MASSACHUSETTS INSTITUTE OF TECHNOLOGY
6.265/15.070J
Homework 4 Solution
1
Fall 2013
Problem 1
√
Proof. We aim to bound |E[Ln ] − mn | by O( mn ). We have that
∞
P(|Ln − mn | > x)dx
|E[Ln ] − mn | ≤ E[|Ln − mn |] =
0
By Talagrand’s inequality, we have that
P(|Ln − mn | > x) ≤ 4 exp(−
√
Let x = β mn log mn , we have
P(|Ln − mn | > x) ≤ 4 exp(−
x2
)
4(mn + x)
β 2 mn log mn
√
)
4(mn + β mn log mn )
n
for some c ≥ 1, we have for β ≤ β0
Let β0 = c logmm
n
2
P(|Ln − mn | > x) ≤ 4 exp(−
− β
β 2 log mn
) = 4mn 4(1+c)
4(1 + c)
√
For β > β0 , let β = y mn log mn where y > c ≥ 1, and then
P(|Ln − mn | ≥ β
y 2 mn
y 2 m2n
) = 4 exp(−
)
4(mn + ymn )
4(1 + y)
ymn
)
≤ 4 exp(−
8
mn log mn ) ≤ 4 exp(−
Thus,
β0
∞
∞
β2
− 4(1+c)
ymn
)dy
P(|Ln − mn | > x)dx ≤ mn log mn
4mn
dβ + mn
4 exp(−
8
0
c
0
8 exp(− cm8 n )
π(1 + c)
+ 4mn
= 4 mn log mn
log mn
mn
≤ 4 π(1 + c)mn + 32 = O(mn )
Namely,
√
|E[Ln ] − mn | ≤ O( mn )
√
√
Thus, we have mn = α n + o( n).
1
1
2
Problem 2 (Kaji’s Solution)
Proof. Define g by g(E) = g(E1 , ..., Edn ) = log Zn . For notational purposes,
denote E−i := (E1 , ..., Ei−1 , Ei+1 , ..., Edn ). Then we will prove the following
equation:
|g(E, E−i − g(Ei , E−i ))| ≤ log 2 for all i.
(1)
Consider the change from g(Ei , E−i ) to g(E−i ). Omitting the edge Ei , we have
that the sets that are independent under E are still independent under E−i , and
that for an independent set I under E, a new set I ∪Ei may become independent.
Hence, the maximum possible change is Zn → 2Zn . Now, by adding a new edge
Ei = {a, b}, observe the following:
• The independent sets that do not contain Ei are still independent.
• If I is an independent set containing Ei , then I\{a} and I\{b} are still
independent.
• So the number of independent sets eliminated by adding Ei cannot exceed
1
3 of that of independent sets under E−i .
Thus, Zn cannot go below 23 Zn , and we have (1). Let di = log 2. Since g is
symmetric in i, Theorem 12.2 gives that for any t > 0,
P(|g(E) − E[g(E)]| ≥ t) ≤ 2 exp(−
t2
)
2dn(log 2)2
Thus, we have
P(log Zn − E[log Zn ] ≥ t) ≤ P(|g(E) − E[g(E)]| ≥ t) ≤ 2 exp(−
3
t2
)
2dn(log 2)2
Problem 3
Proof. Let Xt = Xt+ −Xt− where Xt+ = max{Xt , 0} and Xt− = max{−Xt , 0},
then we have that
Xt = Xt+ − Xt−
Similar to the proof of Proposition 3, define two increasing nonnegative sequences Xtn+ and Xtn− : Xtn+ = Xt+ when 0 ≤ Xt+ ≤ n and Xtn+ = n
2
2
o.w.; Xtn− = Xt− when 0 ≤ Xt− ≤ n and Xtn− = n o.w.. Thus, a.s. we have
that
Xtn+ → Xt+ , Xtn− → Xt−
Let Xn = Xtn+ − Xtn− . By |Xn | ≤ |X| and |Xn | ≤ n, we have that Xn is a.s.
bounded and in L2 . Then,
(Xt − Xtn )2 = ((Xt+ − Xtn+ ) − (Xt− − Xtn− ))2
≤ 2(Xt+ − Xtn+ )2 + 2(Xt− − Xtn− )2
We have that
T
(Xt+ − Xtn+ )2 dt] = E[
E[
0
T
0
(Xt+ )2 dt] + E[
T
0
(Xtn+ )2 dt] − 2E[
(2)
T
0
Xt+ Xtn+ dt]
As n → ∞, by MCT the r.h.s. of the equation above converges to
E[
T
0
(Xt+ )2 dt]
+ E[
0
T
(Xt+ )2 dt]
− 2E[
T
0
(Xt+ )2 dt] = 0
T
Likewise, we have that E[ 0 (Xt− − Xtn− )2 dt] → 0 as n → ∞. By (2), we have
that
T
T
T
(Xt − Xtn )2 dt] ≤ 2E[
(Xt+ − Xtn+ )2 dt] + E[
(Xt− − Xtn− )2 dt]
E[
0
0
0
→ 0 as n → ∞.
4
Problem 4
t
Proof. Since E[ 0 (Xsn − Xs )2 ds] → 0 as n → ∞, given > 0, there exists a
positive integer N such that for any n > N we have
t
E[ (Xsn − Xs )2 ds] < 0
3
3
For any n > N , we have that
t
t
n
2
E[ (Xs + Xs ) ds] = E[ (Xsn − Xs + 2Xs )2 ds]
0
0
t
= E[ (Xsn − Xs )2 + 4Xs (Xsn − Xs ) + 4Xs2 ds]
0
t
≤ E[ 2(Xsn − Xs )2 + 8Xs2 ds]
0
t
(3)
≤ 2 + 8E[ Xs2 ds] < ∞
0
For any n ≤ N , we have that
t
t
t
n
2
n 2
E[ (Xs + Xs ) ds] ≤ 2E[ (Xs ) ds] + 2E[ Xs2 ds]
0
0
<∞
(4)
Thus, (3) and (4) give that
t
sup E[ (Xsn + Xs )2 ds] < ∞
n
0
0
4
4
Problem 5.
5
6
7
8
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15.070J / 6.265J Advanced Stochastic Processes
Fall 2013
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