How long do you expect to live?

While discussing conditional probability, someone said the following:

Problem1

The expected life expectancy of some country is 70, and there exist people who die at every age before 70. What is the expected life expectancy for a 60 year old?

Most people would answer 10. However, he continuous:

It could be 10, but for many distributions, it's likely more than that. You can convince yourself by thinking about the expected life expectancy for a 80 year old.

The quote above would follow directly from the proof of the following theorem:

Theorem2

For any real random variable $X$ , if $\Pr(X\geq a)>0$ , $E[X|X\geq a] \geq E[X]$ .

Proof

Let $c = \Pr(X\leq a)$ $\displaystyle \begin{aligned} E[X] &= \int_{-\infty}^\infty x \Pr(X=x) dx\\ &=\int_{-\infty}^a x \Pr(X=x) dx + \int_a^\infty x \Pr(X=x) dx\\ &=\int_{-\infty}^\infty x \Pr(X=x|X\leq a)\Pr(X\leq a) dx + \int_{-\infty}^\infty x \Pr(X=x|X\geq a)\Pr(X\geq a) dx\\ &=c\int_{-\infty}^\infty x \Pr(X=x|X\leq a) dx + (1-c)\int_{-\infty}^\infty x \Pr(X=x|X\geq a) dx \\ &=cE[X|X\leq a] + (1-c)E[X|X\geq a] \\ \end{aligned}$

If $a = \lambda b + (1-\lambda) c$ , where $\lambda \in [0,1]$ , then $a \leq \max(b,c)$ . Because $E[X|X\leq a]\leq a \leq E[X|X\geq a]$ , $E[X] \leq E[X|X\geq a]$ .

In fact, one can easily modify the above proof and prove the next theorem:

Theorem3

For any real random variable $X$ , if $x\geq y$ and $\Pr(X\geq x)>0$ , then $E[X|X\geq x] \geq E[X|X\geq y]$ .

A heuristics conclusion: The longer you lived, you expect to live longer.

Posted by Chao Xu on 2011-06-15.

Tags: probability.