Stat 310B/Math 230B Theory of Probability
Homework 6 Solutions
Andrea Montanari
Due on 2/26/2014
Exercise [5.4.14]
Recall that ξk are i.i.d. random variables taking values in {A, B, . . . , Z} such that P(ξk = x) = 1/26 if
x ∈ {A, B, . . . , Z} and 0 otherwise.
Pn
(a). The amount of money the gamblers have collectively earned by time n is Mn = i=1 Zi,n , where Zi,n
is the amount of money that the gambler who enters the game just before time step i earns by time step n.
For example, if the letter the monkey types at time i is P then Zi,i = $25, since this gambler paid $1 to bet
and received $26 upon successfully betting on P.
Clearly, Mn is measurable on Fnξ = σ(ξk , k ≤ n) and since each Zi,n is bounded, so is Mn . To show that
Mn is a MG with respect to Fnξ observe that
E(Mn+1 |Fnξ ) =
n
X
E(Zi,n+1 |Fnξ ) + E(Zn+1,n+1 |Fnξ ) .
i=1
The gambler who enters the game at time n + 1 is independent of all the events through time n, hence
E(Zn+1,n+1 |Fnξ ) = E(Zn+1,n+1 ) = 25
25
1
+ (−1)
= 0.
26
26
The earnings of a gambler who entered the game at time i ≤ n and has either lost by the end of time step
n, or left the game upon wining all 11 rounds, are unchanged from time n to n + 1. Any other gambler
who entered the game at time i ≤ n bets all he has, that is 1 + Zi,n , on the outcome of Monkey’s typing at
step n + 1. With probability 1/26 he will have as a result 26(Zi,n + 1) and otherwise, he loses everything
and departs. Since he initially paid $1 for his first gamble, his total earnings after time step n + 1 are
Zi,n+1 = 26(Zi,n + 1) − 1 with probability 1/26 and −1 with probability 25/26. Thus,
E(Zi,n+1 |Fnξ ) =
1
25
E[26(Zi,n + 1) − 1|Fnξ ] −
= Zi,n .
26
26
Consequently, E(Mn+1 |Fnξ ) = Mn with M0 = 0.
(b). Let ξij = (ξi , ξi+1 , . . . , ξj ) for j > i noting that the monkey first types PROBABILITY at the stopping
time
n
τb = inf{n : ξn−10
= (P, R, O, B, A, B, I, L, I, T, Y )} with respect to the filtration {Fnξ }. Setting a = 2611 ,
at time τb exactly one player have won and all the others have lost, from which it follows that Mτb = a − τb.
As each gambler quits after he wins the entire word, we have that always |Mn+1 − Mn | ≤ a, so {Mn } is
a MG of bounded increments. Since P(b
τ ≤ n + r | Fnξ ) ≥ ε for ε = 26−11 and r = 11 we further know
that Eb
τ is finite (see part (c) of Exercise 5.1.15). Consequently, {Mn∧bτ } is U.I. (see part (a) of Proposition
5.4.4), and by Doob’s optional stopping theorem, EMτb = EM0 = 0. Equivalently, Eb
τ = a. Since the word
ABRACADABRA has partial repeats (ABRA and A), the answer for the stopping time τ corresponding
to the first time that the monkey produces ABRACADABRA, is different. Indeed, following the preceding
analysis in case of gambling scheme for ABRACADABRA, at time τ we have that three gamblers won, betting
on ABRACADABRA, ABRA, and A (with all others losing), leading to Mτ = a0 − τ for a0 = 2611 + 264 + 26,
and consequently, to Eτ = a0 > a.
(c). Let T0 = 0 and Tk = inf{n ≥ 11 : Ln = k}, k ≥ 1, denote the k-th time the monkey typed
PROBABILITY. These are almost surely finite {Fnξ }-stopping times and their increments τk = Tk − Tk−1
1
are independent (iteratively apply part (a) of Exercise 5.1.38 to get the independence of Tk ∈ mFTξk and
τk+1 ∈ mσ(ξTk +r , r ≥ 1)). Further, the word PROBABILITY has no partial repeats, so τk ≥ 11 for all k.
That is, at each time Tk the monkey has to start afresh and type by chance this word, from which we deduce
that τk are identically distributed as τb of part (b). We are thus in the renewal theory setting of Example
a.s.
2.3.7 with Ln = sup{k ≥ 0 : Tk ≤ n} and from Exercise 2.3.8 we know that n−1 Ln = 1/Eb
τ . Further, recall
part (b) and Exercise 5.1.15 that P(b
τ > kr) ≤ (1 − ε)k for some r finite and ε > 0, hence by part (a) of
Lemma 1.4.31,
∞
∞
X
X
Eb
τ2 = 2
yP(b
τ ≥ y) ≤ 2r2
(k + 1)P(b
τ > kr)
y=1
k=0
√ D
is finite. We thus deduce from the renewal theory CLT that (Ln − n/a)/ vn→G for the finite, positive
constant v = a−3 Var(b
τ ) (c.f. part (b) of Exercise 3.2.9, applied to Yk = τk /a).
Exercise [5.5.19]
Since Yn → Y−∞ a.s. when n → −∞, clearly |Y−∞ | ≤ supn |Yn | = Z is integrable by our assumption that Z
is integrable. Further, Wr = supn≤r |Yn − Y−∞ | is bounded by 2Z hence integrable. Consequently, by Lévy’s
downward theorem E[Wr |Fn ] → E[Wr |F−∞ ] as n → −∞. Combining this with (cJensen) and monotonicity
of the C.E. we deduce that for any r finite,
lim sup |E[Yn |Fn ] − E[Y−∞ |Fn ]|
≤ lim sup E[|Yn − Y−∞ | | Fn ]
n→−∞
n→−∞
≤
lim E[Wr |Fn ] = E[Wr |F−∞ ] .
n→−∞
The sequence {Wr } is dominated by the integrable 2Z and by its construction, Wr ↓ 0 as r ↓ −∞. Hence, by
(cDOM) E[Wr |F−∞ ] ↓ 0 for r ↓ −∞. We have thus shown that |E[Yn |Fn ] − E[Y−∞ |Fn ]| → 0 as n → −∞.
We are now done, since from Lévy’s downward theorem also E[Y−∞ |Fn ] → E[Y−∞ |F−∞ ] as n → −∞.
Exercise [5.5.29]
Since m(m − 1)W2−m is the sum of finitely many random variables of the form h(ξi , ξj ), each of which is
assumed to be in L1 , it follows by the triangle inequality (in L1 ) that each Wn is also in L1 .
1. Note that W2−m is merely the random variable Sbm (h) we have encountered in Lemma 5.5.25. While
proving this lemma we have shown that Sbm (h) = E[h(ξ1 , ξ2 )|Em ] for all m ≥ 2. With Em a decreasing
sequence of σ-algebras, this implies that FnW ⊆ E2−n and by the tower property, that for any n ≤ 0,
E[h(ξ1 , ξ2 )|FnW ] = E[E(h(ξ1 , ξ2 )|E2−n )|FnW ] = E[Wn |Gn ] = Wn .
W
Thus, Wn = E[h(ξ1 , ξ2 )|FnW ] is a RMG that converges a.s. and in L1 to E[h(ξ1 , ξ2 )|F−∞
]. Further,
W
F−∞ is contained in the exchangeable σ-algebra E for the i.i.d. sequence {ξk }, hence by the HewittW
W
Savage 0-1 law F−∞
is P-trivial. Consequently, E[h(ξ1 , ξ2 )|F−∞
] is a.s. a constant and this constant
must be Eh(ξ1 , ξ2 ). That is, Wn → Eh(ξ1 , ξ2 ) as n → −∞.
2. Direct computation shows that
2
E[W2−m
]=
1
− 1)2
m2 (m
X
X
E(h(ξi , ξj )h(ξk , ξl )) .
1≤i6=j≤m 1≤k6=`≤m
By Cauchy-Schwartz, |E(h(ξi , ξj )h(ξk , ξl ))| ≤ v for all i 6= j and k 6= `. Moreover, if i 6= j 6= k 6= `
then by independence
E[h(ξi , ξj )h(ξk , ξl )] = E[h(ξi , ξj )]E[h(ξk , ξ` )] = [Eh(ξ1 , ξ2 )]2
2
2
There are m(m − 1)(m − 2)(m − 3) such terms in the expression for E[W2−m
]. So, with m(m −
2
2
1)(m − 2)(m − 3)/m (m − 1) → 1, and all other terms bounded uniformly by v, it follows that
E[Wn2 ] → [Eh(ξ1 , ξ2 )]2 when n → −∞.
Exercise [5.5.9]
1. The random tree Tn belongs to a finite setPof all trees of depth n and maximal degree ` + 1 (the vertices
n
of such a tree are a subset of the Ln = k=0 `k vertices of the full `-ary tree, so there are no more
then 2Ln possible trees). Let Fn = σ(Tn ), i.e. all events that can be computed once Tn is given. In
particular, Fn is a filtration, and our definition of C(v) (not counting v among the possible branch
points on the path to v) made sure that C(v) is measurable on Fn for all v ∈ Tn , hence that {Xn } is
0
adapted to this filtration. Next, for each v ∈ ∂Tn let Yv be the sum of e−λ[C(v )−C(v)] over all offspring
v 0 of v. Note that C(v 0 ) = C(v) if v has one offspring and C(v 0 ) = C(v) + 1 otherwise. Since the
number of offspring of v ∈ ∂Tn is independent of Fn it follows that for any n ≥ 0 and v ∈ ∂Tn
X
E[Yv |Fn ] = P(N = 1) + e−λ
kP(N = k) = M (λ) .
k≥2
Consequently,
X
Xn+1 = M (λ)−(n+1)
0
e−λC(v ) = M (λ)−(n+1)
v 0 ∈∂Tn+1
X
e−λC(v) Yv ,
v∈∂Tn
implying that
E[Xn+1 |Fn ] = M (λ)−(n+1)
X
e−λC(v) E[Yv |Fn ] = Xn
v∈∂Tn
as claimed.
2. From the definition of Xn and Bn it follows that
X
1 + Xn = 1 + M (λ)−n
e−λC(v) ≥ M (λ)−n e−λBn .
v∈∂Tn
Since λ > 0 it thus follows that
n−1 Bn ≥ −λ−1 [log M (λ) + n−1 log(1 + Xn )] .
With Xn a non-negative martingale, by Doob’s martingale convergence theorem, with probability one
Xn (ω) → X∞ (ω) ≥ 0 hence n−1 log(1 + Xn (ω)) → 0, from which we deduce that for any λ > 0,
lim inf n−1 Bn (ω) ≥ −λ−1 log M (λ) .
n→∞
Recall that Zn is super-critical, hence M (λ) → P(N = 1) < 1 when λ → ∞. Consequently, taking λ
large enough we have that δ = −λ−1 log M (λ) is positive, as claimed.
Exercise [5.5.15]
1. Let vk = 1 −
√
q k pk −
p
(1 − qk )(1 − pk ) which are in [0, 1). In this problem the likelihood ratios are
Yk =
qk
1 − qk
Iω =1 +
Iω =0 ,
pk k
1 − pk k
√
hence P( Yk ) = 1 Q
− vk . From Proposition 5.5.13 we knowPthat Q is absolutely continuous with
respect to P iff
α
=
k (1 − vk ) is positive, or equivalently iff
k log(1 − vk ) is finite, which in turn is
P
equivalent to k vk finite (as both require that all but finitely many of the vk -s be in [0, 1/2] in which
case − log(1 − vk ) is between vk and 2vk ).
3
2. It is easy to check that
vk =
p
1 √
1 p
√
( pk − qk )2 + ( 1 − pk − 1 − qk )2 .
2
2
√
√ 2
Since
P ( x − y) ≤ |x − y| for any x, y ≥ 0, it follows that vk ≤ |pk − qk | and by part (a) the finiteness
of k |pk − qk | implies the absolute continuity of Q with respect to P.
3. Using the alternative formula for vk given in part (b) it is easy to verify that
p
−2
1 √
vk
√ −2 1 p
=
pk + q k
+
1 − pk + 1 − q k
2
(pk − qk )
2
2
is between one and 1/(4ε)
P when pk , qk ∈ [ε, 1 − ε], hence for such parameters the finiteness of
equivalent to that of k (pk − qk )2 .
P
k
vk is
4. Consider the event A = {ω : ωk = 1 infinitely often }. By Borel-Cantelli I we have that Q(A) = 0,
whereas by Borel-Cantelli II we see that P(A) = 1. Consequently,P
P⊥Q as claimed. In particular, this
happens when qk = 0 for all k and pk = 1/(k + 1), in which case k (pk − qk )2 is finite.
Exercise [5.5.30]
1. In view of Lévy’s downward theorem, it suffices to show that X−k = E[ξ1 |F−k ]. To this end note that
F−k = σ(Sj , j ≥ k) = σ(Sk , ξj , j > k), so with ξj for j > k independent of σ(ξ1 , Sk ), we have from
Proposition 4.2.3 and part (a) of Exercise 4.4.8 that E[ξ1 |F−k ] = E[ξ1 |Sk ] = k −1 Sk , as claimed (a
more direct, albeit longer solution consists of verifying that E[X−k |F−(k+1) ] = X−(k+1) ). Next, for
2 ≤ k ≤ n we have that τ = −k iff {Sj < j for all k < j ≤ n, and Sk ≥ k}, which clearly is in F−k .
Similarly, τ = −1 iff {Sj < j for all 2 ≤ j ≤ n}, an event in F−1 . Thus, τ is a stopping time for the
filtration {Fk , k ≤ −1}.
2. When Sn ≤ n, tracing the monotone non-decreasing path between (0, 0) and (n, Sn ) backwards from
(n, Sn ) we find that it necessarily first hits the set {(x, y) : y ≥ x} on the diagonal y = x. The event Γcn
then corresponds to the path hitting this diagonal at some ` = −τ ∈ {1, . . . , n} with Xτ = 1, while on
Γn it does so only at (0, 0), in which case by definition τ = −1 and S1 < 1 implying that Xτ = S1 = 0.
Consequently, IΓn = 1 − Xτ when Sn ≤ n.
With F−n = σ(Sn ), τ ≥ −n a bounded stopping time and (Xk , Fk ) a RMG, it then follows that for
Sn ≤ n,
P(Γn |Sn ) = 1 − E[Xτ |F−n ] = 1 − X−n = 1 − n−1 Sn .
To get the stated formula for P(Γn |Sn ) simply note that {Sn > n} ⊆ Γcn .
4