate predictive mean *This part is where I mess up usually* So, our goal is E(X|S). We only have info E(X|coin). This will be used in weighted average multiplied by associated posterior probability. Ex) E(X|c1-4) = 0.5(1)+0.5(0)=0.5. This is because the coin toss can either take on value of 1 or 0 for number of heads. Step 6 - Predictive Mean E(X|S) = 0.5*.6808+... = .5638

SOA STAM 215 The conditional distribution of the number of claims per policyholder is Poisson with mean λλ. The variable λλ has a gamma distribution with parameters αα and θθ. For policyholders with 1 claim in Year 1, the credibility estimate for the number of claims in Year 2 is 0.15. For policyholders with an average of 2 claims per year in Year 1 and Year 2, the credibility estimate for the number of claims in Year 3 is 0.20. Calculate θθ.

Buhlmann Credibility Algebra sucks here!!! ewwwwwwwwwww. The conditional mean and variance are: E[X∣λ]=Var[X∣λ]=λ The overall mean, EPV and VHM are: μ=E[E[X∣λ]]=E[Var[X∣λ]]=E[λ]=αθμ v=μ=αθ a=Var[E[X∣λ]]=Var[λ]=αθ^22 The Bühlmann kk and credibility factor are: k=v/a=1/θ (so far, i am good!) Z=n/n+k = n/n+(1/theta) ->n/[ (thetan+1) / theta ] -> ntheta/(ntheta+1) this was done by getting common denominator of theta within denominator and flipping the fraction underneath to match the numerator so there arent double denominators (hopefully that makes sense) *n refers to # of years here. how many years of data* 0.15 = zx+(1-z)u x bar = 1 bc we estimate 1 claim in 1 yr 1/1 =1 u = ntheta/ntheta+1 okay what the HECK OKAY AGAIN ITS COMBINING THE DENOMINATOR TO SIMPLIFY! n=1 bc 1 yr data replace the Z with n=1 0.15=Zx¯+(1−Z)μ <- plug in 1 for n θ / θ+1(1) + 1 / θ+1(μ) <- combined the RHS of + for common denominator θ+μ / θ+1 <- combined bc common denominator repeat for bullet iv 4θ+μ / 2θ+1 solve for theta = 1/55

SOA STAM 251 For a group of policies, you are given: The annual loss on an individual policy follows a gamma distribution with parameters α=4α=4 and θθ. The prior distribution of θθ has mean 600. A randomly selected policy had losses of 1400 in Year 1 and 1900 in Year 2. Loss data for Year 3 was misfiled and unavailable. Based on the data in (iii), the Bühlmann credibility estimate of the loss on the selected policy in Year 4 is 1800. After the estimate in (v) was calculated, the data for Year 3 was located. The loss on the selected policy in Year 3 was 2763. Calculate the Bühlmann credibility estimate of the loss on the selected policy in Year 4 based on the data for Years 1, 2, and 3.

Buhlmann Credibility Idea with this problem is that we use our knowns to solve for unknown, and use that unknown to solve for other unknown. Step 1- Use basic formulas to solve for what was given u = E(4theta) = 4*600 = 2400 <- manual premium xbar = 1400+1900 / 2 yrs = 1650 <- avg loss yrs 1 and 2 Step 2- Use Yr 4 estimate to solve for credibility factor 1800=zx+1-z*u solve z=0.8 *next part is where i missed!!* Step 3- Use Z to solve for k! we know that n=2 bc we have 2 years of data z= n/n+k 0.8 = 2/2+k solve k=0.5 Step 4- Estimate new xbar with the updated info of yr 3 data xbar = 2021 Step 5- recalculate new z z=n/n+k 3/3+0.5 = 6/7 Step 6- Solve P4 P4 = 2075 WE DONT NEED TO SOLVE FOR V OR A BC K=V/A LOL

SOA STAM 41 You are given: Annual claim frequency for an individual policyholder has mean λλ and variance σ2σ2. The prior distribution for λλ is uniform on the interval [0.5, 1.5]. The prior distribution for σ2σ2 is exponential with mean 1.25. A policyholder is selected at random and observed to have no claims in Year 1. Using Bühlmann credibility, estimate the number of claims in Year 2 for the selected policyholder.

Buhlmann Credibility This one is tricky for me! READ CAREFULLY U NUT "Annual claim frequency for an individual policyholder has mean λλ and variance σ2σ2." This tells us we have E(lambda,sigmasq) = lambda and V(l,s)=sigma squared Step 1- Use the given info to solve for EHM, VHM, EPV u = E(E(l,s)) = E(lambda) = 1 v= E(V(l,s)) = E(s) = 1.25 a=V(u(l,s)) = V(l) = .0833 Step 2- Solve for k k=1.25/.0833 = 15 Step 3- Solve for Z n=1 bc we have one year of data to observe Z=1/1+15 = 1/16 Step 4- Solve Pc P2 = 1/16*0+15/16*1 = .9375 x bar is zero bc 0 claims over 1 year

Variance of MLE

For exam purposes, you can say that the information based on n observations is equal to n times the information based on 1 observation.

SOA STAM 47 You are given the following observed claim frequency data collected over a period of 365 days: Fit a Poisson distribution to the above data, using the method of maximum likelihood. Regroup the data, by number of claims per day, into four groups: 0 1 2 3+ Apply the chi-square goodness-of-fit test to evaluate the null hypothesis that the claims follow a Poisson distribution. Determine the result of the chi-square test.

Hypothesis Tests: Chi-Square Goodness-of-Fit KEY FORMULAS: Ej = n*pi Poisson MLE = xbar (sample mean) ************ lambda mle = 1.6438 we use this estimate for the poisson distribution to fit with the data. we will use this to calculate the probabilities of being within each group once these probabilities are calculated, multiply by n=365 to get the Ej Plug these into the chi square formula chi square test statistic is 7.56 after, calculate df df = k-1-r k=4 bc 4 groups r=1 because we estimated one parameter (lambda) df=2 use tables to conclude that 7.56 is bw critical values bw .025 and .01 sig level. Hence, reject null at 2.5% sig level, but not at 1% sig level. use tables

SOA STAM 172 A random sample of five observations from a population is: You use the Kolmogorov-Smirnov test for testing the null hypothesis, H0H0, that the probability density function for the population is: Determine the result of the test.

Hypothesis Tests: Kolmogorov-Smirnov Set up a table to make solving these easier. j | xj | F*(xj) | F(xj_) | F(xj) | D(xj_) | D(xj) | Max Diff Key formulas: Dxj_ = Fxj_ - F*xj (lower limit of empirical - cdf) Dxj = Fxj - F*xj (upper value empirical - cdf) We select the max difference of all values and combos as our KS test statistic It is important to recognize cdf as pareto. Use FX in table After creating table, conclude that TS = 0.680 Next, use the given critical values and plug in n. See where TS falls between. Conclude falls between the critical values at alpha=2.5% and alpha=1.0% Null rejected at 0.025 and not 0.01 bc doesnt exceed 1.0%

SOA STAM 101 Calculate the mean excess loss assuming a deductible of 100 applies.

Insurance Applications First, you gotta know what the mean excess loss is: e(100) E(x)-E(x^100)/S(100) F1000=1, can never exceed. E(X^1000) = EX. limiting value of x to 1000 does nothing E(X^100) = 91 given S100 = 1-0.2 = 0.8 e100 = 300 SHOULDNT HAVE GOTTEN THIS WRONG!

SOA STAM 309 The random variable XX represents the random loss, before any deductible is applied, covered by an insurance policy. The probability density function of XX is: f(x)=2x,0<x<1f(x)=2x,0<x<1 Payments are made subject to a deductible, dd, where 0<d<10<d<1. The probability that a claim payment is less than 0.5 is equal to 0.64. Calculate the value of d

Insurance Applications This is basically Exam P! Use substitution we know Y= X-d for x>d. (Y is payment) we are given that P(Y<0.5) = 0.64 Cant we technically just replace y with x-d? HELL YEAH P(X-d<0.5) -> P(x<0.5+d)=0.64 See this is amazing because we are given x distribution! integral of fx from 0 to 0.5+d x^2 eval at 0.5+d - 0 (0.5+d)^2 = d^2 - d + 0.25 = 0.64 d^2 + d-.39=0 d=0.3 to do it the STAM way, we know that w/ policy limits, the insurer pays the minimum of x or u. ex) policy limit 1000. if loss is 500, they pay 500 if loss is 1500, they pay 1000. in math notation, Y = X ^ u

SOA STAM 87 Calculate the loss elimination ratio for an ordinary deductible of 20.

Insurance Applications You cant really use the concept of area under of a curve. You need to use x and y axes to define fx and the domain. fx= 0.01 for 0 <x<80 fx = c(120-x) for 80<x<120 this is because x can take on any value between 120 and 80, and th loss amount will be removed from 120 multiplied by c because FX=1. solve c say x=80, the first piecewise is capped at 0.01 for x<=80 @ 80, pdf changes from constant to a downward sloped line. the equation defines what the pdf changes to when 80 reached. this equation necessary to solve problem, so slope is needed to be accurate. 0.01 = c(120-80) c=0.00025 (lines slope) LER = E(X^d) / EX EX is integral x*fx piecewise.. = 50.667 EX^20 = min(x,20) = avg x up to 20, 20 thereafter EX^20 using survival func is = 18 LER = 18/50.667 = .3553

SOA STAM 199

Maximum Likelihood Estimators for me, I get these problems wrong bc algebra rules. how nuts. since we are given only info that 2 exceed 1000, its likelihood is the survival function. ∝θ^−0.8exp[−θ^−0.2(1300.2+2400.2+3000.2+5400.2+2⋅10000.2)] this part makes sense bc we only keep in anything that depends on theta. ALGEBRA REFRESHER when adding the exponents in the natural log that are raised to the same exponent, we must raise each individual constant and then add together. x^k⋅y^k=(xy)^k However, ∑x^2≠(∑x)^2 In other words, 1300.2+2400.2+...+10000.2≠3210^0.2 look this up later. taking u way too long lol. Anyways, doing algebra right and then following rest of steps will lead you to right answer ltheta = -0.8lntheta - 20.2505theta^-0.2 derivative = 0 theta = 3325.67 SHORTCUT: IDK HOW TO PASTE IT JUST SEARCH THIS SOA NUMBER ON CA this can be found on ca summary for MLE section under "Other Shortcuts"

SOA STAM 310 Calculate the percentage reduction in loss costs by moving from a 100 deductible to a 250 deductible.

Other Topics / Loss Elimination Ratio? (i think) make it intuitive. why would we want to solve a problem like this? so, insurers put deductibles on their policies because it could reduce their total costs and also takes trouble away w paperwork for really minor losses. the purpose of comparing reduction in loss costs is to see if there is a huge difference in applying a 250 deductible vs 100 deduc, vs no deduc at all. w/ no deductible, we see the ground-up total losses is 598,500. this is our starting point. In general, we know that total payment/loss amount with a deductible is Y= 0 for x<d and Y= X-d for x>d this can be written as E(X)-E(X^d) where E(X) is 598,500 the idea of the EX-EX^d formula is to remove EX^d for the insurer because the insurer will no longer pay for any losses that are less than d. E(x^100) we see that for losses 0-99, the total losses is 58,500. the rest of these losses that are 100 or greater, we are responsible for paying the xs of 100. the insured will pay 100 for all of these losses. Ex-Ex^100 598500 - [ 58500 + 100(2100 - 1100) ] = 440,000 2100=total number of claims observed 1100 = total number of claims that are < 100 (exclude from product bc those are fully paid for by the insured) E(x^250) same idea. this time, we have 2100-1100-400 = 600 claims that will have $250 covered by insured. insurer responsible for rest. 598,500 - [58500+70000+250(600)] = 320,000 to calculate percentage reduction in increasing deductible, do new-old/old. 27%

SOA STAM 97 A group dental policy has a negative binomial claim count distribution with mean 300 and variance 800. Ground-up severity is given by the following table: You expect severity to increase 50% with no change in frequency. You decide to impose a per claim deductible of 100. Calculate the expected total claim payment after these changes.

Aggregate Loss Models 2 ways to calculate expected total claim payment: 1. E(# losses) * E( payment per loss) 2. E(# payment) * (E payment per payment) increase severity and apply deductible the difference between the two options is that losses less than deductible are viewed as losses but NOT payments. Step 1 - Apply the severity increase Step 2 - Apply the deductible Step 3- Calculate E(payment per loss) 0.25(0+20+80+200) = 75 DONT FORGET THIS: Expected total claim payment = Expected # losses * Expected payment per loss EN = 300 (given) EPayment = 75*300 = 22500 by payemnts: E(# claims) * prop losses resulting in payments 300* .75 .75 = 1-P(no payments) E(# payments) = 225 exp payment/payment = (20+80+200)/3 = 100 notes: * its important to recognize that the deductible is on a per claims basis, not an aggregate basis. *"Expected number of payments is often derived from the number-of-loss distribution by making a coverage modification. Expected payment per loss and expected payment per payment only differ by a factor, which equals the probability of a loss resulting in a payment." - Coaching Actuaries How to solve the way you were trying to do so: E(N)*[EX-EX^100] this set up works because deductible only applied to severity mistake: use Y=1.50X EY-EY^100 EY = 60+120+180+200/4 = 165 EY^100 = 60+100(3)/4 = 90 165-90=75 EN=300 300*75 = 22500

SOA STAM 99 For a certain company, losses follow a Poisson frequency distribution with mean 2 per year, and the amount of a loss is 1, 2, or 3, each with probability 1/3. Loss amounts are independent of the number of losses, and of each other. An insurance policy covers all losses in a year, subject to an annual aggregate deductible of 2. Calculate the expected claim payments for this insurance policy. A

Aggregate Loss Models Step 1 - Define the variables that will be used to solve N= number of losses X= severity of losses S= aggregate losses before deductible Step 2- E(S) ES = EN*EX EN = 2 EX = 2 ES=4 Step 3 - E(S ^ 2) *THIS IS WHERE I GET CONFUSED!!!!* Coach's Remarks on Coaching Actuaries - Discrete Limited Expected Value: E[(X∧u)k]=[∑x≤uxkp(x)]+uk⋅S(u) In words, the expected value of min X and policy limit U is the expected value of all x < u times the probability of that x plus the policy limit for all values the policy limit is reached. This makes sense because a policy limit maxes out the payment for insurer. If policy limit is 100, the insurer will pay 100 for all losses 100 or greater. E(S^2) = ps(0)*(0)+ps1(1)+2*(1-ps0-ps1) S=0 when n=0 because no loss amounts if no loss pn0 = 0.1353 using poisson formula with lambda = 2. this is because frequency is poisson! S=1 when n=1 and x=1 ps(1)=pn1*px1 = (e^-2*2^0/1)*(1/3) = 0.0902 E(s^2) = 0 + 1 * 0.0902 + 2 *(1-.1353-0.0902) = 1.6392 ES-ES^2 = 4-1.6392 = 2.36

SOA STAM 210 Each life within a group medical expense policy has loss amounts which follow a compound Poisson process with λ=0.16λ=0.16. Given a loss, the probability that it is for Disease 1 is 1/16. Loss amount distributions have the following parameters: Disease 1Other diseasesMean per loss510Standard Deviation per loss5020Mean per lossStandard Deviation per lossDisease 1550Other diseases1020 Premiums for a group of 100 independent lives are set at a level such that the probability (using the normal approximation to the distribution for aggregate losses) that aggregate losses for the group will exceed aggregate premiums for the group is 0.24. A vaccine which will eliminate Disease 1 and costs 0.15 per person has been discovered. Define:A = the aggregate premium assuming that no one obtains the vaccine, andB = the aggregate premium assuming that everyone obtains the vaccine and the cost of the vaccine is a covered loss. Calculate A/B.-

Aggregate Loss Models This is difficult for me!! N1- number losses disease 1. There are 100 lives in group N1~poisson(lambda = 100*0.16*1/16 = 1) bc n=100, lambda = 0.16, and there is a 1/16 probability of the loss being for disease 1. S1 - total losses from disease 1 ES1 = lambda* E(x1), x1 is given in table as mean per loss for disease 1 ES1 = 1*5 = 5 VS1 FUN FACT: SIMPLIFY AGG VARIANCE FORMULA W POISSON n is poisson VS= EnVx + VnEx^2 VS = λVx + λEx^2 VS = λ ( Vx + (Ex)^2) By definition of VX, we can conclude: VS = λ(E(X^2)) tbh, i think it was kinda weird to use this in the solution because we were already put in form to use aggregate as is. cool i guess. ANYWAYS VS1 = (1)(50^2) + (1)(5^2) = 2525 now, lets repeat these definitions for other diseases N2~poisson(lambda = 100*0.16*15/16 = 15) ES2~ EN2*EX2 = 15*10 = 150 VS2 = 7500 Now, lets use the A and B definitions. A= no vaccine. If no vaccine, the total losses are gonna be the sum of total losses from 1 and 2. ES = 5+150=155 VS = 10025 Given info: P(S>A) = 0.24 plug in known values and solve for A *Just by this solution so far, I see that we must redefine mean and variance based on the definition of A and B, not directly solving for A and B. Otherwise, there really is no difference bw A and B bc the loss info would be same.* plugging in new mean and var, we get z1=.71. we use this z to solve for a, which is the premium. rearrange: x-u/sigma = z zsigma+u = x .71*rad 10025 + 155 = 226.09 Now, B B has vaccine for disease 1 only redefine S = cost providing vaccine + S2 we only look at the cost bc there will be no losses for D1 if the disease is eliminated entirely. cost vaccine = 100*0.15 = 15 S= 15 + S2 ES = 15 + 150 = 165 VS = 0 + VS2 = 7500 plugging in and using p(s>b)=0.24, solve z2= 0.71 solve B = 226.49 A/B = .998 = 1.00

SOA STAM 212 For an insurance: The number of losses per year has a Poisson distribution with λ=10λ=10. Loss amounts are uniformly distributed on (0, 10). Loss amounts and the number of losses are mutually independent. There is an ordinary deductible of 4 per loss. Calculate the variance of aggregate payments in a year. A

Aggregate Loss Models you got this wrong bc of a totally wrong approach. when we are given that we want variance of aggregate payments, we should convert our N to N* to reflect payments. There is a table on CA formula sheet to represent the conversion from N to N*. Formula: VS = EN*VX+VN*EX^2 however, we are given that there is an ordinary deductible of 4 *per loss* in other words, only x, the loss amount, has the deductible applied. also, n is n* bc number of payments. Updated/actual formula: VS = EN'*V(X-4|X>4) + VN'*E(X-4|X>4)^2 In words, X-4|X>4 translates to the expected loss after the deductible is applied, given that the loss exceeds the deductible. This should make intuitive sense because N does not have a deductible applied. because we have N', apply the coverage modification and multiply by P(X>4). Although I stated that N does not have a deductible applied, the reason we multiply by S4 is because payments will only occur when losses > deductible. N* ~ poisson lambda* = 6 EN* = VN* = 6 Payment per Payment: X-4|X>4 ~ uniform(0,6) 0---4-6---10 deduc 4 applied, ignore everything below 4-6---10 number of terms on this line is 10-4=6 since per payment approach, a loss of 4 will have a payment of 0 bc x-4. hence, x-4|x>4 ~ uniform(0,6) *Refer to special distribution shortcuts on formula sheet* E(X-4|X>4) = 3 V = 3 Apply aggregate variance formula VS = 72 Alternate Setup: Var[S]=E[N]Var[(X−d)+]+Var[N]E[(X−d)+]2

SOA 113

Aggregate Loss Models you need to learn how to think critically you nut!! love u tho you were good up to here: ES=1.4 VS=16.04 SS=4.005 ES+2SS = 1.4+2(4.005) = 9.41 WHAT KEEPS TRICKING YOU: We are given that the only possible benefit amounts are 0 or 10. So, can we actually calculate the probability of the agg benefits > 9.41? technically yes, but 9.41 is IMPOSSIBLE in this case. we can look at P(S>=10) now, this becomes your usual aggregate loss models problem. consider scenarios P(S>=10) = 1 - P(S=0) when is S=0? (think of ALL using the given info that X could be 0 or 10) 1. S=0, N=0 2. S=0, N=1, X=0 3. S=0, N=2, X=0,X=0 4. S=0, N=3, X=0,X=0,X=0 calculate the aggregate probabilities of these 4 scenarios answer is 0.12

SOA STAM 86

Aggregate Loss Models - Aggregate Payments you were on the right track! THEY TRYNA TRICK THO X* - conditional pmt amt given pmt occurs since pmt must occur, X>d, the pmt is uniform bw 0 and c(b-d) (drawing a timeline helps this make sense) (1:06 on CA Video soln helps this) idea- modified frequency N* represents number of losses resulting in payment. the deductible eliminates losses <d N* for poisson = lambda* P(x>d) Sx = 1-FX Sx = b-x/b-a Sd= b-d/b. IDK HOW UGH At 1:52 in ca video, "bc severity is uniform, we know the fraction of claims to eliminate id d/b. after eliminating d/b, we have b-d/b between d and b. the payment follows uniform (0, c(b-d)). When modifying frequency, we need to multiply parameter by the probability of a loss BELOW the deductible. We use the LOSS distribution, which is (0,b). Fd = b/d and Sd = b-d/d "losses above d result in payment" - why we use loss dist.

SOA STAM 5 The annual number of claims for a policyholder has a binomial distribution with probability function: p(x∣q)=(2x)qx(1−q)2−x,x=0,1,2 The prior distribution is: π(q)=4q3,0<q<1 This policyholder had one claim in each of Years 1 and 2. Calculate the Bayesian estimate of the number of claims in Year 3.

Bayesian Credibility Follow the basic steps to solve this My mistake is that I accidentally made (1-q)^4 when it should have been 1-q^3. silly lol. Anyways, also important to understand why we use law of total expectation. We use law of total expectation within the conditional X|data because we do not have unconditional X. law of total expectation is a tool for us to get to unconditional x.

SOA STAM 260 You are given: Claim sizes follow an exponential distribution with mean θθ. For 80% of the policies, θ=8θ=8. For 20% of the policies, θ=2θ=2. A randomly selected policy had one claim in Year 1 of size 5. Calculate the Bayesian expected claim size for this policy in Year 2.

Bayesian Credibility Step 1- Likelihood using experience f(x=5|theta) Step 2- Multiply by prior f(x=5|theta)*pi(theta) (used for numerators of posterior) Step 3- Sum these products to get unconditional probability of data P(x=5) Step 4- Calculate posterior probabilities using the updated "After the fact" observations P(theta=8 | data) = .8670 Step 5- Calculate hypothetical means According to Coaching Actuaries, "The Hypothetical Means are the expected claim size for different values of theta." E(x2|theta=8) = theta = 8 Step 5 - Predicted expected value of next claim size for different values of theta given the prior experien answer is 7.2

SOA STAM 11 Losses on a company's insurance policies follow a Pareto distribution with probability density functionf(x∣θ)=θ(x+θ)2,0<x<∞f(x∣θ)=θ(x+θ)2,0<x<∞ For half of the company's policies, θ=1θ=1, while for the other half, θ=3θ=3. For a randomly selected policy, losses in Year 1 were 5. Calculate the posterior probability that losses for this policy in Year 2 will exceed 8. A

Bayesian Credibility The idea of bayesian credibility is to use historical information to update our probabilities of an event occurring, because we now have more information to assume an event is more/less likely of occurring. each theta has a probability of 0.5 (pi data) start with f(data|theta) multiply these by 0.5 and sum products to get unconditional probability of data unconditional probability of data is what we condition the probability of theta on because 0.5 is no longer accurate given that we know the losses in yr 1 are 5. posterior: f(theta=1|data) = 0.5*(1/1+5^2)/f(data) = 16/43 f(theta=3|d) = 27/43 note how the posterior probabilities sum to 1 because these are weights use tables to get survival pareto multiply these by the appropriate weights product is .2126

SOA STAM 203 You are given: The annual number of claims on a given policy has the geometric distribution with parameter ββ. One-third of the policies have β=2β=2, and the remaining two-thirds have β=5β=5. A randomly selected policy had two claims in Year 1. Calculate the Bayesian expected number of claims for the selected policy in Year 2.

Bayesian Credibility These problems become easier to solve once you intuitively understand the law of total expectation, law of total probability, and posterior probabilities. Our goal: E(x2|x1=2) notice how we do not have unconditional x distribution. we only have E(x|b). This is a sign that we need to use law of total expectation. E(E(x|b) | x1=2) understanding posterior probabilities and why they are used here: we are originally given the probabilities of beta=2 and beta=5. However, we now have the experience that policy had 2 claims in year 1. We can use this information to update our probabilities, because it might be more likely to have b=2 rather than b=5 or vice versa. these posterior probabilities look like p(data | b)*p(b) in words, this translates to the probability of x1=2, given that the policy beta was b. times the likelihood/probability that the policy beta was b. expanded, we are asked to solve E(x2|b=2)*p(b=2|x1=2) + E(x2|b=5)*p(b=5|x1=2) in words, expected value of x2, given that the policy has a beta of 2, times the posterior probability/likelihood of beta=2 being the policy beta, conditioned on the fact that x1=2. My mistake before is that I thought k in pk geometric formula represented moments. THIS IS A PROBABILITY FORMULA. so, P(X=2) = p2. plug 2 into k ex of geom = b use bayes theorem to calculate the posterior probabilities p(b=2 | x1=2) = .39024 p(b=5|x1=2) = .60976 E(x2|x1=2) = 3.83

SOA STAM 45 You are given: The amount of a claim, XX, is uniformly distributed on the interval [0,θ][0,θ]. The prior density of θθ is π(θ)=500θ2,θ>500π(θ)=500θ2,θ>500. Two claims, x1=400x1=400 and x2=600x2=600, are observed. You calculate the posterior distribution as: f(θ∣x1,x2)=3(6003θ4),θ>600f(θ∣x1,x2)=3(6003θ4),θ>600 Calculate the Bayesian premium, E[X3∣x1,x2]E[X3∣x1,x2].

Bayesian Credibility You can save SO much time by trying to see if the posterior distribution resembles any distribution from the table. posterior represents SPP a=3, theta=600 our goal is E(x|d) = E(E(x|theta|d)) E(x|theta) = theta/2 0.5* E(x|d) use posterior tables to calculate E(x|d) = 900 900/2= 450 First principles approach: E(x|theta) = theta/2 E(x|data) = integral of 600 to infinity of E(x|theta)*f(theta|data) wrt theta you can directly solve equals 450 this is just the basic moments formula E(gx) = integral gx*fx dx

SOA STAM 76 You are given: The annual number of claims for each policyholder follows a Poisson distribution with mean θθ. The distribution of θθ across all policyholders has probability density function: f(θ)=θe−θ,θ>0f(θ)=θe−θ,θ>0 ∫∞0θe−nθdθ=1n2∫0∞θe−nθdθ=1n2 A randomly selected policyholder is known to have had at least one claim last year. Calculate the posterior probability that this same policyholder will have at least one claim this year.

Bayesian Credibility biggest mistake i make with these problems is forgetting law of total probability and how it translates into these problems goal: P(N2>0|N1>0) E(P(N2>0 | theta) | N1 > 0 ) think EX = integral x*fx dx P(n2>0|n1>0) = integral P(n2>0|theta) * pi(theta | n1>0) dtheta think pi(theta|n1) as updated posterior probability used as weights. this makes sense when you think of law of total probability and expected values CA Coach response: "Before observations, we use prior probs as weights. after obs, we update to use posterior probs as weights, bc we know more about the observed exposure." main idea- always find way to apply law of total probability when asked to find unconditional probability or unconditional expected value

SOA STAM 29 Each risk has at most one claim each year One randomly chosen risk has three claims during Years 1-6. Calculate the posterior probability of a claim for this risk in Year 7.

Bayesian Credibility Goal- use the data to update the probability of claims for a given selected risk. Step 1- Use the experience to calculate the likelihood. data= 3 claims yrs 1-6 since there are two possible outcomes (claim or no claim), and 6 years observed (max 6 claims 6 years), use binomial dist with the annual claim probability given risk class. P(3|class 1) = (6 ncr 3) (0.1)^3(1-0.1)^3 = 0.01458 Step 2 - Identify the prior probabilities Since our model distribution is X|class, our prior distribution is p(class). prior probabilities are given Step 3- Multiply likelihood and prior and sum to get unconditional probability of the data (law of total probability) P(data) = 0.054238 Step 4- Calculate the posterior probabilities What we have: model dist p(data|class) prior dist p(class) to update prior to posterior, we condition on the experience that has occurred. p(class|data) = p(data|class)*p(class)/p(data) AHHH OK ITS MAKING SENSE!! WE ARE CONDITIONING!! BECAUSE WE KNOW MORE!! :) P(Class 1 | data) = (0.7)(0.01458) / .054238 = .18817 Step 5 - Calculate the posterior probability of 1 CLAIM remember, claim can be 0 or 1. We use p(x=1|class) and multiply by our updated class probability (posterior) posterior prob is 0.28313

SOA STAM 78 You are given: Claim size, XX, has mean μμ and variance 500. The random variable μμ has a mean of 1000 and variance of 50. The following three claims were observed: 750, 1075, 2000 Calculate the expected size of the next claim using Bühlmann credibility.

Buhlmann Credibility This has a zero difficulty on CA lol. This can be solved by using basic formulas. From coaching actuaries: μ+Z(X¯−μ)μ+Z(X¯−μ): Theorethical mean, plus credibility times adjustment. ZX¯+(1−Z)μZX¯+(1−Z)μ: Weighted average between experience and theoretical mean.

SOA STAM 35 Calculate the Bühlmann credibility estimate of the second observation, given that the first observation is 1.

Buhlmann as least squares estimate of bayesian *Must memorize for exam (KEY CONCEPT): buhlmann estimate is the least squares approximation to the bayesian estimate.* this makes sense because the historical data and wording of the problem seemed more like a bayesian problem. Buhl-Bayesian ^2 times probability sum over all obs if we minimize this expression, we get least squares estimate, which will give us buhlmann parameters. Using this fact, we determine that Z and u must be selected to minimize the weighted squared errors. minimize: sum all x(px(Yx-Yhatx)^2) Yx is bayesian estimate x1=x Yhatx is buhlmann estimate x-x we start by calculating the yhatx, buhlmann estimate. zx+(1-z)u set partial derivatives equal to zero to get values, but it should be clear that u is the average of bayesian estimates. u=2 minimize contains only 1 var, Z Z=0.75 plug into cred estimate formula = 1.25 ADVANCED QUESTION!!! this isnt very common :)

SOA STAM 72 Calculate the Bühlmann-Straub credibility premium per employee for this policyholder.

Buhlmann-Straub credibility the hardest part w these problems is understanding the exposure basis and how to calculate x bar and n accordingly. columns = avg loss, rows = year exposure basis: per employee per year when calculating n, we should have 3 numbers of employees summed together because it is a per year basis. (3 YEARS SUMMED TOGETHER TO GET THE SUM OF POLICYHOLDERS) most of the info is given yay! our goal: Pc k=v/a = 200 n= 800+600+400=1800 Z=1800/2000 = 0.9 xbar? 800(15)+600(10)+400(5) / 1800 = 100/9 Pc = 12 CA notes: "For buhlmann-straub cred, n is ALWAYS the number of exposures. Buhlmann n is always equal to the n in the calculation of xbar." overall mean = mean of hypothetical mean xbar = average that is calculated from the experience the idea behind credibility premium is weighted avg between sample mean and overall mean Remember: Buhlmann straub is an application of buhlmann credibility when you do not have the same number of exposures for each period. if u have buhlmann straub, your n will be the total number of POLICYHOLDERS!!!

SOA STAM 23 For a sample of 15 losses, you are given: Losses follow the uniform distribution on (0,θ)(0,θ) Estimate θθ by minimizing the function ∑j=13(Ej−Oj)2Oj∑j=13(Ej−Oj)2Oj, where EjEj is the expected number of losses in the jthjth interval and OjOj is the observed number of losses in the jthjth interval. A

Chi Square THERE IS A SHORTCUT: using the fact that the losses follow a uniform distribution between 0,theta, we can calculate the width of the intervals. Defns: Oj- observed value in jth interval Ej- expected number of claims in interval j we would test this with all answer choices to see smallest theta. Ex) choice a=6, theta=6 losses uniform 0,theta 1/theta-0 = 1/theta = 1/6 = fx Widths: 2-0=2 5-2=3 theta-5 = 6-5 = 1 Ejs E1 = 2/6 * 15 = 5 E2 = 3/6 * 15 = 7.5 E3 = 1/6 * 15 = 2.5 Our width is 1 if we use theta=6. Use these to calculate the values of function at each answer choice this is barnacles

SOA STAM 273 A company has determined that the limited fluctuation full credibility standard is 2000 claims if: The total number of claims is to be within 3% of the true value with probability pp. The number of claims follows a Poisson distribution. The standard is changed so that the total cost of claims is to be within 5% of the true value with probability pp, where claim severity has probability density function: f(x)=110,000,0≤x≤10,000f(x)=110,000,0≤x≤10,000 Using the limited fluctuation credibility, calculate the expected number of claims necessary to obtain full credibility under the new standard.

Classical/Limited Fluctuation Credibility The first line tells us that nc=2000. There is no information about severity, so we only have information on NUMBER of claims. since poisson, sigma^2n/un = =1 2000=(z/k)^2(1) solve z2=1.8 or z=1.34 Next part tells us that the standard changes. new k=.05 same p Now, we have information on severity x~uniform (0,10,000) notice how we are asked to get nc for TOTAL COST OF CLAIMS. total cost of claims tells us we must include both frequency and severity. nc= (1.34/.05)^2(1 + (10,0000^2/12)/5000^2) nc = 960 Wording is tricky here! "Standard is regarding total cost of claims, which means both frequency and severity are considered." "If we were asked for cost of an INVIDIUAL claim, we would just use the new severity info." "If asked only for the number of claims within 5%, we would look at just frequency."

SOA STAM 22 You fit a Pareto distribution to a sample of 200 claim amounts and use the likelihood ratio test to test the hypothesis that α=1.5α=1.5 and θ=7.8θ=7.8. You are given: The maximum likelihood estimates are α^=1.4α^=1.4 and θ^=7.6θ^=7.6. The natural logarithm of the likelihood function evaluated at the maximum likelihood estimates is −817.92−817.92. ∑ln(xi+7.8)=607.64∑ln⁡(xi+7.8)=607.64 Determine the result of the test.

Hypothesis Tests: Likelihood Ratio Background Summary: LRT TS T= 2(l1-l0) l0- log likelihood of null model l1- log likelihood of alternative model (evaluated using MLEs) df = # free par H1 - # free par H0 Step 1- Look at the given information to define the variables l0 = unknown l1 = -817.92 T= 2(-817.92-l0) Step 2 - Solve for l0 using Pareto dist L(a,theta) = product of fx over all 200 data points l(a,thea) = 200lna+200lntheta-(a+1)sumln(xi+theta) l(1.5,7.6) = -821.77 Step 3 - Plug back into T T = 2(-817.92 - (-821.77)) = 7.7 Step 4- Solve for df # freeparH1 = 2 #freeparH0 = 0 df=2-0=2 Step 5- Use chi sq table to solve for the test result df=2, t=3.85 3.85is between alpha=1-.975=.025 and 1-.990=.01 Answer: "Test statistic falls between 97.5 percentile and 99th percentile. Reject the null hypothesis at the 0.025 sig elvel, but not at the 0.010 sig level"' you need to improve with calculus! L(a,th) = product of 200 pareto fx log is 200*ln(a) + 200aln(theta) - (a+1)sum ln(xi+theta) ln(a^200) = 200*ln(a) ln(theta^200a) = 200aln(theta) ln((x+theta)^200(a+1))=(α+1)∑i=1200ln(xi+θ) LOG PROPERTY: ln(pi xi) = sumln(xi) l(a,theta) = ln(product fx) ln(athetaa)-ln(product x+theta^a+1) -(a+1)ln(product x+theta) using ln(pi xi) = sum ln(xi) -(a+1)*summation (x+theta)

Understanding zero-modified & zero truncation Ex) C_TC0042 (zero modified) Claim size, X, follows a zero-modified negative binomial distribution with r=3.5r=3.5 and β=1.5β=1.5. You are given pM0=0.4p0M=0.4. Calculate E[X∧3]E[X∧3].

Insurance Applications According to CA: In the context of this question, we think the negative binomial distribution is a good fit for claim size, but there's a mismatch in the probability at 0 between what we've observed and what the negative binomial distribution implies. Everything else looks good. So, we ignore the probability at 0 that the negative binomial distribution computes, and change it to 0.4. This is the zero-modified probability given in the question. Of course, this means the probabilities for all possible values no longer sum to 1. So, we need to scale the probabilities for all other values (1, 2, ...). We do this by multiplying each probability after zero by 1−pM01−p01−p0M1−p0. Why this factor? The division by 1−p01−p0 scales up the probabilities after zero so that they sum to 1. Then, the multiplication by 1−pM01−p0M scales them back down so that they sum to 1−pM01−p0M. This creates just the amount of room we need for the zero-modified probability. Side note: You can think of the zero-truncated distribution as a special case of the zero-modified where the zero-modified probability is 0 instead of some other number. Then, the factor becomes 11−p011−p0. Zero Truncation Assume probability of zero is impossible scale by factor 1/1-p0 to get the truncated probabilities to sum to 1 Zero Modification Assume probability of zero is a modified probability based on the distribution in table. scale by factor 1-p0m/1-p0 to sum probabilities to 1

SOA STAM 209 In 2005 a risk has a two-parameter Pareto distribution with α=2α=2 and θ=3000θ=3000. In 2006 losses inflate by 20%. An insurance on the risk has a deductible of 600 in each year. PiPi, the premium in year ii, equals 1.2 times the expected claims. The risk is reinsured with a deductible that stays the same in each year. RiRi, the reinsurance premium in year ii, equals 1.1 times the expected reinsured claims. R2005P2005=0.55R2005P2005=0.55 Calculate R2006P2006R2006P2006.

Insurance Applications This one is tricky and you have to be really solid with formulas and concepts! in second line, we are given that there is a deductible. for any d and given severity, E(x-d)+ = EX-EX^d Use pareto in table Ex-d+ = 3000*(3000/3000+d) (My mistake here is that I just assumed the payment P5 would be Ex^600) We also are given that Pi has inflation 1.2*Expected claims Expected claims is Ex-d+ BECAUSE WE ARE LOOKING AT THE EXPECTED CLAIMS WITH THE IMPACT OF THE DEDUCTIBLE!!!! P2005= 1.2*3000*3000/3600 = 3000 R2005/P2005 = 0.55 R2005 = 1650 Coaching Actuaries Solution: "let r denote reinsurer's deductible relative to losses (not relative to reinsured claims). if r=1000, then on a loss of 4000, the insured collects 4000-600=3400 (insured responsible to pay 600 dollar deductible). The primary insurer is responsible to pay for the remaining 400. Another way to express that reinsurance is that the primary company pays the insured 3400. The reinsurer reimburses the primary company for its claims less deductible 400 applied to claims. the reinsurer pays 3400-400=3000." ^?????? kinda makes sense?????????? We have R2005=1650, but we need the formula for R Ri = 1.1*EXPECTED REINSURED CLAIMS i guess we can view it like reinsurer has deductible.... like no coverage comes in until this is reached....???? ERi = 1.1 * 3000* (3000/3000+r) r= 3000 In 2006 after 1.2 inflation, APPLY THE SCALED FACTOR ONLY TO THETA!!!!!!!!!!!!!!!!!!!!!!!!! MEMORIZE FROM FORMULA SHEET ITS A SCALING PROPERTY losses ~ 2 par pareto, a=2 theta=1.2*3000 = 3600 General formula for claims Ex-d+ = 3600(3600/3600+d)) 1.2* that formula using d=600 = 3703 = P6 1.1 * that formula using r=3000 = 2160 = R6 P6/R6 = .5833 WHATTTTTTTTTTTTTTTTTTTT OH Ok "R is the premium the insurer pays to reinsurer" not only does insurer cover losses > d=600, but also is responsible to reach deductible with reinsurer so reinsurer can cover some of these losses. we dont know r, but know it exceeds d. "when loss > r, insurer pays policyholder loss-d and collects loss-r from reinsurer. think of this as capping the payments at r-d" reinsurers expected claims is the expected amount in excess of reinsurer deductible (r). Reinsurance deductible = retention limit

SOA STAM 84 A health plan implements an incentive to physicians to control hospitalization under which the physicians will be paid a bonus BB equal to cc times the amount by which total hospital claims are under 400 (0≤c≤1)(0≤c≤1). The effect the incentive plan will have on underlying hospital claims is modeled by assuming that the new total hospital claims will follow a two-parameter Pareto distribution with α=2α=2 and θ=300θ=300. E[B]=100E[B]=100 Calculate cc.

Insurance Applications/Aggregate Loss Models? By reading this problem, we know that a bonus is only paid if the claims are under 400. Otherwise, what's the point of a bonus? Be careful when reading this. By reading this, we should be able to define bonus as c* the difference between 400 and claim amount. This is because we know we will only receive a bonus if the max claim is <400. B=c(400-x) X<400 THE MAIN IDEA OF THESE PROBLEMS IS TO BE ABLE TO REARRANGE THE EXPRESSION TO GET TO SOMETHING THAT WE RECOGNIZE AND CAN HOPEFULLY USE THE TABLES OR MEMORIZED FORMULAS. Let's demonstrate with a few examples: lets say a claim is $0. Bonus = c(400-0) = c(400) what if claim is $250? Bonus = c(400-250) = c(150) what if claim is 395? Bonus = c(400-395) = c5 Okay! Let's view this like deductibles in the perspective of the insured. If you have insurance coverage with a deductible of 100, you know that your max payment will be 100. You also know that for any losses <=100, you will be responsible to pay. Hence, your payment is 100-x for any x 0<x<100 In math terms, our payment is E(100) - E(x^100) Let's take this knowledge and apply it to our example! We know bonus is max 400c. We also know that any claims 0<x<400 will reduce our bonus. Therefore, our bonus is c(400-x) c(400-E(x^400)) for any claim 0<x<400 We are given E(B)=100. Use this to solve for c E(B) = c[ E(400) - E(X^400)] 100 = c[ 400 - E(x^400) ] Use pareto table 100=c (400- 171.43) c=100/228.57 = 0.44

SOA STAM 180 The time to an accident follows an exponential distribution. A random sample of size two has a mean time of 6. Let YY denote the mean of a new sample size of two. Calculate the delta method approximation of the variance of the maximum likelihood estimator of FY(10)FY(10).

Maximum Likelihood Estimators Special Cases for Exponential/Gamma For x~gamma where a is fixed, MLE of theta is sample mean divided by alpha. (log-likelihood derivation is on coaching actuaries derivations tab) we are given that a random sample size of two has a mean time of 6. x bar definition is x1+x2/2. we use this language to define x bar = 6 = MLE we are also given that Y is the mean of a new sample size of 2. Y= x3+x4/2 x3 and x4 measure time to accidents of exponential. we dont know the times yet, so we cant determine the mean. rearrange: 2Y=X3+x4 MUST REMEMBER: "all continuous distributions listed on the exam table, except for lognormal, inverse gaussian, log-t are parameterized such that theta is a SCALE PARAMETER. When a RV is scaled, the new RV will follow the same distribution with the same parameters except THETA will be scaled by the scaling factor." Since 2y=x3+x4, and 2y is a sum of expoentials, we can conclude 2y follows a gamma distribution!! 2y~gamma a=2, theta=theta scale by factor 1/2 to get y distribution y~gamma a=2, theta* = theta/2 *THIS IS THE PART YOU MESS UP. MAKE SURE YOU UNDERSTAND THE SCALING* We are asked to solve for Variance of Fy10 using delta method. we can utilize the poisson shortcut! 1- P(N<2) = 1-P(N<a) x=x/theta* x=10 x=10/theta/2 x=20/theta ^use this x in the poisson shortcut Fy10 = 1-P(n<2) plug in 1 and 0 and remove from 1 using x=20/theta Fy10 = 1-(1+20/theta)*e^-20/theta ^THIS IS THE MLE WE JUST CALCULATED Last step, take the variance of this MLE. Key formula: Var[g(θ^)]Varˆ[g(θ^)]=[g′(θ)]2Var[θ^] where gtheta = Fy10 solve and get .0786 PRACTICE THIS PLEASE!!!

SOA STAM 196 You are given the following 20 bodily injury losses (before the deductible is applied): Past experience indicates that these losses follow a Pareto distribution with parameters αα and θ=10,000θ=10,000. Calculate the maximum likelihood estimate of αα. A

Maximum Likelihood Estimators This problem requires both truncation and censoring. the first loss of 750 has a deductible of 200. In math terms, we know that a loss of x=750 occurs, but we need to condition that on the fact that the loss exceeds the deductible. Left truncation at d means the dataset does not include data below d. We have to condition the likelihood of every observation in this dataset being above d. We do the same conditioning for the loss of 400 with the deductible of 300. 200 and 300 have a policy limit that has not been reached. we do not need to truncate or censor. the loss > 10,000 needs to be right censored because all we know is that the policy limit of 10,000 has been met. in other words, we dont know if the loss is 10,001 or 20,000, we just know that the payment is capped at 10,000. to right censor, just multiply by S(10,000). Right Censored at 10,000 means that the value is known to be at least 10,000, but is only recorded as 10,000. Likelihood by definition is S10000. Right censoring is common with policy limits. La = product of the likelihoods raised to the power of the number of losses experienced. take log likelihood, derivative, set equal to zero solve a = 3.089 It is the calculus, algebra, and log rules that really gets to you La, how can we simplify this? it is best to plug in the numbers right away into the variables. α^14*10,200^−3*10,000^13α*10,300^−4*10,750^−3α−3 *20,000^−6α*10,400^−4α−4 anything that is like 10750^a+1 in denominator is proportional to just 10750^a bc we can write it as 10^a+1= 10^a*10^1 and we only want things wrt a in our La. the 10200 and 10300 in numerator comes from the fraction of dividing by Sx. Like exponents, so bases are combined and added. 10,750^-3a-3 comes from the fact that denominator is raised to a+1 and is within parentheses that are raised to 3 power. this makes sense. looks like numerators were grouped first, and then multiplied by denominators to negative power. constants removed. JUST BE CAREFUL W THESE!! if these come on exam, which they will, probably best to save them for last so ur not messy and wasting time.

SOA STAM 179 The time to an accident follows an exponential distribution. A random sample of size two has a mean time of 6. Let YY denote the mean of a new sample of size two. Calculate the maximum likelihood estimate of Pr(Y>10)

Maximum Likelihood Estimators (& Poisson Shortcut!) MUST RECALL: 1. Sum of exponential RVs with the same mean is gamma, where alpha is the number of exp RV in the sum and theta is the theta from the source exponential. 2. Poisson Shortcut can be used when alpha is a POSITIVE INTEGER. Gamma(a;x) = 1 - P(N<a) This is useful because incomplete gamma function is messy and hard to integrate Step 1- Find theta hat of the given info. x1 and x2 both exp theta=6. x1+x2 = gamma w alpha=2 and theta=6 its helpful if we define w=x1+x2 w/2 = xbar in other words, the sample mean of the two exponential random variables is xbar. Since we are solving for Y, which is the mean of a new sample size of 2, we can isolate Y to identify the distribution. Y=W/2 If w is gamma w a=2 and theta=6 then Y is also gamma w a=2 and theta= 6/2 = 3 (theta is the theta of source) (still kinda confused...) Anyways, knowing we have gamma, either integrate incomplete gamma function or use poisson shortcut poisson shortcut: N~poisson (lambda=x/theta) 10/3 = lambda p(y>10) = 1-gamma(2;3) 1-(1-p(n<2)) .1546

SOA STAM 194 You are also given a^=651.03a^=651.03. Calculate the nonparametric empirical Bayes credibility factor for Group 1.

Non-parametric Empirical Bayes Step 1- Check if exposures are uniform Since the claims have not been captured for the same amount of years between the two groups, we conclude non-uniform exposures. One has data in yr 3 and not yr 1, the other has 1 and not 3. Step 2- Define variables and see what information is already given (This could save a lot of time!!) x11=0 x12=200 x13=250 x21=160 x22=200 m12=50 m13=60 m21=100 m22=90 x1=227.27 m1=110 n1=2 (2 out of 3 years observed) x2=178.95 m2=190 n2=2 x=196.67 m=300 r=2 #groups a=651.03 (given) Step 3- Solve for v using non-uniform formula to solve for k=v/a. *Hint- I like to do this using data function in ti30xs by setting x1=x and x2=y. For v, we divide by (2-1)+(2-1)=2 v=71,985.647 k=v/a = 110.57 Step 4- Plug into z=n/n+k n1= 110 110/(110+110.57)= 0.5!

SOA STAM 123 Annual prescription drug costs are modeled by a two-parameter Pareto distribution with θ=2000θ=2000 and α=2α=2. A prescription drug plan pays annual drug costs for an insured member subject to the following provisions: The insured pays 100% of costs up to the ordinary annual deductible of 250. The insured then pays 25% of the costs between 250 and 2250. The insured pays 100% of the costs above 2250 until the insured has paid 3600 in total. The insured then pays 5% of the remaining costs. Calculate the expected annual plan payment.

ey Insurance Applications idea is to get this into a piecewise function that allows you to use the tables. Y= pmt from insured Y = x when x< 250 (insurer ignores losses < 250) Y= 250 + 0.25(X-250) *THIS IS WHERE I GET STUCK. ITS PIECEWISE! Take the extra time to go "when x=250, the insured will pay 250. When x=255, insured will pay 250 + 0.25(5). Hence, we define Y as 250 + 0.25(x-250). this is the equation used if the loss happens to exceed 250.* Y=750+1(X-2250) 750 comes from 250 from the first layer and including the max insured payment from a loss of 2250 y= 250 + 0.25(2250-250) = 750. *this is a cumulative payment, so 750 included in next layer.* y= 3600 + 0.05(x-5100) * we are given that 100% of costs are paid until insured max payment of 3600 is reached.* where does 5100 come from? 5100 is the point at which the insured has paid 3600 in total. 750 is the payment that has been made by insured for losses up to 2250. after 2250, we are told that the insured is responsible for 100% until 3600 is reached. max payment = payment made to date + (m - 2250), where m is the max loss that has to be incurred for the max payment of 3600 to be met. m=5100 *Key Fact: the sum of payments from insured and insurer must equal total loss* next steps: re-write piecewise func in terms of insurer payment Y = X - X Y=X- (250+0.25(X-250)) Y=X-(750+(X-2250)) Y=X-(3600+0.05(X-5100)) Get these into a form with a lot of manipulation and removal of coinsurance to get in terms of limited loss variables =0.75[(X∧2250)−(X∧250)]+0.95[X−(X∧5100)] ugh wtf tho lol answer is 1162.6