Independence of two random variables. We are given two independent random variables $X_1$ and $X_2$. We need to check whether $X_1$ and $X_1 X_2$ are independent or not.
Probability functions are given as : $P(X_i = -1) = P(X_i = 1) = \frac{1}{2}$ for $i=1,2$
Simply by looking at the random variables $X_1$ and $X_1 X_2$ can't we just conclude both are dependent, since the value of $X_1 X_2$ clearly depends on $X_1$ ? Or this isn't the way to proceed ?
 A: Functional dependence and stochastic dependence are two different things. Think about a random variable $X$ which takes the values $1$ and $-1$ with probability $1/2$. Then $X^2=1$ which is constant, so $X$ is stochastically independent on $X^2$, even though they are functionally dependent. So the only way to check for stochastic independence in your example is to verify whether $$P(X_1=i,X_1X_2=j)=P(X_1=i)P(X_1X_2=j)$$ for $i,j\in\{-1,1\}$.
A: That's in general not the way to proceed. For example, if $X_3$ is a constant function, $X_3$ and $X_3$ are independent, so you cannot say just by looking at it, that $X_1$ and $X_1X_2$ are dependent. You have to look at the probiablities. We have
\begin{align*}
  \def\P{\mathbf P}\P(X_1 = 1, X_1X_2 = 1) 
   &= \P(X_1 = 1, X_2 = 1)\\
   &= \P(X_1= 1) \P(X_2 = 1)\\
   &= \frac 14
\end{align*}
On the other hand
\begin{align*}
  \P(X_1X_2 = 1) &= \P(X_1=1, X_2=1) + \P(X_1=-1, X_2=-1)\\
   &= \frac 14 + \frac 14\\
   &= \frac 12
\end{align*}
So, $$\P(X_1 = 1, X_1X_2=1) = \P(X_1 = 1)\P(X_1X_2 = 1) $$
Along the same lines, we see that for any $i,j$: 
$$ \P(X_1 = (-1)^i, X_1X_2 = (-1)^j) = \P(X_1 = (-1)^i) \P(X_1X_2 = (-1)^j) $$
So, $X_1$ and $X_1X_2$ are independent.
A: It's a bit counterintuitive, but $X_1 X_2$ does not depend on $X_1$.
If $X_1=1$ and $P(X_2 = -1) = P(X_2 = 1) = \frac{1}{2}$, then
$P(X_1X_2 = -1) = P(1X_2 = -1) = P(X_2 = -1) = \frac{1}{2}$
In the same way, $P(X_1X_2 = 1)=\frac{1}{2}$
Second case:
If $X_1=-1$ and $P(X_2 = -1) = P(X_2 = 1) = \frac{1}{2}$, then
$P(X_1X_2 = -1)= P(-1X_2 = -1)= P(X_2 = 1) = \frac{1}{2}$
In the same way, $P(X_1X_2 = 1)=\frac{1}{2}$
So no matter what $X_1$ is, $P(X_1X_2 = -1)=P(X_1X_2 = 1)=\frac{1}{2}$
Intuitively, you could bescribe it as you being $X_1$ and picking $1$ or $-1$ at random and giving someone else ($X_2$) the possibility to negate what ou have chosen. As you can't know whether the other person will flip your choice, your pick and the result after the flip are independant.
