Section 28

Infinitely Divisible Distributions*

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Problem 28.1

Show that μn→ιμ\mu_{n} \rightarrow_{\iota } \mu implies μ(R1)≤lim inf⁡nμn(R1)\mu \left(R^{1}\right) \leq \liminf_{n} \mu_{n}\left(R^{1}\right). Thus in vague convergence mass can "escape to infinity" but mass cannot "enter from infinity."

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Problem 28.2
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(a)

Show that μn→ιμ\mu_{n} \rightarrow_{\iota } \mu if and only if (28.2) holds for every continuous ff with bounded support.

(b)

Show that if μn→ιμ\mu_{n} \rightarrow_{\iota } \mu but (28.1) does not hold, then there is a continuous ff vanishing at ±∞\pm \infty for which (28.2) does not hold.

Problem 28.3

23.7↑23.7 \uparrow Suppose that N,Y1,Y2,…N, Y_{1}, Y_{2}, \ldots are independent, the YnY_{n} have a common distribution function FF, and NN has the Poisson distribution with mean α\alpha. Then S=Y1+⋯+YNS=Y_{1}+\cdots +Y_{N} has the compound Poisson distribution.

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(a)

Show that the distribution of SS is infinitely divisible. Note that SS may not have a mean.

(b)

The distribution function of SS is Σn=0∞e−ααnFn∗(x)/n!\Sigma_{n=0}^{\infty } e^{-\alpha } \alpha^{n} F^{n *}(x) / n!, where Fn∗F^{n *} is the nn-fold convolution of FF (a unit jump at 0 for n=0n=0 ). The characteristic function of SS is exp⁡α∫−∞∞(eitx−1)dF(x)\exp \alpha \int_{-\infty }^{\infty }\left(e^{i t x}-1\right) d F(x).

(c)

Show that, if FF has mean 0 and finite variance, then the canonical measure μ\mu in (28.6) is specified by μ(A)=α∫Ax2dF(x)\mu (A)=\alpha \int_{A} x^{2} d F(x).

Problem 28.4
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(a)

Let ν\nu be a finite measure, and define

φ(t)=exp⁡[iγt+∫−∞∞(eitx−1−itx1+x2)1+x2x2ν(dx)],(28.12) \varphi (t)=\exp \left[i \gamma t+\int _{-\infty }^{\infty }\left(e^{i t x}-1-\frac{i t x}{1+x^{2}}\right) \frac{1+x^{2}}{x^{2}} \nu (d x)\right], \tag {28.12}

where the integrand is −t2/2-t^{2} / 2 at the origin. Show that this is the characteristic function of an infinitely divisible distribution.

(b)

Show that the Cauchy distribution (see the table on p. 348) is the case where γ=0\gamma =0 and ν\nu has density π−1(1+x2)−1\pi^{-1}\left(1+x^{2}\right)^{-1} with respect to Lebesgue measure.

Problem 28.5

Show that the Cauchy, exponential, and gamma (see (20.47)) distributions are infinitely divisible.

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Problem 28.6

Find the canonical representation (28.6) of the exponential distribution with mean 1:

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(a)

The characteristic function is ∫0∞eitxe−xdx=(1−it)−1=φ(t)\int_{0}^{\infty } e^{i t x} e^{-x} d x=(1-i t)^{-1}=\varphi (t).

(b)

Show that (use the principal branch of the logarithm or else operate formally for the moment) d(log⁡φ(t))/dt=iφ(t)=i∫0∞eitxe−xdxd(\log \varphi (t)) / d t=i \varphi (t)=i \int_{0}^{\infty } e^{i t x} e^{-x} d x. Integrate with respect to tt to obtain

11−it=exp⁡∫0∞(eitx−1)e−xxdx(28.13) \frac{1}{1-i t}=\exp \int _{0}^{\infty }\left(e^{i t x}-1\right) \frac{e^{-x}}{x} d x \tag {28.13}

Verify (28.13) after the fact by showing that the ratio of the two sides has derivative 0.

(c)

Multiply (28.13) by e−ite^{-i t} to center the exponential distribution at its mean: The canonical measure μ\mu has density xe−xx e^{-x} over (0,∞)(0, \infty ).

Problem 28.7

↑\uparrow If XX and YY are independent and each has the exponential density e−xe^{-x}, then X−YX-Y has the double exponential density 12e−∣x∣\frac{1}{2} e^{-\left|x\right|} (see the table on p. 348). Show that its characteristic function is

11+t2=exp⁡∫−∞∞(eitx−1−itx)1x2∣x∣e−∣x∣dx \frac{1}{1+t^{2}}=\exp \int _{-\infty }^{\infty }\left(e^{i t x}-1-i t x\right) \frac{1}{x^{2}}\left|x\right| e^{-\left|x\right|} d x
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Problem 28.8

↑\uparrow Suppose X1,X2,…X_{1}, X_{2}, \ldots are independent and each has the double exponential density. Show that ∑n=1∞Xn/n\sum_{n=1}^{\infty } X_{n} / n converges with probability 1. Show that the distribution of the sum is infinitely divisible and that its canonical measure has density ∣x∣e−∣x∣/(1−e−∣x∣)=∑n=1∞∣x∣e−∣nx∣|x| e^{-|x|} /\left(1-e^{-|x|}\right)=\sum_{n=1}^{\infty }|x| e^{-|n x|}.

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Problem 28.9

26.8↑26.8 \uparrow Show that for the gamma density e−xxu−1/Γ(u)e^{-x} x^{u-1} / \Gamma (u) the canonical measure has density uxe−xu x e^{-x} over (0,∞)(0, \infty ).

The remaining problems require the notion of a stable law. A distribution function FF is stable if for each nn there exist constants ana_{n} and bn,an>0b_{n}, a_{n}>0, such that, if X1,…,XnX_{1}, \ldots , X_{n} are independent and have distribution function FF, then an−1(X1+⋯+Xn)+bna_{n}^{-1}\left(X_{1}+\cdots +X_{n}\right)+b_{n} also has distribution function FF.

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Problem 28.10

Suppose that for all a,a′,b,b′a, a^{\prime }, b, b^{\prime } there exist a′′,b′′a^{\prime \prime }, b^{\prime \prime } (here a,a′,a′′a, a^{\prime }, a^{\prime \prime } are all positive) such that F(ax+b)∗F(a′x+b′)=F(a′′x+b′′)F(a x+b) * F\left(a^{\prime } x+b^{\prime }\right)=F\left(a^{\prime \prime } x+b^{\prime \prime }\right). Show that FF is stable.

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Problem 28.11

Show that a stable law is infinitely divisible.

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Problem 28.12

Show that the Poisson law, although infinitely divisible, is not stable.

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Problem 28.13

Show that the normal and Cauchy laws are stable.

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Problem 28.14

28.10↑28.10 \uparrow Suppose that FF has mean 0 and variance 1 and that the dependence of a′′,b′′a^{\prime \prime }, b^{\prime \prime } on a,a′,b,b′a, a^{\prime }, b, b^{\prime } is such that

F(xσ1)∗F(xσ2)=F(xσ12+σ22) F\left(\frac{x}{\sigma _{1}}\right) * F\left(\frac{x}{\sigma _{2}}\right)=F\left(\frac{x}{\sqrt{\sigma _{1}^{2}+\sigma _{2}^{2}}}\right)

Show that FF is the standard normal distribution.

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Problem 28.15
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(a)

Let YnkY_{n k} be independent random variables having the Poisson distribution with mean cnα/∣k∣1+αc n^{\alpha } /|k|^{1+\alpha }, where c>0c>0 and 0<α<20<\alpha <2. Let Zn=n−1∑k=−n2n2kYnkZ_{n}=n^{-1} \sum_{k=-n^{2}}^{n^{2}} k Y_{n k} (omit k=0k=0 in the sum), and show that if cc is properly chosen then the characteristic function of ZnZ_{n} converges to e−∣t∣αe^{-\left|t\right|^{\alpha }}.

(b)

Show for 0<α≤20<\alpha \leq 2 that e−∣t∣αe^{-\left|t\right|^{\alpha }} is the characteristic function of a symmetric stable distribution; it is called the symmetric stable law of exponent α\alpha. The case α=2\alpha =2 is the normal law, and α=1\alpha =1 is the Cauchy law.