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Let us consider $\xi\sim\sum\limits_{i=1}^v\text{Bin}(1,p):=\sum\limits_v\eta$.

Where $v$ is random variable with unknown distribution. Let $\mathbb{E}v^m<\infty \quad\forall m$.

We know, that $$\mathbb{E}\xi= \mathbb{E}\sum\limits_v\eta = \mathbb{E}v \cdot \mathbb{E}\eta $$ $$\text{Var}\xi= \text{Var}\sum\limits_v\eta = \mathbb{E}v \cdot\text{Var}\eta + \text{Var} v \cdot (\mathbb{E}\eta)^2 $$

Let us concider sample $(\xi_1,\dots,\xi_n)$.

For given $\gamma$ and $k$ I need to obtain $a$ and $b$ such that $$\Pr(a\lt\sum\limits_{j=1}^k\xi_j \lt b)=\gamma.$$

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  • $\begingroup$ do you really want confidence interval on the sum of $\xi$ or $p$? Use the fact that the sum of binomials is another binomial. $\endgroup$ – papgeo Aug 29 '18 at 18:12
  • $\begingroup$ Yes, for the sum. It will be again random sum of binomials, that's the point. $\endgroup$ – Kess Aug 29 '18 at 18:27
  • $\begingroup$ Do you know the distribution of $v$? $\endgroup$ – jbowman Aug 29 '18 at 20:11
  • $\begingroup$ @jbowman, no, that's the point too :D $\endgroup$ – Kess Aug 29 '18 at 21:33
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    $\begingroup$ @whuber Thanks! Both answers are "yes" $\endgroup$ – Kess Aug 31 '18 at 22:27

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