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Let $N$ be a positive integer. I consider $N$ random variables $X_1^{(N)}, X_2^{(N)}, \dots, X_N^{(N)}$, all independent and identically distributed, each taking values $\pm 1$ with probabilities $p/(2N)$ and $0$ with probability $1-p/N$. Consider the sum:

$$Z_N = \sum_{i=1}^N X_i^{(N)}$$

I am interested in the limiting probability distribution of $Z_N$, as $N\rightarrow\infty$. Denote by $Z$ a random variable following this limit distribution. Note that $Z$ should have a well-defined distribution. I can compute its first few moments, since:

$$\langle Z_N\rangle = \langle Z\rangle = 0, \quad \langle Z_N^2\rangle = \langle Z^2\rangle = p$$

However I don't think I can apply the central limit theorem, because the distribution of the $X_i^{(N)}$ is changing with $N$.

So how can I compute the limiting form of the distribution of $Z$?

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    $\begingroup$ $X_N,$ being a discrete random variable, has no density, whence its density has no limiting form: it sounds like you want to compute the limiting distribution function, should it exist at all. Are you willing to standardize the $X_N$ if necessary? $\endgroup$
    – whuber
    Apr 2, 2020 at 13:21
  • $\begingroup$ @whuber Yes, I want the distribution. Using "density" was incorrect. Here standardizing would just mean dividing by $\hat p$, which I think will not help much. $\endgroup$
    – a06e
    Apr 2, 2020 at 13:26
  • $\begingroup$ Please note, writing $p_N =\hat{p}/N$ for all indexes $N,$ that the variance of $X_N$ is $$\operatorname{Var}(X_N)=\sum_{i=1}^N\operatorname{Var}(x_i) = \sum_{i=1}^N p_i\approx \sum_{i=1}^N \frac{\hat p}{i}=\hat{p}H_i\approx \hat{p}\log(N)$$ where $H_i$ are the harmonic numbers. $\endgroup$
    – whuber
    Apr 2, 2020 at 13:41
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    $\begingroup$ @whuber Ah, I see the confusion. Thanks for being explicit. Maybe my notation was not clear enough. In the problem I am formulating, for a given $N$, the variables $x_1,x_2,\dots,x_N$ are all independent and identically distributed. I'll try to rewrite the question and answer with a notation that makes this dependence more explicit. $\endgroup$
    – a06e
    Apr 2, 2020 at 13:44
  • $\begingroup$ @whuber Please see edited question. $\endgroup$
    – a06e
    Apr 2, 2020 at 13:49

1 Answer 1

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Here is a partial answer.

The moment generating function of $Z_N$ is given by:

$$\psi_N (\zeta) = \langle e^{\zeta Z_N} \rangle = \left( 1 + \frac{p}{N} [\cosh (\zeta) - 1] \right)^N$$

Taking the limit $N\rightarrow\infty$, we obtain the moment-generating function of $Z$:

$$\psi(\zeta) = \lim_{N \rightarrow \infty} \psi_N (\zeta) = \lim_{N\rightarrow \infty} \left( 1 + \frac{p}{N} [\cosh(\zeta) - 1] \right)^N = e^{p \cosh (\zeta) - p}$$

Sow now I just need to take the inverse Fourier transform of $\psi(i\zeta)$, but I can't get the integral done.

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    $\begingroup$ This answer looks inconsistent with the question: the question supposes $p$ varies with $N$ but you don't introduce any such variation. It would help to use clearer notation in both the question and the answer so that it explicitly shows how both $p$ and $X$ vary. This answer looks more like part of the question: I suggest you move it into the question itself. $\endgroup$
    – whuber
    Apr 2, 2020 at 13:23
  • $\begingroup$ @whuber Note that $p = \hat p / N$, and I assume that $\hat p$ remains finite. That answers your question? $\endgroup$
    – a06e
    Apr 2, 2020 at 13:24
  • $\begingroup$ No: I don't have any questions. My comments were aimed at showing why your calculations are incorrect. $\endgroup$
    – whuber
    Apr 2, 2020 at 13:38
  • $\begingroup$ @whuber I don't see what's incorrect. I do consider the variation of $p$ with $N$ in this answer. $\endgroup$
    – a06e
    Apr 2, 2020 at 13:40
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    $\begingroup$ @whuber I also updated the notation of this answer. $\endgroup$
    – a06e
    Apr 2, 2020 at 13:51

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