Proving probability inequality

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SUMMARY

The discussion centers on proving the probability inequality involving a non-negative function X and a positive constant a. The inequalities presented are Ʃ_{n∈N} P(X>an)≥(1/a)(E[X]-a) and Ʃ_{n∈N} P(X>an)≤E[X]/a. Participants explore the use of indicator functions and the expectation operator, with a suggestion to redefine the variable by setting Y = X/a to simplify the proof. The conversation highlights the importance of correctly applying the properties of expectations and indicator functions in probability theory.

PREREQUISITES
  • Understanding of probability theory, specifically inequalities involving expectations.
  • Familiarity with indicator functions and their properties.
  • Knowledge of the expectation operator E[X] and its applications.
  • Basic skills in manipulating summations and series in mathematical proofs.
NEXT STEPS
  • Study the properties of indicator functions in probability theory.
  • Learn about the application of the expectation operator in inequalities.
  • Research techniques for manipulating series and summations in proofs.
  • Explore advanced topics in probability inequalities, such as Markov's and Chebyshev's inequalities.
USEFUL FOR

Students and researchers in mathematics, particularly those focusing on probability theory and mathematical statistics, will benefit from this discussion.

Lily@pie
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Homework Statement



Prove the following
a>0, X is a non-negative function

Ʃ_{n\in N} P(X>an)≥\frac{1}{a}(E[X]-a)

Ʃ_{n\in N} P(X>an)≤\frac{E[X]}{a}

The Attempt at a Solution



I know that
\sum_{n\in N} P(X>an)=\sum_{k \in N} kP((k+1)a≥X>ka)=\sum_{k \in N} E[k1_{[(k+1)a,ka)}(X)]
where 1_{[(k+1)a,ka)} is the indicator function.

I'm not sure how to preceed from here.

All I know is that for all ε>0, E[ε1_{[ε,∞)}(X)]≤E[X]
 
Last edited:
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The second is not true. Is there a factor 1/a missing on the right?
 
haruspex said:
The second is not true. Is there a factor 1/a missing on the right?

Sorry, my bad... I've edited it...
 
You can eliminate a by defining Y = X/a. Does that help?
 

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