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There are $n$ IID experiments all with a success rate of $p$. Let $a_i=1$ denote a success in experiment $i$ and $a_i=0$ denote a failure. Is there a way to calculate the following: $E(\sum a_i|\sum a_i>B)$ for any $B\in\{1\cdots n\}$?

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  • $\begingroup$ If $np$ and $n(1-p)$ are large enough, a normal approximation works well, unless if you really need an exact solution. $\endgroup$ – jdods Dec 4 '17 at 14:13
  • $\begingroup$ I wish to avoid using approximations, although it will prob. be good enough for my asymptomatic analysis. I guess there is no ``closed form" for this, is there? $\endgroup$ – MoRkO Dec 4 '17 at 15:44
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Can you do something with the formula for conditial expectation?

Given $X$ a discrete random variable and $\mathbb{P}(B) > 0$, then is the conditional expectation of $X$ given $B$ defined by

$\mathbb{E}(X|B) = \sum_{x\in Im(X)} x\mathbb{P}(X=x|B)$, with

$\mathbb{P}(X=x|B) = \frac{\mathbb{P}(X=x \cap B)}{\mathbb{P}(B)}$

Note that if the sum is smaller than $B$, the conditional probability is zero.

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Yes.

The summation of successes is a random variable $\sim\mathsf{Bin}(n,p)$

If it is denoted by $X$ then to be calculated is: $$\mathbb E[X\mid X>B]$$

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  • $\begingroup$ Thanks, but I am afraid this is obvious, I am looking for the formula for $E(X|X>B)$. I know it is above $B*Pr(X>B)$ and below $n*Pr(x>B)$, is there anything more I can say on it? $\endgroup$ – MoRkO Dec 4 '17 at 15:47
  • $\begingroup$ There is not a closed form for this. $\endgroup$ – drhab Dec 4 '17 at 17:51
  • $\begingroup$ Thanks, I suspected this was the answer.... $\endgroup$ – MoRkO Dec 4 '17 at 22:07

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