Are there any known formulas that approximate the expected value of the maximum of $N$ i.i.d. lognormal random variables?

I am looking for something similar to: Approximate order statistics for normal random variables

But in general any useful info on the expected value will be appreciated.

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    $\begingroup$ The logarithm is a monotonic transformation, so you could simply work on the logged data, which are normally distributed (so your link applies), then exponentiate the order statistics. Or what am I missing? $\endgroup$ Commented May 23, 2016 at 13:08
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    $\begingroup$ @StephanKolassa That is what I thought, but I'm wondering how to get at the expected value as $E[e^{X}]$ does not equal $e^{E[X]}$ generally. $\endgroup$
    – Ron
    Commented May 23, 2016 at 13:27
  • $\begingroup$ Yes, of course, you are right. Perhaps the beta-F distribution would help, and maybe there is something about its exponential... $\endgroup$ Commented May 23, 2016 at 14:14
  • $\begingroup$ Do you only care about the maximum, or are you interested in higher order statistics from the lognormal in general? $\endgroup$
    – jbowman
    Commented May 25, 2016 at 17:19
  • $\begingroup$ @jbowman At this point I care only about the maximum. Others would be nice, of course, but even maximum will be helpful. $\endgroup$
    – Ron
    Commented May 26, 2016 at 1:25

1 Answer 1


I do not know of any standard approximate formulas for this. So I tried the following standard technique.

Let $X_1,...,X_N$ be $N$ iid log-normal random variables. Let $Y=\max_i X_i$ be their maximum. Now we need to find $E[Y].$ Let $F_Y(y)$ be the cdf of $Y$ and $F_X(x)$ be the cdf of any $X_i$'s. Then,

\begin{eqnarray} F_Y(y) &=& P(Y \leq y) = P(\max_i X_i \leq y) \\ &=& P(\cap_i \{X_i \leq y\}) \\ &=& \prod_i P(X_i \leq y) = F_X(y)^N \end{eqnarray}

Since $X_i$'s are iid. Now note that $X_i$'s are non-negative random variables and so is $Y.$ For any non-negative RV we can express its expectation in terms if its cdf as follows. You can find its derivation here.

$$ E[Y] = \int_0^\infty (1-F_Y(y))dy = \int_0^\infty (1-F_X(y)^N)dy $$

Now for any log-normal random variable $F_X(x) = \Phi\left(\frac{\ln x - \mu}{\sigma} \right),$ where $\Phi$ is the cdf of the standard normal distribution. Therefore

$$ E[Y] = \int_0^\infty \left(1-\Phi\left(\frac{\ln y - \mu}{\sigma} \right)^N\right)dy $$

Till now we have an exact formula for $E[Y].$ All I can think of now is a numerical integration to approximately compute $E[Y].$ The function $\Phi$ is well approximated and there are library functions for it. For a given $\mu,\sigma,N$ we can approximately compute $E[Y]$ using standard numerical integration techniques.

  • $\begingroup$ Thanks - I am looking for an analytical closed form approximation (it'll go into a game theory model). The numerical approach is quite obvious. $\endgroup$
    – Ron
    Commented Jun 1, 2016 at 13:38
  • $\begingroup$ @Ron but what do you want from closed form? If you want to calculate derivatives you still can do it by differentiating the integral by parameter and then calculating numerically the resulting integral. $\endgroup$ Commented Jun 1, 2016 at 15:10
  • $\begingroup$ I guess I don't see the benefit of an approximate closed form solution either, as opposed to an accurate numeric value. As @a-rodin suggested you can always approximate using a Taylor series, by taking derivative of the integral and numerically computing the Taylor coefficients. $\endgroup$
    – A. Ray
    Commented Jun 1, 2016 at 15:46
  • $\begingroup$ I will need derivatives with respect to N, which indeed I can take. The issue is the error function/normal PDF. Taking taylor approximations is a good point - I'll just do that. Thanks guys! $\endgroup$
    – Ron
    Commented Jun 1, 2016 at 16:28

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