Question

An insurance company offers its policyholders a number of different premium payment

options. For a randomly selected policyholder, let X = number of months between

successive payments. The cdf of X is as follows:

0  x < 1

0.30 1 ≤ x< 3

f(x)= 0.40 3 ≤ x < 4

0. 45 4 ≤ x < 6

0.60 6 ≤ x ≤ 12

1 12 ≤ x .

a) What is the pmf of X ?

b) Using just the cdf, compute P(3 ≤ X ≤ 6) and P(4 ≤ X).

Answer 1

Solution:

We are given with cumulative distribution function:

$F(x)-\left\{\begin{array}{cc}0 & x<1 \\ 0.3 & 1 \leq x<3 \\ 0.4 & 3 \leq x<4 \\ 0.45 & 4 \leq x<6 \\ 0.60 & 6 \leq x<12 \\ 1 & 12 \leq x\end{array}\right.$

 (a)

At every step, by subtracting subsequent cumulative distribution function, we can find probability distribution function.

$p(0)=0 \\p(1)=F(1)-F(0)=0.3-0=0.3 \\p(2)=0 \\p(3)=F(3)-F(2)=0.4-0.3=0.1 \\p(4)=F(4)-F(3)=0.45-0.40=0.05 \\p(5)=0 \\p(6)=F(6)-F(5)=0.60-0.45=0.15 \\p(6)=0 \\p(7)=0 \\p(8)=0 \\p(9)=0 \\p(10)=0 \\p(11)=0 \\p(12)=F(12)-F(11)=1-0.60=0.40$

(b)

We have to find probability between x=3 and x= inclusive. This is given by:

$\begin{aligned} &p(3 \leq x \leq 6)=F(6)-F(2) \\ &=0.60-0.30 \\ &=0.30 \end{aligned}$

Therefore, $p(3 \leq x \leq 6)=0.30$

Also, we have to find probability for $x$ greater than or equal to 4 . This is given by:

$\begin{aligned} &p(4 \leq x)=F(12)-F(3) \\ &=1-0.40 \\ &=0.60 \end{aligned}$

Therefore,

$p(4 \leq x)=0.60$