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Hi,

(I hope this is not too basic)

I basically have a set of data from the same underlying distribution (which I would like to estimate), but I only have available the mean and N from partitions of the data. I can estimate the mean using a weighted average, and, I thought, the sample variance using weighted variance (e.g: https://stat.ethz.ch/pipermail/r-help/2008-July/168762.html)

However, I can't make much sense of the results. And I suspect I'm misunderstanding how/what weighted sample average should work.

Edit: From a set of (unobserved) data points all drawn from some distribution with some $\mu$ and $\sigma$, I can observe only the average and cardinality for partitions of the data. I.e. my observed values $X_i$ are each the average of some partition of size $N_i$, and thus have, $\mu_i=\mu$ and a $\sigma_i=\sigma/\sqrt{N_i}$.

Question: how can I estimate $\mu$ and $\sigma$ from the observed $X_i$s? Note that I can't observe the variance (or any other parameters) for the partitions.

The goal is to identify partitions that do not fit the expected distribution, i.e. identify partitions where $X_i$ is too many $\sigma_i$s from $\mu$.

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it does appear too basic for this site, although like mikitov I'm not 100% sure I understand what you mean. Did you consider asking your question on stats.stackexchange.com –  Thierry Zell May 6 '11 at 13:39
    
Like Thierry I don't quite understand what exactly you are asking for. For one possible interpretation ("you are looking for a numerical formula") I'd recommend you to read infoserve.sandia.gov/sand_doc/2008/086212.pdf and for another possible interpretation ("you are speaking about quantization") to read statistik.uni-dortmund.de/fileadmin/user_upload/Lehrstuehle/… –  Someone May 6 '11 at 14:24
    
I didn't know about stats.stackexchange.com, but I can try there. I don't know how to clarify it or explain better - I looked at the links but it is not quanization, and I don't understand the other link enough to see any relevance. –  Ketil May 6 '11 at 20:26
    
Maybe the same question as this?stats.stackexchange.com/questions/10441/… –  Ketil May 6 '11 at 20:30
    
The <b>Question</b> may be pretty basic, but I wonder if the part after that starting with "The goal is..." might be a research question. –  Michael Hardy May 7 '11 at 0:55
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3 Answers

up vote 1 down vote accepted

Note: New answer.

If you don't know the variances $Var X_1,...,VarX_m$, you have to make parametric assumptions about the distribution of the data. For example, under the assumption of a Poisson distribution, the the maximum likelihood estimator of the variance would simply be the aggregated mean $EX$. Under the assumption of the normal distribution, the intuitive answer

$$VarX = (1/m)\sum_i N_i (EX_i-\mu)^2$$ is the maximum likelihood estimator.

Assuming your distirbution is normal $(\mu, \sigma^2)$, then the maximum likelihood estimate of $\mu$ is the aggregated mean $(1/N)\sum_i N_i EX_i$. The profile likelihood as a function of $\sigma^2$ is then

$$L(\sigma^2) = \prod_i \frac{1}{\sqrt{2\pi\sigma^2/N_i}} exp\left(\frac{-(EX_i - \mu)^2}{2\sigma^2/N_i}\right).$$

The maximum likelihood estimate of $\sigma^2$ is the maximizing argument of the above. For computational convenience we take the logarithm of the likelihood as it is easier to maximize.

$$\log L(\sigma^2) = \sum_i -\frac{1}{2}\log(2\pi/N_i) - \frac{1}{2}\log(\sigma^2) - \frac{(EX_i - \mu)^2}{2\sigma^2/N_i}$$

Setting the derivative to zero,

$$0=\frac{\partial L(\sigma^2)}{\partial \sigma^2} = \sum_i -\frac{1}{2\sigma^2} + \frac{(EX_i-\mu)^2}{2/N_i}\frac{1}{\sigma^4}$$

multiplying by $\sigma^4$,

$$ 0 = \sum_i -\frac{1}{2}\sigma^2 + \frac{N_i(EX_i-\mu)^2}{2}$$

which yields the maximum likelihood estimator

$$\sigma^2 = (1/m)\sum_i N_i (EX_i-\mu)^2.$$

For general parametric $f(\theta)$ you need the densities of the convolutions $f * \cdots * f(\theta)$, which makes the problem much more difficult outside of a few special parametric families. However if all of your $N_i$ are large, you can use the normal distribution as an approximation by the Central Limit Theorem.

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@charles: You're misusing the word "sample" in the usual way; you should say two samples of sizes $N_1$ and $N_2$.

However (and this is the occasion for making this an "answer" rather than a comment), I think a conceptually more efficient way to think about this is to recall that $\text{var}(X) = E(\text{var}(X\mid Y)) + \text{var}(E(X \mid Y))$. I.e. the total variance is the explained variance plus the unexplained variance. See Law of total variance.

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As far as I understood, you have a set of, let's say $M$ partitions of different data with cardinality $N_i$, mean $\mu_i$ mean and variance $\sigma^2_i$ for all $i = 1, \ldots , M$. Is it right?

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Any clearer now? –  Ketil May 6 '11 at 13:42
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