# Given $X_1, X_2..$ independent real r.v., say if $\{X_1 X_2, X_1 X_3, X_1 X_4, \ldots \}$ are always independent

I have a set of problems about independence of RV which I'm having quite a hard time solving. I think I know the theory, but when it's time to get the hands dirty, I do not know where to begin.
Here is one of them

Given $$X_1, X_2,\ldots$$ independent real r.v., say if the following proposition is always true or not $$\{X_1 X_2, X_1 X_3, X_1 X_4, \ldots\}$$ are independent.
I know that my objective is to prove that
$$P(\{X_1 X_2, X_1 X_3, X_1 X_4, \ldots\}) = P(X_1 X_2) \cdot P(X_1 X_2) \cdot P(X_1 X_3)\cdots$$ As the definition of independence. But I do not find the way.
What I tried so far:\

1. Conditioning on $$X_1$$ like $$P(a X_2\mid X_1=a, aX_3\mid X_1=a,\ldots) P(X_1=a)$$, given the $$X_is$$ are independent that factorizes, but i cannot put the "times $$X_1$$" back into each probability.\
2. Thinking as a cartesian product $$P(\{X_1\} \times \{X_2,X_3 \dots \})$$, and I couldn't actually elaborate much here.\

I am looking to both: A hint to advance with this one, and maybe some insight on how to address these kind of problems, since I have a few left.

• I assume the $X_i$ are independent to begin with, no? Commented Aug 24, 2020 at 23:30
• Also, by the way you are conditioning the $X_i$ it seems that these are discrete. Is that the case? Otherwise $P(X_i = a)$ will always be zero. (to clarify: I didn't downvote you) Commented Aug 24, 2020 at 23:31
• $P(X_1X_2)$ has no meaning. Why do you assume that all random variables are discrete? Sorry to say that there is too much of nonsense in what you have written. The question itself is absurd since there is no assumption of $X_i$'s. Commented Aug 24, 2020 at 23:33
• Let $X$ be a random variable with, say $N(0,1)$ distribution, and $X_i=X$ for all $i$. Do you see that this is a counter-example? Commented Aug 25, 2020 at 0:09
• omg, yes sorry, they are all independent. Commented Aug 25, 2020 at 0:49

As a hint, here is a particular random variable to think about: Let $$X_1$$ be Bernoulli(1/2), that is, it is $$0$$ with probability 1/2, and $$1$$ with probability 1/2. Now, construct some $$X_i$$s for yourself, and then ask questions like "What is the probability that $$X_1 X_i = 0$$ given $$X_1 X_j = 0?$$"
• Thanks a lot for your comment, really clear, and I take your recommendations and advice. I still have one final doubt:\ Sometimes I think I'm overthinking the problems. When you say $P(X_1 X_i =0 | X_1 X_j = 0)$ you mean think the $\frac{P(X_1 X_i =0 | X_1 X_j = 0)}{P(X_1 X_j = 0)}$ in some easy cases to have a counter-example, or is there a more direct way of thinking it? Commented Aug 25, 2020 at 13:04
This is obviously untrue. Consider the case of $$X_2 = X_3 = X_4 = ... = 1$$ and $$X_1 = N(0,1)$$. Then $$X_1, X_2, ...$$ are jointly independent, but the products are not.