SOLUTIONS MANUAL for Stochastic Modeling: Analysis and ...

20 CHAPTER 3. BASICS
30. (a) F −1
X (q) =− ln(1 − q)/λ for the exponential. To obtain the median we set q =0.5.
Then η = − ln(1 − 0.5)/λ ≈ 0.69/λ. This compares to E[X] =1/λ.
Notice that
F X (1/λ) = 1− e −λ(1/λ)
= 1− e −1 ≈ 0.63
showing that the expected value is approximately the 0.63 quantile for all values
of λ.
(b) FX
−1 (q) =α +(β − α)q for the uniform distribution. For q =0.5, η = α +(β −
α)0.5 =(β + α)/2 which is identical to the E[X].
31. We want Pr{Y = b} =1/5 forb =1, 2,...,5.
Pr{Y = b} = Pr{X = b | X ≠6}
=
=
Pr{X = b, X ≠6}
Pr{X ≠6}
Pr{X = b}
Pr{X ≠6} = 1/6
5/6 =1/5
(definition)
32. (a)
Pr{Y = a} =
{
Pr Z = a
∣ V ≤ p }
Y (Z)
cp Z (Z)
= Pr { Z = a, V ≤ p }
Y (Z)
cp Z (Z)
Pr { } = (1)
V ≤ p Y (Z)
(2)
cp Z (Z)
(2) = ∑ {
Pr V ≤ p }
Y (a)
a∈B
cp Z (a) | Z = a p Z (a)
= ∑ a∈B
{
(1) = Pr
( )
pY (a)
cp Z (a)
p Z (a) = 1 ∑
p Y (a) = 1 c
a∈B
c
V ≤ p Y (a)
cp Z (a) | Z = a }
p Z (a)
= p Y (a)
cp Z (a) p Z(a) = p Y (a)
c
Therefore, (1)
(2) = p Y (a).