Multiple Decrement Models

1 Multiple Decrement ModelsLecture: Weeks 8-9 Lecture: Weeks 8-9 (STT 456) Multiple Decrement ModelsSpring 2015 - Valdez1 / 25 Multiple Decrement modelsLecture summaryMultiple Decrement model - expressed in terms of Multiple statemodelMultiple Decrement Tables (MDT)several causes of decrementprobabilities of decrementforces of decrementThe Associated Single Decrement Tables (ASDT)Uniform distribution of decrementsin the Multiple Decrement contextin the associated single Decrement contextChapter 8 (DHW), Sections : Weeks 8-9 (STT 456) Multiple Decrement ModelsSpring 2015 - Valdez2 / 25 Multiple Decrement Models examplesExamples of Multiple Decrement modelsMultiple Decrement Models are extensions of standard mortalitymodels whereby there is simultaneous operation of several causes life fails because of one of these include: life insurance contract is terminated because of death/survival orwithdrawal (lapse).

2 An insurance contract provides coverage for disability and death, whichare considered distinct insurance contract pays a different benefit for different causes ofdeath ( accidental death benefits are doubled).pension plan provides benefit for death, disability, employmenttermination and : Weeks 8-9 (STT 456) Multiple Decrement ModelsSpring 2015 - Valdez3 / 25 Multiple Decrement Models notationIntroducing notationageno. of livesheart diseaseaccidentsother causesx`( )xd(1)xd(2)xd(3)x504,832,5555,1681,1574, 293514,821,9275,3631,2065,162524,810,206 5,6181,4435,960534,797,1855,9291,6796,84 0544,782,7276,2772,1527,631 Conventional notation.

3 `( )xrepresents the surviving population present at exact (j)xrepresents the number of lives exiting from the population betweenagesxandx+ 1due to is also conventional to denote the total number of exits by all modesbetween agesxandx+ 1byd( ) ( )x=m j=1d(j)xwheremis the total number of possible decrements, and therefore,d( )x=`( )x `( )x+ : Weeks 8-9 (STT 456) Multiple Decrement ModelsSpring 2015 - Valdez4 / 25 Multiple Decrement Models probabilitiesProbabilities of decrementThe probability that a life (x)will leave the group within one year asa result of decrementj:q(j)x=d(j)x/`( ) probability that(x)will leave the group (regardless ofdecrement):q( )x=d( )x/`( )x=m j=1d(j)x/`( )x=m j=1q(j) probability that(x)will remain in the group for at least one year:p( )x= 1 q( )x=`( )x+1/`( )x= (`( )x d( )x)/`( ).

4 Weeks 8-9 (STT 456) Multiple Decrement ModelsSpring 2015 - Valdez5 / 25 Multiple Decrement Models probabilities- continuedWe also have the probability of remaining in the group afternyearsp( )n x=`( )x+n/`( )x=p( )x p( )x+1 p( )x+n the complementq( )n x= 1 p( )n number of failures due to decrementjover the interval(x,x+n]isd(j)n x=n 1 t=0d(j)x+ relationships should be straightforward to follow:d(j)n x=`( )x q(j)n xd( )n x=`( )x q( )n xLecture: Weeks 8-9 (STT 456) Multiple Decrement ModelsSpring 2015 - Valdez6 / 25 Multiple Decrement Models MDTI llustration of Multiple Decrement TableExpand Multiple Decrement Table (MDT) into:x`( )xd(1)xd(2)xd(3)xq(1)xq(2)xq(3)xq( )xp( )x504,832,5555,1681,1574, ,821,9275,3631,2065, ,810,2065,6181,4435, ,797,1855,9291,6796, ,782,7276,2772,1527.)

5 Weeks 8-9 (STT 456) Multiple Decrement ModelsSpring 2015 - Valdez7 / 25 Multiple Decrement Models illustrative problemsIllustrative problemsUsing the previously given Multiple Decrement table, compute andinterpret the following:1d(3)2 512p( )3 503q(1)2 534q(2)2|2 50 Lecture: Weeks 8-9 (STT 456) Multiple Decrement ModelsSpring 2015 - Valdez8 / 25 The continuous case force of decrementTotal force of decrementThe total force of Decrement at agexis defined as ( )x= limh 01hq( )h x= 1`( )xddx`( )x= ddxlog`( )xTherefore, analogous to the single Decrement table, we havep( )t x= exp ( t0 ( )x+sds)andq( )x= 10p( )s x ( )x+sdsor, more generallyq( )t x= t0p( )s x ( )x+.

6 Weeks 8-9 (STT 456) Multiple Decrement ModelsSpring 2015 - Valdez9 / 25 The continuous case force of single decrementForce of a single decrementThe force of Decrement due to decrementjis defined as: (j)x= 1`( )xddx`(j) that the denominator is NOT`(j)xbut is rather`( ) a consequence, we see that ( )x=m j=1 (j) total force of Decrement is (indeed) the sum of all the otherpartial forces of can also show thatq(j)x= 10p( )s x (j)x+ : Weeks 8-9 (STT 456) Multiple Decrement ModelsSpring 2015 - Valdez10 / 25 The continuous case illustrative exerciseIllustrative exerciseSuppose that in a triple- Decrement model , you are given constant forces ofdecrement, for a person now agex, as follows: (1)x+t=b, fort 0, (2)x+t=b, fort 0, (3)x+t= 2b, fort are also given that the probability(x)will exit the group within 3years due to decrement1is the length of time a person now agexis expected to remain inthe triple Decrement (to be discussed in lecture):83 1.

7 Weeks 8-9 (STT 456) Multiple Decrement ModelsSpring 2015 - Valdez11 / 25 The ASDTThe associated single- Decrement table (ASDT)For each of the causes of Decrement in an MDT, a single-decrementtable can be defined showing the operation of that decrementindependent of the the associated single- Decrement table (ASDT)Each table represents a group of lives reduced continuously by onlyone Decrement . For example , a group subject only to death, but notto other decrements such as associated probabilities in the ASDT are called absolute rates ofdecrements.

8 For example , the absolute rate of Decrement due todecrementjover the interval(x,x+t]isq (j)t should be able to explain intuitively why the following alwayshold true:q (j)t x q(j)t : Weeks 8-9 (STT 456) Multiple Decrement ModelsSpring 2015 - Valdez12 / 25 Link between the MDT and the ASDTLink between the MDT and the ASDTIf given the absolute rates of decrements, sayq (1)x,q (2)x,..,q (m)x,how do we derive the probabilities of decrementsq(1)x,q(2)x,..,q(m)xinthe MDT? And vice fundamental link: (j)x= (j)xfor allj= 1,2,.., , it follows thatp( )t x=p (1)t x p (2)t x p (m)t , we note thatq(j)t x= t0p( )s x (j)x+sds= t0p( )s x (j)x+sds= t0p( )s xp (j)s xp (j)s x (j)x+ : Weeks 8-9 (STT 456) Multiple Decrement ModelsSpring 2015 - Valdez13 / 25 Link between the MDT and the ASDTIn the Multiple Decrement contextWe assume the following UDD assumption:q(j)t x=t q(j)x,for0 t leads us to the following result.)

9 P (j)t x= (1 t q( )x)q(j)x/q( ) to be done in result allows us to compute the absolute rates of decrementsq (j)xgiven the probabilities of decrements in the Multiple decrementmodel. In particular, whent= 1, we haveq (j)x= 1 (1 q( )x)q(j)x/q( ) : Weeks 8-9 (STT 456) Multiple Decrement ModelsSpring 2015 - Valdez14 / 25 Link between the MDT and the ASDTI llustrative exampleIn a double Decrement table where causedis death and causewiswithdrawal, you are given:both deaths and withdrawals are each uniformly distributed over eachyear of age in the double Decrement table.

10 `( )x= 1000q(w)x= (d)x= (w)xCalculateq (d)xandq (w) : One way to check your results make sense is to ensure the inequalityq (j)x q(j)xis : Weeks 8-9 (STT 456) Multiple Decrement ModelsSpring 2015 - Valdez15 / 25 Link between the MDT and the ASDTIn the associated single Decrement contextWe assume the following UDD assumption:q (j)t x=t q (j)x,for0 t implies:p (j)t x (j)x+t=p (j)t x (j)x+t=q (j) the previous link, one can deriveq(j)t x= t0p( )s x (j)x+sds= t0 i6=jp (i)s xp (j)s x (j)x+sds=q (j)x t0 i6=j(1 s q (i)x) this integration to derive the probabilities of Decrement given theabsolute rates of : Weeks 8-9 (STT 456) Multiple Decrement ModelsSpring 2015 - Valdez16 / 25 Link between the MDT and the ASDTThe case of two decrementsWhen we havem= 2, we can deriveq(1)t x=q (1)x t0(1 s q (2)x)ds=q (1)x(t 12t2q (2)x),and similarly,q(2)t x=q (2)x(t 12t2q (1)x).