Independence of a random vector and a random variable

Question

I have a random variable $X$ and a random vector $Y=(Y_1,Y_2,...,Y_n)$.

Let $X$ be stochastically independent of the elements of $Y$, i.e. $X$ is independent of $Y_1$, of $Y_2$,..., of $Y_n$. Can I then conclude that $X$ is also independent of vector $Y=(Y_1,Y_2,...,Y_n)$?

So I tried to show that the following relationship holds (in terms of density functions): $$f(X,Y_1,Y_2,...,Y_n) = f(X) \cdot f(Y_1,Y_2,...,Y_n)$$

given that $$f(X,Y_i) = f(X) \cdot f(Y_i), i=1,...,n.$$

Unfortunately, I have no idea of how to proceed.

Answer 1

Consider the following distribution for $X$ and $\{Y_1,Y_2\}$:

$$ f(y_1) = \left\{ \begin{array}{cc} 1 & 0 < y_1 < 1 \\ 0 & \mbox{ otherwise} \end{array} \right. \\ f(y_2) = \left\{ \begin{array}{cc} 1 & 0 < y_2 < 1 \\ 0 & \mbox{ otherwise} \end{array} \right. \\ f(x| Y_1 = y_1, Y_2 = y_2) = \left\{ \begin{array}{cc} \delta\left(x-(y_1-y_2) \right) & \mbox{if }y_1 > y_2 \\ \delta\left(x-(y_1-y_2+1) \right) & \mbox{otherwise }\end{array} \right. $$ The distribution of $X$ for any fixed value of $Y_1$ is simply that of a uniform random on $(0,1)$. Similarly, the distribution of $Y$ for any fixed value of $Y_1$ is simply that of a uniform random on $(0,1)$.

Yet $X$ is clearly not independent of the vector $\{Y_1,Y_2\}$. Indeed, for a given value of the vector, the value of $X$ is completely determined, and will be different for different values of the vector.

Answer 2

It turns out the answer is false. It is somewhat difficult to explain the intuition behind coming up with counterexamples like these, but here is one (it is similar in spirit to Mark Fischler's answer). I leave it to you to prove it is correct.

Let $n=2$, and let $Y_1,Y_2$ be fair, independent coin flips. Let $$ X = \begin{cases} 1 & {Y_1\neq Y_2}\\ 0 & {Y_1=Y_2} \end{cases} $$