Proving Bounds on Moments of a Random Variable

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The discussion centers on proving the inequality P{X ≥ 0} ≤ inf[E[φ(t) : t ≥ 0]] ≤ 1 for any random variable X, where φ(t) = E[exp(tX)]. Participants clarify that E[φ(t)] is not valid as φ(t) is a function, not a random variable. The correct interpretation involves comparing expectations of different functions of X, specifically using the indicator function H(X) to express P{X ≥ 0} as E[H(X)]. This approach allows for a structured proof of the stated inequality.

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Homework Statement


For any random variable X, prove that
P{X\geq0}\leqinf[ E[ phi(t) : t \geq 0] \leq 1

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


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\geq0}\leqinf[ E[ phi(t) : t \geq 0] \leq 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\geq0}\leqinf[phi(t) : t \geq 0] \leq 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\geq0}\leqinf[ E[ phi(t) : t \geq 0] \leq 1

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


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
 

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