How to show that the distribution of $X=V+W$ is infinitely divisible?

Question

The question is :

Suppose that $V$ and $W$ are independent random variables such that $V$ is uniformly distributed on the interval $[-1,1]$ and $W$ has the standard Cauchy distribution. Define the random variable $X$ by $X=V+W$. Is the distribution of $X$ infinitely divisible?

By definition, I know that a distribution of $X$ is infinitely divisible if for every $n \in \mathbb{N}$ there exist a sequence of i.i.d random variables $(X_1,X_2,..,X_n)$ has the same distribution as $X$.

For this question, I proved that the variable $W$ is infinitely divisible by using the characteristics function.

I am not sure how to solve this for sum of two random variable with different distribution?

Can anyone help me with this one?

There are zero-points for the characteristic functions of $\phi_V(t), \phi_{V+W}(t)$ — JGWang, Mar 04 '22 at 03:19
@JGWang you mean $\phi(t) \neq 0 $ for all $t \in \mathbb{R}$, where $\phi(t)$ is the characteristic function of an infinitely divisible distribution. — User124356, Mar 04 '22 at 12:22

score 1 · Accepted Answer · answered Mar 04 '22 at 12:45

1

This is my attempt:

$\phi_{(X)}(t) = E(exp(iXt))= E(exp(i(V+W)t)$

Since, $V$ and $W$ are independent. So,

\begin{eqnarray*} \phi_{(X)}(t) &=& E(exp(iVt)). E(exp(iWt))\\ &=& \frac{e^{it}-e^{-it}}{2it} . e^{-|t|}\\ &=& \frac{sin(t)}{t} . e^{-|t|}\\ \end{eqnarray*}

For $t=n\pi$, where $n$ is a non-zero integer , $\phi_{(X)}(t) = 0$. So the distribution of $X$ is not infinitely divisible.

Is this solution correct?

answered Mar 04 '22 at 12:45

User124356

1,617

Yep, that is correct. Here is a proof that characteristic functions of infinitely divisibly distributions are nonzero everywhere. – Mike Earnest Mar 04 '22 at 15:03

How to show that the distribution of $X=V+W$ is infinitely divisible?

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