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

AI Thread Summary
The discussion centers on whether the equation E(e^{-X}) = 0 implies that X = ∞ almost surely for non-negative X. Participants agree that intuitively, this makes sense, but seek mathematical validation. A key point raised is that if E[X] = 0, it leads to the conclusion that X = 0 almost surely. This result can then be applied to the function e^{-X} to support the initial claim. The conversation highlights the connection between expectations and almost sure outcomes in probability theory.
osprey
Messages
3
Reaction score
0
Does the following make sense:

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

(Intuitively yes, but mathematically?)

Thank you in advance for your help! :-)

/O
 
Last edited:
Physics news on Phys.org
Hi osprey! :smile:

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

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

Then you just need to apply this result for e^{-X}...
 
Back
Top