Is X(\omega) = \frac{1}{\omega} a Random Variable?

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SUMMARY

The discussion centers on whether the function X(ω) = 1/ω qualifies as a random variable within the probability space (Ω, ℱ, P) = ([0,1], ℬ([0,1]), λ), where λ represents the Lebesgue measure. It is concluded that X(ω) is not a random variable because it lacks a set of measure zero on which the function is bounded. Consequently, X cannot be considered almost surely bounded due to the unbounded nature of the function as ω approaches 0.

PREREQUISITES
  • Understanding of probability spaces, specifically (Ω, ℱ, P).
  • Familiarity with Lebesgue measure and its properties.
  • Knowledge of random variables and the concept of almost sure boundedness.
  • Basic calculus, particularly limits and behavior of functions near singularities.
NEXT STEPS
  • Study the properties of Lebesgue measure in detail.
  • Learn about the definition and properties of random variables in probability theory.
  • Explore the concept of almost sure convergence and boundedness in probability.
  • Investigate functions with singularities and their implications in probability spaces.
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Mathematicians, statisticians, and students of probability theory who are examining the properties of random variables and their definitions within measure theory.

wayneckm
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Hello all,

I have the following question:

Assume [tex](\Omega, \mathcal{F},P) = ([0,1],\mathcal{B}([0,1]),\lambda)[/tex], where [tex]\lambda[/tex] is Lebesgue mesure, so is [tex]X(\omega) = \frac{1}{\omega}[/tex] a random variable defined on this probability space?

If yes, then can I say that [tex]X[/tex] is bounded a.s. because the set for unboundedness is [tex]{0}[/tex] which is of measure 0?

Thanks.

Wayne
 
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wayneckm said:
Hello all,

I have the following question:

Assume [tex](\Omega, \mathcal{F},P) = ([0,1],\mathcal{B}([0,1]),\lambda)[/tex], where [tex]\lambda[/tex] is Lebesgue mesure, so is [tex]X(\omega) = \frac{1}{\omega}[/tex] a random variable defined on this probability space?

If yes, then can I say that [tex]X[/tex] is bounded a.s. because the set for unboundedness is [tex]{0}[/tex] which is of measure 0?

Thanks.

Wayne

No.

Because there is no set of measure 0 on the compliment of which the function is bounded.
 

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