I am wondering if the following holds/can be proven: Let $X_1, \ldots, X_n$ be iid copies of a random variable $X$, and let $Y_1, \ldots, Y_n$ be iid copies of a random variable $Y$. $X$ and $Y$ are also assumed to be independent. Then, $$A=\Big(\sum_{i=1}^n X_i\Big)\,\cdot\,\Big(\sum_{i=1}^n Y_i\Big)$$ is identically distributed as $$B=n\sum_{i=1}^n X_i Y_i.$$ Perhaps a look at characteristic functions can be conclusive?
It holds $A=\sum_{i=1}^n \sum_{j=1}^n X_i Y_{i+j\text{ mod }n}$, where $\sum_{i=1}^n X_i Y_{i+j\text{ mod }n}$ for $j=1,\ldots, n$ clearly follow the same distribution, but are not independent from each other. Any suggestions? Thanks!
This is false. The two random variables have the same mean, namely $n^2 \mathbb{E}(X) \mathbb{E}(Y)$, so assume WLOG that $\mathbb{E}(X) = \mathbb{E}(Y) = 0$ and let's compute the variance. Then
$$\mathbb{E}(A^2) = \mathbb{E} \left( \left( \sum X_i \right)^2 \right) \mathbb{E} \left( \left( \sum Y_i \right)^2 \right) = n^2 \text{Var}(X) \text{Var}(Y)$$
but
$$\mathbb{E}(B^2) = n^2 \mathbb{E} \left( \left( \sum X_i Y_i \right)^2 \right) = n^3 \text{Var}(X) \text{Var}(Y).$$