Let X be a random variable with finite expectation, and suppose that E[e^{-X}] =< 1. Prove that E[X] >= P(X >= 2).

Josie Kennedy 2022-11-18 Answered
Let X be a random variable with finite expectation, and suppose that E [ e X ] 1. Prove that E [ X ] P ( X 2 ).<br<My first thought was to use Markov's inequality, but I can only use that if X is non-negative, and in any case, the resulting inequality doesn't seem very helpful. Even if I assume that X c and apply it to Y = X + c, I'm left to show P ( X 2 ) c c + 1 , which doesn't seem like it's always true.
I'm primarily unsure of how to use the E [ e X ] condition; Jensen's inequality tells me that the expectation is non-negative but there's presumably more to it.
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Answers (1)

Antwan Wiley
Answered 2022-11-19 Author has 13 answers
Step 1
I'm not entirely sure how the hypothesis about the expectation comes into play, but here's my solution. I'll simply use the Markov inequality for a non-negative r.v. and a > 0 in the form
E [ X ] a P ( X a )
with a = 2. Remembering that probabilities are non-negative and using Markov we have:
E [ X ] P ( X 2 ) + P ( X 2 ) P ( X 2 )
Edit after the comments about non-negativity:
I think I saw the light :D I use the extended form of Markov inequality for monotonically increasing function:
Step 2
Theorem
Let φ be a monotonically increasing function for x > 0 and let X be a r.v., a > 0 , φ ( a ) > 0. Then
P ( X a ) E [ φ ( X ) ] φ ( a )
In our case I'll consider the (somehwat artificial) function φ ( x ) := x + e x 1, which we can prove satisfy all the hypothesis with a = 2. Hence, since 1 / φ ( a ) 1 and using the bound for e X , we have
P ( X a ) E [ X + e X 1 ] φ ( a ) E [ X ] + E [ e X ] 1 E [ X ]
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