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Let $f$ be some probability density function with undefined central moments. For example, suppose $f$ is a Cauchy distribution.

Say I draw two samples of size $N=100$ from that distribution. The mean of the first sample could be $1.5$ and the mean of the second sample $-2.3$.

However, since the population distribution $f$ does not have the central moments, calculating the mean values should get me into some trouble? Am I correct?

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You can show that CLT does not apply, so that you can compute sample mean, but it does not provide a good estimator of the first moment.

a <- rcauchy(1e5,0,1)
b <- rnorm(1e5,0,1)

plot(seq(1,1e5), cumsum(a)/cumsum(as.numeric(rep(1,1e5))), type = 'l')
plot(seq(1,1e5), cumsum(b)/cumsum(as.numeric(rep(1,1e5))), type = 'l')

See figure below. Cauchy's partial sums do not converge to the true mean $0$ as $n \to \infty$.

enter image description here

See figure below. Normal's partial sums converge to the true mean $0$ as $n \to \infty$. enter image description here

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  • $\begingroup$ How can we tell that the upper plot simply isn't converging more slowly to a constant? There's a logical gap in your argument: the conclusions of a theorem can happen to be true even when its assumptions do not hold. In this case, although the CLT says nothing about the mean of a large number of iid Cauchy variates, that in itself does not show the mean will not converge. One way to demonstrate the lack of convergence is to compute the distribution of the mean explicitly (which isn't hard). $\endgroup$
    – whuber ♦
    Aug 4, 2015 at 13:15
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    $\begingroup$ In principle I agree with you, however the OP was asking "calculating the mean values should get me into some trouble?". The simulations I provided show that the speed of convergence is someway low (we can say at least that), hence the use of the sample mean can cause some troubles. $\endgroup$
    – stochazesthai
    Aug 4, 2015 at 13:52
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    $\begingroup$ OK--but what kind of "trouble," exactly? And under what circumstances? I see that your post and its illustrations hint at answers to these questions but perhaps you could state your answers a little more explicitly. $\endgroup$
    – whuber ♦
    Aug 4, 2015 at 14:15
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    $\begingroup$ @stochazesthai there is similar problem in your argument as is in mine, but your answer is more advanced (+1). $\endgroup$
    – Analyst
    Aug 5, 2015 at 5:25
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Sample statistics of course exists but they are not estimators for the population moments which are not meaningfully defined.

Here is a R code example drawing 800 drawings from the cauchy (100,1000) distribution.

a=rcauchy(800,100,1000);
print(mean(a));
[1] -555.4276
b=rcauchy(800,100,1000);
print(mean(b));
[1] -262.3275

These values differ a lot.

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    $\begingroup$ Maybe they differ a lot for means of $800$ iid samples, but perhaps by the time you draw $10^{800}$ samples they will stabilize. This possibility is an inherent flaw in trying to use simulations to demonstrate lack of convergence. Such simulations can be successful only insofar as they are supported by theoretical analysis. $\endgroup$
    – whuber ♦
    Aug 4, 2015 at 13:17
  • $\begingroup$ @whuber of course you are right (+1) but question was about sample statistics and not necessarily about converge/divergence of certain integrals.. $\endgroup$
    – Analyst
    Aug 5, 2015 at 5:23

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