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Following the ideas from this post and, especially, this post, i was wondering if the a sum of two independent groups of Bernoulli distributed variables whose probabilities are know a priori is a Poisson-Binomial distribution (according to Le Cam's theorem), and a few other questions. To add some context, have:

Let $X_1, ..., X_n$ follow a Bernoulli distribution with probability $p = 1$. So: $$X_i \sim Bern(p_i = 1), \forall{i} \in 1,...,n$$

Also let $X_{n+1}, ..., X_N$ follow a Bernoulli distribution with probability $p = 0.5$. So: $$X_j \sim Bern(p_j = 0.5), \forall{j} \in n+1,...,N$$

Then, does the following statement still apply? $$S_n = \sum_{i=1}^{N}X_i \sim \text{Poisson-Binomial}$$

Given that the probabilities are known, the only "parameters" are N and n. Suppose that N is known as well, and n is not known. Is there any way to estimate n, for a given large enough N (N > 100). If it is possible, how sure can i be of the estimate?

Pedro Henrique
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    You can substantially simplify your question with a tiny bit of analysis. Your $X_1,\ldots, X_n$ are *constantly* equal to $1,$ and so they introduce a *constant* additive term of $n$ in the sum. The sum of the $X_{n+1},\ldots, X_N$ has a Binomial$(N-n,1/2)$ distribution. Thus, $S_n$ has a *shifted Binomial* distribution. Because $N$ is not terribly small, Maximum Likelihood will perform well. It is so strange to add a value of $n$ by prescribing a set of Bernoulli variables, though, I have to ask: is this how you intended your question to be interpreted? – whuber May 23 '20 at 14:49
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    Yes, @whuber. The main question was if $S_n$ followed a Poisson-Binomial (now i know it doesn't, thanks to you). To give some context, the problem i'm trying to model is basically two groups of people answering a question with two alternatives. $n$ know the answer ($p_i$ = 1) and $N-n$ dont know ($p_j$ = 0.5). The size of $n$ determines the "mean probability" of the two groups added. That is why $n$, odd as it may seem, is important. – Pedro Henrique May 23 '20 at 18:14

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