Question 387180 discusses the pdf of the log-log-normal distribution. I'd like to know if there's an expression for the mean of this distribution. I'm trying to work it out with pencil and paper, but it's hard (at least for me).


1 Answer 1


When $Y$ has a "log-lognormal" distribution, $\log(\log(Y)) = X$ has a Normal distribution of (say) mean $\mu$ and variance $\sigma^2.$ The expectation of $Y=\exp\left(e^X\right)$ therefore is obtained from

$$\begin{aligned} E[Y] &= E\left[\exp\left(e^X\right)\right] = \frac{1}{\sigma\sqrt{2\pi}}\int_\mathbb{R} \exp\left(e^x - \frac{(x-\mu)^2}{2\sigma^2}\right)\,\mathrm{d}x. \end{aligned}$$

There are many ways to prove that when $x$ is sufficiently large, $e^x \ge (x-\mu)^2/(2\sigma^2).$ Let's leave this as a Calculus exercise and just apply it. It implies there is some number $N$ for which this inequality holds, whence when $x\ge N,$ $e^x - (x-\mu)^2/(2\sigma^2) \ge 0.$ Plugging this into the right hand side gives

$$\int_\mathbb{R} \exp\left(e^x - \frac{(x-\mu)^2}{2\sigma^2}\right)\,\mathrm{d}x \gt \int_N^\infty \exp\left(e^x - \frac{(x-\mu)^2}{2\sigma^2}\right)\,\mathrm{d}x \ge \int_N^\infty \exp\left(0\right)\,\mathrm{d}x = \int_N^\infty \mathrm{d}x $$

which diverges. Therefore the expectation of $Y$ is infinite.

If, on the other hand, you seek the expectation of $1/Y = \exp\left(-e^X\right),$ you won't find a simple expression in terms of familiar mathematical functions. But it does exist, because necessarily $Y\ge 1,$ whence $1/Y \le 1.$ If you would like a formal demonstration, note that $ \exp\left(-e^x\right) \le 1$ for all $x,$ whence

$$\begin{aligned} E\left[\frac{1}{Y}\right] &=E\left[\frac{1}{|Y|}\right] = \frac{1}{\sigma\sqrt{2\pi}}\int_\mathbb{R} \exp\left(-e^x\right) \exp\left( - \frac{(x-\mu)^2}{2\sigma^2}\right)\,\mathrm{d}x\\ &\le \frac{1}{\sigma\sqrt{2\pi}}\int_\mathbb{R} \exp\left(- \frac{(x-\mu)^2}{2\sigma^2}\right)\,\mathrm{d}x = 1 \end{aligned}$$

demonstrates $E[1/Y]$ converges absolutely to a finite value. Compute this value numerically.

  • $\begingroup$ Thanks Richard for pointing that out. @whuber...thanks for the very quick answer. That all makes sense. I realize now that I should have asked my real question, which, using your notation, is "what is the expectation of Y=exp(-exp(X)), where X has a normal distribution." I had assumed that if I had the expectation of exp(exp(X)) then I could just transform one to the other, but I realize now that's wrong. I think this new expectation probably is finite...at least my code converges. Sorry for not asking the right question in the first place. $\endgroup$ Commented Aug 2, 2021 at 10:31
  • $\begingroup$ @whuber...thanks for part II. $\endgroup$ Commented Aug 2, 2021 at 13:28

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