How to calculate expected value of normal distribution with the condition that value is higher than x

Question

I have following problem. Let assume that lifespan in the population has normal distribution with certain mean, variance and skewness.

When the baby is born, its average lifespan will be equal to mean - this is obvious. But let's assume that the person did not die for x years and now its age is x. What is expected lifespan for this person if we know that he or she lives already x years?

How we calculated it if the distribution was e.g. Lévy stable distribution (or any other)?

Answer 1

In general, you are asking the following question. If we know the distribution of $Y$, what is $E(Y|Y > x).$ The fundamental idea is to take the part of the distribution of $Y$ to the right of $x,$ re-scale that part so that it sums (or integrates) to unity, and then find the mean of the new conditional distribution. Methods and results differ greatly depending on the distribution.

If $Y$ is exponential, then (by the no-memory property), the conditional mean of additional life, starting at $x$ is the same as the mean of the original distribution. Some electronic devices that "die" exclusively by random accident have this property. A used one is as good as a new one. If the device had a 10-year expected lifetime when made, and it survives to age 5, then its expected additional lifetime is still 10 years. Obviously, this distribution does not apply to human lifetimes--but neither does a normal distribution.

If $Y$ has a uniform distribution on $(0, 100)$ starting at time $0$ and survives to age $x = 40,$ then the remaining part of the distribution is uniform on $(40, 100).$ Its expected lifetime is now 30 more years or a total of 70.

Suppose $Y$ has a normal distribution with mean 50 and and standard deviation 10. (As in the Comment by @Michael Hardy, this is a symmetrical distribution with zero skewness.) Suppose we are given that $Y > 60,$ then the remaining part of the distribution is not normal but considerably skewed, it would require numerical methods to find the mean of the corresponding conditional distribution. I got about 65.2, or about 5.2 additional years.

An actual lifetime distribution of humans typically has a decreasing death rate over the early months of life, as babies survive many special hazards of newborns and then an increasing death rate as bodies begin to wear out. That is why most computations on actual lifetimes use life tables based on actual data rather than theoretical distributions. The principle is the same, but you would have to look at the life table for specifics. Many life tables include projections for additional years of life at each age. (In practice, human lifetime distributions are not exponential, uniform, or normal.)

Addendum: Below are simple simulations in R, each based on a million cases, illustrating the statements above about exponential, uniform, and normal distributions. The method is not optimal, but very easy to program and adequate for present purposes. Accuracy is to the nearest integer, maybe a little more.

 # Exponential
 y1 = rexp(10^6, 1/10)  # death rate 1/10 per year
 mean(y1)
 ## 10.00546  # check that original mean is 10
 mean(y1[y1 > 5])  # survives to age 5
 ## 15.01287  # conditional mean 15;  10 years additional life

 # Uniform
 y2 = runif(10^6, 0,  100)
 mean(y2)
 ##  49.95337 # check that original mean is 50
 mean(y2[y2 > 40])  # survives to age 40
 ## 70.00422  # conditional mean 70,  30 years additional life

 # Normal  
 y3 = rnorm(10^6, 50, 10)
 mean(y3)
 ## 50.01684  # check that original mean is 50
 mean(y3[y3>60])  # survives to age 60
 ## 65.26052  # conditioal mean 65.2.  5.2 years additional life