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 $c = Pr (X \leq a)$ $E [X] = \int_{- \infty}^{\infty} x Pr (X = x) d x = \int_{- \infty}^{a} x Pr (X = x) d x + \int_{a}^{\infty} x Pr (X = x) d x = \int_{- \infty}^{\infty} x Pr (X = x ∣ X \leq a) Pr (X \leq a) d x + \int_{- \infty}^{\infty} x Pr (X = x ∣ X \geq a) Pr (X \geq a) d x = c \int_{- \infty}^{\infty} x Pr (X = x ∣ X \leq a) d x + (1 - c) \int_{- \infty}^{\infty} x Pr (X = x ∣ X \geq a) d x = c E [X ∣ X \leq a] + (1 - c) E [X ∣ X \geq a]$

If $a = λb + (1 - λ) c$ , where $λ \in [0, 1]$ , 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.