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Hej,

I am studying a proof for the Markov inequality, and there is a certain step, which I don't understand:

$\mathbb{E}(X \cdot \mathbb{I}_A) \ge \mathbb{E}(a \mu \mathbb{I_A})$

where $\mathbb{I}_A$ is the indicator function, $\mu = \mathbb{E}(X)$ and $A = [X \ge a \cdot \mathbb{E}(X)]$.

Thanks for any help.

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    Sorry, that was a typing mistake. I just have in the proof the following inequalities: $\mu = \mathbb{E}(X) \ge \mathbb{E}(X \cdot \mathbb{I}_A) \ge \mathbb{E}(a \cdot \mu \cdot \mathbb{I}_A) = a \mu \mathbb{P}[X \ge a \cdot \mu]$, which proofs the Markov inequality. But the step above is unclear. Why is this inequality correct?2012-12-03

2 Answers 2

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Use the pointwise inequality $X\mathbf 1_A\geqslant Y$ with $Y=a\mathbb E(X)\,\mathbf 1_A$. Deduce that $\mathbb E(X\mathbf 1_A)\geqslant\mathbb E(Y)$ and note that $\mathbb E(Y)=a\mathbb E(X)\mathbb P(A)$.

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We have $X\mathbb I_A\geqslant aE(X)\mathbb I_A= a\mu \mathbb I_A,$ because either we are in $A$ and it's by definition, or we aren't and we have equality. Now integrate on both sides.

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    That's not what I said. I mean the equality when we have the characteristic function of $A$ on both sides.2012-12-03