Are X and Y independent random variables?
$\bullet$ Let $Z$ be uniformly distributed on $[-1,1]$.
$\bullet$ $X$ is a random variable such that $X=1$ when $Z>0$ and $X=-1$ otherwise.
$\bullet$ $Y$ is a random variable such that $Y=ZX$
Are $X$ and $Y$ independent??
From reading the question my first conclusion was that they are not independent but I am having difficulty finding a counterexample such that $$\mathbb{P}(A \cap B) \neq \mathbb{P}(A)\mathbb{P}(B)$$
Any help would be appreciated
2 Answers
$\begingroup$Consider two borelian $>0$ functions $f,g$.
$$ \begin{align} E(f(X)g(Y)) &= \frac 12\left(\int _{-1}^0 f(-1) g(-x) dx + \int_0^1 f(1)g(x)dx\right) \\&= \frac 12\left(\int _{0}^1 f(-1) g(x) dx + \int_0^1 f(1)g(x)dx\right) \\&= \frac 12\left(f(-1) + f(1)\right) \int _{0}^1 g(x)dx \\&= \left[\frac 12\left( \int_{-1}^0 f(-1) dx + \int_0^1 f(1) dx \right)\right] \left[\frac 12\left( \int _{-1}^0 g(-x) dx + \int_0^1 g(x) dx \right)\right] \\&= E(f(X))\times E(g(Y)) \end{align} $$
hence $X,Y $ are independant.
$\endgroup$ $\begingroup$Random variables $X : \Omega_X \rightarrow \mathbb R$ and $Y : \Omega_Y \rightarrow \mathbb R$ are independent random variables if events ${X \in [a,b]}$ and $Y \in [c,d]; \forall a,b,c,d \in \mathbb R$ are independent events. We can see that $Y$ equals $|Z|$, while $X = \frac{Z}{|Z|}$. Now try to understand whether $Y$ distribution is altered if we know the value of $X$. Let $X = -1$. Then $Z \in [-1,0]$ uniformly and $Y \in [0,1]$ uniformly. If $X = 1$, then $Z \in (0,1]$ uniformly and $Y \in (0,1]$ uniformly. Thus, we have the only value for $Y = 0$ which can't happen when $X = 1$. However, as this is one point, the measure of set containing such outcomes is $0$ and $P(Y=0|X=x) = 0$ regardless of distribution. Thus, $Y$ is uniformly distributed on $[0,1]$ independent from $X$ random variable.
$\endgroup$ 2
