Proving Bounds on Moments of a Random Variable

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silentone
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Homework Statement


For any random variable X, prove that
P{X[itex]\geq[/itex]0}[itex]\leq[/itex]inf[ E[ phi(t) : t [itex]\geq[/itex] 0] [itex]\leq[/itex] 1

where phi(t) = E[exp(tX)] o<phi(t)[itex]\leq[/itex]∞


Homework Equations





The Attempt at a Solution


I am not sure how to begin this. Any hints to get started would be greatly appreciated.
 
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silentone said:

Homework Statement


For any random variable X, prove that
P{X[itex]\geq[/itex]0}[itex]\leq[/itex]inf[ E[ phi(t) : t [itex]\geq[/itex] 0] [itex]\leq[/itex] 1
E[ phi(t)] doesn't mean anything. E[] requires a r.v., whereas phi(t) is just an ordinary function. So I guess you mean
P{X[itex]\geq[/itex]0}[itex]\leq[/itex]inf[phi(t) : t [itex]\geq[/itex] 0] [itex]\leq[/itex] 1

Write out E[phi()] as an integral and consider the positive and negative ranges of X separately.
 
silentone said:

Homework Statement


For any random variable X, prove that
P{X[itex]\geq[/itex]0}[itex]\leq[/itex]inf[ E[ phi(t) : t [itex]\geq[/itex] 0] [itex]\leq[/itex] 1

where phi(t) = E[exp(tX)] o<phi(t)[itex]\leq[/itex]∞


Homework Equations





The Attempt at a Solution


I am not sure how to begin this. Any hints to get started would be greatly appreciated.

Broad hint: P{X ≥ 0} = E H(X), where H(x) = 0 for x < 0 and H(x) = 1 for x ≥ 0. So, you are really comparing expectations of two different functions of X.

RGV