Does E(e^{-X}) = 0 imply X = \infty almost surely for X \geq 0?

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The discussion confirms that if E(e^{-X}) = 0, then X must equal infinity almost surely (a.s.) for X ≥ 0. This conclusion is supported by the established result that E[X] = 0 implies X = 0 a.s. The reasoning follows that since e^{-X} is a non-negative function, the expectation being zero necessitates that X approaches infinity almost surely. This mathematical relationship is crucial for understanding the behavior of random variables in probability theory.

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Does the following make sense:

[itex] <br /> E(e^{-X}) = 0 \Rightarrow X = \infty\quad a.s. ?<br /> [/itex]

(Intuitively yes, but mathematically?)

Thank you in advance for your help! :-)

/O
 
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Hi osprey! :smile:

Yes, that makes sense! Can you show in general for [itex]X\geq 0[/itex] that

[tex]E[X]=0~\Rightarrow~X=0~\text{a.s.}[/tex]

Then you just need to apply this result for [itex]e^{-X}[/itex]...
 

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