The Art Gallery Guardian

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 \(c = \Pr(X\leq a)\) \begin{align*} 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{align*}

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: ###### {type=Theorem index=3 name=} 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.