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Suppose I have a family of $n$ independent Bernoulli random variables described by a vector of parameters $(p_i)_{i=1}^n$. As it is well known, the number of successes within this family is a random variable $S_n$ which follows a Poisson-binomial distribution of parameter $(p_i)_{i=1}^n$ supported in $\{0,\dots,n\}$.

I would like to generalize this situation to a ``unit mass" population of independent Bernoullis. To describe this more formally, let $F$ be the cdf of an (absolutely continuous) probability distribution, supported in $[0,1]$. Take $F$ to describe the frequency of the possible values of the Bernoulli parameters within the population.

Define $S$ to be the fraction of successes within the population. Clearly,should take values in $[0,1]$. Is $S$ a well defined random variable? If yes, what is its distribution? If no, is there something close to what I am after?

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    $\begingroup$ I'm not really able to parse the question, but will attempt an interpretation and an answer to that interpretation. What I think you're saying is you want to have a family of Bernoulli's $(X_t)$ indexed by $[0,1]$. For each $t\in [0,1]$, you want a probability $f(t)$ of success. You are then interested in $S$, which is the integral of the $X_t$'s. Leaving aside the measurability issues (which are likely insurmountable), I think your r.v. $S$ would be a.e. constant, just $\int f(t)\,dt$. This should be basically the strong law of large numbers $\endgroup$ Commented Jun 19 at 16:50
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    $\begingroup$ Sorry if my question is vague. It is so on purpose as I don't know how to make it precise. I know the Law of Large Numbers. Indeed, I think you can generalize to a continuum of random variables, asking convergence in $L2$ . See for instance this paper. home.uchicago.edu/~huhlig/papers/uhlig.et.1996.pdf What I am asking is if the same idea of the Poisson binomial can be captured in this case. Probably one should not focus on $S$ as I defined it. $\endgroup$ Commented Jun 19 at 17:35
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    $\begingroup$ @AnthonyQuas : That was also my guess, but if so, then why must $F$ be a nowhere decreasing function? $\endgroup$ Commented Jun 20 at 1:18
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    $\begingroup$ Not sure if that's what you have in mind, but this reminds me a bit of the construction by Yeneng Sun (NUS), see for example the paper "The exact law of large numbers via Fubini extension and characterization of insurable risks". $\endgroup$ Commented Jun 20 at 20:30

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$\newcommand\de\delta$Let $(X_p)_{p\in[0,1]}$ be a family of uncorrelated random variables (r.v.'s) on a probability space $(\Omega,\mathcal F,P)$ such that for each $p$ the r.v. $X_p$ has the Bernoulli distribution with parameter $p$, and let $F$ be the c.d.f. of a non-atomic distribution over the interval $[0,1]$, so that $F$ is continuous on $[0,1]$.

Then $$\int_0^1 X_p\,dF(p)=\mu:=\int_0^1 p\,dF(p), \tag{1}\label{1}$$ where $$\int_0^1 X_p\,dF(p):=\lim\sum_{i=0}^{n-1}X_{p_i^*}\,(F(p_{i+1})-F(p_i))$$ and the limit is in $L^2(P)$ under the condition that $$\de:=\max_{i=0}^{n-1}(p_{i+1}-p_i)\to0, \tag{2}\label{2}$$ $0=p_0<\cdots<p_n=1$, and $p_i^*\in[p_i,p_{i+1}]$ for all $i$.

Indeed, $$\Big|\sum_{i=0}^{n-1}p_i^*\,(F(p_{i+1})-F(p_i))-\mu\Big| =\Big|\sum_{i=0}^{n-1}\int_{p_i}^{p_{i+!}}(p_i^*-p)\,dF(p)\Big| \\ \le \sum_{i=0}^{n-1}\int_{p_i}^{p_{i+!}}|p_i^*-p|\,dF(p) \le\de\sum_{i=0}^{n-1}\int_{P_i}^{p_{i+!}}dF(p)=\de\to0. \tag{3}\label{3}$$ On the other hand, $$E\Big(\sum_{i=0}^{n-1}X_{p_i^*}\,(F(p_{i+1})-F(p_i)) -\sum_{i=0}^{n-1}p_i^*\,(F(p_{i+1})-F(p_i))\Big)^2 \\ =E\Big(\sum_{i=0}^{n-1}(X_{p_i^*}-p_i^*)\,(F(p_{i+1})-F(p_i))\Big)^2 \\ =\sum_{i=0}^{n-1}E(X_{p_i^*}-p_i^*)^2\,(F(p_{i+1})-F(p_i))^2 \\ =\sum_{i=0}^{n-1}p_i^*(1-p_i^*)\,(F(p_{i+1})-F(p_i))^2 \\ \le\frac14\max_{i=0}^{n-1}(F(p_{i+1})-F(p_i))\to0 \tag{4}\label{4}$$ given \eqref{2}, because $F$ is contiunous and hence uniformly continuous on $[0,1]$.

Thus, \eqref{1} follows by \eqref{3} and \eqref{4}.

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  • $\begingroup$ Thanks, a lot! I edited the title because I think it better describes the direction of the discussion. I will post a new and clearer question describing what I have in mind. $\endgroup$ Commented Jun 20 at 7:28

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