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Actuarial: An Introduction to Mortality
Posted:
Nov 21, 2012 11:14 AM


In the cohort lifetable 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 lifetable 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 lifetables 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 lifetable 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).



