Let $X$ be a random variable, and $\mathbf{Y}=<Y_1,Y_2,...,Y_n>$ be a vector of random variables. If $X$ is independent of $Y_i \forall i=1,2,3,...,n $, will this $X$ be independent of any linear combination of $\mathbf{Y}?$ that is, $X \perp C^T \mathbf{Y}$ ?
I can only show $\text{cov}(X,C^T\mathbf{Y})=\sum c_i (X,Y_i)=0$. Any other suggestion? thanks,
I don't think so. Suppose $Y_1$ and $Y_2$ are independent coin tosses, and $X$ is $0$ if both coins are the same, and $1$ if they are different. Then $X$ is independent of $Y_1$ and $Y_2$, but not independent of $Y_1 + Y_2$.
On the other hand, if $X$ is independent of the set $\{Y_1,\ldots,Y_n\}$ of random variables, then the result is true. Independence here means that for any vector $\mathbf{v}$ and real number $x$, we have $\{X \leq x\}$ and $\{\mathbf{Y} \leq \mathbf{v}\}$ are independent events. (The inequality of vectors is just component-wise inequality).
More generally, this means that for any measurable subset $U$ of $\mathbb{R}^n$, $\{X \leq x\}$ and $\{\mathbf{Y} \in U\}$ are independent. Then if $y \in \mathbb{R}$, the set of vectors $\mathbf{v} \in \mathbb{R}^n$ such that $C^{T}v \leq y$ is measurable. So $\{X \leq x\}$ and $\{C^{T}\mathbf{Y} \leq y\}$ are independent events.
