The Art Gallery Guardian

How long do you expect to live?

While discussing conditional probability, someone said the following:


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:


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


Let c=Pr(Xa)c = \Pr(X\leq a) E[X]=xPr(X=x)dx=axPr(X=x)dx+axPr(X=x)dx=xPr(X=xXa)Pr(Xa)dx+xPr(X=xXa)Pr(Xa)dx=cxPr(X=xXa)dx+(1c)xPr(X=xXa)dx=cE[XXa]+(1c)E[XXa]\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=λb+(1λ)ca = \lambda b + (1-\lambda) c, where λ[0,1]\lambda \in [0,1], then amax(b,c)a \leq \max(b,c). Because E[XXa]aE[XXa]E[X|X\leq a]\leq a \leq E[X|X\geq a], E[X]E[XXa]E[X] \leq E[X|X\geq a].

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


For any real random variable XX, if xyx\geq y and Pr(Xx)>0\Pr(X\geq x)>0, then E[XXx]E[XXy]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 .
Tags: probability.