Confidence interval for sum of random subsequence generated by coin tossing This question is related to Sum of random subsequence generated by coin tossing. Here is the corresponding problem description as given by Memming:

Let $(\pi_1, \pi_2, \cdots)$ be an infinite sequence of real numbers such that $\forall i\; \pi_i > 0$ and $\sum_i \pi_i = 1$. This can be thought of as a probability over natural numbers.
Let $(z_1, z_2, \ldots)$ be a sequence of independently and identically distributed Bernoulli random variables such that $P(z_i = 1) = p$ and $P(z_i = 0) = (1-p)$.
What can we say about the distribution of $X = \sum_i \pi_i z_i$?
$X$ is the sum of a random subsequence of $(\pi_i)$ generated by coin tossing.

Since $E[X] = p$, $X$ can be used to get an estimation for $p$. Given the sequence $\pi_i$, how does the corresponding confidence interval look like? I am especially interested in the case, where $\pi_i$ is a geometric sequence $\pi_i := (1-\rho) \rho^{i-1}$.
Edit: More precisely, I would like to know a method to calculate the optimal (smallest) confidence interval. The corresponding lower and upper bounds are functions of the given sequence $(\pi_1, \pi_2, \cdots)$, $L_\alpha=L_\alpha(\pi_1, \pi_2, \cdots)$ and $U_\alpha=U_\alpha(\pi_1, \pi_2, \cdots)$, respectively, which fulfill $P(X<L_\alpha)=P(X>U_\alpha)\leq\frac{\alpha}{2}$ for given confidence level $\alpha$. I would also be satisfied with an efficient numerical procedure.
Edit: Changed ...how do the corresponding confidence intervals look like? to ...how does the corresponding confidence interval look like? to make this question more clearly.
 A: Since $\mathbb E(X)=p$ and $\mathrm{var}(X)=p(1-p)\vartheta$ with $\vartheta=\sum\limits_i\pi_i^2$, iterating $n$ times the experience and denoting by $S_n$ the sum of these $n$ results yields $S_n$ of mean $np$ and variance $np(1-p)\vartheta$. Thus,
$$
Z_n=\frac{S_n-np}{\sqrt{np(1-p)\vartheta}}\longrightarrow Z,
$$
where $Z$ is standard normal. Hence
$$
\mathbb P\left(\left|p-\frac{S_n}n\right|\geqslant\frac{z_\alpha}{n\sqrt{n}}\sqrt{S_n(n-S_n)\vartheta}\right)\longrightarrow\mathbb  P(|Z|\geqslant z_\alpha)=2(1-\Phi(z_\alpha)).
$$
If $\pi_i=\rho(1-\rho)^{i-1}$, then $\vartheta=\dfrac{\rho}{2-\rho}$. 
Edit: Nonasymptotic bounds are that, for every $z\gt0$ and every $n\geqslant1$,
$$
\mathbb P(|Z_n|\geqslant z)\leqslant\frac1{z^2}.
$$
In other words, considering the domain
$$
D_{n,z}(s)=\{u\in[0,1]\mid (s-nu)^2\leqslant nzu(1-u)\vartheta\},
$$
one gets
$$
\mathbb P(p\in D_{n,z}(S_n))\geqslant1-\frac1{z^2}.
$$
Note that if $S_n/n\approx p$, $D_{n,z}(S_n)$ is approximately the interval
$$
\left[p-\frac{z}{n\sqrt{n}}\sqrt{S_n(n-S_n)\vartheta};p+\frac{z}{n\sqrt{n}}\sqrt{S_n(n-S_n)\vartheta}\right],
$$
hence the loss in the apparent quality of the approximation this surplus of rigor entails is mainly to replace the asymptotic upper bound $2(1-\Phi(z))$ by $1/z^2$.
