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 X, if \Pr(X\geq a)>0, E[X|X\geq a] \geq E[X].
Let c = \Pr(X\leq a) [\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:
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.