Download APM 504 - PS7 Solutions 3.4) Suppose that X1 and X2 are

APM 504 - PS7 Solutions
3.4) Suppose that X1 and X2 are independent normal random variables with mean 0 and variances σ12 and σ22 . Noting that the characteristic function of Xi is φi (t) = exp(−σi2 t2 /2), it follows
that the characteristic function of X1 + X2 is
1
2
2
2
φ(t) = φ1 (t)φ2 (t) = e− 2 (σ1 +σ2 )t
which shows that X1 + X2 is normally distributed with mean 0 and variance σ12 + σ22 .
3.8) Now suppose that X1 , X2 , · · · are independent Cauchy random variables. In this case the
characteristic function of Xi is exp(−|t|) and so the characteristic function of Tn = (X1 + · · · +
Xn )/n is
n
E etTn = e−|t/n| = e−|t| .
This shows that Tn also has the Cauchy distribution.
3.11) Let X1 , X2 , · · · be independent random variables with characteristic functions φi (t) and
suppose that Sn = X1 + · · · + Xn → S almost surely. Then Sn also converges to S in probability
and in distribution, and so it follows from the continuity theorem that
∞
n
Y
Y
tS
tSn φk (t).
E e ] = lim E e
= lim
φk (t) =
n→∞
n→∞
k=1
k=1
3.12) Let Xn be a sequence of independent random variables such that
1
2
n ). If S = X + · · · + X ,
and notice that the characteristic
function
of
X
is
φ
(t)
=
cos(t/2
n
n
n
1
n
P
P∞
∞
−2n < ∞, Theorem (1.8.3) tells us that
2
Var(X
)
=
then because
EX
=
0
and
n
n
n=1
n=1
P
Sn → S ≡ ∞
n=1 Xn almost surely. I claim that S is uniformly distributed on (−1, 1). Let D
be the collection of dyadic rationals in (−1, 1) and let Id be the collection of half open intervals with endpoints in D. Since D is dense in (−1, 1), an argument analogous to that given
in exercises (1.1.3) and (1.1.5) shows that the Borel σ-algebra on (−1, 1) is generated by Id .
However, Id is also a π-system and so by Theorem (2.2) of the Appendix it is enough to show
that P(S ∈ (a, b]) = 21 (b − a) for every interval (a, b] ∈ Id .
P(Xn = 2−n ) = P(Xn = −2−n ) =
To do so, observe that each dyadic rational x ∈ D has a unique decomposition
x=
n
X
xk 2−k
i=k
where xk ∈ {−1, 1} and n ≥ 1. Since
it follows that
P∞
k=n+1 |Xn |
≤ 2−n and since P(X = y) = 0 for every y,
P S ∈ (x − 2−n , x + 2−n ] = P(X1 = x1 , X2 = x2 , · · · , Xn = xn )
n
Y
=
P(Xk = xk ) = 2−n .
k=1
1
The claim then follows upon noting that every interval (a, b] ∈ Id has a unique decomposition
into disjoint intervals of this form and that the total length of the interval is equal to the sum
of the lengths of the sub-intervals.
Combining the fact that the characteristic function of S is sin(t)/t with the result of exercise
(3.11) allows us to conclude that
∞
Y
sin(t)
=
cos t/2n .
t
n=1
3.14) We proceed by induction. For the case n = 0, the claim amounts to the requirement
that φ(t) is continuous atR all t, which we know to be true. Suppose that the claim is true for
some n and assume that |x|n+1 µ(dx) < ∞. To show that the claim is also true for n + 1, we
will use Theorem (9.1) in the Appendix which comes with four conditions that need to be verified.
We first define the function f (t, x) = (ix)n eitx and observe that the induction hypothesis asserts
that
Z
φ(n) (t) = f (t, x)µ(dx)
and that
Z
Z
|f (t, x)|µ(dx) =
|x|n µ(dx) < ∞.
This proves condition (i). For (ii), we note that the partial derivative
∂t f (t, x) = (ix)n+1 eitx
exists and is continuous in both arguments. Since |∂t f (t, x)| = |x|n+1 is an integrable function,
the dominated convergence theorem implies that
Z
Z
n+1 isx
lim (ix)
e µ(dx) = (ix)n+1 eitx µ(dx)
s→t
at every t and so (iii) is verified. Finally, for (iv), let δ > 0 and observe that
Z Z δ
Z Z δ
Z
n+1
∂t f (t + s, x)dsµ(dx) =
|x|
dsµ(dx) = 2δ |x|n+1 µ(dx) < ∞.
−δ
−δ
Since Theorem (9.1) allows us to interchange differentiation and integration, we have
Z
φ(n+1) (t) =
(ix)n+1 eitx µ(dx),
which completes the induction.
3.15) Let X be a standard normal random variable and observe that
2 /2
φ(t) = e−t
=
∞
X
(−1)n
n=0
2
t2n
2n n!
is the characteristic function of X. Since X has exponentially-decaying tails, it is clear that X
has moments of all orders. In particular, by using (3.14) with t = 0, we can calculate
EX 2n = i2n φ(2n) (0) =
(2n)!
.
2n n!
Of course, symmetry considerations show that all of the odd moments vanish: EX 2n+1 = 0.
3.17) Let X1 , X2 , · · · be i.i.d. with characteristic function φ and let Sn = X1 + · · · + Xn .
(i) Suppose that φ0 (0) = ia. By the definition of the derivative, this means that
φ(t) = 1 + ia · t + o(t)
as t → 0.
Since Theorem (2.4.2) implies that
EeitSn /n = φ(t/n)n =
1 + iat/n + o (t/n)
n
→ eiat
for all t ∈ R, we can use the continuity theorem along with exercise (2.2.9) to conclude that
Sn /n → a in probability.
(ii) Now suppose that Sn /n → a in probability and let
δn (t) = n φ(t/n) − 1 .
In this case, the continuity theorem gives
EeitSn /n = φ(t/n)n =
1 + δn (t)/n
n
→ eiat ,
which then (using (2.4.2) again) implies that limn→∞ δn (t) = iat, i.e. for all t 6= 0,
lim
n→∞
φ(t/n) − φ(0)
= ai.
t/n
Since φ(·) is continuous, an application of the following lemma to the function
ψ(t) ≡
φ(t) − 1
t
shows that φ is differentiable at t = 0 with φ0 (0) = ai.
Lemma: Suppose that ψ : R/{0} → R is a continuous function with the property that for all
t 6= 0,
lim ψ(t/n) = c (?).
n→∞
Then
lim ψ(h) = c.
h→0
Proof: Suppose that the conclusion does not hold. Without loss of generality, we may assume
that
lim sup ψ(h) > c,
h↓0
3
in which case there is a sequence hn ↓ 0 and numbers e > d > c such that ψ(hn ) > e for all
n. Furthermore, since ψ is continuous on R/{0}, there exist numbers ln < rn with ln → 0 such
that for every n ≥ 1, ψ(t) > d for all t ∈ In ≡ (ln , rn ).
In light of (?), there can be no t > 0 such that the set O(t) ≡ {t/m; m ≥ 1} intersects infinitely
many of the open intervals In . So, for n ≥ 1, let Un be the collection of t ∈ [1, 2] such that
O(t)∩In 6= ∅ and O(t)∩In+k = ∅, and let U0 be the collection of t ∈ [1, 2] such that O(t)∩Im = ∅
for all m ≥ 1. Since [1, 2] is a complete metric space and since
∞
[
[1, 2] =
Un ,
n=0
the Baire category theorem tells us that the Un cannot all be nowhere dense. In other words,
there exists n ≥ 0 and some interval (a, b) ⊂ [1, 2] such that (a, b) is contained in the closure of
Un .
To arrive at a contradiction, let M be the smallest integer greater than a/(b − a) and notice
that for all m ≥ M , a/m ≤ b/(m + 1) so that the successive intervals (a/m, b/m) and (a/(m +
1), b/(m + 1)) overlap and consequently
(0, M ) =
∞
[
(a/m, b/m).
m=M
Then, either In+k ∩(0, M ) = ∅ for all k ≥ 1, which would contradict our assumption that ln → 0,
or there exists an t ∈ (a, b) and integers m, k ≥ 1 such that t/m ∈ In+k . If the latter is true,
then since Un is dense in (a, b), there is also a sequence (tj ; j ≥ 1) ⊂ Un such that tj → t. Since
In+k is an open interval, this implies that tj ∈ In+k for all j sufficiently large, which contradicts
the definition of Un . This completes the proof of the lemma.
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