# Age expectation if a person is immortal if he's not murdered

1. Jun 19, 2012

### moonman239

A while back I posted a thread about the probability that someone who is alive at 12 years of age will be alive at age 82. This person could only die if he were murdered or if he died when he was greater than 82 years old.

Now, let's assume this person will live forever if he's not murdered. As before, we assume that the probability of him dieing in a one-year period is 1/18690. My question is, on average, how long could he expect to live?

I know that there's about a 50/50 chance he'd live to be 140 years old - is this the answer to my question?

2. Jun 20, 2012

### willem2

Where do you get that from? The person has a probability of (1-1/18690)^128 to survive for 128 years (past 12) This is about 99.3%. If you start at 0, the survival rate rounded to 3 significant figures is still 99.3%

Suppose the average survival age is a.
Someone will survive in a year with prob 18689/18690, and will still live a time a after that, so there's a probability of 18689/18690 of living a+1 years, and 1/18690 of living 0 years.
This means that ( 18689/18690)(a+1) = a, wich means a = 18690.
The probability of living to the average age is ( 18689/18690)^18690, wich is about 36%. (or 1/e)

3. Jun 20, 2012

### moonman239

My apologies. He has about 50/50 chance of living to be 12,967 years old.

Clearly, assuming someone is in the US, there's a very small chance they'll be murdered.

4. Jun 23, 2012

### SW VandeCarr

There is no need to assume people are immortal unless they're murdered. A simple arithmetic model can be used which is more appropriate to this kind of problem and does in fact give a similar answer for the problem you're posing for age 82 but can also answer the question I posed in that thread. That is: Given a constant annual murder rate of 0.00005 and 1/3 mortality at age 82, what fraction of the cohort (people born in the same year) will have been murdered when the entire cohort has died?

We can use some simplifying assumptions because of the lack of data. First we assume a straight line model from 1 to 0 for the fraction of the original cohort that survives at any point. At age 82 0.67 survive. The murder rate only applies to survivors in the cohort and must be adjusted against the rate in the general population. Second, because this is a straight line model, we can use the median/mean age to calculate the average annual survival in the interval. Thirdly, we can simply add the annual murder rate intervals to obtain the murder rate over a larger interval.

The mean/median age between 0 an 82 is 41. At this point 5/6 of the cohort survive. So we adjust the murder rate for survivors over the interval using median/mean value:

(0.00005)(0.833)=0.00004165; (0.00004165)(82)=0.00342, fairly close to other estimates.

Since your survival rate is optimistic, we get a maximum lifetime estimate from the straight line model of 246. Using the same method I get a fraction of murdered when the entire cohort has died of 0.00615. In the real world, this figure would be a bit lower than the rate obtained for age 82. (The actuarial survival at age 82 in the US (2003) is about 0.46.)

Last edited: Jun 24, 2012
5. Jun 24, 2012

### awkward

If the probability that a person will die each year has a fixed value, say p, then his age at death follows a geometric distribution, and its expected value is 1/p.

6. Jun 25, 2012

### SW VandeCarr

Did you mean expectation or point at which the last member of a cohort has died using the straight line method? It's the latter. The expectation is the median/mean of this "curve" which is 123. Half the members will have died at this point given a constant 1/246 death rate per year of the original cohort, that population taken as unity.

While simple, this method doesn't fit the data for the real human mortality curve. (It does fit many invertebrates pretty well.) However, if you connect the point x=0 with x=t by a straight line with t:39<x<71 you get reasonable answers if you're just interested in the mortality at some end point t.

The actual curve is better approximated by the logistic model:

$\mu(x)=\alpha e^{\beta x}$ where alpha and beta are positive real numbers. There are more elaborate refinements to this model.

Last edited: Jun 26, 2012
7. Jun 25, 2012

### awkward

I mean the expected value:
$E[X] = \sum_x x \; p(x)$

I agree that the geometric distribution model is over-simplified, but that's what the OP asked-- at least the one in this thread. I gather from some of the posts that there may be a history I am not privy to.