Proof: If $X_1$ and $X_2$ are independent random variable, are $X_1$ and $X_1X_2$ independent? I am working through a set of example problems.
I am stuck on this on which defines the distribution are:
$P(X_i=-1)=P(X_i=1)=\frac{1}{2}$ for $i=1,2$.
I am trying to show whether $X_1$ and $X_1X_2$ are independent.
So far I know that if $X$ and $Y$ are independent then the joint mass function should be equivalent to $P(X_1)\cdot P(X_1X_2)$
So from this I worked out $P(X_1=1)=\frac{1}{2}$
And the $P(X_1X_2=1)=P(X_1=1)\cdot P(X_2=1)=1$ (as they are independent random variables).
However I can't seem to proceed from here.
I think I'm missing something obvious that the probabilities tell me about the distribution but I can't seem to think what it could be. Instinctively, I'm thinking that this should be a uniform distribution but not sure how to go from there.

Let's try to answer as quickly as possible, exploiting the structure of the problem (also enumeration would work) and the relation between independence and conditional distributions.
Let $Z=X_1X_2$.
We observe that:
$P(Z=z|X_1=x_1)=1/2$
because given $X_1$, $X_2$ will just change randomly its sign in the multiplication $Z=X_1X_2$.
This is enough to declare independence since the distribution of $Z$ conditioned on $X_1$ does not depend on $X_1.$
I think this approach "tells us" also "why" the variables are independent.
If we have Bernoulli variables, as in the example of BruceET (i.e. taking values in {0,1}), than the conclusion would be different since in that case:

*

*$P(Z=z|X_1=0)=1$ if $z=0$ and $0$ otherwise

*$P(Z=z|X_1=1)=1/2$ for every $z$
(can you see why?). And therefore in this case the variables would not be independent. Intuitively, knowing one gives us information about the other since the conditional distribution changes.
