Tables for Exam STAM - SOA

1 Tables for exam stam The reading material for exam stam includes a variety of textbooks. Each text has a set of probability distributions that are used in its readings. For those distributions used in more than one text, the choices of parameterization may not be the same in all of the books. This may be of educational value while you study, but could add a layer of uncertainty in the examination. For this latter reason, we have adopted one set of parameterizations to be used in examinations. This set will be based on Appendices A & B of Loss Models: From Data to Decisions by Klugman, Panjer and Willmot.

2 A slightly revised version of these appendices is included in this note. A copy of this note will also be distributed to each candidate at the examination. Each text also has its own system of dedicated notation and terminology. Sometimes these may conflict. If alternative meanings could apply in an examination question, the symbols will be defined. For exam stam , in addition to the abridged table from Loss Models, sets of values from the standard normal and chi-square distributions will be available for use in examinations. These are also included in this note. When using the normal distribution, choose the nearest z-value to find the probability, or if the probability is given, choose the nearest z-value.

3 No interpolation should be used. Example: If the given z-value is , and you need to find Pr(Z < ) from the normal distribution table , then choose the probability for z-value = : Pr(Z < ) = When using the normal approximation to a discrete distribution, use the continuity correction. The density function for the standard normal distribution is 2121()2xxe =. Excerpts from the Appendices toLoss Models: From Datato Decisions, 5th editionSeptember 12, 2018 Appendix AAn Inventory of Continuous incomplete gamma function is given by ( ;x) =1 ( ) x0t 1e tdt, >0, x >0,with ( ) = 0t 1e tdt, > , defineG( ;x) = xt 1e tdt, x > times we will need this integral for nonpositive values of . Integration by parts producesthe relationshipG( ;x) = x e x +1 G( + 1;x).

4 This process can be repeated until the first argument ofGis +k, a positive it can be evaluated fromG( +k;x) = ( +k)[1 ( +k;x)].The incomplete beta function is given by (a,b;x) = (a+b) (a) (b) x0ta 1(1 t)b 1dt, a >0, b >0,0< x < Beta Generalized Pareto , , f(x) = ( + ) ( ) ( ) x 1(x+ ) + , F(x) = ( , ;u), u=xx+ ,E[Xk] = k ( +k) ( k) ( ) ( ), < k < ,E[Xk] = k ( + 1) ( +k 1)( 1) ( k)ifkis a positive integer,E[(X x)k] = k ( +k) ( k) ( ) ( ) ( +k, k;u) +xk[1 F(x)], k > ,Mode= 1 + 1, >1, else Burr , , (Burr Type XII, Singh Maddala)f(x) = (x/ ) x[1 + (x/ ) ] +1, F(x) = 1 u , u=11 + (x/ ) ,VaRp(X) = [(1 p) 1/ 1]1/ ,E[Xk] = k (1 +k/ ) ( k/ ) ( ), < k < ,E[(X x)k] = k (1 +k/ ) ( k/ ) ( ) (1 +k/ , k/.)

5 1 u) +xku , k > ,Mode= ( 1 + 1)1/ , >1,else Inverse Burr , , f(x) = (x/ ) x[1 + (x/ ) ] +1, F(x) =u , u=(x/ ) 1 + (x/ ) ,VaRp(X) = (p 1/ 1) 1/ ,E[Xk] = k ( +k/ ) (1 k/ ) ( ), < k < ,E[(X x)k] = k ( +k/ ) (1 k/ ) ( ) ( +k/ ,1 k/ ;u) +xk[1 u ], k > ,Mode= ( 1 + 1)1/ , >1,else Pareto , (Pareto Type II, Lomax)f(x) = (x+ ) +1, F(x) = 1 ( x+ ) ,VaRp(X) = [(1 p) 1/ 1],E[Xk] = k (k+ 1) ( k) ( ), 1< k < ,E[Xk] = kk!( 1) ( k),ifkis a positive integer,E[X x] = 1[1 ( x+ ) 1], 6= 1,E[X x] = ln( x+ ), = 1,TVaRp(X) = VaRp(X) + (1 p) 1/ 1, >1,E[(X x)k] = k (k+ 1) ( k) ( ) [k+ 1, k;x/(x+ )]+xk( x+ ) , k > 1, k6= ,E[(X x) ] = (xx+ ) [1 + n=0[x/(x+ )]n+1 +n+ 1],Mode= Inverse Pareto , f(x) = x 1(x+ ) +1, F(x) =(xx+ ) ,VaRp(X) = [p 1/ 1] 1,E[Xk] = k ( +k) (1 k) ( ), < k <1,E[Xk] = k( k)!

6 ( 1) ( +k)ifkis a negative integer,E[(X x)k] = k x/(x+ )0y +k 1(1 y) kdy+xk[1 (xx+ ) ], k > ,Mode= 12, >1,else Loglogistic , (Fisk)f(x) = (x/ ) x[1 + (x/ ) ]2, F(x) =u, u=(x/ ) 1 + (x/ ) ,VaRp(X) = (p 1 1) 1/ ,E[Xk] = k (1 +k/ ) (1 k/ ), < k < ,E[(X x)k] = k (1 +k/ ) (1 k/ ) (1 +k/ ,1 k/ ;u) +xk(1 u), k > ,Mode= ( 1 + 1)1/ , >1,else Paralogistic , This is a Burr distribution with = .f(x) = 2(x/ ) x[1 + (x/ ) ] +1, F(x) = 1 u , u=11 + (x/ ) ,VaRp(X) = [(1 p) 1/ 1]1/ ,E[Xk] = k (1 +k/ ) ( k/ ) ( ), < k < 2,E[(X x)k] = k (1 +k/ ) ( k/ ) ( ) (1 +k/ , k/ ; 1 u) +xku , k > ,Mode= ( 1 2+ 1)1/ , >1,else Inverse Paralogistic , This is an inverse Burr distribution with =.

7 F(x) = 2(x/ ) 2x[1 + (x/ ) ] +1, F(x) =u , u=(x/ ) 1 + (x/ ) ,VaRp(X) = (p 1/ 1) 1/ ,E[Xk] = k ( +k/ ) (1 k/ ) ( ), 2< k < ,E[(X x)k] = k ( +k/ ) (1 k/ ) ( ) ( +k/ ,1 k/ ;u) +xk[1 u ], k > 2,Mode= ( 1)1/ , >1,else Gamma Gamma , (When =n/2 and = 2, it is a chi-square distribution withndegrees of freedom.)f(x) =(x/ ) e x/ x ( ), F(x) = ( ;x/ ),E[Xk] = k ( +k) ( ), k > ,E[Xk] = k( +k 1) ifkis a positive integer,E[(X x)k] = k ( +k) ( ) ( +k;x/ ) +xk[1 ( ;x/ )], k > ,E[(X x)k] = ( + 1) ( +k 1) k ( +k;x/ )+xk[1 ( ;x/ )] ifkis a positive integer,M(t) = (1 t) ,t <1/ ,Mode= ( 1), >1,else Inverse Gamma , (Vinci)f(x) =( /x) e /xx ( ), F(x) = 1 ( ; /x),E[Xk] = k ( k) ( ), k < ,E[Xk] = k( 1) ( k)ifkis a positive integer,E[(X x)k] = k ( k) ( )[1 ( k; /x)] +xk ( ; /x)= kG( k; /x) ( )+xk ( ; /x),allk,Mode= /( + 1).

8 Weibull , f(x) = (x/ ) e (x/ ) x, F(x) = 1 e (x/ ) ,VaRp(X) = [ ln(1 p)]1/ ,E[Xk] = k (1 +k/ ), k > ,E[(X x)k] = k (1 +k/ ) [1 +k/ ; (x/ ) ] +xke (x/ ) , k > ,Mode= ( 1 )1/ , >1,else Inverse Weibull , (log-Gompertz)f(x) = ( /x) e ( /x) x, F(x) =e ( /x) ,VaRp(X) = ( lnp) 1/ ,E[Xk] = k (1 k/ ), k < ,E[(X x)k] = k (1 k/ ){1 [1 k/ ; ( /x) ]}+xk[1 e ( /x) ],= kG[1 k/ ; ( /x) ] +xk[1 e ( /x) ],allk,Mode= ( + 1)1/ . Exponential f(x) =e x/ , F(x) = 1 e x/ ,VaRp(X) = ln(1 p),E[Xk] = k (k+ 1), k > 1,E[Xk] = kk! ifkis a positive integer,E[X x] = (1 e x/ ),TVaRp(X) = ln(1 p) + ,E[(X x)k] = k (k+ 1) (k+ 1;x/ ) +xke x/ , k > 1,E[(X x)k] = kk! (k+ 1;x/ ) +xke x/ ifk > 1 is an integer,M(z) = (1 z) 1,z <1/ ,Mode = Inverse Exponential f(x) = e /xx2, F(x) =e /x,VaRp(X) = ( lnp) 1,E[Xk] = k (1 k), k <1,E[(X x)k] = kG(1 k; /x) +xk(1 e /x),allk,Mode = Lognormal , ( can be negative)f(x) =1x 2 exp( z2/2) = (z)/( x), z=lnx ,F(x) = (z),E[Xk] = exp(k +12k2 2),E[(X x)k] = exp(k +12k2 2) (lnx k 2 )+xk[1 F(x)],Mode = exp( 2).

9 Inverse Gaussian , f(x) =( 2 x3)1/2exp( z22x), z=x ,F(x) = [z( x)1/2]+ exp(2 ) [ y( x)1/2], y=x+ ,E[X] = ,Var[X] = 3/ ,E[Xk] =k 1 n=0(k+n 1)!(k n 1)!n! n+k(2 )n, k= 1,2,.. ,E[X x] =x z [z( x)1/2] yexp(2 / ) [ y( x)1/2],M(z) = exp[ (1 1 2 2 z)],z < 2 Log-t r, , ( can be negative) LetYhave atdistribution withrdegrees of freedom. ThenX=exp( Y+ ) has the log-tdistribution. Positive moments do not exist for this as thetdistribution has a heavier tail than the normal distribution, this distributionhas a heavier tail than the lognormal (x) = (r+ 12)x r (r2)[1 +1r(lnx )2](r+1)/2,F(x) =Fr(lnx )withFr(t) the cdf of atdistribution withrdf,F(x) = 12 r2,12;rr+(lnx )2 ,0< x e ,1 12 r2,12;rr+(lnx )2 , x e.

10 Single-Parameter Pareto , f(x) = x +1, x > , F(x) = 1 ( x) , x > ,VaRp(X) = (1 p) 1/ ,E[Xk] = k k, k < ,E[(X x)k] = k k k ( k)x k, x , k6= ,E[(X x) ] = [1 + ln(x/ )],TVaRp(X) = (1 p) 1/ 1, >1,Mode= .Note:Although there appear to be two parameters, only is a true parameter. Thevalue of must be set in with Finite SupportFor these two distributions, the scale parameter is assumed Generalized Beta a,b, , f(x) = (a+b) (a) (b)ua(1 u)b 1 x,0< x < , u= (x/ ) ,F(x) = (a,b;u),E[Xk] = k (a+b) (a+k/ ) (a) (a+b+k/ ), k > a ,E[(X x)k] = k (a+b) (a+k/ ) (a) (a+b+k/ ) (a+k/ ,b;u) +xk[1 (a,b;u)]. Beta a,b, The case = 1 has no special name but is the commonly used version of this (x) = (a+b) (a) (b)ua(1 u)b 11x,0< x < , u=x/ ,F(x) = (a,b;u),E[Xk] = k (a+b) (a+k) (a) (a+b+k), k > a,E[Xk] = ka(a+ 1) (a+k 1)(a+b)(a+b+ 1) (a+b+k 1)ifkis a positive integer,E[(X x)k] = ka(a+ 1) (a+k 1)(a+b)(a+b+ 1) (a+b+k 1) (a+k,b;u)+xk[1 (a,b;u)].