# Approximate Order Statistics for lognormal variables

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.

• 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? – Stephan Kolassa May 23 '16 at 13:08
• @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. – Ron May 23 '16 at 13:27
• Yes, of course, you are right. Perhaps the beta-F distribution would help, and maybe there is something about its exponential... – Stephan Kolassa May 23 '16 at 14:14
• Do you only care about the maximum, or are you interested in higher order statistics from the lognormal in general? – jbowman May 25 '16 at 17:19
• @jbowman At this point I care only about the maximum. Others would be nice, of course, but even maximum will be helpful. – Ron May 26 '16 at 1:25

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.

• Thanks - I am looking for an analytical closed form approximation (it'll go into a game theory model). The numerical approach is quite obvious. – Ron Jun 1 '16 at 13:38
• @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. – Alexander Rodin Jun 1 '16 at 15:10
• 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. – A. Ray Jun 1 '16 at 15:46
• 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! – Ron Jun 1 '16 at 16:28