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


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*$\frac{1}{1-it}$

*$\frac{1}{1+t^2}$

*$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 $X \stackrel{D}{=} X_{1,n}+\cdots+X_{n,n}$
 A: I would prefer to give hints when possible:


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*$\frac1{1-it}$ 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.

*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 $X_1$.

*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. 
