First of all, I apologize for not being rigorous, but I am not a statistitian by background.

Imagine you have $N$ i.i.d. positive random variables $X_1...X_N$ and you are trying to compute a logarithmic sum of expectation values: $$S = \ln\textrm{E}[X_1]+...+\ln\textrm{E}[X_N]$$

This is of course equal to: $$S = \ln\Big(\prod_{n=1}^{N}\textrm{E}[X_n]\Big)$$

Since the random variables are independent: $$S = \ln\Big(\textrm{E}\Big[\prod_{n=1}^{N}X_n\Big]\Big) \equiv \ln\textrm{E}[Z]$$

where $Z$ is a product of $N$ random variables.

For very large finite $N$ the distribution of $Z$ approaches the lognormal distribution and this $S$ can be estimated by:

$$S \approx \mu_{\ln Z} + \frac{1}{2}\sigma^2_{\ln Z} = N\Big(\mu_{\ln X_1} + \frac{1}{2}\sigma^2_{\ln X_1}\Big)$$

But we also know that, since the the random variables have identical probability distributions:

$$S = N\ln\textrm{E}[X_1]$$

However, since the PDF of $X$ need not be lognormal, the following is not in general true:

$$N\ln\textrm{E}[X_1] \neq N\Big(\mu_{\ln X_1} + \frac{1}{2}\sigma^2_{\ln X_1}\Big)$$

It seems that we have reached a paradox, which stems from a limiting behaviour I am most probably not handling correctly. On one hand, the higher-order cumulants in the limiting case approach zero but in the other case they always contribute the same amount to the logarithmic sum. Any suggestions where this paradox comes from?

  • $\begingroup$ I just assume that $X\sim N(\mu,\sigma)$ then $log(X)$ is not defined! you are missing an assumption in your definition ... $\endgroup$ – TPArrow Jun 13 '19 at 9:51
  • $\begingroup$ Thanks for that, I completely missed that one out, I will edit it. $\endgroup$ – Godzilla123 Jun 13 '19 at 9:54

It appears that you are confusing the random variables $X_1,X_2,...,X_N$ with their expected values. As presently defined, the value $S$ is not a random variable at all; it is a constant (and so the notation of using an upper-case letter is not really appropriate). If we let $\mu \equiv \mathbb{E}(X_i)$ be the common mean of your IID values, then you have the constant value:

$$S = \ln \mathbb{E}(X_1) + \cdots + \ln \mathbb{E}(X_N) = N \ln \mu.$$

It follows that you also have the constant value $Z = e^S = \mu e^N$. This is a constant value, so it is not true that $Z$ is normally distributed for large $N$.

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