SOLUTIONS MANUAL for Stochastic Modeling: Analysis and ...

124 CHAPTER 9. TOPICS IN SIMULATION OF STOCHASTIC PROCESSES
∑ ki=1
X i Z i = 644.99
∑ kj=1
X j =80.3
∑ kh=1
Z h =80.3
̂σ XZ ≈ 0.02
16.
k∑
(V i − ¯V ) 2 =
i=1
=
=
=
k∑
(X i − Z i − ( ¯X − ¯Z)) 2
i=1
k∑
((X i − ¯X) − (Z i − ¯Z)) 2
i=1
k∑ {
(Xi − ¯X) 2 − 2(X i − ¯X)(Z i − ¯Z)
i=1
+(Z i − ¯Z)
2}
k∑
(X i − ¯X)
k∑
2 + (Z i − ¯Z) 2
i=1
i=1
k∑
− 2 (X i − ¯X)(Z i − ¯Z)
i=1
= (k − 1)(̂σ X 2 + ̂σ Z 2 − 2̂σ XZ )
17. There are mostly disadvantages.
• Unless we understand the bias quite well, we have no idea what overload approximately
compensates the underload. Therefore, the bias may still be quite significant,
and we will still have to study it.
• When we attempt to study the bias, we must either study the underload and overload
individually, which is twice as much work, or study them together, in which case the
combined bias process may have very unusual behavior.
An advantage is that we might be able to bound the bias by looking at convergence
from above and below.
18. If ̂θ is an estimator of θ, then the bias of ̂θ is E[̂θ − θ] where the expectation is with
respect to the distribution of ̂θ.
Let S 0 be the initial state, as in the Markov chain. Then, if we sample the initial state
from the steady-state distribution, we proved
E[̂θ] = ∑
E[̂θ | S 0 = x]Pr{S 0 = x} = θ (1)
x∈M