Seminar 2012 - Counterexamples in Probability Presenter : Joung In Kim

Seminar 2012- Counterexamples in Probability

Presenter : Joung In KimSeminar | 19.11.2012 |

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Seminar – Counterexamples in Probability

Ch8. Characteristic and Generating Functions

Ch9. Infinitely divisible and stable distributions

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Notation and Abbreviations

• r.v. : random variable

• ch. f. : characteristic function (ϕ(t))

• d.f. : distribution function (F)

• i.i.d. : independent and identically distributed

• : equality in distributiond =

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Definition (Characteristic function)

Let X be a r.v.defined on ሺΩ,ℱ,Ρሻ with distribution function F∶ ℝ →[0,1] The 𝐜𝐡𝐚𝐫𝐚𝐜𝐭𝐞𝐫𝐢𝐬𝐭𝐢𝐜 𝐟𝐮𝐧𝐜𝐭𝐢𝐨𝐧 ϕ:ℝ→ℂ of X is defined by ϕሺtሻ= Ε ൣ �eitX ൧ = eitx dFሺxሻ∞−∞

= ൞

eitx fሺxሻdx,∞−∞ if X is 𝒂𝒃𝒔𝒐𝒍𝒖𝒕𝒆𝒍𝒚 𝒄𝒐𝒏𝒕𝒊𝒏𝒖𝒐𝒖𝒔 with density f σ eitxnn pn, if X is 𝒅𝒊𝒔𝒄𝒓𝒆𝒕𝒆 with PሾX= xnሿ= pn, pn > 0,σ pn = 1n

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Properties of a characteristic function

i) ϕሺ0ሻ = 1, ϕሺ−tሻ= ϕሺtሻതതതതതത ,aFϕሺtሻaF ≤ 1, t ∈ℝ ii) If ΕaFXaFn < ∞,then ϕሺnሻሺ0ሻ exists and ΕሾXn ሿ= i−nϕሺnሻሺ0ሻ ൬ϕሺnሻሺtሻ= dndnt ϕሺtሻ൰ iii) If ϕሺnሻሺ0ሻ exists and n is even, then ΕaFXaFn < ∞ If n is odd ,then ΕaFXaFn−1 < ∞ iv) If ΕaFXaFn < ∞ ൫and hence ΕaFXaFk < ∞ for k < 𝑛 ൯then

ϕሺtሻ= (it)kk! Eൣ �Xk൧+ ο(tn)nk=0 ሺt →0ሻ

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Properties of a characteristic function v) If X1 and X2 are independent random variables with ch.f.s ϕ1 and ϕ2. ⇒ the ch.f.of X1+ X2 is given by ϕሺtሻ= ϕ1ሺtሻϕ2ሺtሻ,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: If ϕ is absolutely integrable over ℝ, it follows that X is absolutely continuous with density: fሺxሻ= 12π e−itxϕ(t)dt∞−∞ (inverse Fourier-transformation)

Pሺa < 𝑋< 𝑏ሻ= limc→∞ 12πන e−ita − e−itbitc−c ϕሺtሻdt

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Fourier expansion of a periodic function

g(t+T)=g(t) => 𝑔ሺtሻ= (an cos(nw0t) + bn sin(nw0t))∞

n=0 where, w0 = 2πT ,T:Period, an,bn ∶ Fourier coefficient 𝐚𝟎 = 1Tන gሺtሻdt,T2

−T2 𝐚𝐧 = 2Tන gሺtሻcosሺnw0tሻdtT2−T2 , 𝐛𝐧 = 2Tන gሺtሻsinሺnw0tሻdtT2

−T2

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Example 1. Discrete and absolutely continuous distributions with the same characteristic functions on [-1, 1]

ϕ1ሺtሻ= ൜1− aFtaF, if aFtaF≤ 1 0 , otherwise. 𝑓ሺ𝑥ሻ= 12πන e−it𝑥ϕ1ሺtሻdt = 1− cos𝑥π𝑥2∞

−∞ , 𝑥∈ℝ

ΡሾY = 0ሿ= 12 , PሾY = ሺ2𝑘− 1ሻπሿ= 2ሺ2k−1ሻ2π2 ,k = 0,±1,±2,∙∙∙

ϕ2(t) = 12+ 4π−2 cos(ሺ2k− 1ሻπt)(2k− 1)2∞

k=1

continuous

discrete

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Example 1. Discrete and absolutely continuous distributions with the same chacteristic functions on [-1, 1]hሺtሻ= aFtaF= a0 + σ an∞n=1 cosnπt (Fourier series)

൬ aFtaF≤ 1, a0 = 12, an = 2(cosnπ− 1)n2π2 ൰

= 12+ −4cos(ሺ2k− 1ሻπt)(2k− 1)2π2∞

k=1

ϕ1ሺtሻ= 1− aFtaF= 12+ 4cos(ሺ2k− 1ሻπt)(2k− 1)2π2∞

k=1

⇒ ϕ1ሺtሻ= ϕ2ሺtሻ, t ∈[−1,1]

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Example 2. The absolute value of a characteristic function is not necessarily a characteristic function.

ϕሺtሻ= 18൫1+ 7eit൯, t ∈ℝ , aFϕ(t)aF= 18൫50+ 7e−it + 7eit൯1/2 If aFϕ(t)aF is a ch.function, then ψ(t) ∶= aFϕ(t)aF must be of the form ψሺtሻ= p∙eitx1 + (1− p) ∙eitx2 Where 0 < 𝑝< 1 𝑎𝑛𝑑 x1 ,x2 are different real numbers. Comparing ψሺtሻ2 and aFϕ(t)aF2 => p2 = (1− p)2 = 764 , 2p(1− p) = 5064 => contradiction! 𝐂𝐨𝐧𝐜𝐥𝐮𝐬𝐢𝐨𝐧∶ ϕ is a ch.function, but not aFϕaF.

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Decomposable and Indecomposable

We say that a ch.f. ϕ is decomposableif 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.

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Example 3. The factorization of a characteristic function into indecomposable factors may not be unique.

(i) discrete caseX : discrete uniform distribution on the set {0, 1, 2, 3, 4, 5}.Characteristic function of X :

We can factorize the ch. f. in the following way :ϕሺtሻ= E൫eitX൯= P(X= k)eitk5

k=0 16 = eitk5k=0

ϕ1ሺtሻ= 13൫1+ e2it + e4it൯, ϕ2ሺtሻ= 12൫1+ eit൯ ψ1ሺtሻ= 13൫1+ eit + e2it൯, ψ2ሺtሻ= 12൫1+ e3it൯ ϕሺtሻ= ϕ1ሺtሻ∙ϕ2ሺtሻ= ψ1ሺtሻ∙ψ2ሺtሻ, t ∈ℝ

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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 ψ1ሺtሻ= ψ11ሺtሻ∙ψ12ሺtሻ, ψ11ሺtሻ, ψ12ሺtሻ: non− trivial Let ψ11ሺtሻ= peitx1 +ሺ1− pሻeitx2, ψ12ሺtሻ= qeity1 +ሺ1− qሻeity2 ሺ0 < p < 1, 0 < 𝑞 < 1ሻ then pq = ሺ1− pሻሺ1− qሻ= pሺ1− qሻ+ qሺ1− pሻ= 13 => contradiction ⇒ ψ1 is indecomposable. ⋅ ϕ1ሺtሻ= ψ1ሺ2tሻ ⇒ ϕ1 is also indecomposable.∎

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(ii) Continuous case∙ Let X be a r.v. which is uniformly distributed on (-1,1). ∙ Ch. f. of X :ϕሺtሻ= t−1 sint, t ∈ℝ Factorization 1.

ϕሺtሻ= t−1 sint = �ෑ cos൬t2k൰n

k=1 ൩൬t2n൰−1 sin൬t2n൰

=> limn→ ∞ ϕሺtሻ= ෑ� cos൬t2k൰∞

k=1

Is cosቀ t2kቁ an indecomposable ch.f.? cos൬t2k൰= 12(eit/2k + e−it/2k)

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Factorization 2.

t−1 sint = ൬t3൰−1 sin൬t3൰2cos൬2t3൰+ 1൨/3

ϕሺtሻ= t−1 sint = 13[2cos൬2t3൰+ 1]ෑ� cos൬ t3 ∙2k൰∞

k=1

⇒ We have two different factorizations.

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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

X Xn1 + ∙∙∙ + Xnn

Equivalent :

∙ Ǝ d.f. Fn with F=(Fn)*n

∙ Ǝ ch.f. ϕn with ϕ =(ϕ n)n

d =

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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. : b1X1+b2X2 bX + γ Equivalent :

d =

ϕሺb1tሻϕሺb2tሻ= ϕሺbtሻeiγt

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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.

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Example 4. A non-vanishing characteristic function which is not infinitely divisible

random variable X

=>

=> ϕ does not vanish.

X -1 0 1

P(X=x) 1/8 3/4 1/8

ϕሺtሻ= 18e−it + 34eit0 + 18eit = 14(3+ cost) ϕሺtሻ> 0,∀𝑡 ∈ℝ

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Example 4. A non-vanishing characteristic function which is not infinitely divisible

X1+X2 2a a+b 2b

P(X1+X2=x) p2 2p(1-p) (1-p)2

d =

d =

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Example 5. Infinitely divisible distribution, but not stable

i) X ~ Poi(λ) , n=0, 1, 2, ∙∙∙, λ>0 Characteristic funtion of X : Characteristic funtion of Xn ~Poi(λ/n) :

=> => X is infinitely divisible

en

nXn

!][

ϕሺtሻ= expൣ �𝜆൫𝑒𝑖𝑡 − 1൯൧, 𝑡 ∈ℝ ϕnሺtሻ= exp𝜆𝑛൫𝑒𝑖𝑡 − 1൯൨ ϕሺtሻ= [ϕn(t)]n

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Is X a stable distribution?If yes, for any b1 and b2 >0, there exist b>0 and γ∈ s.t.ϕሺb1tሻϕሺb2tሻ= ϕ(bt)eiγt ∙Ch.f.of X ∶ ϕሺtሻ= expൣ �𝜆൫𝑒𝑖𝑡 − 1൯൧

∙ϕሺb1tሻϕሺb2tሻ= exp �ൣλ൫eib1t − 1൯൧expൣ �λ൫eib2t − 1൯൧= exp [λ൫eib1t + eib2t − 2൯] ∙ϕሺbtሻeiγt = expൣ�λ൫eibt − 1൯൧expሾ𝑖𝛾𝑡ሿ= expൣ�λeibt + iγt− λ൧ => ϕሺb1tሻϕሺb2tሻ ≠ ϕሺbtሻeiγt ,i.e. X is not stable.

ℝ

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ii) Let see the gamma distribution with parameter θ=1, k=1/2gሺxሻ= xk−1e−xθΓሺkሻθk = 1ξπx−12e−x , x > 0

ϕሺtሻ= (1− itθ)−k = (1− it)−12 ϕnሺtሻ= ሺ1− itሻ− 12n ∶ ch.f.of gamma distr.with θ= 1,k = 12n ϕሺtሻ= [ϕn(t)]n ,i.e. X is infinitely divisible

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Is X a stable distribution?If yes, for any b1 and b2 >0, there exist b>0 and γ ∈ s.t.ϕሺb1tሻϕሺb2tሻ= ϕ(bt)eiγt

∙Ch.f.of X ∶ ϕሺtሻ= (1− it)−12 ∙ϕሺb1tሻϕሺb2tሻ= (1− ib1t)−12 (1− ib2t)−12 = (1− iሺb1 + b2ሻt− b1b2t2)−12 ∙ϕሺbtሻeiγt = (1− ibt)−12eiγt => 𝜙ሺb1tሻϕሺb2tሻ ≠ ϕሺbtሻeiγt , i.e. X is not stable.

ℝ

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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 !!!

Seminar 2012 - Counterexamples in Probability Presenter : Joung In Kim

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