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Seminar 2012
- Counterexamples in Probability
Seminar | 19.11.2012 |
Presenter : Joung In Kim
Seite 2
Seminar – Counterexamples in
Probability
Ch8. Characteristic and Generating
Functions
Ch9. Infinitely divisible and stable
distributions
Seite 3
Seminar – Counterexamples in
Probability
Ch8. Characteristic and Generating
Functions
Ch9. Infinitely divisible and stable
distributions
Seite 4
Notation and Abbreviations
•
r.v. : random variable
•
ch. f. : characteristic function (ϕ(t))
•
d.f.
•
i.i.d. : independent and identically distributed
•
d
=
: distribution function (F)
: equality in distribution
Seite 5
Definition (Characteristic function)
Let X be a r. v. defined on Ω, ℱ, Ρ with distribution function F ∶ ℝ → [0,1]
The 𝐜𝐡𝐚𝐫𝐚𝐜𝐭𝐞𝐫𝐢𝐬𝐭𝐢𝐜 𝐟𝐮𝐧𝐜𝐭𝐢𝐨𝐧 ϕ: ℝ → ℂ of X is defined by
ϕ t = Ε eitX
=
∞ itx
e
−∞
dF x
∞ itx
e
−∞
f x dx, if X is 𝒂𝒃𝒔𝒐𝒍𝒖𝒕𝒆𝒍𝒚 𝒄𝒐𝒏𝒕𝒊𝒏𝒖𝒐𝒖𝒔 with density f
=
n
eit x n pn , if X is 𝒅𝒊𝒔𝒄𝒓𝒆𝒕𝒆 with P X = xn = pn , pn > 0,
n
pn = 1
Seite 6
Properties of a characteristic function
i) ϕ 0 = 1, ϕ −t = ϕ t , ϕ t
ii) If Ε X
ϕ
n
iii) If ϕ
n
< ∞, then ϕ n 0 exists and Ε X n = i−n ϕ n 0
t =
n
dn
dn t
ϕ t
0 exists and n is even, then Ε X
If n is odd , then Ε X
iv) If Ε X
≤ 1, t ∈ ℝ
n
n−1
ϕ t =
k=0
<∞
<∞
< ∞ and hence Ε X
n
n
k
< ∞ for k < 𝑛 then
(it)k
E X k + ο(t n )
k!
t→0
Seite 7
Properties of a characteristic function
v) If X1 and X 2 are independent random variables with ch. f. s ϕ1 and ϕ2 .
⇒ the ch. f. of X1 + X 2 is given by ϕ t = ϕ1 t ϕ2 t , t ∈ ℝ
vi) If we know the ch.f. ϕ of a r.v. X then the d.f. F of X
is given by the inversion formula:
1
P a < 𝑋 < 𝑏 = lim
c→∞ 2π
c
e−ita − e−itb
ϕ t dt
it
−c
If ϕ is absolutely integrable over ℝ, it follows that
X is absolutely continuous with density:
f x =
∞ −itx
e ϕ(t)dt
2π −∞
1
(inverse Fourier-transformation)
Seite 8
Fourier expansion of a periodic function
g(t+T)=g(t)
∞
=> 𝑔 t =
(an cos(nw0 t) + bn sin(nw0 t))
n=0
where,
𝐚𝟎 =
1
T
T
2
T
−
2
2π
w0 =
, T: Period, an , bn ∶ Fourier coefficient
T
g t dt,
𝐚𝐧 =
2
T
T
2
T
−
2
g t cos nw0 t dt ,
𝐛𝐧 =
2
T
T
2
T
−
2
g t sin nw0 t dt
Seite 9
Example 1. Discrete and absolutely continuous distributions with
the same characteristic functions on [-1, 1]
continuous ϕ t =
1
1
𝑓 𝑥 =
2π
discrete
ΡY=0 =
1− t ,
0
,
∞
−it𝑥
e
−∞
1
2
if t ≤ 1
otherwise.
1 − cos 𝑥
ϕ1 t dt =
, 𝑥∈ℝ
2
π𝑥
, P Y = 2𝑘 − 1 π =
1
ϕ2 (t) = + 4π−2
2
∞
k=1
2
2k−1 2 π 2
cos( 2k − 1 πt)
(2k − 1)2
, k = 0, ±1, ±2,∙∙∙
Seite 10
Example 1. Discrete and absolutely continuous distributions with
the same chacteristic functions on [-1, 1]
∞
n=1 an
h t = t = a0 +
t ≤ 1,
1
= +
2
∞
k=1
1
a0 = ,
2
2(cos nπ − 1)
an =
n2 π2
−4cos( 2k − 1 πt)
(2k − 1)2 π2
1
ϕ1 t = 1 − t = +
2
⇒ ϕ1 t = ϕ2 t ,
cos nπt (Fourier series)
∞
k=1
4cos( 2k − 1 πt)
(2k − 1)2 π2
t ∈ [−1,1]
Seite 11
Example 2. The absolute value of a characteristic function is
not necessarily a characteristic function.
ϕ t =
1
8
1 + 7eit , t ∈ ℝ ,
ϕ(t) =
1
8
50 + 7e−it + 7eit
1/2
If ϕ(t) is a ch. function, then ψ(t) ∶= ϕ(t) must be of the form
ψ t = p ∙ eit x 1 + (1 − p) ∙ eit x 2
Where 0 < 𝑝 < 1 𝑎𝑛𝑑 x1 , x2 are different real numbers.
Comparing ψ t
2
and ϕ(t)
=> p2 = (1 − p)2 =
7
64
2
, 2p(1 − p) =
50
64
𝐂𝐨𝐧𝐜𝐥𝐮𝐬𝐢𝐨𝐧 ∶ ϕ is a ch. function, but not ϕ .
=> contradiction!
Seite 12
Decomposable and Indecomposable
We say that a ch.f. ϕ is decomposable
if it can be represented as a product of two non-trivial ch.f.s. ϕ1 and ϕ2, i.e.
ϕ(t) = ϕ1(t) ϕ2(t)
and neither ϕ1 nor ϕ2 is the ch.f. of a probability measure which is
concentrated at one point.
Otherwise ϕ is called indecomposable.
Seite 13
Example 3. The factorization of a characteristic function into
indecomposable factors may not be unique.
(i) discrete case
X : discrete uniform distribution on the set {0, 1, 2, 3, 4, 5}.
Characteristic function of X :
5
itX
ϕ t =E e
itk
=
P(X = k)e
k=0
1
=
6
5
eitk
k=0
We can factorize the ch. f. in the following way :
ϕ1 t =
ψ1 t =
1
3
1
3
1 + e2it + e4it , ϕ2 t =
1
1 + eit + e2it , ψ2 t =
1
2
2
1 + eit
1 + e3it
 ϕ t = ϕ1 t ∙ ϕ2 t = ψ1 t ∙ ψ2 t , t ∈ ℝ
Seite 14
Example 3. The factorization of a characteristic function into
indecomposable factors may not be unique.
Need to check :
− Are ϕ1 , ϕ2 , ψ1 and ψ2 ch. f. s?
− Are ϕ1 , ϕ2 , ψ1 and ψ2 indecomposable?
⋅ ϕ2 and ψ2 correspond to two − point distribution. => 𝑖𝑛𝑑𝑒𝑐𝑜𝑚𝑝𝑜𝑠𝑎𝑏𝑙𝑒
⋅ Suppose that ψ1 t = ψ11 t ∙ ψ12 t ,
ψ11 t , ψ12 t : non − trivial
Let ψ11 t = peitx 1 + 1 − p eitx 2 , ψ12 t = qeity 1 + 1 − q eity 2
0 < p < 1, 0 < 𝑞 < 1
then
pq = 1 − p 1 − q = p 1 − q + q 1 − p =
1
=> contradiction
3
⇒ ψ1 is indecomposable.
⋅ ϕ1 t = ψ1 2t ⇒ ϕ1 is also indecomposable. ∎
Seite 15
Example 3. The factorization of a characteristic function into
indecomposable factors may not be unique.
(ii) Continuous case
∙ Let X be a r.v. which is uniformly distributed on -1,1).
∙ Ch. f. of X :
ϕ t = t −1 sin t , t ∈ ℝ
Factorization 1.
n
ϕ t = t −1 sin t =
k=1
∞
=>
Is cos
cos
lim ϕ t =
n→ ∞
t
2k
t
cos k
2
cos
k=1
t
2n
−1
t
2k
an indecomposable ch.f.?
t
1 it/2k
−it /2k
=
(e
+
e
)
2
2k
sin
t
2n
Seite 16
Example 3. The factorization of a characteristic function into
indecomposable factors may not be unique.
Factorization 2.
−1
t
t
t −1 sin t =
sin
3
3
ϕ t = t −1 sin t =
2 cos
2t
+ 1 /3
3
1
2t
[2 cos
+ 1]
3
3
∞
cos
k=1
⇒ We have two different factorizations.
t
3 ∙ 2k
Seite 17
Seminar – Counterexamples in
Probability
Ch8. Characteristic and Generating
Functions
Ch9. Infinitely divisible and stable
distributions
Seite 18
Definition (infinitely divisible distribution)
∙ X : a r.v. with d.f. F
∙ ϕ : ch.f. of X
∙ X is called infinitely divisible if
for each n≥1 there exist i.i.d. r.v.s Xn1, ..., Xnn such that
d
X = Xn1 + ∙∙∙ + Xnn
Equivalent :
∙ Ǝ d.f. Fn with F=(Fn)*n
∙ Ǝ ch.f. ϕn with ϕ =(ϕ n)n
Seite 19
Definition (stable distribution)
∙ X : a r.v. with d.f. F
∙ ϕ : ch.f. of X
∙ X is called stable if
for X1 and X2 independent copies of X and any positive numbers b1 and b2,
there is a positive number b and a real number γ s.t. :
d
b1X1+b2X2 = bX + γ
Equivalent :
ϕ b1 t ϕ b2 t = ϕ bt eiγt
Seite 20
Properties of infinitely divisible and stable distributions
• The ch.f. of an infinitely divisible r.v. does not vanish.
• If a r.v. X is stable, then it is infinitely divisible.
Seite 21
Example 4. A non-vanishing characteristic function which is
not infinitely divisible
random variable X
X
-1
0
1
P(X=x)
1/8
3/4
1/8
1 −it 3 it0 1 it
1
ϕ t = e + e + e = (3 + cos t)
8
4
8
4
=>
ϕ t > 0, ∀𝑡 ∈ ℝ
=>
ϕ does not vanish.
Seite 22
Example 4. A non-vanishing characteristic function which is
not infinitely divisible
Is X infinitely divisible?
d
Assume X = X1+X2, (X1, X2 are iid r.v.s)
Since X has three possible values, each of X1 and X2 can take only two
values, say a and b, a<b.
Let P[Xi=a]=p, P[Xi=b]=1-p for some p, 0<p<1, i=1,2
X1+X2
2a
a+b
2b
P(X1+X2=x)
p2
2p(1-p)
(1-p)2
 2a= -1, a+b=0, 2b=1, p2=1/8, 2p(1-p)=3/4, (1-p)2=1/8 =>
contradiction!
d
 X= X1+X2 is not possible. => X is not infinitely divisible.
Seite 23
Example 5. Infinitely divisible distribution, but not stable
i) X ~ Poi(λ)
[ X  n] 
n
n!
e  , n=0, 1, 2, ∙∙∙, λ>0
Characteristic funtion of X :
ϕ t = exp 𝜆 𝑒 𝑖𝑡 − 1 , 𝑡 ∈ ℝ
Characteristic funtion of Xn ~Poi(λ/n) :
ϕn t = exp
𝜆 𝑖𝑡
𝑒 −1
𝑛
=>
ϕ t = [ϕn (t)]n
=>
X is infinitely divisible
Seite 24
Example 5. Infinitely divisible distribution, but not stable
Is X a stable distribution?
If yes, for any b1 and b2 >0, there exist b>0 and γ∈ℝ s.t.
ϕ b1 t ϕ b2 t = ϕ(bt)eiγt
∙ Ch. f. of X ∶ ϕ t = exp 𝜆 𝑒 𝑖𝑡 − 1
∙ ϕ b1 t ϕ b2 t = exp λ eib 1 t − 1 exp λ eib 2 t − 1
= exp⁡
[λ eib 1 t + eib 2 t − 2 ]
∙ ϕ bt eiγt = exp λ eibt − 1 exp 𝑖𝛾𝑡 = exp λeibt + iγt − λ
=> ϕ b1 t ϕ b2 t ≠ ϕ bt eiγt
X is not stable.
, i. e.
Seite 25
Example 5. Infinitely divisible distribution, but not stable
ii) Let see the gamma distribution with parameter θ=1, k=1/2
g x =
x
k−1 −θ
x e
Γ k θk
ϕ t = (1 − itθ)
ϕn t = 1 − it
1
=
−k
π
1
− −x
x 2e
= (1 −
1
−
2n
,
x>0
1
−
it) 2
∶ ch. f. of gamma distr. with θ = 1, k =
ϕ t = [ϕn (t)]n , i. e.
X is infinitely divisible
1
2n
Seite 26
Example 5. Infinitely divisible distribution, but not stable
Is X a stable distribution?
If yes, for any b1 and b2 >0, there exist b>0 and γ ∈ ℝ s.t.
ϕ b1 t ϕ b2 t = ϕ(bt)eiγt
∙ Ch. f. of X ∶ ϕ t = (1 −
∙ ϕ b1 t ϕ b2 t = (1 −
iγt
∙ ϕ bt e
= (1 −
1
−
it) 2
1
−
ib1 t) 2
(1 −
1
−
ib2 t) 2
1
− iγt
ibt) 2 e
=> 𝜙 b1 t ϕ b2 t ≠ ϕ bt eiγt , i. e.
X is not stable.
= (1 − i b1 + b2 t −
1
2 −2
b1 b2 t )
Seite 27
REFERENCES
[1] J. Stoyanov. Counterexamples in probability (2nd edition). Wiley 1997
[2] G. Samorodnitsky, M. S. Taqqu. Stable Non-Gaussian Random Processes.
Chapman&Hall, 1994
[3] K. L. Chung. A course in probability theory. Academic Press, 1974
[4] E. Lukacs. Characteristic functions. Griffin, 1970
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Thank you very much !!!