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I have got this homework to check which distributions with the following characteristic functions are infinitely divisible:

  • $\frac{1}{1-it}$
  • $\frac{1}{1+t^2}$
  • $e^{-t^2}\cos t $

I literally have no idea how to approach it. All I know is the definition of infinite divisibility. Could you show me how to deal with tasks like that?

The definition I was given at the lecture:

Distribution of random variable $X$ is infinitely divisible if for every $n \in N$ there exist $X_{1,n},.., X_{n,n}$ i.i.d such that $X \stackrel{D}{=} X_{1,n}+\cdots+X_{n,n}$

StubbornAtom
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ראָזעווע וואַלפיש
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    "All I know is the definition of infinite divisibility" Then you know more than I do. Please share. – fleablood Jun 09 '20 at 16:21
  • Sory I should have done that in the first place. I've edited my question. – ראָזעווע וואַלפיש Jun 09 '20 at 16:32
  • Do you know the characteristic function of the sum of $n$ i.i.d. random variables? Can you reverse this and check whether you still have a characteristic function? – Henry Jun 09 '20 at 16:32
  • So the characteristic function of the sum of n i.i.d. random variables is the characteristic function of $X_1$ to the nth power, but what do you mean by reversing? – ראָזעווע וואַלפיש Jun 09 '20 at 16:38
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    Reversing the $n$th power would be taking the $n$th root. So for example is $\sqrt[n]{ \frac{1}{1-it}}$ a characteristic function? If it is for all $n$ then the first example is infinitely divisible, and if not then not. It may help if you know the characteristic functions of some common distributions – Henry Jun 09 '20 at 16:58
  • In general, determining whether a function is a characteristic function is a hard problem, Maybe some of the methods in this answer are helpful: https://math.stackexchange.com/a/2417928/177399 – Mike Earnest Jun 10 '20 at 18:26

1 Answers1

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I would prefer to give hints when possible:

  1. $\frac1{1-it}$ is the c.f. of a famous distribution. Furthermore, that distribution is part of a $2$-parameter family of distributions with nice additive properties, which allows you to show it is infinitely divisible (i.d.). For example, any normal distribution is i.d. because $N(\mu_1+\mu_2,\sigma_1^2+\sigma_2^2)\sim N(\mu_1,\sigma_1^2)\oplus N(\mu_2,\sigma_2^2)$. A similar relation holds for family containing the random variable with this ch.f.

  2. You can leverage the answer for $1$ to get an answer for $2$. Note that $\phi_2(t):=1/(1+t^2)$ is the complex modulus of $\phi_1(t):=1/(1-it)$. This means that $\phi_2(t)=\phi_1(t)\overline{\phi_1(t)}$, which further implies that $X_2\stackrel{d}=X_1-X_1'$, where $X_1$ has ch.f. $\phi_1$, $X_2$ has ch.f. $\phi_2$, and $X_1'$ is an iid copy of $X_1$.

  3. There is a well-known theorem that the characteristic function of an infinitely divisible distribution cannot have any zeroes, which rules out this function. For a proof, see this MSE annswer.

Mike Earnest
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