Math 180C, Spring 2016
Homework 1 Solutions
Ex. 6.1.1. P0 (t) = e−t . For the rest I use the recursion (formula (6.5) on page 280 of the text)
Z
t
Pn (t) = λn−1
e−λn (t−s) Pn−1 (s) ds
0
for n = 1, 2, 3:
Z t
e−3(t−s) P0 (s) ds = e−3t
e3s e−s ds
0
0
Z t
= e−3t
e2s ds = e−3t (e2t − 1)/2
Z
t
P1 (t) =
0
1
1
= e−t − e−3t .
2
2
Z
3 t −2(t−s) −s
e
(e − e−3s ) ds
e−2(t−s) P1 (s) ds =
2
0
0
Z
3 −2t t s
3 −2t t
−s
= e
(e − 1) − (1 − e−t )
(e − e ) ds = e
2
2
0
3 −t
= (e − 2e−2t + e−3t ).
2
Z t
Z t
−5(t−s)
−5t
P3 (t) = 2
e
P2 (s) ds = 3e
e5s (e−s − 2e−2s + e−3s ) ds
0
0
Z t
−5t
= 3e
(e4s − 2e3s + e2s ) ds
0
2 3t
1 2t
−5t 1 4t
(e − 1) − (e − 1) + (e − 1)
= 3e
4
3
2
3
3
1
= e−t − 2e−2t + e−3t − e−5t .
4
2
4
Z
t
P2 (t) = 3
Ex. 6.1.2. (a) W3 = S0 + S1 + S2 , so
E[W3 ] = E[S0 ] + E[S1 ] + E[S2 ]
−1
−1
= λ−1
0 + λ1 + λ2
1 1
11
=1+ + =
.
3 2
6
(b) Similarly, E[W1 ] = 1 and E[W2 ] = 4/3, so E[W1 + W2 + W3 ] = 1 + 4/3 + 11/6 = 25/6.
(c) The variance of W3 is the sum of the variances of S0 , S1 , and S2 . We know that the variance of an
exponentially distributed random variable with parameter λ is 1/λ2 . Therefore,
Var[W3 ] = 1 +
1 1
49
+ =
.
9 4
36
Ex. 6.1.5. According to formula (6.10) (page 282 of the text), if X(0) = 1 then X(t) has the geometric
distribution with parameter p = e−βt . From known formulas for the mean and variance of a geometric
random variable we deduce that
E[X(t)] = eβt
1
and
Var[X(t)] = e2βt (1 − e−βt ).
(Cf. the formulae for the moments of Z 0 on page 21 of the text.)
Pr. 6.1.3. If, at time t, there are X(t) infected individuals, then at that time there are N −X(t) susceptible
individuals in the population. Since the individual infection rate (for each infected/susceptible pair) is α,
the total infection rate, when X(t) = k, is αk(N − k). That is, λk = αk(N − k) for k = 0, 1, 2, . . . , N .
Pr. 6.1.8. Evidently,
P0 (t + h) = P0 (t)(1 − βh) + P1 (t)αh + o(h),
h→0+.
Consequently,
o(h)
P0 (t + h) − P0 (t)
= −βP0 (t) + αP1 (t) +
,
h
h
whence
P00 (t) = −βP0 (t) + αP1 (t).
(6.1.8.1)
Similarly,
P10 (t) = −αP1 (t) + βP0 (t).
(6.1.8.2)
Because P1 (t) = 1 − P0 (t), equation (6.1.8.1) can be rewritten as
P00 (t) = −(β + α)P0 (t) + α.
Thus,
P00 (t) + (β + α)P0 (t) = α,
so
d (α+β)t
e
P0 (t) = αe(α+β)t .
dt
Integrating we find that
e(α+β)t P0 (t) − 1 =
α
(e(α+β)t − 1),
α+β
because P0 (0) = 1. It follows that
P0 (t) =
β
α
+
e−(α+β)t .
α+β
α+β
Because P1 (t) = 1 − P0 (t), we also have
P1 (t) =
β
β
−
e−(α+β)t .
α+β
α+β
Pr. 6.1.9. [This problem, as stated in the text, is much harder than the authors may have intended. To
make the problem do-able, as discussed in class, we assume that N (t) is a pure birth process N (t) with birth
rates λk = α if k is odd and λk = β if k is even.]
Let Pk (t) = P[N (t) = k] for k = 0, 1, 2, . . . and M (t) = E[N (t)]. Let us also write η(t) for P[N (t) is
even]. (This was called P0 (t) in Problem 6.1.8.) We know that
0
P2k
(t) = −βP2k (t) + αP2k−1 (t),
2
k = 1, 2, . . . ,
and
0
P2k+1
= −αP2k+1 (t) + βP2k (t),
k = 0, 1, 2, . . . .
Therefore,
M 0 (t) =
∞
X
0
2kP2k
(t) +
k=1
= −β
∞
X
0
(2k + 1)P2k+1
(t)
k=0
∞
X
2kP2k (t) + α
k=1
∞
X
2kP2k−1 (t) − α
k=1
∞
X
(2k + 1)P2k+1 (t) + β
k=0
∞
X
(2k + 1)P2k (t)
k=0
= βη(t) + α(1 − η(t))
= α + (β − α)η(t).
Using the formula for η(t) from problem 6.1.8, we see that
β
α
−(α+β)t
0
+
e
.
M (t) = α + (β − α)
α+β
α+β
Since M (0) = 0, we must have
M (t) =
β(β − α)
2αβ
t+
(1 − e−(α+β)t ).
α+β
(α + β)2
Ex. 6.2.1. Using the formulas on page 287 of the text (and a little patience), we find that
P3 (t) = e−5t
5 −2t
P2 (t) =
e
− e−5t
3
5 −2t
P1 (t) =
2e
− 3e−3t + e−5t ,
3
and of course P0 (t) = 1 − P1 (t) − P2 (t) − P3 (t), so that
P0 (t) = 1 − 5e−2t + 5e−3t − e−5t .
As a check, note that P3 (0) = 1, P2 (0) = P1 (0) = P0 (0) = 0, and
P30 (t) = −5P3 (t)
P20 (t) = −2P2 (t) + 5P3 (t)
P10 (t) = −3P1 (t) + 2P2 (t), and
P00 (t) = 3P1 (t),
as expected.
Alternatively, you can use the formula P3 (t) = e−µ3 t , and the recursion discussed in class:
Z
Pk (t) = µk+1
t
e−µk (t−s) Pk+1 (s) ds,
k = 2, 1, 0,
0
to compute (in succession) P2 (t), P1 (t), and P0 (t).
Ex. 6.2.2. (a) W3 = S3 + S2 + S1 , so
E[W3 ] = E[S3 ] + E[S2 ] + E[S1 ] =
3
1 1 1
31
+ + =
.
5 2 3
30
(b) Similarly, E[W2 ] = 5/6 and E[W1 ] = 1/3, so E[W1 + W2 + W3 ] = 11/5.
(c) Var[W3 ] = Var[S3 ] + Var[S2 ] + Var[S1 ] = 1/25 + 1/4 + 1/9 = 361/900 = .4011 . . ..
Pr. 6.2.2. Since the death rates are all the same (namely θ), the sojourn times in the various states
all have the same exponential distribution, and their sums have gamma distributions. More precisely, for
k = 1, 2, . . . N , the random variable Wk = SN + SN −1 + · · · + SN −k+1 has the gamma distribution with
parameters θ and k; that is, the density function of Wk is
fWk (t) =
θk tk−1 e−θt
,
(k − 1)!
t > 0.
Therefore
P[X(t) = n] = P[WN −n ≤ t < WN −n+1 ] = P[WN −n ≤ t] − P[WN −n+1 ≤ t],
n = 1, 2, . . . , N.
But we know from studying Poisson processes that
P[Wk ≤ t] =
∞
X
e−θt
j=k
(θt)j
.
j!
Combining this with the last-displayed equation we find that
P[X(t) = n] = e−θt
(θt)N −n
,
(N − n)!
Similarly,
P[X(t) = 0] = P[WN ≤ t] =
n = 1, 2, . . . N.
∞
X
j=N
e−θt
(θt)j
.
j!
Pr. 6.2.3. A glance at the picture on page 287 of the text should be enough to convince you that the area
under the trajectory of the pure death process is
N
X
k · Sk .
k=1
Consequently, the desired expectation is
N
X
k
.
µk
k=1
4