Let X be a random variable with finite expectation, and suppose that E [ e − X ] ≤ 1. Prove that E [ X ] ≥ P ( X ≥ 2 ).&lt;br&lt;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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Let X be a random variable with finite expectation, and suppose that   E  [      e          −      X        ]  ≤  1. Prove that   E  [  X  ]  ≥  P  (  X  ≥  2  ).&amp;lt;br&amp;lt;My first thought was to use Markov&#039;s inequality, but I can only use that if X is non-negative, and in any case, the resulting inequality doesn&#039;t seem very helpful. Even if I assume that   X  ≥  −  c and apply it to   Y  =  X  +  c, I&#039;m left to show   P  (  X  ≥  2  )  ≥      c          c      +      1      , which doesn&#039;t seem like it&#039;s always true.I&#039;m primarily unsure of how to use the   E  [      e          −      X        ] condition; Jensen&#039;s inequality tells me that the expectation is non-negative but there&#039;s presumably more to it.

Antwan Wiley · Accepted Answer

Step 1I&#039;m not entirely sure how the hypothesis about the expectation comes into play, but here&#039;s my solution. I&#039;ll simply use the Markov inequality for a non-negative r.v. and   a  &amp;gt;  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 2TheoremLet   φ be a monotonically increasing function for   x  &amp;gt;  0 and let X be a r.v.,   a  &amp;gt;  0  ,  φ  (  a  )  &amp;gt;  0. Then  P  (  X  ≥  a  )  ≤                    E            [      φ      (      X      )      ]              φ      (      a      )      In our case I&#039;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        ]

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

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