MATH 403A, fall 98, CHAPTER 3

X = age-at-death. Distribution: F_{X}(x) = P (X ≤ x) is the distribution function of X. Survival: s(x) = 1 − F_{X}(x) = P (X > x) the survival function.

P (x < X ≤ z|X > x) = \frac{F_{X}(z) − F_{X}(x)}{1 − F_{X}(x)} = \frac{s(x) − s(z)}{s(x)}

is the probability that a newborn will die between the ages of x and z given survival to age x. To simplify notation, it is customary to use the following variables:

• (x) denotes life-age-x and

• T (x) = X − x is the future life of (x). Thus, in the above

P (x < X ≤ z|X > x) = P (T (x) ≤ z − x)

Other notation:

tq_{x} = P (T (x) ≤ t), _{t}p_{x} = 1 −_{t} q_{x} = P (T (x) > t)
q_{x} =_{1} q_{x}, p_{x} =_{1} p_{x}

t|uq_{x} = P (t < T (x) ≤ t_{u}) =_{t+u} q_{x} −_{t} q_{x} =_{t} p_{x} −_{t+u} p_{x} =_{t} p_{x} _{u}q_{x+t}

Note: The notation can get a little confusing here. The survival function s(x) means that the age-at-death is at least x, however, for most of the book, (x) is the current age. denotes

Another important variable is the curtate-future-lifetime of (x) K(x) = number of future years complete by (x) prior to deaath.

P (K(x) = k) = P (k ≤ T (x) < k + 1) =_{k} p_{x}q_{x+k} =_{k|} q_{x}

The distribution function of K(x) is

F_{K(x)}(y) = ∑_{h=0}^{k} _{h|}q_{x} =_{k+1} q_{x}

Force of mortality: The distribution function F_{X} has density f_{X}(x) = F′_{X}(x). The force of mortality represents the conditional density (given survival to age (x):

μ(x) = \frac{f_{X}(x)}{1 − F_{X}(x)} = \frac{−s′(x)}{s(x)}

Integrating we obtain

− ∫_{x}^{x+n} μ(y)dy = ln(\frac{s(x + n)}{s(x)} = ln_{n} p_{x}

or

np_{x} = e^{− ∫_{x}^{x+n} μ(y)dy}.

In particular

F_{X}(x) = 1 − s(x) = 1 −_{x} p_{0} = 1 − e^{− ∫_{0}^{x} μ(y)dy}

hence

f_{X}(x) = F′_{X}(x) = e^{− ∫_{0}^{x} μ(y)dy}μ(x) =_{x} p_{0}μ(x).

Now, F_{T (x)}(t) =_{t} q_{x} so its density is

f_{T (x)}(t) =_{t} p_{x}μ(x + t).

Note: since _{n}p_{x} → 0 as n → ∞, ∫_{x}^{x+n} μ(t)dt → ∞ as n → ∞. Properties of Survival functions and Force of mortality

Requirements  Restrictions on s(x) Restrictions on μ(x)
I. Survival function s(0) = 1 μ(x) = 0 for x < 0

must start at 1

II. Survival function  s′(x) ≤ 0 (if it is  μ(x) ≥ 0
must be non–decreasing  differentiable)  for x ≥ 0

III. Survival function

must approach 0 as s(∞) = 0 ∫_{0}^{∞} μ(x)dx = ∞

age approaches ∞

Homework

1. Complete the following table:

This  s(x)  f_{T (x)}(t) F_{T (x)}(t) μ(x + t)  _{n}p_{x}  _{n}q_{x}

In terms of

s(x)  —

f_{T (x)}(t)  —  —

F_{T (x)}(t)  —  —

μ(x + t)  —  —

np_{x}  —  —

nq_{x}  —  —  —

2. s(x) = e^{−x^{2}/500} for x ≥ 0. Find _{t}p_{20} μ(20 + t).

3. The force of mortality is constant for all ages. The probability that a 35-year old dies within the next 10 years is .10. Find the force of mortality.

4. You are given that _{t}q_{35} = \frac{70t+t^{2}}{8775} for 0 ≤ t ≤ 65. Find μ(60).

5. You are given that f_{T (50)}(t) = \frac{t/3+1/2}{t^{2}+3t+7} for 0 ≤ t ≤ 51.60. Find 10|20q_{50}.

6. You are given a survival function of s(x) = 1 − \frac{x^{2}}{10,000} for 0 ≤ x ≤ 100. Let y be the age at which the force of mortality is equal to .1. Find s(y).

7. Determine if s(x) = \frac{1+2/(x+2)}{2} is a legitimate survival function.

8. Determine if μ(x) = 1/(x + 1) for x ≥ 0 is a legitimate force of mortality.

9. Textbook problems: 3.1, 3.4-3.7, 3.9.