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I'm working on a problem where I have $n^2$ real numbers $x_{11},...,x_{nn}$, all drawn i.i.d. from the same distribution $F$. I don't observe each $x_{ij}$, but I do observe the $n$ means:

$$\bar{x}_j = \frac{1}{n} \sum_{i=1}^n x_{ij}$$

As $n$ grows large, it is clear that I can get estimates for both the mean and variance of $F$. I can estimate the mean as

$$\hat{\mu} = \frac{1}{n} \sum_{j=1}^n \bar{x_j}$$

And I can use the central limit theorem to estimate variance as $\hat{\sigma^2} = \sum_{j=1}^n (\bar{x}_j - \hat{\mu})^2.$

If I could recover more moments (and assumed F was nice enough) I could recover the original distribution from $F$ using the moment generating functions (https://math.stackexchange.com/questions/353490/deducing-a-probability-distribution-from-its-moment-generating-function).

My intuition is that it is hard to recover more moments, since the observed $\bar{x_1},...,\bar{x_n}$ behave as if they were drawn from a normal distribution $N(\mu,\frac{\sigma^2}{n})$ when $n$ is large and $\mu,\sigma^2$ are the mean and variance of $F$.

My question is, is it possible to recover higher-order moments given just $\bar{x_1},...,\bar{x_n}$?

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  • $\begingroup$ Here's a possible answer $\endgroup$
    – Asterix
    Commented Apr 7, 2017 at 17:45

1 Answer 1

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I figured this out right after I posted it.

From observing $\bar{x_1},...,\bar{x_n}$ (for large enough $n$) I can compute the characteristic function

$$\psi(\bar{X},t) = \mathbb{E}[e^{it\bar{X}}].$$

where $\bar{X} = \frac{1}{n} \sum_{i=1}^n X $ and $X$ is the original random variable (i.e. the random variable form which each $x_{ij}$ is drawn).

But the characteristic function of a sum of random variables is the product of the individual random variable's characteristic functions. So, I can recover the original characteristic function

$$\psi(X,t) = \psi(\bar{X},nt)^{\frac{1}{n}}.$$

From this characteristic function I can recover the original random variable $X$.

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    $\begingroup$ I think this is very confused. I don't agree that you can calculate the cf from some sample means; even if you had the actual theoretical moments, I don't see how you could reconstruct a closed form cf from the moments (without additional information). One cannot recover anything in the manner set out here. $\endgroup$
    – wolfies
    Commented Apr 9, 2017 at 8:07
  • $\begingroup$ Thanks for your comment. My intuition is that 1. I can recover an empirical characteristic function for $\overline{X}$ 2. I can recover an ``empirical cdf'' for $X$ from this empirical characteristic function 3. As long as everything is nice and continuous ( a big assumption) my empriical cdf for $X$ should be close to the true cdf for $X$. Why do you think this should not work? $\endgroup$
    – Asterix
    Commented Apr 10, 2017 at 18:32
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    $\begingroup$ Well, if the central limit theorem applies and $n$ is large enough that you are close to the limit,. This will be a badly cpnditioned problem. The name of the problem is deconvolution, so you can search for that $\endgroup$
    – kjetil b halvorsen
    Commented Apr 21, 2017 at 13:12

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