Solutions to Problems from Chapter 4 of Ross
Exercise 1 (Ross 4.17). Suppose that the distribution function of X is given by


0
b<0



b


0≤b<1
4
1
b−1
F (b) = 2 + 4
1≤b<2


11

2≤b<3

2


1
3 < b.
(a) Find P (X = i) for i = 1, 2, 3.
(b) Find P (1/2 < X < 3/2).
Solution. Recall that P (X < t) = lims↑t F (s). We denote this limit-from-the-left by F (t−).
We have
P (X = i) = P (X ≤ i) − P (X < i) = F (i) − f (i−) .
So
1
1 1
− = ,
2 4
4
11 3
1
P (X = 2) = F (2) − F (2−) =
− = ,
12 4
6
1
11
=
P (X = 3) = F (3) − F (3−) = 1 −
12
12
P (X = 1) = F (1) − F (1−) =
For the second part,
P (1/2 < X < 3/2) = P (X < 3/2) − P (X ≤ 1/2) = F (3/2−) − F (1/2) =
Exercise 2 (Ross 4.19). If the distribution

0




1


2


3
F (b) = 45


5


9


 10


1
calculate the pmf of X.
1
function of X is given by
b<0
0≤b<1
1≤b<2
2≤b<3
3 ≤ b < 3.5
b ≥ 3.5
5 1
1
− = .
8 8
2
Solution. The pmf satisfies
p(a) = P (X = a) = F (a) − F (a−) .
Thus p(a) > 0 only at the values t where F has a jump. So
1
2
1
3 1
P (X = 1) = F (1) − F (1−) = − =
5 2
10
4 3
1
P (X = 2) = F (2) − F (2−) = − =
5 5
5
4
1
9
− =
P (X = 3) = F (3) − F (3−) =
10 5
10
9
1
P (X = 3.5) = F (3.5) − F (3.5−) = 1 −
=
.
10
10
P (X = 0) = F (0) − F (0−) =
At all other values a, we have p(a) = 0.
Exercise 3 (Ross 4.T3). Express P (X ≥ a) in terms of the d.f. of X.
Solution.
P (X ≥ a) = 1 − P (X < a) = 1 − F (a−) ,
where again F (a−) = lims↑a F (s).
Exercise 4 (Ross 4.T4). If X has d.f. X, what is the d.f. of eX .
Solution. Let Y = eX . The question asks for the d.f. of Y . We will find this in terms of
the d.f. FX of X.
FY (t) = P (Y ≤ t)
= P (eX ≤ t)
= P (X ≤ log(t))
= FX (log(t)) .
Exercise 5 (Ross 4.T5). Find the d.f. for αX + β, where α 6= 0.
2
Solution.
FαX+β (t) = P (αX + β ≤ t)
(
P (X ≤ t−β
) if α > 0
α
=
t−β
P (X ≥ α ) if α < 0
(
)
if α > 0
FX ( t−β
α
=
t−β
1 − FX ( α −) if α < 0
3