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Let $X$ and $Y$ are two independent binomial random variables where $X\sim B(K, q), Y\sim B(K, p)$. I am wondering how to compute or estimate the following expectations: $$ \ \mathbb{E}[|pX-qY|]\ \ \text{and}\ \ \mathbb{E}[\text{ReLU}(pX-qY)].$$

where ReLU($x$) = max($x, 0$) Furthermore, what does the distribution of $pX-qY$ look like?

Update: Actually I'm more interested in $\mathbb{E}[pX\mathbb{1}(pX-qY>0)]\ \text{and}\ \mathbb{E}[qY\mathbb{1}(pX-qY>0)]$, where $\mathbb{1}$ is the indicator function. We know that $\mathbb{E}[pX\mathbb{1}(pX-qY>0)]-\mathbb{E}[qY\mathbb{1}(pX-qY>0)]=\mathbb{E}[\text{ReLU}(pX-qY)]$.

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    $\begingroup$ The title doesn't reflect the question. The question in the title is trivial due to the linearity of expectation, whereas the question in the body of the text leads to a double sum for which I don't think there's a closed form. $\endgroup$
    – joriki
    Mar 14 '20 at 7:32
  • $\begingroup$ It does not look pretty to me, though I think you can say $\mathbb{E}[|pX-qY|] = \mathbb{E}[\text{ReLU}(pX-qY)] + \mathbb{E}[\text{ReLU}(qY-pX)] =2 \mathbb{E}[\text{ReLU}(pX-qY)]$. If $Z=pX-qY$ then I think $-qK \le Z \le pK$ and $\mathbb{E}[Z]=0$ and $\text{Var}(Z) = (p+q-2pq)pqK$ $\endgroup$
    – Henry
    Mar 14 '20 at 7:37
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    $\begingroup$ Also asked at stats.stackexchange.com/q/454103/119261. $\endgroup$ Mar 14 '20 at 7:46
  • $\begingroup$ @Henry Thank you! How do you deduce $\mathbb{E}[\text{ReLU}(pX-qY)]=\mathbb{E}[\text{ReLU}(qY-pX)]$ and the value of Var($Z$)? $\endgroup$
    – luw
    Mar 14 '20 at 14:38
  • $\begingroup$ @joriki Thank you! I have modified the title. $\endgroup$
    – luw
    Mar 14 '20 at 14:45

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