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Math 171 Homework #3 – due Friday 4/
π(x) =
y∈S
π(y)p(y, x)
= π(0)px + π(x + 1) π(x + 1) = π(x) − π(0)px
Taking π(0) to be known, we compute
π(1) = π(0) π(2) = π(0)(1 − p 1 ) π(3) = π(0)(1 − p 1 − p 2 ) .. .
π(x) = π(0)
∑^ x−^1
j=
pj
Which explicitly defines π(x). Now, we require that the sum of π(x) over all states be unity:
∑^ ∞
x=
π(x) = 1 = lim N →∞
∑^ N
x=
π(x)
= lim N →∞
π(0)
[
1 + 1 + 1 − p 1 + 1 − p 1 − p 2 +...
]
= lim N →∞
π(0)
[
1 + N − (N − 1)p 1 − (N − 2)p 2 −... − 2 pN − 2 − 2 pN − 1
]
= π(0) + lim N →∞
π(0)
[
N (1 −
N∑ − 1
x=
px) +
N∑ − 1
x=
xpx
]
1 = π(0)
∑^ ∞
x=
xpx
π(0) =
x=
xpx
We see that if the expectation value of the return time E(T ) is infinite, then π(0) = 0 and thus π(x) = 0 for all x. So, for the system to be positive recurrent, we require that the expectation value of the return time to be finite.
The limiting probability distribution is then given by
π(x) =
μ
[
∑^ x−^1
j=
pj
]
P{x 1 ≥ x|x 0 = 0} μ
where μ = E(T ).
a. a = 0.25+0. 4 a+0. 35 a^2 has roots 1 and 0.71429. So we have approximately a 71% chance of extinction.
b. a = 0.5 +. 1 a +. 4 a^2 has roots 1 and 1.25, so the system goes extinct with certainty.
c. μ < 1, so the system goes extinct with certainty.
d.
φ(a) =
∑^ ∞
i=
Pkai
∑^ ∞
i=
(1 − q)qiai
1 − q 1 − qa
Then we solve
q = φ(a)
=
1 − q 1 − qa qa^2 − a + (1 − q) = 0
yielding a = 1 and a = (^1) q − 1. So, if q > 12 , then the extinction probability is
- Otherwise, the extinction probability is (^1) q − 1.
a.
P{X 2 = 0|X 1 6 = 0} = P{X 2 = 0|X 1 = 1}p 1 + P{X 2 = 0|X 1 = 3}p 3 = p 0 p 1 + p^30 p 3
