conditional probability question from sheldon ross

This question is quite spectacularly devious, but actually leads to some really satisfying results.

HINT:

Suppose WLOG that $p_m>p_f$ - men are more likely to have accidents than women. The intuition behind your problem is that, if you have an accident in year one, it's more likely that you're a man, so it's more likely that you'll go on to have an accident in year 2. This should be enough of a hint to get you started, so spend some time thinking along these lines before you read on. You don't actually need this assumption for the proof to work out.

SPOILER:

So, we'll use $M$ and $F$ to denote the events that the policyholder is male and female respectively, and let $\widetilde{\alpha} = (1-\alpha)$. Conditioning on gender, the probability that you have an accident in year 1 is

\begin{align*} \mathbb{P}(A_1) &= \mathbb{P}(A_1 | M)\mathbb{P}(M) + \mathbb{P}(A_1|F)\mathbb{P}(F) \\ &= \alpha p_m + \widetilde{\alpha}p_f \end{align*}

And, using the same conditioning, the probability you have an accident in year 2 given that you had one in year 1 is

\begin{align*} \mathbb{P}(A_2|A_1) &= \mathbb{P}(A_2 | A_1 \cap M)\mathbb{P}(M|A_1) + \mathbb{P}(A_2|F\cap A_1)\mathbb{P}(F |A_1) \\ &= \frac{\alpha p_m^2 + \widetilde{\alpha}p_f^2}{\mathbb{P}(A_1)} \end{align*}

Now, we're trying to prove that $\mathbb{P}(A_2|A_1)>\mathbb{P}(A_1)$. We'll instead show that $\mathbb{P}(A_2|A_1)\mathbb{P}(A_1)> \mathbb{P}(A_1)^2$, which is true iff

$$ \alpha p_m^2 + \widetilde{\alpha}p_f^2 > (\alpha p_m + \widetilde{\alpha}p_f)^2 $$

A quick bit of fiddling, recalling that $\widetilde{\alpha} = (1-\alpha)$ and $(1- \widetilde{\alpha}) = \alpha$, shows that this is true if and only if

$$ (p_m - p_f)^2 > 0 $$

And since $p_m \neq p_f$, this is clearly true, and the result follows.

This sort of thing will probably give you the same result regardless of what we take into account. If we don't consider gender, but instead consider some function of every property a human might have, you'll probably get the same kind of result, providing a sound mathematical basis to the idea that past performance is a good indicator of future performance, without having to rely on any data gathered by experiment.