3.46 - A First Course in Probability

Question

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 $\ne$}p_m. The fraction of the policyholders that are male is α, 0 <α< 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(A₂|A₁) > P(A₁) 

Give an intuitive explanation of why the preceding inequality is true. 

Step-by-Step Solution

Answer

Probabilities are weighted averages of p_m and p_f. Starting weights are $\alpha$ and$1-\alpha$ .If a accident occur there is more probability for category with high risk of accidents.

Step 1: Given information

Events  

 M = A male automobile policyholder

 F =  A Female automobile policyholder

A_i = A person participate in an accident  in an i -th  year 

  The percentage of male  is given 

              P(M) = $\alpha$

Hence the percentage of female is 

   P(F) =(1 - $\alpha$)

Men and women have different Probabilities 

   p_m = probability for the female policyholder

    p_m = P( $\frac{A_i}{M}$)

     p_f  = P($\frac{A_i}{F}$)

 For every i

Step 2: Solution

pmαP(A1) >α ⇔pmαP(A1)=α +∈ for Intuitively we can say that probability of an accident for a person is somewhere between the probabilities for male and female. It is an average of being p_m and p_f which is the probability that a person is male or female.

If an accident occurs, men are more likely to have accidents than women .there is more chance in the category with a higher risk of accidents so the probability that another accident happens is greater.

   Formaly

Conditioning on M and F gives firstly

        P(A₁) =P$\left(\frac{A_1}{F}\right)$P(F) + P$\left(\frac{A_1}{M}\right)$P(M)

        P(A1) = P_f . (1 - $\alpha$)

For P$\left(\frac{A_2}{A_1}\right)$ also condition on F and M:

P$\left(\frac{A_2}{A_1}\right)$ = $\frac{P(A_2.A_1)}{P\left(A_1\right)}$=

$\frac{P{\displaystyle \left(\frac{A_2{A}_1}{M}\right)}P\left(M\right)+P{\displaystyle \left(\frac{A_2{A}_1}{F}\right)}P\left(F\right)}{P\left(A_1\right)}$

Since A₁ and A₂ are conditionally independent given the condition F or M 

   P($\left(\frac{A_2}{A_1}\right)$) = $\frac{P\left({\displaystyle \frac{A_2}{M}}\right)P\left({\displaystyle \frac{A_1}{M}}\right)P\left(M\right) +P\left({\displaystyle \frac{A_2}{F}}\right)P\left({\displaystyle \frac{A_1}{F}}\right)P\left(F\right)}{P\left(A_1\right)}$

P$\left(\frac{A_2}{A_1}\right)$=$\frac{{p^2}_m \alpha + {p^2}_f (1 - \alpha )}{P\left(A_1\right)}$ +

P $\left(\frac{A_2}{A_1}\right)$ = p_m $\frac{p_m\alpha }{P\left(A_1\right)}$ +p_f $\frac{p_{f }. (1 -\alpha )}{P\left(A_1\right)}$

without loosing abstraction p_m $>$ p_f ,

         $\frac{p_m}{P\left(A_1\right)}>1$

Because 

  p_m$>p_f .(1- \alpha )+p_m .\alpha$

  p_m$$\begin{gathered} (1-\alpha ) > p_f(1-\alpha ) \\ from that \\ \frac{p_m\alpha }{P\left(A_1\right)}>\alpha \iff \frac{p_m\alpha }{P\left(A_1\right)}=\alpha +\in for some \in >0 \end{gathered}$$ and

   $\frac{p_{f }.(1-\alpha )}{P\left(A_1\right)}=1-\frac{p_m\alpha }{P\left(A_1\right)} \iff \frac{p_m(1-\alpha )}{P\left(A_1\right)}=1-\alpha -\in$ So

       $$\begin{gathered} P\left(\frac{A_2}{A_1}\right)-P\left(A\right)= p_m(\alpha +\in )+p_f(1-\alpha -\in )-p_m\alpha -p_f(1-\alpha ) \\ =p_m\in +p_f(-\in ) \\ =\in (p_m-p_f) \end{gathered}$$

 

since we assumed firstly that p_m$>p_f$

 so $\in >0$

     hence p$\left(\frac{A_2}{A_1}\right)$ - P(A₁)$>0$ 

         Or P$\left(\frac{A_2}{A_1}\right)$ > P(A₁)

This concludes the proof.

Step 3: conclusion

both probabilities are weighted averages of p_m and p_f . If an accident occurs The person mostly included in the category with higher risk of accidents.