Proving probability inequality

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Homework Help Overview

The discussion revolves around proving a probability inequality involving a non-negative function X and a positive constant a. The inequalities to be proven relate to the expected value of X and its probabilities over a sequence of natural numbers.

Discussion Character

  • Exploratory, Assumption checking

Approaches and Questions Raised

  • The original poster attempts to manipulate the summation of probabilities using an indicator function and expresses uncertainty about the next steps. Some participants question the validity of the second inequality and suggest there may be a missing factor. Another participant proposes a transformation of the variable to simplify the problem.

Discussion Status

The discussion is ongoing, with participants exploring different interpretations of the inequalities and considering potential transformations to aid in the proof. There is no explicit consensus yet, but suggestions for re-evaluation and alternative approaches have been provided.

Contextual Notes

Participants are navigating the complexities of the inequalities and the implications of the variable a, with some uncertainty regarding the correctness of the original statements. The original poster's approach involves specific mathematical expressions that may require further clarification.

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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