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Let $X_1, X_2, \ldots, X_n \in \mathbb{R}^n$ be continuous random variables in $\mathbb{R}^n$ that are pairwise and elementwise independent. By continuous random variables, it is meant that their distributions do not include any Dirac delta fnc $\delta_x$.

My question is:

in this general situation, are $X_1, \ldots, X_n$ linearly independent (LI) with propbability 1? That is, is it true that \begin{equation} \operatorname{Prob}\big ( \operatorname{span} \{ X_1, \ldots, X_n\} = \mathbb{R}^n \big ) = 1\text{?} \end{equation}


This is a general version of the topic:

Probability that $n$ vectors drawn randomly from $\mathbb{R}^n$ are linearly independent

since here we consider a general independent prob. distribution, rather than uniform (or Gaussian) distribution. By referring the related posts above and below:

The probability that two vectors are linearly independent.

I think the statement above is true as shown in the short proof below.

First, the probability of "$X_1 = 0$" is zero since the singleton $\{ 0 \}$ is a set of measure zero. Hence, $X_1$ is LI with probability 1. Next, assume that $X_1,\ldots,X_r$ are LI with probability 1 for some $r \in \{1,2,\ldots, n-1\}$. Then, with probability 1, they span an $r$-dimensional subspace of $\mathbb{R}^n$ which is, however, a set of measure zero, too. Hence, the next vector $X_{r+1}$ lies on $\mathbb{R}^n$ outside the subspace with prob. $1$. By this process, $\operatorname{span}\{X_1,\ldots,X_n\} = \mathbb{R}^n$ with prob. $1$.

But, I'm just not sure that this proof is correct and has no problem.

Many thanks in advance for your comments and discussions!


Edited: I think this can be extended to a more general case where the RVs are stochastically dependent. The proof seems to be also true in this dependent case as well---it is relevant to the fact that the distributions include no Dirac delta fnc, not to the stochastic independence.

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