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Michael Hardy
Michael Hardy
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Let $\xi, \eta$ be a discrete random values and $\mathbb E| ξ |$, $\mathbb E | η |$ < $+\infty$$\mathbb E | η | < +\infty$, and any value of these values ​​are accepted with a non-zero probability. How to prove that from $\mathbb E (ξ | η) ≥ η$$\mathbb E (ξ \mid η) ≥ η$, $\mathbb E (η | ξ) ≥ ξ$$\mathbb E (η \mid ξ) ≥ ξ$ follows $ξ = η$?

Let $\xi, \eta$ be a discrete random values and $\mathbb E| ξ |$, $\mathbb E | η |$ < $+\infty$, and any value of these values ​​are accepted with a non-zero probability. How to prove that from $\mathbb E (ξ | η) ≥ η$, $\mathbb E (η | ξ) ≥ ξ$ follows $ξ = η$?

Let $\xi, \eta$ be a discrete random values and $\mathbb E| ξ |$, $\mathbb E | η | < +\infty$, and any value of these values ​​are accepted with a non-zero probability. How to prove that from $\mathbb E (ξ \mid η) ≥ η$, $\mathbb E (η \mid ξ) ≥ ξ$ follows $ξ = η$?

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Iosif Pinelis
Iosif Pinelis
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Lisa
Lisa
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Сoincidence of discrete random variables

Let $\xi, \eta$ be a discrete random values and $\mathbb E| ξ |$, $\mathbb E | η |$ < $+\infty$, and any value of these values ​​are accepted with a non-zero probability. How to prove that from $\mathbb E (ξ | η) ≥ η$, $\mathbb E (η | ξ) ≥ ξ$ follows $ξ = η$?