## Life Expectancy

This is not the case. We cannot extrapolate to forecast an infinite life. Look more closely at the figures and one very interesting fact emerges. For very old people, those over one hundred years old, the probability of dying in the next years approches a limit. No matter how old you are, it is not certain that you will die in the next year. In fact, for old people, the probability of dying in the next year is about 0.6. The oldest person in the worl has only a probability of about 0.6 of dying in the next year.

If a person is a hundred, they can expect to die in the next year with probability 0.6 If a person is a hundred, to live to 102, they must survive the first year with probability 0.4 and die in the second year with probability 0.6 ie 0.4*0.6. If a person is a hundred, to live to 103, they must survive the first two years with probability 0.4

^{2}and die in the third year with probability 0.6 ie 0.4

^{2}*0.6. If a person is a hundred, to live to 104, they must survive the first three years with probability 0.4

^{3}and die in the fourth year with probability 0.6 ie 0.4

^{3}*0.6 and so on. The life expectancy of a one hundred year old is the sum of these:

\[1 \times 0.6+ 2 \times 0.4 \times 0.6 + 3 \times 0.4^2 \times 0.0.6 + 4 \times 0.4^3 \times 0.6 + ...= \sum 0.6 \times n \times 0.4{^n-1} = \sum_{n=1} \frac{d}{dx}_{p=0.4} {0.6 p^n}\]

The summation is of a geometric series with

\[a=0.6, r=p\]

The sum of the series is

\[\frac{a}{1-r} = \frac{0.6}{1-p}\]

Now differentiate and put

\[p=0.4\]

\[ \frac{d}{dx}_{p=0.4} \frac{0.6}{1-p} = [ \frac{0.6}{(1-p)^2}]_{p=0.4} = \frac{0.6}{0.6^2} = \frac{5}{3} years\]