Prediction of random sum of binomial variables

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 $$\Pr(a\lt\sum\limits_{j=1}^k\xi_j \lt b)=\gamma.$$

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