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\begin{eqnarray*} E\left[\left.\left(X+k\right)\right|\left(X+k\right)>0\right] & = & E\left[k\left|\left(X+k\right)>0\right.\right]+E\left[X\left|\left(X+k\right)>0\right.\right] \end{eqnarray*}

$k$ is a constant and $X$ is a random variable that could be discrete, continuous and having any distribution.

Does the above equality hold and if so, please provide the proof.

QUESTION ORIGIN

Please note, I have assumed the above equality is correct in the proof to the related question here, but would be keen to know if there a formal proof or any cases where this would not hold.

Conditional Expected Value of Product of Normal and Log-Normal Distribution

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  • $\begingroup$ Is this a self-learning question? Can you show us where you got stuck in the proof? $\endgroup$ Jul 21, 2015 at 15:05

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This is just linearity of expectation. \begin{align} \text{E}(X + k \mid X + k > 0) &= \frac{\text{E}(X + k \, ; \, X + k > 0)}{P(X + k > 0)} \\ &= \frac{\text{E}(X\, ; \, X + k > 0) + k P(X + k > 0)}{P(X + k > 0)} \\ &= \text{E}(X \mid X + k > 0) + k. \end{align}

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  • $\begingroup$ What if the sum involves two random variables , i.e. $E[X+Y | X+Y>0]$? Does the linearity also hold here? Thank you $\endgroup$
    – Confounded
    Oct 19, 2020 at 11:26

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