Let $X_1,X_2,\ldots,X_n\in\mathbb{N}_0$ be independent and identically distributed random variables. The corresponding cumulative sums $S_k$ are defined as $S_k:=\sum_{i=1}^k X_i$. Let $(T_1,T_2,\ldots,T_n)$ be a random permutation of the cumulative sums $(S_1,S_2,\ldots,S_n)$. How do $X_1,X_2,\ldots,X_n$ have to be distributed such that $T_1, T_2, \ldots, T_n$ are mutually independent? Particulary, is the independence fulfilled if $X_1,X_2,\ldots,X_n$ are geometrically distributed?


Consider the case $n=2$: you have $T_1 = X_1 + \epsilon X_2$ and $T_2 = X_1 + (1-\epsilon) X_2$, where $\epsilon$ has Bernoulli($1/2$) distribution and is independent of $X_1$ and $X_2$. It's easy to compute that the covariance of $T_1$ and $T_2$ is

$$ \text{Cov}(T_1, T_2) = \text{Var}(X_1) - \frac{ \mathbb E[X_2]^2}{4}$$

For $T_1$ and $T_2$ to be independent, it's necessary (but not sufficient) for this to be $0$.

In particular if $X_1$ and $X_2$ are geometric random variables with parameter $p$, that turns out to be $$ \frac{(1-p)(3+p)}{4p^2}$$ and in particular is never $0$ except in the trivial case $p=1$. So we do not have independence for the geometric case.


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