X is a random variable so is |X|?

AI Thread Summary
To prove that |X| is a random variable when X is a random variable, it is essential to understand the definitions involved. A random variable is defined as a measurable function from a probability space to the real numbers. Since the absolute value function is continuous and measurable, the composition of measurable functions, in this case, |X|, remains measurable. The discussion highlights that the pre-image of a Borel set under |X| can be shown to be in the sigma-algebra F, confirming that |X| is indeed a random variable. The inquiry reflects a common concern in understanding the properties of measurable functions in probability theory.
BoogieE
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Howdy guys. Given that X is a random variable how would you prove |X| to be one too? Thanks for any suggestions!
 
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What is your definition of random variable?
 
Just mapping from (S,F) to (R, B(R))
 
X is measurable, | | is continuous hence measurable. And the composition of measurables is measurable.
 
I am expecting that from the fact that X(-1)(G) = {w in S such that X(w) is in G for all G in B(R) } is in F you can somehow show that |X|(-1)(G') = {w in S such that |X|(w) is in G' for all G' in B(R)} is also if F
 
micromass said:
X is measurable, | | is continuous hence measurable. And the composition of measurables is measurable.
I asked my math professor and she said this is ok. I probably overthought the problem. Thank you very much!
 

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