An insurance company offers its policyholders a number of different premium payment options. For a randomly selected policyholder, let X = the number of months between successive payments. The cdf of X is as follows:

F (x) =

0 x < 1

0.33 1 < x < 3

0.44 3 < x < 4

0.48 4 < x < 6

0.86 6 < x < 12

1 12 < x

(a) What is the pmf of X?

x 1 3 4 6 12

p (x)

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

P (3 < X < 6) =

P (4 <

X) =

a)

The pmf of x is

x p (x)

1 0.33

3 0.11

4 0.04

6 0.38

12 0.14

b)

P (3≤X≤6) = 0.53

P (X≥4) = 0.56

Step-by-step explanation:

a)

We have to find pmf of x.

We know that

F (x) = P (X≤ x)

F (0) = P (X≤ 0) = P (X=0) = f (0)

F (1) = P (X≤ 1) = P (X=0) + P (X=1) = f (0) + f (1)

As, f (0) = F (0), so

f (1) = F (1) - F (0)

F (2) = P (X≤ 2) = P (X=0) + P (X=1) + P (X=2) = F (1) + f (2)

f (2) = F (2) - F (1)

So, we can say that

f (3) = F (3) - F (2)

f (4) = F (4) - F (3)

f (6) = F (6) - F (5)

f (12) = F (12) - F (11).

We are given that

F (0) = 0, F (1) = 0.33, F (2) = 0.33, F (3) = 0.44, F (4) = 0.48, F (5) = 0.48, F (6) = 0.86, F (7) = 0.86, F (8) = 0.86, F (9) = 0.86, F (10) = 0.86, F (11) = 0.86, F (12) = 1.

We have to find f (1), f (3), f (4), f (6) and f (12).

f (1) = F (1) - F (0) = 0.33-0=0.33

f (3) = F (3) - F (2) = 0.44-0.33=0.11

f (4) = F (4) - F (3) = 0.48-0.44=0.04

f (6) = F (6) - F (5) = 0.86-0.48=0.38

f (12) = F (12) - F (11) = 1-0.86=0.14

The pmf of x is

x p (x)

1 0.33

3 0.11

4 0.04

6 0.38

12 0.14

b)

P (3≤X≤6) = ?

We know that P (a
P (3≤X≤6) = P (2
P (3≤X≤6) = 0.53

P (X≥4) = ?

P (X≥4) = 1-P (X<4) = 1-P (X≤3) = 1-F (3) = 1-0.44=0.56

P (X≥4) = 0.56