In any given year a male automobile policyholder will make a claim with probability $p_{m}$, and a female policyholder will make a claim with probability $p_{f}$, where $p_{f} \neq p_{m}$. The fraction of the policyholders that are male is $\alpha, 0 < \alpha < 1$. A policyholder is randomly chosen. If $A_{i}$ denotes the event that this policyholder will make a claim in year $i$, show that $P(A_{2}\mid A_{1}) > P(A_{1})$.

It appears that $A_{2}$ and $A_{1}$ are independent events, hence I don't understand why the inequality should hold. Can anybody please help? Thanks a lot.

  • $\begingroup$ They are only independent conditioned on the gender. As a check, suppose that the probability that $p_f = \varepsilon$ is close to $0$, and $p_m = 1-\varepsilon$, and $\alpha = 1/2$. Then $P(A_1) = 1/2$, but $P(A_2 \mid A_1)$ is very close to $1$. (I think in fact it is $\frac{(1-\varepsilon)^2}{\varepsilon^2+(1-\varepsilon)^2}$.) ETA: Oops, I just realized this question is ancient. Oh, well, it's worth making the short observation. $\endgroup$
    – Brian Tung
    May 29, 2019 at 20:37

3 Answers 3


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


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.


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.

  • $\begingroup$ +1. // Why "quite spectacularly devious"? $\endgroup$
    – Did
    Sep 21, 2013 at 16:46
  • $\begingroup$ Because it initially looks like a very simple and slightly odd question, but leads to some useful conclusions. I'm fairly sure this was intended by the question's author, and I don't use "devious" in any kind of disparaging way. $\endgroup$
    – ymbirtt
    Sep 21, 2013 at 17:16
  • $\begingroup$ OK. (I would not say it is, but nevermind.) $\endgroup$
    – Did
    Sep 21, 2013 at 17:20
  • 2
    $\begingroup$ @ymbirtt: How did you calculate $P(A_{2}\mid A_{1} \cap M)$ in the above derivation? $\endgroup$
    – akhil
    Nov 5, 2013 at 10:10
  • $\begingroup$ @akhil Implicit from the first part of the question, accidents occur independently each year for each person - $A_2$ and $A_1$ are conditionally independent given the occurrence or non-occurrence of $M$. This means that $\mathbb{P}(A_2 | A_1 \cap M) = \mathbb{P}(A_2 | M)$ $\endgroup$
    – ymbirtt
    Nov 5, 2013 at 12:11

Intuitively, if a policy holder makes a claim in year $1$ then that event suggests that they are more likely to have been in the group which is more likely to make a claim than that group's share of the general population.

In turn, being more likely to be to the group which is more likely to make a claim suggests that they are more likely than the general population to make a claim in year $2$.

You could formalise this into conditional probability expressions, but I doubt that doing so would increase understanding by much.

  • $\begingroup$ +1. I have explicitly formalized your intuition into an answer. Please review it. $\endgroup$
    – Hans
    May 29, 2019 at 19:41

I put the intuitive answers of ymbirtt and Henry into mathematical expression as follows. WLOG, we assume $p_m>p_f$.

\begin{align} P(A_2|A_1)&=\frac{P(A_1\cap A_2)}{P(A_1)} \\ &=\frac{P(\text{male}\cap A_1\cap A_2)}{P(A_1)}+\frac{P(\text{female}\cap A_1\cap A_2)}{P(A_1)} \\ &=\frac{P(A_1\cap A_2|\text{male})P(\text{male})+P(A_1\cap A_2|\text{female})P(\text{female})}{P(A_1)} \\ &=\frac{P(A_2|\text{male})P(A_1|\text{male})P(\text{male})+P(A_2|\text{female})P(A_1|\text{female})P(\text{female})}{P(A_1)} \\ &= \frac{\alpha p_m^2+(1-\alpha)p_f^2}{\alpha p_m+(1-\alpha)p_f} \end{align} as $P(A_1)=\alpha p_m+(1-\alpha)p_f$. \begin{align} (\alpha p_m+(1-\alpha) p_f)^2&=(\sqrt{\alpha} \sqrt{\alpha}p_m+\sqrt{1-\alpha}\sqrt{1-\alpha} p_f)^2 \\ &\le (\alpha p_m^2+(1-\alpha)p_f^2)(\alpha+(1-\alpha)), \end{align} with the equality achieved only at $p_m=p_f$, by the Cauchy-Schwartz inequality. So $P(A_2|A_1)\ge P(A_1)$ with the quality achieved only at $p_m=p_f$.

  • $\begingroup$ Why is $P(male \cap A_{2})$ not equal to $\alpha p_{m}$? From what I understand, $P(male \cap A_{2}) = P(male \cap A_{1}) = \alpha p_{m}$. $\endgroup$
    – Ankit Seth
    Aug 14, 2021 at 9:21
  • $\begingroup$ @AnkitSeth: You are right. $P(\text{male}\cap \text{male})\ne P(\text{male})P(\text{male})$. I have modified and simplified my proof. Check it out. $\endgroup$
    – Hans
    Aug 16, 2021 at 6:49

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