STAT 333 Assignment 2 SOLUTIONS
1.
Consider a sequence of independent tosses of a fair coin. Each toss results in a head H or a tail T.
Let
λ
be the event that the total number of H equals exactly onethird the number of tosses, i.e.
we say
λ
occurs on the n
th
toss if the total number of H up to and including the n
th
toss equals n/3.
a.
Give an argument that
λ
is a renewal event
The first time we are waiting for this event to occur, if we want it to happen on trial 3n, we
will need
exactly n Heads and 2n Tails
.
Once the event occurs, say on trial 3k, we know there are exactly k Heads and 2k Tails so far.
Now if we want it to occur 3n trials later (on trial 3k+3n), we need a total of k+n Heads and
2(k+n) Tails in the total 3(k+n) flips, which means we need an
additional n Heads and 2n
Tails
in the next 3n trials. In other words, we need the exact same conditions for the event to
occur on trial 3n the very first time as after an earlier occurrence.
b.
What is the period of
λ
?
It can only occur on a multiple of three trials, so the period is d=3
c.
Prove that lambda is transient. Why does this make logical sense when you consider the law
of large numbers? (that with a large number of trials, the percentage of successes will
approach the probability of a success)
There are a couple of ways to do this.
We can see that the renewal sequence is r
0
=1, r
3n+1
= 0, r
3n+2
= 0, and r
3n
= (3n C n) (½)
3n
for
all integer n.
Using Stirling’s approximation,
3nC
n = (3n)!/(2n!)n! ≈
(3n)
3n
e
3n
√
(2
π
3n) / [ (2n)
2n
e
2n
√(
2
π2n)
(n)
n
e
n
√(
2
πn)
] = c (27/4)
n
/√n
So E[V
λ
] = c Σ ((27/4)*(½)
3
)
n
/√n = c Σ 0.84375
n
/ √n < ∞ since 0.84375 < 1.
Since the expected number of occurrences is finite, λ is transient.
This makes sense because as the number of trials approaches infinity, the proportion of heads
will approach ½. It would be impossible for the proportion to hit 1/3 infinitely often since it
is converging to ½.
2.
Suppose a sequence of independent trials X
1
,X
2
, . . . is generated by randomly picking digits
(with replacement) from the set {0, 1, 2, . . ., 9}. Prove that
the event “2 3 5”
is recurrent by:
a.
Finding r
n
= P(“2 3 5” occurs on trial n) and show
ing that
Σ
r
n
=
∞
r
0
= 1, r
1
= r
2
= 0, and r
n
= (1/10)
3
= 0.001 for n ≥ 3
E[V
235
] = Σ
∞
n=1
r
n
= 0 + 0 + Σ
∞
n=3
0.001 = ∞
Hence “235” will occur infinitely often and is recurrent.
b.
Obtaining the generating function R
235
(s) of the renewal sequence {r
n
}, finding the pgf
F
235
(s) of the first waiting time T
235
and evaluating F
235
(s) at s = 1 to prove recurrence.
R
235
(s) = Σ
∞
n=0
r
n
s
n
= 1 + 0 + 0 + Σ
∞
n=3
0.001s
n
= 1 + 0.001s
3
/(1s)
F
235
(s) = 1
–
1/R
235
(s) = 1
–
(1s)/(1
–
s + 0.001s
3
) = (0.001s
3
)/(1
–
s + 0.001s
3
)
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f
235
= F
235
(1) = 0.001/(1
1+0.001) = 1 so “235” is recurrent.
c.
Using F
235
(s) to obtain E(T
235
). Is “2 3 5” positive recurrent? Expand F
235
(s) in a power series
(you may use mathematical software) to find the probability f
7
that the first time “2 3 5”
occurs is on trial 7. Find a shortcut to get this probability using a logical argument.
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 Winter '08
 Chisholm
 Probability theory

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