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.

Comment: The following simulation of a similar problem may give some clues, if you are
willing to deal with a few results from R statistical software.
This is for visualization; it is neither a solution nor a proof.
We simulate 10,000 realizations of each random variable.
set.seed(2021);  m = 10^4
x1 = rbinom(m, 1, .5)
x2 = rbinom(m, 1, .5)
y = x1*x2

table(x2)
x2
   0    1 
5011 4989 

table(y)
y
   0    1 
7494 2506

What does that tell you about the distributions of $X_1$
and of $Y$ taken separately? (The 'marginal' distributions.)
If you know about correlation, you may know that the correlation
of independent random variables must be $0.$ However, the
the correlation of $X_1$ and $Y$ seems far from $0.$
[Also, the product of the marginal distributions does not yield the joint distribution.]
cor(x1, y)
[1] 0.5744699

In order to plot the joint distribution of $(X_1, Y)$
it is helpful to 'jitter' their values, so that the
values won't all coincide on three points. Jittering
is done by adding a bit of random noise.
j1 = runif(m,-.2,.2)
jy = runif(m,-.2,.2)
plot(x1+j1, y+jy, pch=".")

When you do a tabulation as suggested by @whuber, you
should be able to explain the plot below. Notice that the three
squares (randomly expanded points) have different numbers
of dots (are not of the same 'darkness'). Altogether,
there are 10,000 dots in the figure.

sum(x1 == 1  &  y == 0)
[1] 2527                # dots at lower right
um(x1 == 1  &  y == 1)
[1] 2506                # dots at upper right


Note: In order to simulate your problem, you might start
with R code:
set.seed(2022);  m = 10^4
x1 = sample(c(-1,1), m, rep=T)
x2 = sample(c(-1,1), m, rep=T)
y = x1*x2
cor(x1, y)
...

