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Topic: Actuarial: An Introduction to Mortality
Replies: 1   Last Post: Nov 21, 2012 11:32 AM

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 richardskaa@hotmail.co.uk Posts: 1 Registered: 11/21/12
Actuarial: An Introduction to Mortality
Posted: Nov 21, 2012 11:14 AM

In the cohort life-table model, imagine a number of individuals born simultaneously and followed until death. Death is assumed to occur by an identical ?mechanism? of failure for these individuals.
In this cohort life-table model, the number of people living at age x is denoted by lx and this number of people living at age x expressed as a fraction of the initial number of people is then equal to lx/l0. This is denoted by S(x) meaning the survival function of x. It is a non increasing function and has no units.
The number of people living at age x who die before reaching age x + t is then equal to lx - lx+t and is denoted by tdx.
The number of people living at age x who die before reaching age x + t expressed as a fraction of the initial number of people is then equal to tdx/l0 which then equals (lx - lx+t)/l0 which then equals lx/l0 - lx+t/l0 which then equals S(x) ? S(x + t).
And the number of people living at age x who die before reaching age x + t expressed as a fraction of the number of people living at age x is then equal to tdx/lx which then equals (lx - lx+t)/lx which then equals (lx/l0 - lx+t/l0)/(lx/l0) which then equals (S(x) ? S(x + t))/S(x). This is denoted by tqx.
The number of people living at age x + t expressed as a fraction of the number of people living at age x is then equal to lx+t/lx which then equals (lx+t/l0)/(lx/l0) which then equals S(x + t)/S(x). This is denoted by tpx.
Note: Because tpx + tqx = S(x + t)/S(x) + (S(x) ? S(x + t))/S(x), tpx + tqx = (S(x + t) + S(x) ? S(x + t))/S(x) and so tpx + tqx = S(x)/S(x) and so tpx + tqx = 1. tpx and tqx have no units.
Note: Ages are in a chosen unit of time and if x is 0 or an integer, i.e. 0 or an integer number of time units, AND t = 1, i.e. 1 time unit, then tdx, tqx and tpx can be written as dx, qx and px respectively rather than having to be written as 1dx, 1qx and 1px respectively. However, I will still refer to them as 1dx, 1qx and 1px respectively.
Because S(x) is the number of people living, i.e. SURVIVING, at age x expressed as a fraction of the initial number of people, it is the number of people who die, i.e. FAIL, anytime after reaching age x expressed as a fraction of the initial number of people.
?
And S(x) = ?x f(T)dT where f(T) is the failure density function of T. f(T) is in units of reciprocal time units. Note:* the upper limit of this integral is ? because the terminal age here = ? (see later).
This means that
?
S(x + t) = ?x+t f(T)dT and S(x) ? S(x + t) =
? ?
?x f(T)dT - ?x+t f(T)dT and so S(x) ? S(x + t) =
? ?
[F(T)]x ? [F(T)]x+t and so S(x) ? S(x + t) =
F(?) ? F(x) ? (F(?) ? F(x + t)) and so S(x) ? S(x + t) =
F(x + t) ? F(x) which equals
x+t
?x f(T)dT.
This means that lim ??0 (S(x) ? S(x + ?))/?, equal to - lim ??0 (S(x + ?) ? S(x))/?, equals lim ??0 (F(x + ?) ? F(x))/?, and so equals ? S?(x), which equals F?(x) i.e. f(x).
Note: because lim ??0 (S(x) ? S(x + ?))/? = f(x), lim ??0 S(x) ? S(x + ?) = lim ??0 ?f(x).
And because tqx = (S(x) ? S(x + t))/S(x), lim ??0 ?qx/? = lim ??0 (S(x) ? S(x + ?))/?S(x) and because lim ??0 (S(x) ? S(x + ?))/? = f(x) = ? S?(x), this means that lim ??0 ?qx/? = f(x)/S(x) which then equals ? S?(x)/S(x).
This quantity is called the force of mortality function of x and is denoted by µ(x), i.e. µ(x) = lim ??0 ?qx/? = f(x)/S(x) = ? S?(x)/S(x). µ(x) is in units of reciprocal time units.
Note: because lim ??0 ?qx/? = lim ??0 (S(x) ? S(x + ?))/?S(x) = µ(x), lim ??0 ?qx = lim ??0 (S(x) ? S(x + ?))/S(x) = lim ??0 ?µ(x).
Looking at the equation µ(x) = ? S?(x)/S(x), i.e. the equation µ(x) = (-d/dx)(lnS(x)), ?µ(x)dx then equals - lnS(x) + constant and so
x
?0 µ(T)dT = - lnS(x) + constant ? (- lnS(0) + constant) and so
x
?0 µ(T)dT = lnS(0) - lnS(x).
And because S(x) = lx/l0, S(0) = l0/l0 and so S(0) = 1 and so this means that
x
?0 µ(T)dT = ln1 - lnS(x) and so
x
?0 µ(T)dT = - lnS(x) and so lnS(x) =
x
- ?0 µ(T)dT and so S(x) =
x
exp (- ?0 µ(T)dT).
Because - lnS(x) =
x
?0 µ(T)dT, - lnS(x + t) =
x+t
?0 µ(T)dT and so - lnS(x + t) ? (- lnS(x)) =
x+t x
?0 µ(T)dT - ?0 µ(T)dT and so lnS(x) ? lnS(x + t) =
x+t
?x µ(T)dT and so lnS(x + t) ? lnS(x) =
x+t
- ?x µ(T)dT and so ln(S(x + t)/S(x)) =
x+t
- ?x µ(T)dT and so S(x + t)/S(x) =
x+t
exp(- ?x µ(T)dT), i.e. tpx =
x+t
exp(- ?x µ(T)dT).
Obviously lx can?t be recorded continuously and is instead recorded for each integer value of the chosen time unit ages are in. This means that where x is 0 or an integer and 0 < t < 1, lx+t, i.e. l0S(x + t), has to be assumed i.e. interpolated. S(x) is then treated as a piecewise continuously differentiable non increasing function. Note: the lowest integer for which lx = 0 is denoted by ? and because S(x) = lx/l0, this means that ? is also the lowest integer for which S(x) = 0. While ? is finite in real life-tables and in some analytical survival models, most theoretical forms have no finite age, ?, at which l? and S(?) equal 0. And in these forms, ? = ? by convention. ? = ? in *this cohort life-table model. ? is known as the terminal age of the table.
The 3 main types of interpolation are as follows;
Type 1, interpolation under the assumption that f(x + t) = f(x) where x is 0 or an integer and 0 ? t < 1;
Because S(x) ? S(x + t) =
x+t
?x f(T)dT, under this assumption, S(x) ? S(x + t) will equal
x+t
?x f(x)dT and so will equal f(x)(x + t) ? f(x)x and so will equal f(x)(x + t ? x) and so will equal f(x)t.
And so, under this assumption, S(x + t) will equal S(x) ? f(x)t. Note: this means that, under this assumption, S(x + t) will be linear where x is 0 or an integer and 0 ? t < 1.
And this means that, under this assumption, S(x + 1) will equal S(x) ? f(x).
Because µ(x) = f(x)/S(x), µ(x + t) = f(x + t)/S(x + t) and so this means that, under this assumption, µ(x + t) will equal f(x + t)/(S(x) ? f(x)t). And because this assumption is that f(x + t) = f(x), this means that, under this assumption, µ(x + t) will equal f(x)/(S(x) ? f(x)t). This means that, under this assumption, µ(x + 1) will equal f(x)/(S(x) ? f(x)).
Because tpx = S(x + t)/S(x) and because, under this assumption, S(x + t) will equal S(x) ? f(x)t, under this assumption, tpx will equal (S(x) ? f(x)t)/S(x) and so will equal 1 ? f(x)t/S(x).
And because µ(x) = f(x)/S(x), this means that, under this assumption, tpx will equal 1 - µ(x)t.
And because tqx = 1 - tpx, this means that, under this assumption, tqx will equal f(x)t/S(x) and will equal µ(x)t.
This means that, under this assumption, 1qx will equal µ(x) and so, under this assumption, tqx will equal t1qx.
And because 1qx = 1 - 1px, this means that, under this assumption, tqx will equal (1 - 1px)t.
And because tpx = 1 - tqx, this means that, under this assumption, tpx will equal 1 - t1qx and will equal 1 ? (1 - 1px)t.
Because tdx/l0 = S(x) ? S(x + t) and so tdx = l0(S(x) ? S(x + t)) and because, under this assumption, S(x + t) will equal S(x) ? f(x)t, under this assumption, tdx will equal l0(S(x) ? (S(x) ? f(x)t)) and so will equal l0(S(x) ? S(x) + f(x)t) and so will equal l0tf(x), i.e. tdx will be uniformly distributed under this assumption.
Note: if one were to assume that S(x + t) is linear where x is 0 or an integer and 0 ? t ? 1, i.e. that S(x + t) = S(x) + t(S(x + 1) ? S(x)) where x is 0 or an integer and 0 ? t ? 1, then tpx would still equal 1 ? (1 - 1px)t. This is because tpx, i.e. S(x + t)/S(x), would equal 1 + t(S(x + 1)/S(x) -1) and so would equal 1 ? t(1 ? S(x + 1)/S(x)), i.e. 1 ? (1 - 1px)t. Because tdx/l0 = S(x) ? S(x + t) and so tdx = l0(S(x) ? S(x + t)), under this assumption, tdx would equal l0(S(x) ? (S(x) +t(S(x + 1) ? S(x)))) and so would equal l0(S(x) ? S(x) - t(S(x + 1) ? S(x))) and so would equal - l0t(S(x + 1) ? S(x)), i.e. tdx would be uniformly distributed under this assumption.
Type 2, interpolation under the assumption that µ(x + t) = µ(x) where x is 0 or an integer and 0 ? t < 1;
Because tpx =
x+t
exp(- ?x µ(T)dT), under this assumption, tpx will equal
x+t
exp(- ?x µ(x)dT) and so will equal exp(-(µ(x)(x + t) - µ(x)x)) and so will equal exp(-(µ(x)(x + t ? x))) and so will equal exp(-µ(x)t). Note: this means that, under this assumption, lnS(x + t) will be linear where x is 0 or an integer and 0 ? t < 1. This is because, under this assumption, ln(tpx), i.e. ln(S(x + t)/S(x)) equal to lnS(x + t) ? lnS(x), will equal -µ(x)t and so lnS(x + t) will equal lnS(x) - µ(x)t.
This means that, under this assumption, 1px will equal exp(-µ(x)) and so, under this assumption, tpx will equal (1px)^t. ^ means to the power of.
Because tqx equals 1 - tpx, this means that, under this assumption, tqx will equal 1 - exp(-µ(x)t) and will equal 1 - (1px)^t.
And because 1px = 1 - 1qx, this means that, under this assumption, tpx will equal (1 - 1qx)^t and, under this assumption, tqx will equal 1 - (1 - 1qx)^t.
Because tpx = S(x + t)/S(x) and so S(x + t) = tpxS(x), the above means that, under this assumption, S(x + t) will equal exp(-µ(x)t)S(x). And this means that, under this assumption, S(x + 1) will equal exp(-µ(x))S(x).
Note: if one were to assume that lnS(x + t) is linear where x is 0 or an integer and 0 ? t ? 1, i.e. that lnS(x + t) = lnS(x) + t(lnS(x + 1) ? lnS(x)) where x is 0 or an integer and 0 ? t ? 1, then tpx would still equal (1px)^t. This is because lnS(x + t) would then equal lnS(x) + tlnS(x + 1) ? tlnS(x) and so would equal lnS(x) + ln(S(x + 1))^t ? ln(S(x))^t and so would equal ln(S(x)(S(x + 1))^t/(S(x))^t) and so S(x + t) would equal S(x)(S(x + 1))^t/(S(x))^t and so S(x + t)/S(x), i.e. tpx, would equal (S(x + 1)/S(x))^t, i.e. (1px)^t.
Type 3, interpolation under the assumption that 1/S(x + t) = 1/S(x) + t(1/S(x + 1) ? 1/S(x)) where x is 0 or an integer and 0 ? t ? 1;
Note: this means that, under this assumption, 1/S(x + t) will be linear where x is 0 or an integer and 0 ? t ? 1.
And so, under this assumption, S(x + 1)/S(x + t) will equal S(x + 1)/S(x) + t(1 ? S(x + 1)/S(x)), i.e. will equal 1px + t(1 - 1px) which equals t + (1 ? t)1px. And so, under this assumption, (S(x + 1)/S(x + t))(S(x)/S(x + 1)), equal to S(x)/S(x + t), i.e. 1/tpx, will equal (t + (1 ? t)1px)/1px and so, under this assumption, tpx will equal 1px/(t + (1 ? t)1px).
And because tqx equals 1 - tpx, this means that, under this assumption, tqx will equal 1 - 1px/(t + (1 ? t)1px).
And because 1px = 1 - 1qx, this means that, under this assumption, tpx will equal (1 - 1qx)/(t + (1 ? t)(1 - 1qx)) and so tpx will equal (1 - 1qx)/(t + 1 - 1qx - t + t1qx) and so tpx will equal (1 - 1qx)/(1 - 1qx + t1qx) and so tpx will equal (1 - 1qx)/(1 + (t ? 1)1qx) and that, under this assumption, tqx will equal 1 - (1 ? 1qx)/(1 + (t ? 1)1qx).

Date Subject Author
11/21/12 richardskaa@hotmail.co.uk
11/21/12 Jesse F. Hughes