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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.