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.

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(role="math" localid="1646394713725" $\frac{A_i}{M}$)

p_f= P($\frac{A_i}{F}$)

For every i

pmαP(A1)>α⇔pmαP(A1)=α+∈forIntuitively 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 Plocalid="1646415690917" $\left(\frac{A_2}{A_1}\right)$also condition on F and M:

Plocalid="1646415702142" $\left(\frac{A_2}{A_1}\right)$= $\frac{P(A_2.A_1)}{P\left(A_1\right)}$=

localid="1646415716744" $\frac{P{\displaystyle \left(\frac{A_2{A}_1}{M}\right)}P\left(M\right)+P{\displaystyle \left(\frac{A_2{A}_1}{F}\right)}P\left(F\right)}{P\left(A_1\right)}$

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

P(localid="1646415741605" $\left(\frac{A_2}{A_1}\right)$) = localid="1646415756577" $\frac{P\left({\displaystyle \frac{A_2}{M}}\right)P\left({\displaystyle \frac{A_1}{M}}\right)P\left(M\right)+P\left({\displaystyle \frac{A_2}{F}}\right)P\left({\displaystyle \frac{A_1}{F}}\right)P\left(F\right)}{P\left(A_1\right)}$

Plocalid="1646415781783" $\left(\frac{A_2}{A_1}\right)$=$\frac{{p^2}_m\alpha +{p^2}_f(1-\alpha )}{P\left(A_1\right)}$+

P localid="1646415798376" $\left(\frac{A_2}{A_1}\right)$ = 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_mlocalid="1646402594084" $$\begin{gathered} (1-\alpha )>p_f(1-\alpha ) \\ fromthat \\ \frac{p_m\alpha }{P\left(A_1\right)}>\alpha \iff \frac{p_m\alpha }{P\left(A_1\right)}=\alpha +\in forsome\in >0 \end{gathered}$$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

localid="1646415846120" $$\begin{gathered} P\left(\frac{A_2}{A_1}\right)-P\left(A\right)=p_m(\alpha +\in )+p_f(1-\alpha -\in )-p_m\alpha -p_f(1-\alpha ) \\ =p_m\in +p_f(-\in ) \\ =\in (p_m-p_f) \end{gathered}$$

since we assumed firstly that p_m$>p_f$

so $\in >0$

hence plocalid="1646415830743" $\left(\frac{A_2}{A_1}\right)$ - P(A₁)$>0$

Or Plocalid="1646415814589" $\left(\frac{A_2}{A_1}\right)$ > P(A₁)

This concludes the proof.

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.