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.

  • $\begingroup$ The Edit modified drastically the question... The new version roughly asks for the full distribution of $X$. This is not computable theoretically in full generality. Hence one can rely on Monte Carlo simulations to generate a sample of size $n$ and decide that $L_{\alpha}$ is the $\alpha/2$-empirical quantile (and similarly for $U_\alpha$). To check whether $L_\alpha\leqslant x$ for some given $x$, one needs only to decide if each sample $X$ is $\leqslant x$ or not. To do that, a finite (random) number of $z_i$s suffices. $\endgroup$ – Did Jan 7 '13 at 10:26
  • $\begingroup$ Do not understand me wrong, I appreciate your answer below (although my reputation is still too low to rate it as useful). It is the best I have so far. I agree that a general solution might not exist. However, there may be a better solution for the special case $\pi_i = (1-\rho) \rho^{i-1}$. At least for another special case ($\pi_i=\frac{1}{n}$ if $i\leq n$ and $\pi_i=0$ else) there exists a better one: Since $n X$ is binomial, an optimal confidence interval can be constructed easily and calculated using numerical means. $\endgroup$ – otmar Jan 7 '13 at 12:04
  • $\begingroup$ I understand quite well the situation--and I fully disagree with the way you managed this question, which reflects, either a lack of reflexion about what you really wanted to ask when you posted the question, or a lack of consideration for the answerers, or both. $\endgroup$ – Did Jan 7 '13 at 12:13
  • $\begingroup$ I am sorry for your inconvenience, this is my first question here. I think the formulation of the question was somewhat misleading due to a small typo. See edit. $\endgroup$ – otmar Jan 7 '13 at 12:46
  • $\begingroup$ See edit... Sorry, but no. $\endgroup$ – Did Jan 7 '13 at 12:47

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 $1/z^2$.

  • $\begingroup$ Thanks for your reply. However, I miss some condition, for which this standard normal approximation works well. $\endgroup$ – otmar Jan 7 '13 at 8:23
  • $\begingroup$ Assume $\pi_i = \frac{1}{N}$ for $i\leq N$ and $\pi_i = 0$ else. Then $X$ corresponds to a binomial distribution, for which the standard normal approximation is justified for $\max(Np,N(1-p))\gg 1$. Therefore, due to similar reasons, I doubt that your approximation works well for the geometric sequence, if for example $\rho=0.5$ and $p=0.9$. I would like to know a method, which allows an accurate and efficient numerical evaluation of such confidence intervals for any $\rho$ and $p$. $\endgroup$ – otmar Jan 7 '13 at 8:39
  • $\begingroup$ See Edit. $ $ $ $ $\endgroup$ – Did Jan 7 '13 at 9:03

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