# Linearly independent random variables and independent random variables

Does one of these two assertions imply the other ?

(1) $X_1, X_2, ..., X_n$ are linearly independent random variables (i.e. $\lambda_1 X_1 + \lambda_2 X_2 + ... + \lambda_n X_n = 0$ => $\lambda_1 =\lambda_2=...=\lambda_n=0$)

and

(2) $X_1, X_2, ..., X_n$ are independent random variables (stochastically independent)

If not, is there some special cases for which one implication (1=>2 or 2=>1) is true (Gaussian law? etc.)

• Well I feel like (1) and (2) belong to different mathematical topics. (1) is for algebra (meaning X is a vector), where as (2) is for probability (meaning X is a stochastic variable). Therefore (1) and (2) wouldn't have to do anything with each other. – Bernd Jan 9 '14 at 10:59
• To start with: if $X_i \in \mathbb{R}$ (real scalar) and $n>1$, then 1) is never true. I think you are confusing probabilistic concepts (eg orthogonality) with linear-algebra concepts. see eg stats.stackexchange.com/questions/12128/… – leonbloy Jan 9 '14 at 11:37
• No @leonbloy, I don't confuse anything, I know that linear independence (linear algebra) and stochastic independence (probability theory) are different concepts in different topics (linalg vs proba), but I wanted to know if, in some cases, there are links between them – Basj Jan 9 '14 at 13:15

As a counterexample of $1 \Rightarrow 2$ take $X_1 = X,\ X_2 = X^2$ for some r.v. $X$, say $X\sim U(0,1)$. They are of course not stochastically independent, but should be linearly independent.
For $2 \Rightarrow 1$, just use a.s. constant $X_1, X_2$. The are stochastically independent, but not linearly independent.
• You are right (+1), but for 2)$\rightarrow$1) being a.s. constant is the only counterexample. The only thing needed to get 2)$\rightarrow$1) is the extra condition that $X_1$ or $X_2$ is not a.s. constant. – drhab Jan 9 '14 at 12:41
• Thank you @drhab, do you claim that 2 => 1 is true as long as $X_1$ and $X_2$ are not a.s. constants ? Do you have a proof for that ? – Basj Jan 9 '14 at 13:14
• @Basj: drhab's answer proves that for $n=2$ and I think this can easily be extended. – Jens Jan 9 '14 at 13:49
Let it be that $X$ and $Y$ are not linearly independent. Then some pair $\left(\lambda,\mu\right)\in\mathbb{R}^{2}\backslash\left\{ \left(0,0\right)\right\}$ exists with $\lambda X+\mu Y=0$ leading to $Y=cX$ or $X=cY$ for some constant $c$. So independence is quite far away. If they are independent then they must be a.s. constant.