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Exam LTAM Updated 9/10/2018
Adapt to Your Exam SURVIVAL DISTRIBUTIONS SURVIVAL DISTRIBUTIONS Probability Functions Survival Function : future lifetime or time-to-death of Probability that survives years
d
( ) # () = ( ) # == PrPrP r# >> + | > = /(/+ ) ) / # ()( ) /() #(0∞) = =10 ##() ( ) 6# = == PrP r(()# > ) # () 6# = == Pr (()# ≤ ) # 6# + 6 # = 1 ( ) |6# = and dies within the following years == # −⋅ 6#K = K6 ## −K6## ( ) # # = ⌊#⌋ Pr# = = Q# ∙ #KQ = Q|# 6# = #K6 # − 6# = 6# # = # # #K6 = #K − #KK6 |6# = 6 #K # # 6# #K6 ( ) # #K6 = #() = − 6# = − #K6 #() = #() ⋅ #K6 = 6# ⋅ #K6 W# W W# = expZ−[/ #K6 \ = expZ−[ pZ−[##KW] \ must satisfy:
d
Actuarial Notations Probability that
survives
Probability that
•
Probability that
survives
Force of Mortality
Finding
d
Using Force of Mortality d
d
Properties of Force of Mortality •
≥ 0 ∫/#K6#K6 = ∞ ∗#K6∗ = #K6 + ⇒ W∗#∗ = W# ∙Q QW #K6 = ∙ #K6 ⇒ W# = W# ∞
•
d
Adding/Multiplying a Constant • •
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prior
# = # = [/j ⋅ 6 ##K6 = [/j 6# # = [/j ⋅ 6##K6 = [/j2 ⋅ 6 # # = # − q# d
d
Second Moment
d
•
d
Variance
∘
Curtate Expectation First Moment •
j
j
•
Q/
Qu
Second Moment
j
j
# = ⋅ Q|# = (2 − 1) Q# Q/ Qu # = # − (#) #:W|: = [/W ⋅ 6 # #K6 + W# = [/W 6# Wu W #:W|: = ⋅ Q|# + ⋅ W# = Q# Q/ Qu # # # ≈ # + 0.5 # = #:W|: + W# ⋅ #KW # = #:W|: + W# ⋅ #KW #:zKW|: = #:z|: + z# ⋅ #Kz:W|: #:zKW|: = #:z|: + z# ⋅ #Kz:W|: # = #(1 + #Ku) •
Variance
Temporary Expectation ∘
d
Relationship between
d
∘
and
∘
Recursive Formulas ∘
∘
∘
∘
∘
∘
Special Mortality Laws Constant Force of Mortality
# = {6 6# = 1 # = #:W|: = 1 (1 − {⋅W) # = (1 − ) # = − ⇒ #K6 = − (1 + ) = − − ( + ) ) 6# = #K6 # − ∘
∘
Uniform Distribution
# = ( − )} # = − ⇒ #K6 = − ( + ) } 6 # = #K6# = − − (− + ))Ä # = +− 1 # = # # > 1, > 0 exp Ñ− ln 6# = expÑ ln (6 − 1)Ö # = + # # > 1, 1, > 0, 0, ≥ − exp Ñ− ln 6# = expÑ ln (6 − 1)Ö ⋅ exp(−) (0 ≤ < 1) #K6 == (1⋅ − ) ⋅ # + ⋅ #Ku 6# # #K6 = 1 −# ⋅ # # () = 6# ∙ #K6 = # (0 ≤ < 1)1) #K6 == ((# ))u66 ⋅ (#Ku)6 6#K6(#) = −ln( −#ln(#) {⋅6 # = 6# ∙ #K6 = ⋅ ∘
Gompertz’s Law
Makeham’s Law
# = # = ⋅ Q|# = Q#
years
Life Table
d
•
years
Curtate Future Lifetime : number of completed future years by to death
Beta Distribution
∘
years
dies within
∘
Moments Complete Expectation First Moment
•
is a non-increasing function of t
∘
d
•
•
− 6# = # # #K6 = − − |6# = #K # #KK6 = − # = −2 : #:W| = W# () + W# q 2
j 6# = [6 .# ⋅ #K 6 6# = [/ .# ⋅ #K K6 |6# = [ .# ⋅ #K
Express ’s or ’s in terms of
Fractional Ages UDD Use linear interpolation: interpolation:
Constant Force of Mortality Use exponential interpolation: interpolation:
Select & Ultimate Mortality The age at which a person is selected is is denoted as .
#K6
Select mortality is written as where is the selected age age and is the number of years after selection.
The mortality after the select period is is called the ultimate mortality , where:
#K6 = #K6
Common Approach Read from the left to the right and then continue downwards.
# #Ku #K
#Kà
30 31 32 33
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1
INSURANCE
Type of Insurance
Discrete
j
# = sQKu ⋅ Q|# Whole Life
Qt
Continuous
j ̅# = [ 6 ⋅ 6# #K6 d Discrete Term Life
ANNUITIES
INSURANCE
u#:W| = # − W# ⋅ #KW Continuous
̅ u#∶W| = ̅# − W# ⋅ ̅#KW
Constant Force
Uniform Distribution
̅# = + ̅ #u:W| = + e1 − W# f
̅# = êîd#| − u = êW| ̅ #:W| − − ( + ) W# = d({Kï)W W# = W ⋅ − Calculate l and l ̅ similarly to and ̅, but with double the force of interest, . Equivalently, replace with l , or replace with 2 + l . ] l # − ( # )l l #:W|êêê − e #:W|êêêfl
Discrete
W| # = # − u#:W| = W# ⋅ #KW
Whole Life
W| ̅# = ̅# − ̅#u∶W| = W# ⋅ ̅#KW
Endowment Insurance
Continuous Discrete
Pure Endowment
Endowment Insurance
#:W|u = W# = W W# Continuous N/A Discrete
#:W| = u#:W| + W# Continuous
̅#:W| = ̅ u#:W| + W# j
Varying Insurance
()# = s( + 1) QKu ⋅ Q|# Qt ( ̅ ̅)# = [j6 ⋅ 6##K6 d ( ̅ ̅) #u:W|êêê = [W6 ⋅ 6# #K6 d (í ̅) #uW:W|êêê = [ ( − )6 ⋅ 6# #K6 d u êê +() u êêê = ( () #:W| #:W| +1) u#:W|
u êêê = ⋅ ̅ u êêê ( ̅ ̅) #u:W|êêê + (í ̅) #:W| #:W|
Type of Annuities
Due; Discrete
j
̈ # = sQ ⋅ Q# Qt
Immediate; Discrete
Whole Life
# = ̈ # − 1 Continuous
j
ê# = [ 6 ⋅ 6# d Due; Discrete
̈ #:W|êêê = ̈ # − W# ⋅ ̈ #KW
Variances
Discrete
Deferred Whole Life
ANNUITIES
̅ Recursive Formulas u# = # + # ⋅ #Ku u #:W|êêê = # + # ⋅ #Ku:Wdu| êêêêêêê ()# = # + # ( #Ku + ()#Ku ) 1/mthly Insurance #(z) = 1 ⌊# ⌋ Prô#(z) = ö = õ# ⋅ zu #Kõ = õ | zu # j (z) # = s (QKu)/z ⋅ zQ | zu # Qt () í Relationship between , , and (Under UDD Assumption) ̅# = # u êêê = u êêê ̅ #:W| #:W| W| ̅# = W|# u êêê + êêêu ̅#:W|êêê = #:W| #:W| (z) # = (z) # Replace A with for continuous cases.
Percentiles The 100 th percentile of Z is the value
°
Prô ≤ ° ö =
#
Continuous
W|ê# = ê# − ê#:W|êêê = W# ⋅ ê#KW êêêêêêê #:W|êêê = êW|êêê + W| ê# j (̈ )# = s ( + 1) Q ⋅ Q# Qtj ( ̅ê)# = [ 6 ⋅ 6# d ()#:W|êêê + ()#:W|êêê = ( + 1)#:W|Wêêê ( ̅ê)#:W|êêê = [ 6 ⋅ 6# d (íê)#:W|êêê = [W( − ) 6 6# d ( ̅ê)#:W|êêê + (íê)#:W|êêê = ê#:W|êêê
Certainand-Life
Varying Annuities
Uniform Distribution
Constant Force
To calculate : 1. Draw a graph with Z on y -axis and on x -axis. 2. Identify the parts of the c urve where . Determine the value of that corresponds to those parts. 3. Use the value of from Step 2 to calculate .
#
#:W|êêê = ̈ #:W|êêê −1 + W # Continuous ê#:W|êêê = ê# − W# ⋅ ê#KW Due; Discrete W|̈ # = ̈ # − ̈ #:W|êêê = W# ⋅ ̈ #KW
Deferred Whole Life
°such that:
#
Immediate; Discrete
Temporary Life
≤ °
°
ê# = +1 ê#:W|êêê = +1 e1 − W# f
Integrate directly, or use
̅# = 1 −ê# Integrate directly, or use ̅#:W| = 1 − ê#:W| êêê
Variances
]
Discrete Whole Life Temporary Life
̅
l # − ( # )l l l #:W|êêê − e #:W|êêêfl l
Replace A with and with for continuous cases.
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2
PREMIUMS
Recursive Formula
̈# = 1+ # ⋅ ̈#K
Net Future Loss
Relationship between Insurances and Annuities Discrete
Temporary Life
Continuous
ê# # ̅=# =11−− (2 ) ê# ̅ = 1 − ê ̅ #:W|êêê#:W|ê =ê 1 − (2)#:W|êê ê#:W|êêê
Whole Life Temporary Life
(z)# = 1 − (z)̈#(z) ̈#(z)(z) = ()⋅̈# − () ̈#:W|êê = ()⋅ ̈#:W|êê − ()(1 − W#) W|̈#(z) = ()⋅ W|̈# − ()⋅ W# () = (z)(z) () = (z)− (z)(z) (z) z (z) z UDD Assumption
Z1+ \ = Z1− \ •
•
are
ô /ö = # − ̈# ô /ö = ¨1 + ≠ ô # − ( #) ö ô /ö = ̅# − ê# ô /ö = ¨1 + ≠ ô ̅# − ( ̅#) ö ô /ö (= 0 ) ( ⇒ f.premiums (benefit) = f.benefits) ⇒ Net Premium = (annuity) ô /ö = ⋅ (1# −− ( #)#) ô /ö = ⋅ ( 1̅# −− ( ̅# ̅)#) |í ′ ′ /≤ = (f.benefits) + (f.expenses) − (f.gross premiums) ô /≤ ö (= 0 ⇒ =f.gross premi) u+ms) (f.expenses) (f.benefits Continuous
Add
to
and
•
If the question asks to use the Woolhouse’s formula with two terms, just drop the last ter m. If is not available, approximate as:
# 1 # # ≈ − 2 (ln# + ln#) ̈#(z):W|êê ≈ ̈#:W|êêê − 2− 1 1 − W# − 1 − 12 ô# + − W # (#KW + )ö # ≈ # + 12 − 121 # ° Prô °ö = ° # ° # # ° If the interest rate is 0: ∘
such as
To calculate : 1. Draw a graph with Y on y -axis and on x -axis. 2. Identify the parts of the c urve where . Determine the value of that corresponds to those parts. 3. Use the value of from Step 2 to calculate .
Gross Premium Reserve Prospective Method ben )
6 ≤=
− õ ≠™´K − − õ + (¥ − õ) ô /≤ ö = ¨ + + − õ ≠ ô # − ( # ) ö |í ′ [=] = +⋅ +[⋯+ ∂ ] [] = ⋅ [] 1. Replace and d with their continuous counterparts for fully continuous policies. 2. Add to for endowment insurance. Portfolio Percentile Premium
Using the portfolio percentile premium principle, the premium is set such that t here is a specified probability ( x %) that the total loss is negative:
Pr[ < 0] = % Prô / °ö = / ° ° # / / ° # # °
Percentile of The 100pth percentile of
that
is the value
. To determine
6(f. . + 6(f. − 6 (f. . exp.)
pre )
Expense Reserve
= Gross Premium − Net Premium 6 º = 6 ≤ − 6 W 6 º = 6 (f. exp.) − 6 (f. exp. premium)
Expense Premium
Recursive Formula Net premium reserve •
•
6 + (1 + ) = #K6 ⋅ + #K6 ⋅ 6K 6 ≤ + − (1+ ) = #K6 ⋅ ( + )≤ + #K6 ⋅ 6K 6 = 6 ⋅ 6 + 6 − 6 − 6 + 6 − 6#K6 : − 6 6KΩ 6 = ℎ : − 6 6Ω 6 = ℎ 6 < ) + (1 + ) − #K6 ⋅ ⋅ 6K = 6 #K6 6K = 6 + (1 − )+ ⋅ 6K Gross premium reserve
Thiele’s Differential Equation d
d
Euler’s Method Forward Euler Approximation d •
•
d Backward Euler Approximation d d
For net premium reserve, drop expense-related terms and replace with net premium. Interim Reserves ( • Exact value:
Variance For a fully discrete whole life policy:
/≤ = ¨ + +
premiums)
̈ 6 = 1 − #̈ K6# − # 6 = #K6 1 − #
for endowment insurance.
Gross Premium
efits − 6(f.
Special Formulas For a fully discrete whole life insurance policy:
Variance
Equivalence Principle
− 1 ̈#(z) ≈ ̈# − 2− 1 − 12 (# + )
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Discrete
Woolhouse’s Formula ( 3 terms)
Percentiles The 100 pth percentile of Y is the value that:
Fully Discrete
•
are provided on
The values for and when also given in the LTAM Table.
•
= (1+ ) = (1− )
and
/ = Æ´ − êêêêê Æ´| = ¨1 + ≠ Æ´ −
Equivalence Principle
( ) ( ) () () = 0.05
The formulas for the LTAM Table.
/ = ™´K − ̈êêêêêêêêê ™´K| ™ K = ¨1 + ≠ ´ −
6 = 6 (f.
Fully Continuous
Fully Continuous
Annuities with mthly Payments
Note:
Fully Discrete
RESERVES
Net Premium Reserve Prospective Method ben )
/ = (f.benefits) − (f.premiums)
̈ # = 1# =− (12− − #) ̈# ê ê = 1 − ̈ ê ê #:W|êê =#:W|1 − (2 − #:W|) ê #:W|êê
Whole Life
RESERVES
PREMIUMS
•
Linear approximation:
Modified Reserve Full preliminary term (FPT): The policy is treated as if it were issued one year later, with the first year of the policy being treated as if it were a oneyear term insurance. • FPT net premium
•
1st year modified net premium = #:|í Renewal modified net premium = ̈##KK = 0 #¬√Æ 6 #¬√Æ = 6#K FPT reserve
such
:
1. Graph on y -axis and on x -axis. 2. Identify the parts of the c urve where . Determine the value of that corresponds to those parts. 3. Use the value of from Step 2 to calculate .
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3
MARKOV CHAINS MARKOV CHAINS
6#¥ƒ : 6#¥ :
MULTIPLE DECREMENT MODELS MODELS MULTIPLE DECREMENT
agei+ s ) ageage age + 6 #(ƒ) = Q/6z Q#(Œ) #KQ(ƒ) = Q/6 Q|#(ƒ) 6#(Œ) = ƒ 6 #(ƒ) ¥6# = ∆−[/6 #K¥ƒ » ƒ«¥ (Œ) + 6#(Œ) = 1 6 # 6 6K ¥ ∙ = expZ−[/ #K \ ( ) (Œ)#KQ(ƒ) = 6#(Œ) #K6(ƒ) ƒ = 6| # Q # Q6 6 #¥ƒ = [/6 #¥ ⋅ #K¥ƒ ⋅ 6#Kƒƒ (ƒ ) 6#(ƒ) = 6#(Œ#) W ( ) ( ) ( ) Œ Œ Œ − 6 # # ( ) Œ # K6 ¥ ƒ Qƒ ¥ ƒ ƒQ ¥ Q = = 6 # #(Œ) #(Œ) 6# = Q/Q«ƒ 6# ⋅ #K6 − 6# ⋅ #K6 =−Pr(Pr(StSatratrtinin,m,moveoveinoutto )of ) 6 #(Œ) = #(#(Œ)ŒK6) (ƒ) (ƒ) = (#K6Œ) 6| # ¥ ƒ ¥ ƒ Qƒ W ¥ Q # 6KΩ¥ƒ# ≈ƒQ 6# + ℎ ∑Q/Q«ƒ 6# ⋅ #K6 − 6#¥ƒ ⋅ #K6ℎ #¥ƒ ≠ 6 #(ƒ) = [z/6 #(Œ) #K(ƒ) ⇒ #K6(ƒ) = 66#(Œ#(ƒ)) Ω# ≈ à 1 − ℎ#¥∙ = #K6(Œ) = ƒ#K6(ƒ) 6 ( ) 6#(Œ) = 6 Z−[/ #K(Œ) \ ¥ ̅#ƒ = [/j ï6 6#¥Q ⋅ #K6Qƒ (Œ) = [/ #(Œ) #K(Œ) Q«ƒ 6 # j = [/ ï6 ⋅ Pr(sta)rt in ,move into d (Œ) − 6#(Œ) 6 # ( Œ ) ⇒ #K66K= 6#(Œ) = 6#(Œ) 6|#(ƒ) = [6 #(Œ) #K(ƒ) ê#¥ƒ = [j/jï6 6#¥ƒ ̈#¥ƒ = Q/Q Q#¥ƒ z j QK(ƒ) (Œ)(ƒ) benefits = ƒ Q # #KQ Q/ j Q (Œ) a nnui t y = Q/ Q # 6 (¥) = 6 6W(¥) − 6(¥) −ƒ/ƒ«¥ #K6¥ƒ6¥ƒ + 6(ƒ) − 6(¥) benefits = [/j6 6#(Œ)#K6(ƒ) (ƒ) (Under CF) # b enef i t s = (Œ) + j 6 (¥) = 6(¥) −ℎ6Ω(¥) annuity = [/ 6 6#(Œ)
probability someone in state at state (where may equal at probability someone in state at remains in state until
in
Multiple Decrement Tables (MDT) Decrements are dependent on each other. Discrete Probabilities
•
•
Continuous Probabilities Direct Approach •
exp
d
d
For permanent disability model: d
•
Life Table
Approximation Kolmogorov’s Forward Equations: d d
Euler’s Method: 1.
Continuous Probabilities
2.
d d
d
Premiums For an insurance on currently in state i that pays $1 immediately upon every transition to state j :
exp
d
d
d
For an annuity on currently in state i that pays $1 per year while the person is i n state j : d
d d
d d
d
Insurance Applications Consider a whole life policy:
Reserves Direct Approach Use prospective method. Approximation Thiele’s Differential Equation: d
Discrete
•
•
d
Continuous
d
Euler’s Method: d d
d
0 ≤ < 1) ##((ƒŒ)) == ⋅⋅ ##((ƒŒ)) ##((ƒŒ)) = ##((ƒŒ)) = ##((Œƒ)) (ƒ ) #(ƒ) = ##(Œ) “1 − #(Œ)”
Fractional Ages ( UDD in the multiple decrement table:
Constant forces of decrement:
Associated Single Decrement Tables (ASDT) The associated single decrements are independent.
6#K(ƒ) \ Z−[ / ‘ (ƒ ) (ƒ) ⇒ 6#K6 = − ln 6# ‘(ƒ) 6#‘(ƒ) = [/ #‘(ƒ)#K(ƒ) ⇒ #K6(ƒ) = 66#‘(#ƒ) 6#‘((Œƒ)) + z6#‘(ƒ) =‘(ƒ)1 6# = ’ƒ 6# 6#‘(ƒ) 6#(ƒ) 6÷ #((Œ)Œ) = ∏zƒ ∑z6#‘(ƒ) (ƒ) 6#(Œ) =+ 1 −(Œ) ƒ= 1 6# 6 # 6 #(fi) (#K6(ƒ)) = ‹‹‹°‹››´(´‡fl´()fi) ⎩ #K6ƒ = ‹‹°´‡(fi) 0 ≤‘(ƒ) < 1 (Œ) ›´((fifl)) # = # ›´ 0 ≤ < 1‘() #() = #‘() Z − 2# \ #() = #‘() · − #‘()2 + #‘(à) + à#‘()3 ∙ #‘(à)„ ((fifl)) 0 ≤ < 1 › ´ #‘(ƒ) = #(Œ)›´ 6#‘(ƒ) =
exp
d
d d
d d
d
Key Relationships between No Assumption
and
•
d d
•
d d
UDD in Multiple-Decrement Tables (UDDMDT) ( )
UDD in Associated Single Decrement Tables (UDDASDT)( ) For 2 decrements:
For 3 decrements:
CF in MDT or ASDT (
)
MULTIPLE LIVES MULTIPLE LIVES
6#]#] =+mi6n#]#=, ]1 |6#] == #] ⋅−6#K: ]K #] K6 #] = K6#] − #] 6#] == 6# +⋅ 6] − ⋅ 6#K6:#]]K6 =6 ##K6 +6 ]]K6 6 # 6 ] Joint Life
Independent Lives
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4
Moments
j ∘#] = [ 6#] d j “#] ” = 2[ ⋅ 6 #] j
Contingent Probabilities
6
d
#] = Q#] Q
Last Survivor
#]êêêê = max# ,] 6#] + 6 #] = 1 |6 #] = #] − K6 #] = K6#] − #] 6#] = 6 # ⋅ 6 ] 6#] = 6 # + 6 ] − 6 # ⋅ 6] #]() = 6 # 6#K6# +⋅ 66]] +− 66]# ⋅]K66⋅] 6 # Moments
j
∘ #] = [ 6#] d j #] = Q #] Relationships between Status
+1 6z(ƒ) + ℎ6(ƒ) (1 + )Ω ƒQ ℎ (Q) + ƒQ + (Q) = Ω #K6 6KΩ 6KΩ 6KΩ Q
If lump sum benefit is assumed to be paid in the middle of an interval:
6z(ƒ) + ℎ6(ƒ) (1 + )Ω ƒQ ℎ (Q) + ƒQ (1+ )Ω/ + (Q) = Ω #K6 6KΩ 6KΩ 6KΩ
Relationships
Independent Lives
Q
= [ #] ∙ #K 6#] 6 6#] = [ #] ∙ ]K 6 = [ # ∙ #K ⋅ 1 − ] 6#] 6 6#] = [ ] ∙]K ⋅ 1− #
Reserve Recursion for Policies with Multiple States Assuming there are states and cash flows are made every h years:
+ 6 #] = 6 #] 6#] + 6 #] = 6 # 6#] = 6 #] + 6 # ⋅ 6] 6#] + 6 #] = 6 #] 6#] #] + #] = 1 #] + #] = 1 #] = #]
∞
Q
Activities of Daily Living (ADLs): Bathing •
•
∞
∞
•
∞
∞
•
∞
•
Contingent Insurance
j ̅#] = [ 6 ⋅ 6#] ⋅ #K6 d j ̅#] = [ 6 ⋅ 6# ⋅#K6 ⋅ 6] d Relationships
() Status and
( êêêê) #] + #] = # + ] #] ⋅ #] = # ⋅ ] 6#] + 6 #] = 6 # + 6] ∘ + ∘ = ∘ + ∘ #] #] # ] ∘ + ∘ = ∘ + ∘ #]:W|êêê #]:W|êêê #:W|êêê ]:W|êêê #] + #]êêêê = # + ] ̅#] + ̅#] = ̅# + ̅] ê#] + ê#] = ê# + ê] W#] + W#] = W # + W] Covariance of #] and #] ∘ ∘ ∘ ∘ Cov#] ,#] = Cov# ,] + # − #] ] − #] Cov # ,] = 0 if # and ] are independent Exactly One Life Survives
Pr(exactly one life survivies years) = 6 #] − 6#] = 6 # + 6 ] − 2 ⋅ 6 #] Relationships between Insurance Policies, Annuities, and Premiums
#] = 1 − ̈ #] ê#]êêêê = 1 − ̅êêêê#] ̈ #]:W|êêê = ̈ #] − W#] ∙̈ #KW:]KW #]êêêê = ̈1êêêê#] − #] = 1− ̅# ]̅#]
Note: The list above is not exhaustive; similar relationships can be applied to other forms of insurance/annuities with appropriate adjustments.
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Δ 6 = 6(1+ ) − #K6 6
Make payments to ( y ) after ( x ) has died:
ê#|] = ê] − ê#]
Make payments only when exactly one life is alive:
(annuities) = ê#] − ê#]
LONG-TERM INSURANCE COVERAGE LONG-TERM INSURANCE COVERAGE Disability Income Insurance (DII) Continuous Sojourn Annuity The EPV of an n-year continuous sojourn annuity on ( x ) in state i that pays $1 per year continuously while the life remains in state i is:
¥¥ êêê = [ W 6 #¥¥ ï6 d ê#:W| W
êêê = [ 6 # #K6 êêêêêêê ï6 d ê#K6:W6| ê#:W| With waiting period of w years, the EPV is:
WÁ
êêêêêêê − ê êêêê ï6 d qê#K6:W6| 6# #K6 #K6:Á|
With waiting period of w years and benefit term of m years, the EPV is:
W(zKÁ)
Profit Vector
Pr = (Pr Pr Pr … PrW) Profit Signature Profit per policy issued
Π = Pr6 ⋅Prob“in force −at1 time Î in force at time 0” Profit signature: (Π Π Π … ΠW ) where Π = Pr Π6 = Pr6 ⋅ 6 # , = 1,2,3,…, Profit Measures NPV ∞
EPV of benefit of an n-year DII:
[
Profits for Traditional Products The profit per policy in force at time t is
Change in Reserve
Reversionary Annuities
•
PROFIT TESTS PROFIT TESTS
Pr6 = 6 + 6 − 6(1 + ) − #K66 − #K6 6
̅#] + ̅#] = ̅#] ̅#] + ̅#] = ̅#] ̅#] + ̅#] = ̅# •
•
Dressing Eating Toileting Continence Transferring
êêêêêêêêê − ê#K6:Á| êêêê ï6 d qê#K6:zKÁ| [ 6 # #K6 WÁ êêêêêêê − ê êêêê ï6 d qê#K6:W6| + [W(zKÁ) 6# #K6 #K6:Á|
NPV = Πƒ ⋅ õƒ ƒ where = risk discount or hurdle rate Partial NPV
Q
NPV() = Πƒ ⋅ õƒ ƒ
IRR ∞
NPV = Πƒ ⋅ õƒ = 0 DPP
ƒ
DPP = min:NPV() > 0 Profit Margin NPV Profit margin = (f.premiums )
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5
Zeroized Reserves 1. Begin with the last year an d work backwards 2. Set the profit for the year to zero then solve for the beginning-of-year reserve 3. If the reserve is negative, set to zero and repeat this entire process again until time 0 Gain by Source Gain in the order of expenses, interest, a nd mortality ( actual): Expense: Interest :
′ = ‘6 − 6‘(1 + 6 ) + 6‘ − 6‘#K6 6 − 6 6‘ + 6 − 6 ‘ #K6 − #K6 6 + 6 − 6
Defined Contribution Pension Plans
(pension fund) = (pension benefits) Annual Retirement Benefit = ⋅ ¯˘˙ ⋅ = total number of years of service ¯˘˙ = final average salary = accrual rate Defined Benefit Pension Plans where
•
§
Mortality :
Actual Profit Using the actual experience:
Actual Profit = 6‘ + 6 − 6‘(1+ 6‘)‘ ‘ −#K6(6 + 6 ) − #K6 ⋅ 6 Expected Profit Using the assumed experience:
Expected Profit = 6 + 6 − 6 (1+ 6) −#K6(6 + 6) − #K6 ⋅ 6 Total Gain = Actual Profit − Expected Profit = Gain from Expenses + Gain from Interest + Gain from Mortalities Total Gain
Two methods to calculate the amount of retirement benefit: PUC: projects salary to retirement or exit date TUC: calculates salary based on employee’s current age §
•
•
•
Early retirement
Annual Retirement Benefit = ⋅ ¯˘˙ ⋅ ⋅(1 − pension reduction factor) Annual Retirement Benefit = ⋅ ¯˘˙ ⋅ Annual Retirement Benefit = ⋅ ¯˘˙ ⋅ ˝˛ˇ˘˝˛!˛˙ˇ "˛$˘ˇ%&˝"$"' "˛ ⋅ (1+ COLA) Withdrawal without COLA Withdrawal with COLA
Normal Contribution
•
Two methods of funding benefits: PUC and TUC
•
If there are no mid-year exits: §
PUC:
§
TUC:
6
6 )´*+
6
Retiree Health Benefits Benefit Premium Annuity for age x at time t
Replacement Ratio
j , + )\ ̈,(,) = Q Q # Z( +(, )
=
Value of retiree health benefit at ret irement for a life retiring at age xr in t years: ,
salary in the year before retirement
Salary Projection S : Salary s: Salary scale •
•
•
•
Q
pension income in the year after retirement
̅ ̅ ]K6 = ](1+ %)6 ] = # ⋅ ]# ̅] = #̅ ⋅ ̅̅]# ] = [ ̅]K6 = ̅ #̅# ⋅ [̅]K6 Rate of salary function to salary scale: ] = [ ̅]K6 Salary scale to rate of salary function: ̅] ≈ ].ˆ : Rate of salary : Rate of salary function
Constant percentage of increase Salary Scale
Rate of salary
Relationships
d
6 + 6 = EPV(benefits for mid-year exits)+# 6K
If there are no mid-year exits:
6 = 6
SURVIVAL MODEL ESTIMATION SURVIVAL MODEL E STIMATION Kaplan-Meier and Nelson-Aalen Estimators Empirical Distribution
Pr( = ) = # of data points = W () = Pr( ≤ ) = # of data points ≤ Var2 W() = Var2 W () = W () ⋅ 1− W () Kaplan-Meier Estimator
ƒ
3ƒ = ’¨1 − ¥¥ ≠ ¥ 3 ℎ4ƒ = 1 − 3ƒƒ
Tail Correction Efron’s tail correction: •
3() = 0 for > 3() = 53(), 6 = (all accrued benefits at time ) 0, forfor +# 6K 6 + 6 = EPV ƒ 7ƒ = ¥¥ ¥87] 6 = 6 ⋅ 6 fi 3 = ƒ 6K ) ´ = q ⋅ − 1
Funding the Benefits Actuarial Liability
(benefits for mid-year exits)
MATHEMATICS PENSION PENSION MATHEMATICS Valuation of Benefits Motivations 1. Attract potential employees 2. Provide incentive for employees to stay 3. Facilitate turnover of older employees 4. Provide tax-friendly compensation 5. Pressure from trade unions 6. Reward employees who have contributed to the company’s success
Normal Cost
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Klein and Moeschberger's tail correction:
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Brown, Hollander, and Korwar’s tail correction:
Nelson-Aalen Estimator
Variance of Estimators Greenwood’s Approximation: •
Used for Kaplan-Meier •
(,)̈ (,)
When healthcare premiums increase exponentially with age and at a constant annual inflation rate where:
ƒ Var2 ô3ƒ ö = ô3ƒ ö ¥(¥ −¥ ¥ ) ¥ Klein’s Estimation:
ƒ
Var2 ô7ƒ ö = ¥ (¥ −à ¥) ¥
¥
ƒ
Var2 ô3ƒ ö = ô3ƒ ö ¥ (¥ −à ¥ ) ¥
¥
= ( + 1,)⁄(,) = Var2 ô3ƒ ö = ô3ƒ ö ⋅ Var2 ô7ƒ ö ( + , + ) = Q(1+ )Q(,) 1 + ̈,(,) = ̈#õ|¥∗ where ∗ = (1 + ) − 1 /ˆ#6 = #KQ# Q ( + , + )̈,( + , + ) Q /ˆ# = (,) #KQ# ¥Q∗̈#KQ|¥∗ Q Used for Nelson-Aalen
• •
annual rate of inflation for healthcare costs
In general,
Actuarial Value of Total Health Benefit (AVTHB)
Actuarial Liability at time t, t V
/ˆ#
+ ∙ #KQ# Q ( + , + )̈,( + , + ) Q
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6
Confidence Interval Linear Confidence Interval for
( )
:
3() ± (uK°)/ : V2ar ô3()ö )u/;
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