Can we show $X_1,\ldots,X_k$ are independent iff $\mathcal L(X_1+\cdots+X_k)=\mathcal L(X_1)\ast\cdots\ast\mathcal L(X_k)$?

Question

Let $X_1,\ldots,X_k$ be real-valued random variables on a probability space $(\Omega,\mathcal A,\operatorname P)$ for some $k\in\mathbb N$.

Are we able to show that

1. $X_1,\ldots,X_k$ are independent; i.e. $$\mathcal L(X_1,\ldots,X_k)=\mathcal L(X_1)\otimes\cdots\otimes\mathcal L(X_k)\tag1;$$
2. $\mathcal L(X_1+\cdots+X_k)=\mathcal L(X_1)\ast\cdots\ast\mathcal L(X_k)$

are equivalent?

Let $$\theta:\mathbb R^k\to\mathbb R\;,\;\;\;x\mapsto x_1+\cdots+x_k.$$ Then, if $X_1,\ldots,X_k$ are independent, then \begin{equation}\begin{split}\mathcal L(X_1+\cdots+X_k)&=\mathcal L(\theta(X_1,\ldots,X_k)\\&=\theta(\mathcal L(X_1,\ldots,X_k))=\mathcal L(X_1)\ast\cdots\ast\mathcal L(X_k),\end{split}\tag2\end{equation} where the first and last equality hold by definition and the inner equality holds by $(1)$.

But does the other implication hold as well?

Answer 1

It does not hold. For example, if $X$ and $Y$ are independent and Cauchy distributed, then $(X + Y)/2$ is also Cauchy distributed. Therefore $X+Y$ and $2X$ have the same law, but $2X = X + X$ is the sum of two non-independent Cauchy distributions.

In your notation, let $X_1 = X_2 = X$ with $X$ Cauchy distributed. Then $X_1$ and $X_2$ are not independent, but $\mathcal L(X_1 + X_2) = \mathcal L(X_1) * \mathcal L(X_2)$.