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ShortTerm Actuarial Math Old Exam C Forum 

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#1




Exam C Calculator Tips
I was hoping to get an organized thread going about how the TI30X calculator's table function can be used advantageously (saving time or simplifying calculations) on this exam.
What are specific topics this function is specifically helpful for? How can it be applied to topics such as credibility? Feel free to note anything that comes to mind. If possible, please point to examples from the SOA 306 for practice. Thanks! 
#3




Topics I'm fluent with on my TI30XSII
ChiSquared tests
KolmogovSmirnov test Discrete Bayesian Least Squares Credibility
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Introductory I: Introductory II: Actuarial: Advanced:FAP, PA Seminar: APC 
#5




Quote:
ChiSquared: Put observations in L1, put empirical values in L2, use formula to calculate F*fitted CDF in L3, but then press enter 2x in L3 on a value to copy/paste which removes the formula. Go back to L1 and enter the formula (L2L3)^2/L3 which is (OjEj)^2/Ej. Use stats 1VAR with L1 data, freq=one and find #5 Sum X. This is the ChiSquared stat. By hand you can use one of two formulas: SUM(OjEj)^2/Ej = SUM(Oj)^2/Ejn If there are independent years, we have a different formula by hand: SUM(OjEj)^1/Var(Oj) #1 i.e. When we have Poisson, Ej=mean=Lambda*exposures=Var(Oj) #2 i.e. When we have Binomial, Ej=mean=mq*exposures, but Var(Oj)=mq(1q)*exposures. ***NOTE***degrees of freedom is kp where p=number of estimated parameters and is ***NOT** kp1 as in the standard case for degrees of freedom. KolmogorovSmirnov: (remember you need individual data to use this test) Enter all data observations two times each, but don't repeat a value more than once, so only unique values  all in L1. Then in L2 enter 0, 1/n, 1/n, 2/n, 2/n, ... (k1)/n, (k1)/n, 1. This can get a little trickier, so I'll try to explain. You'll enter 0 for the first data pt. and 1 for the last, but only a single time. All other values of this empirical CDF (that's what this column is) are entered twice unless there's multiple values. i.e. 1 2 3 3 4 there's 2 three's, so we'll have to make an alteration for it and similar methods are used for more then 2. There's 5 data pts. so empirical CDF is k/5 L1 = 1 1 2 2 3 3 4 4 (always enter each unique data pt. only twice) L2 = 0 .2 .2 .4 .4 .8 .8 1 (notice I skipped a value because if you were to graph it you'd see there was twice the jump as the rest, so I added an extra .2 to do that. L3= use a formula for fitted CDF to calculate F*(L1), but then also copy/paste values by pressing enter twice on one value in L3. This removes the formula and just copy/paste numbers as values. revise L1: ((L2L3)2)^.5, so SQRT( (L2L3)^2 ). Yes, two sets of parentheses because you have to square the difference before taking the SQRT, so we get only positive values of L2L3. You can find the max by inspection, but I used 1VAR stats of L1, freq=one and press up arrow once to get the last bottom or last statistics in the list (MAX). select MAX to your home screen and multiply by SQRT(N). This is your test statisitic=MAX*SQRT(n). Calculations by hand are the same as on TI, but do it all by hand. Discrete Bayesian: NLambda is Poisson with NLambda being annual claims. For prior distribution of Lambda: P(Lambda1)=.4 & P(Lambda2)=.6 with the observation that N1=3, so 3 claims made in first year. L1 = Enter known mean from the prior and prior is Lamba1=2 Lambda2=4 with SO, enter 2 4 L2 = Prior probabilities, so enter .4 .6 L3 will be the joint probabilities. Enter a formula=L1^3*exp(L1)/3!*L2. 1VAR status #5 (Sum X) is the P(N=3). For E(N2N1=3) we use 1VAR stats with data=L1 (where we put the means) and freq=L3 where the joint probabilities are at, but you might object because L3 doesn't sum to 1. You would be correct, but the beautiful thing about the TI30XSII is that it renormalizes everything, so #2=average will give you the predictive mean of E(N2N1). I you needed E(3NN1=3), which is the expected number over the next 3 periods, you'd multiply it by 3. You can also find higher moments by using #4  sigma^2 for variance with same L1=data, L3=freq. The tricky thing is you need to calculate using conditional expected moments. #1 i.e. what's the variance of next years' claims? ** use conditional variance formula: Var(N2N1=3)=E[Var(N2N1=3)]+Var[E(N2N1=3)] =E(LambdaN1=3)+Var(LambdaN1=3). Where these values are from 1VAR stats with data=L1/freq=L3, namely #2average and #4^2empirical variance #2 i.e. Given q, m=3 N is Binomialq and is monthly claims where q~(q1=.2, q2=.4, q3=.7 with P(q1)=.5, P(q2)=.3, P(q3)=.2. Observation was 2 claims last month. What's the variance of next month's claims? L1=q1 q2 q3 (.2 .4 .7) L2=Priors (.5 .3 .2) L3=joint probabilities  enter formula: (3 nCr 2)*L1^2*(1L1)*L2 Answer to question, what's variance of next month's expected claims? Var(N2N1=2)=E[Var(N2N1=2)] + Var[E(N2N1=2)] =E[3q(1q)N1=3] + Var[3qN1=3] =3E[q  q^2N1=3] + 9Var(qN1=3) =3E[qN1=3]  3E[q^2N1=3] + 9Var(qN1=3) The first and third terms are automatically calculated with 1VAR stats: 3* #2average is first term, 9* #4^2empirical variance is third term. To get second term, use 3*(#2)^2+(#4)^2=3*second moment. Calculations by hand should be written out in some tabular form with good organization to reduce errors: #1 i.e. cases: Lamda1=2 Lambda2=4 Prior Prob: P(Lambda1)=.4 P(Lambda2)=.4 Joint Prob: f(N=n1)*P(Lambda1) f(N=n1)*P(Lambda2)  Total the two here =========================================== Posterior Prob: Joint Prob(1)/Total Joint Prob(2)/Total = 1Joint Prob(1)/Total Predictive/E[N2N1] = Lambda1*Posterior1 + Lambda2*Posterior2 Least squares credibility: given X values, Pr(X) and Bayesian Estimates X={0 2 8}, P(0)=.5, P(2)=.25, P(8)=.25. E(X2X1)={1,2,6}. find Pc(8), the Bulhmann estimate of X=8. L1=X's  values need to be uniformly distributed, so it's possible SOA gives values that can't be converted to have common denominators, but highly unlikely IMHO and Dave from TIA conferred. Basically, we have X=0 is .5 where X=2 and X=8 is .25, so everything needs to be .25. Enter 0 twice as follows: L1 = 0 0 2 8 L2 = 1 1 2 6. (note I added a double first Bayesian to match the double X value) Use 2VAR stats, Xvalues L1, Yvalues L2. a=Z credibility factor b=Pc(0), alternatively, it's b in this "Least Squares Approximation" formula: Pc(X)=aX+b  this gives all Bulhmann credibility calculations using the "Least Squares" approach. You can generate all values in L3 as such: L3=(formula) a*L1+b 2 approaches by hand: #1 i.e. Z=Cov(Bayes,X)/Var(X)=a/(a+v) *These are from Credibility notation =E(Bayes*X)(EX)^2 = [X1*Bayes(1)*P(X1)+...+Xn*Bayes(n)*P(Xn)  (average of X)^2]/Var(X) a=Z b=Pc(0)=(1Z)average(X) Pc(Xi)=a*Xi + b #2 i.e. X={1 3 5} P(X=1)=.25, P(X=3)=.5, P(X=5)=.25, Bayes={1.2, 3.4, 4.0} Goal is to minimize g(Z)=E(Bulhmann  Bayes)^2. It's not as hard as it looks: Find overall u=EX g(Z)=E(BulhmannBayes)^2 = SUM[Pc(Xi)Bayes]^2/P(Xi) (u+Z(X1u)Bayes(1))^2/P(X1)+...+(u+Z(X3u)Bayes(3)/P(X3) *TIP* simplify (uZu) within each numerator, then take the derivative. It's a simple Chain rule with only power 2 and linear Z. Turns out that Z in the middle case usually zero out and no need to differentiate zero because even I can do that, easy, ZERO g'(Z)=0 Set the derivative equal to zero and solve for Z. It's a linear equation now because it was only 2nd power to begin with. This is the value of a=Z Estimate Bulhmann by hand now with Pc(Xi)=aX + b where b=(1Z)EX, but that's too difficult, just apply the stand Bulhmann: Pc(X)=u+Z(Xu)
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Introductory I: Introductory II: Actuarial: Advanced:FAP, PA Seminar: APC Last edited by dkamka; 12142017 at 10:12 AM.. 
#7




I figured it out! and thought I'd share. Using the table function, I input E(X) in L1. The E(X^D) section will always be 100. In L2, I input 100/L1. In L3, enter (L2.125)^2. Then hit second stat, 1  Var Statistics, select Data L3 and FRQ One and calc, find Xbar is .0029.

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