2
$\begingroup$

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 $\bar{x_1},...,\bar{x_n}$?

$\endgroup$
1
  • $\begingroup$ Here's a possible answer $\endgroup$
    – Asterix
    Apr 7, 2017 at 17:45

1 Answer 1

2
$\begingroup$

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 $X$.

$\endgroup$
3
  • 2
    $\begingroup$ 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
    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
    Apr 10, 2017 at 18:32
  • 2
    $\begingroup$ 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 $\endgroup$ Apr 21, 2017 at 13:12

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.