MAT 902 STOCHASTIK II
HOMEWORK 8
Solution 1. a) Xn is indeed a martingale, since it is integrable and E[Xn+1 |Fn ] = E[Yn+1 Xn |Fn ] =
Xn E[Yn+1 |Fn ] = Xn .
b) (i) ⇒ (ii): If Xn is uniform integrable, then Xn → X∞ a.s. and in L1 as n → ∞. We have
1 = E[Xn ] = E[X∞ ].
(ii) ⇒ (iii): If E[X∞ ] > 0 then P [X∞ > 0] > 0 since X∞ ≥ 0.
√
√
Q
Q
(iii) ⇒ (iv): Suppose that n≥1 E[ Yn ] = 0. Let Mn = ni=1 E[√YYi i ] , so that Mn is also a
non-negative martingale. By the martingale convergence theorem, Mn → M∞ ≥ 0 a.s. We
have
√
1 √
Mn = Qn E[
Xn .
Y
]
i
i=1
√
√
Q
Q
Thus Xn = Mn2 ni=1 (E[ Yi ])2 → 0 as n → ∞ since M∞ is finite and n≥1 E[ Yn ] = 0.
√
Q
Therefore X∞ ≡ 0 and the contradiction is proved. (iv) ⇒ (i): Assume that n≥1 E[ Yn ] > 0.
Then
h
i
Xn√
1√
1√
2
Q
≤ Q (E[
<∞
E[Mn ] = E n (E[ Yi ])2 = Qn (E[
Yi ])2
Yi ])2
i=1
i=1
i≥1
2
Therefore, Mn is bounded in L . Doobs inequality gives that
E[(Mn∗ )2 ] ≤ 4E[Mn2 ] ≤ E[√4Yi ])2 < ∞,
∗
where Mn∗ = max1≤i≤n Mi . Therefore, supn≥0 E[(Mn∗ )2 ] < ∞. Then M∞
= supn≥0
which implies that
∗ 2
∗ 2
) = supn≥1 Qn XE[n√Yi ]2 so supn≥1 Xn ≤ (M∞
(M∞
).
√
Qn
i=1
Xn√
,
E[ Yi ]
i=1
Xn is dominated by an integrable random variable, so (Xn )n≥1 is uniformly integrable.
Solution 2. a) We need to show that E[Xn |F−n−1 ] = X−n−1 . To do this, first note that if
i, j ≤ n + 1, by symmetry we have E[ξi |F−n−1 ] = E[ξj |F−n−1 ]. Hence,
n+1
1 X
E[ξn+1 |F−n−1 ] =
E[ξi |F−n−1 ]
n + 1 i=1
=
On the other hand, X−n =
Sn+1 −ξn+1
.
n
Sn+1
1
E[Sn+1 |F−n−1 ] =
n+1
n+1
It follows that
Sn+1
ξn+1
|F−n−1 ] − E[
|F−n−1 ]
n
n
Sn+1
Sn+1
Sn+1
=
−
=
= X−n−1
n
n(n + 1)
n+1
E[X−n |F−n−1 ] = E[
Therefore, Xn is a backwards martingale.
b) For 2 ≤ n ≤ k, we have that τ = −n if and only if {Sj < j for all n < j ≤ k and Sn ≥ n},
which clearly is in F−n .
Similarly, τ = −1 iff {Sj < j for all 2 ≤ j ≤ k}, an event in F−1 . Thus, τ is a stopping time
for the filtration (Fn )n≤−1 .
c) We want to show that Xτ = 1GCk , when Sk ≤ k. But in this case tracing the monotone nondecreasing path between (0, 0) and (k, Sk ) backwards from (k, Sk ) we find that it necessarily
first hits the set (Sl , l) for some l ≤ k. The event GC
k then corresponds to the path hitting this
diagonal at l = −τ with Xτ = 1. Therefore, Xτ = 1GCk .
2
MAT 902 STOCHASTIK II HOMEWORK 8
With F−k = σ(Sk ), τ ≥ −k a bounded stopping time and (Xn , Fn ) a backwards martingale, it
then follows that for Sk ≤ k,
P [Gk |Sk ] = 1 − E[Xτ |F−k ] = 1 − Xk = 1 − Skk .
Solution 3. We use Proposition 12.1 of the online lecture notes to see that
for any /2
there is a
δ such that for all events A in our sigma-algebra, if P(A) < δ then E(|X| 1A ), E(|Y | 1A )< /2,
for allX ∈ C, Y ∈ D. Hence for all X + Y ∈ C + D, we have E(|X + Y | 1A ) ≤ E(|X| 1A ) +
E(|Y | 1A ) < provided P(A) < δ. Applying Proposition 12.1 again, this implies that C + D is
uniformly integrable.
Solution 4. This is essentially just an exercise in repeating the proof of Corollary 12.2 from the
online lecture notes, replacing the single variable X with elements of a u.i. family of random
variables. The same proof works.