Page 16 - Life Insurance Today March 2016

find out the mortality that its insured lives are likely to l 40 963206 = .9821
experience. Life Insurers depend on the mortality ==
experience of the insured lives observed in the recent past l 30 980776
as the basis for estimating the probabilities of survival and
deaths. If it is observed that out of 10,000 lives all aged 2. The required probability is l10q30
35, 13 die within one year i.e. before attaining age 36; the l 30 - l 40 l 30 l 40
observed mortality rate at age 35 works out to 13/ 10,000
= 0.0013, thus Q35 = .0013. = =-
l 30 l 30 l 30
The mortality rates at various ages are determined in the
above manner, by using the data of recent past by the = 1 - 10 P30 = 1- .9821 = .0179
insurers. The observed rates of mortality, say from age of
15 years till extreme age, are then subjected to a process 3. The required probability that a person aged 30 dies
of graduation. after 10 years
No. of persons dying after age 40 l 40
The process of graduation enables the insurer to find the = = = .9821
limiting values of Q1 when the data is increased infinitely. No. of persons living at age 30 l 30
These graduated values of Q1 are used for constructing a
mortality table which contains seven (7) mortality functions This probability is the same as the probability that a person
for successive ages x like lx, dx, qx, px, Lx, Tx and ex. now aged 30 survives to age 40.

As we proceed to derive values of Ix (i.e. the no. of persons Example: B - Of two persons A aged 35 years and B aged
who survives at the age x - say W) where lw = 0 i.e. all 42 years, find the probability that :-
persons reaching age (w-1) i.e. the values of lw, lw+, etc. (1) A and B both survive 10 years ;
will all be zero. The age w, on and after which, there are
no survivors, is called the limiting age or extreme age of (2) A and B both die within 10 years.
mortality table. In the LIC (1970-73) Ultimate table it will
be seen that l102 = 601 and l103 = 0 - so as per the table (3) One of the two survives 10 years while the other dies
103 years is the extreme age of this table from survival within that period.
point of view.
(4) At least one survives for 10 years.
The mortality table is thus used to calculate the
probabilities of survival and death in the following worked Solution: - (1) The probability that A survives 10 years
out examples by using the LIC (1970-73) Mortality Ultimate
Table as illustrated below:- = 10 P35 = l 45 946656 = .9724
=
Example: A - Find the following probabilities: l 35 973550
1. That a life aged 30 survives 10 years.
The probability that B survives 10 years
2. That a life aged 30 dies within the next 10 years.
l 52 904837 = .9450
3. That a life aged 30 dies after 10 years. = 10 P42 = =
l 42 957541
Solution:
1. The required probability is 10 P30 The probability that both A and B survive 10 years is the
product of the two probabilities, i.e. 10 P35 and 10 P42 =
No. of persons living at age 40 (.9724) X (.9450) = .9189.
= No. of persons living at age 30
Solution: - (2) The probability that both die within 10 years
= (Probability that A dies within 10 years) X (Probability
that B dies within 10 years)

= (1- 10 P35) (1- 10 P42) = (1-.9724) (1-.9450) = (.0276)
(.0550) = .0015.

Solution: - (3) The event that one of them survives 10 years
while the other dies within that period is made up of two
events, viz.

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16 March 2016 Life Insurance Today