Probability 1
Problems for week 12
Exercise 1.
√
(a) Since ( t)0 =
1
√
2 t
we have
Z
ε
P (Xn ≤ ε) =
fXn (x) dx =
√
ε.
0
(b) Clearly FMn (t) = 0 if t < 0 and FMn (t) = 1 if t > 1. For t ∈ (0, 1) we have
FMn (t) = P (Mn ≤ t) = P (Xn ≤ t)n = tn/2 ,
where we used independence. Since this function is piecewise C 1 , it follows that Mn is
absolutely continuous and its density is
0
fMn (t) = FM
(t) =
n
n n/2−1
t
1(0,1) (t).
2
(c) The distribution function of Zn is given by FZn (z) = 0 if z < 0, while for z ≥ 0 we compute
FZn (z) = P (Zn ≤ z) = P (1 − Mn ≤
z
z
z
) = P (Mn ≥ 1 − ) = 1 − P (Mn < 1 − )
n
n
n
and since Mn is absolutely continuous P (Mn < t) = FMn (t) and so for n > z
FZn (z) = 1 − (1 −
z n/2
)
→ 1 − e−z/2
n
which is the distribution function of an exponential random variable with parameter 1/2.
(d) By (b) it follows that FMn (t) → 0 if t < 1 and FMn (t) → 1 if t > 1, that is
lim FMn (t) = 1[1,∞) (t) =: F (t)
n→∞
which is the distribution function of the random variable constantly equal to 1, that is Mn → 1
in law.
Since for every ε > 0
X
X
X
X
P (|Mn − 1| > ε) =
P (Mn < 1 − ε) =
FMn (1 − ε) =
(1 − ε)n/2 < ∞
n∈N
n∈N
n∈N
n∈N
it follows by exercise (1), (iii) of last week that Mn → 1 a.s. and so also in probability.
Finally as |Mn | ≤ 1 we also have convergence in Lp .
Exercise 2. Let us prove the ’if ’ part. By Markov inequality and the fact that φ is increasing
P (|Zn − Z| > ε) = P (φ(|Z − n − Z|) > φ(ε)) ≤
E(φ(|Zn − Z|))
→0
φ(ε
Exercises
2014
for every ε > 0, since φ(ε) > 0.
Viceversa, if Zn → Z in probability, then for every ε > 0 fixed, we can write
E(φ(|Zn − Z|)) = E(φ(|Zn − Z|)1{|Zn −Z|≥ε} ) + E(φ(|Zn − Z|)1{|Zn −Z|>ε} )
≤ φ(ε) + ||φ||∞ P (|Zn − Z| > )
from which, by the boundedness of φ, lim supn→∞ E(φ(|Zn − Z|)) ≤ ψ() for every epsilon. By
right continuity of ψ in 0 we conclude by sending ε → 0.
Exercise 3.
(a) Yn−1 and Xn are independent because Yn−1 is a function of (X1 , . . . , Xn−1 ).
(b) Let us prove by induction that V ar(Yn ) = nA2n , the other conclusion being trivial. The case
n = 1 is immediate, as Y0 = 1 is a constant. By induction and the property of the variance
(recall also point (a))
V ar(Yn ) = A2 V ar(Yn−1 ) + V ar(Xn ) = A2 ((n − 1)A2(n−1) ) + A2n = nA2n .
(c) For every ε > 0 we have
P (|Yn | > ε) ≤
nA2n
V ar(Yn )
=
→∞
ε2
ε2
as n → ∞, that is Yn → 0 in probability. Moreover, since
X
P (|Yn | > ε) ≤
X nA2n
n
n
ε2
<∞
because |A| < 1, the convergence is a.s..
(d) We have
2
V ar(Yn ) = A2 V ar(Yn−1 ) + σn2 = A2 V ar(Yn−2 ) + A2 σn−1
+ σn2 = · · · =
n
X
2
A2k σn−k
.
k=1
2
limn→∞ σn−k
For every fixed k,
= 0 by assumption. Since the sequence (σn2 )n is bounded
(intact it converges to 0), i.e. C := supn σn2 < ∞, it follows that
2
|A2k σn−k
| ≤ CA2k .
By the dominated convergence theorem we can pass the limit n → ∞ inside the sum obtained
above to conclude V ar(Yn ) → 0. Finally, as E(Yn ) = 0, Chebychev inequality proves the
convergence in probability.
*Exercise 4. (a) For a ∈ R, let P
Aan := {Xn > a log n}. The events {Aan }n are independent
1
a
and P (An ) = naλ . Therefore n P (Aan ) < ∞ if and only if a > λ1 , and by Borel-Cantelli,
P (lim supn→∞ Aan ) = 1 if a ≤ λ1 , P (lim supn→∞ Aan ) = 0 if A > λ1 . Set
L := lim sup
n→∞
and observe that
while
Xn
log n
n
n X
1o
1o
n
L≥
⊃ lim sup
>
= lim sup A1/λ
n .
λ
log n
λ
n→∞
n→∞
n
1o
L>
⊂
λ
[
ε∈Q,ε>0
n X
o
1
n
lim sup
> +ε =
log n
λ
n→∞
[
ε∈Q,ε>0
lim sup A1/λ+ε
n
n→∞
from which P (L ≥ λ1 ) = 1 and P (L > λ1 ) = 0, that is P (L = λ1 ) = 1
2
Exercises
2014
(b) We have Zn → 0 in probability, because P (|Zn | > ε) = P (Xn > ε log n) =
gously Zn → 0 in Lp for every p, because
E(|Zn |p ) =
1
nε
→ 0. Analo-
1
cost.
1
E(|Xn |p ) =
E(|X1 |p ) =
→ 0.
(log n)p
(log n)p
(log n)p
Finally by uniqueness of the limit, if Zn had limit a.s. then it should be 0, however lim supn Zn =
1/λ a.s..
*Exercise 5. By the SLLN Sn /n → m a.s., in particular for a.e. ω ∈ Ω there exists n0 = n0 (ω) <
∞ such that for n ≥ n0 we have Sn (ω)/n ≥ m/2. It follows that Sn → ∞ for every such ω, that
is a.s.. Moreover for these ω
X
X
m
e−εSn (ω) ≤
e−ε 2 n < ∞
n≥n0 (ω)
so that
P
n
n≥n0 (ω)
e−εSn (ω) < ∞.
3