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Monte Carlo method, from what I could gather, allows one to obtain observations/draws from a possibly unknown statistical distribution. Let's say $T(X)\sim F$, where $T$ is a statistic, $X$ is a random vector, and $F$ is possibly unknown distribution.

The method is composed of the following steps:

  • Draw ${x_i}$ from a distribution $F_X$.
  • Compute ${T(x_1)=t_1,..., T(x_m)=t_m}$. We have m samples, giving m values from the statistic.
  • Use the set ${t_1,...,t_m}$ to compute histograms, quantiles, etc.

Well, I was trying to apply this to check the Central Limit Theorem. So, I followed the steps of the MC method, and draw an histogram

enter image description here

From one histogram to the other I increased the sample size.The $m$ is number of samples. In each histogram I used $m=20000$. By sample size $n$, I mean each sample has $n$ observations. For the histograms I used $n=10, 100, 1000$

It looks nice.

The problem is that the CLT is a result of convergence of distributions, and what I draw are an approximation to the density(histogram) and the limiting normal density. Having convergence of densities implies convergence of distributions, by Scheffé's theorem. However, we the converse is not generally true. We may have the same distribution, with several different densities. So, although what I did looked nice, it seems wrong.

Should I have, instead of an histogram of a density, tried to plot an 'histogram of the distribution'?

Any help would be appreciated.

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    $\begingroup$ sounds like you are missing one asymptotic dimension: not only the sample size must grow to $\infty$ to recover the distribution of interest, but for a CLT the sample on which each term of the sample is computed must also grow to $\infty$... $\endgroup$ – Xi'an Feb 16 '18 at 9:27
  • $\begingroup$ @Xi'an. The $m$ is number of samples. In the histogram I used 20000. By sample size $n$, I mean each sample has $n$ observations. For the histograms I used $n=10, 100, 1000$ $\endgroup$ – An old man in the sea. Feb 16 '18 at 9:51
  • $\begingroup$ Then you are fine, producing a sample of independent $\bar{X}_n$ which limiting distribution is a $\mathcal{N}(0,\sigma^2)$. Looking at the histogram sounds kosher to me... $\endgroup$ – Xi'an Feb 16 '18 at 9:53
  • $\begingroup$ @Xi'an but the thing is that the TLC is a result for convergence in distribution, while usually convergence of histograms is convergence of densities, which needs convergence in probability if I'm not mistaken. Either way, Shouldn't I have drawn an approximation to the CDF instead of an approximation to a PDF? $\endgroup$ – An old man in the sea. Feb 16 '18 at 9:55
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    $\begingroup$ You are correct that a rigorous treatment of convergence concerns the distribution function (cdf) rather than the density (which for general $F$ doesn't even exist). What prevents you from drawing plots of the distribution functions? $\endgroup$ – whuber Feb 22 '18 at 19:33

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