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I came across this question

Let X and Y be independent random variables with distributions.

enter image description here

Let $Z = XY$

  1. Write down a table giving the probability distribution of $Z$
  2. Are the random variables $X$ and $Z$ independent?

So I managed I already calculated the probability distribution of $Z$ which is:

enter image description here

The part that I am having difficulty of finding out is if Z and X are independent.

I know that if $P(X \cap Z) = P(X)P(Z)$, then $X$ and $Z$ are independent.

How do I use that to prove that X and Z are independent or not independent?

Any help is truly appreciated, thank you in advance.

Bryan Hii
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1 Answers1

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Be very careful when you say "if $P(X \cap Z) = P(X)P(Z)$, then $X$ and $Z$ are independent". $X$ and $Z$ are random variables so a statement like $P(X)$ doesn't have any meaning.

You can only ask what is the probability of an event. A random variable is not an event. An example of an event is $(X=0)$. So you could ask what is $P(X=0)$? (The answer is $\frac{1}{4}$ from what you have above).

Two events $A$ and $B$ are independent if $P(A \ {\rm and} \ B) = P(A)P(B)$. I'm using "and" because I think it's a bit more intuitive but $\cap$ is also good.

Two (discrete) random variables $X$ and $Z$ are independent if $P(X = a \ {\rm and} \ Z = b) = P(X = a)P(Z = b)$ for all possible values $a$ and $b$.

Although more intuitively you could think about it as "if I knew the value random variable $X$ takes, could that influence my belief of what $Z$ is? If yes, then $X$ and $Z$ are dependent. If no for all values $X$ could take, then they're independent.

Brian Lai
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  • so if I put something like $P(X = 2 and Z = 8)$, how do I calculate it? – Bryan Hii Sep 29 '21 at 06:09
  • One way: $P(X = 2 \ {\rm and} \ Z = 8) = P(Z=8|X=2) P(X = 2) $ where $P(Z=8|X=2) $ is the probability that $Z=8$ assuming you already know $X=2$. If you already know $X=2$, what value(s) of $Y$ would get you $Z=8$? How likely are you to get those values? – Brian Lai Sep 29 '21 at 06:13
  • so is it $\frac{1}{16}$ for $P(Z=8|X=2)$? – Bryan Hii Sep 29 '21 at 06:28
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    $P(Z=8|X=2) = P(Y=4) = \frac{1}{4}$ since you're given that $X=2$. For conditional probability, you take the conditional part to already be true, you don't factor in how likely it is to happen. – Brian Lai Sep 29 '21 at 06:56
  • For example, $P($ I get wet $|$ I'm outside in a hurricane $)$ is very high regardless of how small $P($ I'm outside in a hurricane $)$ is. – Brian Lai Sep 29 '21 at 06:58
  • @ohh, I understand now. Thank you for your explanation – Bryan Hii Sep 29 '21 at 07:24