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Consider $X$ that follows a log-normal distribution with parameters $\mu=1$ and $\sigma=1$. The moments are known: $m_n=E[X^n]=e^{n+n^2/2}$. For example, $m_{10}=e^{60}$ (of the order of $10^{26}$). I simulate samples of $X^{10}$ in Mathematica of length $1,000,000$, and I notice that the mean values of such samples are always less than $e^{60}$:

Mean[RandomVariate[LogNormalDistribution[1, 1], 1000000]^10] < Exp[60]
(* Comment: The output seems always to be True. *)

Why is Monte Carlo simulation not working with the log-normal distribution? The tails of the distribution (the extreme values) should be the problem, I think. Is there a modification of the Monte Carlo simulation to tackle this problem?

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    $\begingroup$ Computers do not deal with all real numbers. Instead, a carefully selected finite subset of the rationals. I think you may be exceeding the largest numbers Mathematical can handle--'overflow'. // Why are you interested in the 10th moment of lognmormal? $\endgroup$ – BruceET Sep 16 '19 at 16:46
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    $\begingroup$ There's a reason for this and it's to be found in the fact that the variance of the tenth moment will be on the order of the twentieth moment, which is enormous. The only way that can happen is for the vast majority of random samples to have relatively tiny tenth moments. You can figure this out for yourself by examining the situation with smaller moments: study the (raw) second moment, for instance. $\endgroup$ – whuber Sep 16 '19 at 17:11
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    $\begingroup$ The question is not off-topic, it is about the impact of tails in estimating higher moments by Monte Carlo. It really has nothing to do with software and in particular @BruceET 's comment about overflow is a red herring. $\endgroup$ – Chris Haug Sep 20 '19 at 16:51
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From your expression above for $m_n$, you can derive the following:

$$P[X^n > m_n] = 1 - \Phi(n/2)$$

For $n=10$, that's less than 0.3 per million, so the vast majority of samples will be smaller than $m_n$ (and it's not unusual to not have a single sample out of 1 million turn out to be larger than $m_n$). It's the rare but very large values that pull up the moment, as you mention, and it gets more evident for larger moments.

One relatively simple method for improving a Monte Carlo method that depends critically on rare events is to use importance sampling and "tweak" the distribution to make those events more likely, and then weigh our samples to get back to the distribution we're actually trying to sample from. One such approach is called exponential twisting (or tilting). For example, for a normal distribution $Y\sim \mathcal{N}(\mu,\sigma^2)$, consider the twisted random variable:

$$Y_\theta \sim \mathcal{N}(\mu + \theta\sigma^2,\sigma^2)$$

For some function $f$ of interest, you can show that:

$$E(f(Y)) = E_{\theta}\left[f(Y)p(Y)/p_{\theta}(Y)\right]=E_{\theta}\left[f(Y)e^{(\mu-Y)\theta+\sigma^2\theta^2/2}\right]$$

The expectation $E$ is taken with respect to the distribution of $Y$, while $E_{\theta}$ is with respect to that of $Y_{\theta}$. The parameter $\theta$ can be tuned depending on the problem at hand (e.g. which events we need to make more likely), and $\theta = 0$ corresponds to the regular MC method.

The above identity means that you can then estimate the expectation of interest on the left by Monte Carlo like this:

  1. Draw $Y_i \sim \mathcal{N}(\mu + \theta\sigma^2,\sigma^2)$
  2. Compute the weighted sample average $\frac{1}{M}\sum_{i=1}^M f(Y_i) e^{(\mu-Y_i)\theta+\sigma^2\theta^2/2}$

For your specific problem, you would take $\mu = 10$, $\sigma^2 = 100$ and $f(x) = e^x$.

For a range of values of $\theta$, sampling only 1000 $Y_i$'s for each value, I obtain the following estimates (the horizontal line is the exact value):

enter image description here

We see that the regular MC method ($\theta=0$) performs poorly, and that values near $\theta=1$ produce a result much closer to the true value.

Obviously, this is a somewhat contrived example: if I simply take $\theta = 1$, then $Y_i = e^{\mu+\sigma^2/2}$ for all $i$, and I get the exact result with no sampling variation, which we already knew from the beginning. The method can still be used in other situations where a closed-form solution or other efficient numerical approximation does not exist.

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  • $\begingroup$ Thank you for the answer, that's exactly what I was looking for. What is the definition of $p$ and $p_\theta$? $\endgroup$ – user260042 Sep 17 '19 at 9:25
  • $\begingroup$ @user260042 Sorry, those are just the densities that correspond to $Y$ and $Y_{\theta}$ (normal, with the parameters stated above). $\endgroup$ – Chris Haug Sep 17 '19 at 10:42

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