Probability Qualifying Exam
Solution, Spring 2008 (SKETCH)
Shiu-Tang Li
December 27, 2012
1
R 2π R ∞
R∞
Use polar coordinates: 0 0 g(r)r dr dθ = 1. Thus 0 g(r)r dr =
1
. Let U = X/Y , V = Y , we have X = U V , Y = V . fU,V (u, v) =
2π
√
√
√
R∞
R∞
R∞
2
g(|v| u2 + 1)|v|; −∞ g(|v| u2 + 1)|v| dv = 2 0 g(v u2 + 1)v dv = 1+u
g(z)z dz =
2
0
1
.
π(1+u2 )
2
Sn −nE[X1 ]
(a) CLT: let X1 , · · · , Xn , · · · ∈ L2 (Ω) and be i.i.d RVs, then √
nV ar(X1 )
converges to N (0, 1) weakly.
Kolmogrov’s 0-1 law: Let X1 , · · · , Xn , · · · be i.i.d, if A ∈
then P (A) = 0 or 1.
T∞
n=1
σ(Xn , Xn+1 , · · · ),
(b)
P (An i.o.) = P (
∞ [
∞
\
Am )
n=1 m=n
∞
[
= lim P (
n→∞
Am )
m=n
≥ inf P (Ak ) > 0.
k≥1
(c) We have
√
√
√
P (Sn > 0) ≥ P (Sn > log(n)) = P (Sn / n > log(n)/ n) ≥ P (Sn / n > )
1
for arbitrary > 0 and n = n() large. Therefore by CLT,
1/2 = lim P (Sn > 0) ≥ lim sup P (Sn > log(n)) ≥ lim inf P (Sn > log(n))
n
n
n
√
≥ lim lim sup P (Sn / n > ) = 1/2.
→0+
n
As a result, P (limT
supn T
Sn = S
∞) ≥ limn P (Sn >
log(n))
Since
T∞
T∞ =S1/2.
∞
∞
∞
{lim supn Sn = ∞} = ∞
{S
>
k}
=
{S
m
m−
k=1
n=N
m=n
k=1
n=N
m=n
SN > k} ∈ σ(XN , XN +1 , · · · , ) for every N ∈ N, it follows that {lim supn Sn =
∞} is a tail event with positive probability, so P (lim supn Sn = ∞) = 1.
3
Consider Ω = {w1 , w2 w3 }, with probability 1/3 each. Let X = 1{w1 ,w2 } , Y =
1{w2 ,w3 } , Z = 1{w1 ,w3 } .
4
This is misstated. Consider X1 ≡ 1, X2 ≡ 3, Xn ≡ 2 for n ≥ 3.
5
Note that X1 /Sn , · · · , Xm /Sn are identically distributed, and the result
is straightforward.
6
If k = 1, then P (Tn = k) = 0 for all n. For k > 1,
P (Tn = k) = P (X1,n + · · · + Xk,n > n) − P (X1,n + · · · + Xk−1,n > n)
= P (X1,n + · · · + Xk−1,n ≤ n) − P (X1,n + · · · + Xk,n ≤ n)
n−1 1
n−1
1
−
=
k
nk
k − 1 nk−1
1
1
→
−
as n → ∞.
(k − 1)! k!
7
See 2004 problem 7. for relevant techniques.
n n
2
t2
1
t2
1
2
EX
+
o(
)
=
1
−
+
o(
)
→ e−t /2 .
2n
n
2n
n
2
n
φ( √tn )
= 1 + i √tn EX −
8
P
n
P (Xn > (1 + ) log(n)) =
P
n
e( (1 + ) log(n)) =
9
10
(Def?) Use inclusion-exclusion principle.
3
P
n
n1+ .