Lets say I have a normal distribution $N(\mu, \sigma^2)$ from which I have drawn $n$ i.i.d. samples $x_1, \dots, x_n$.

Now, lets define a random variable $Y = max(x_1, \dots, x_n)$.

When $n=1$, the expected value of $Y$ is $\mu$. I would expect that as $n$ increases, the expected value of $Y$ should increase as well. Is it possible to determine the expected value of $Y$ for any value of $n$, in terms of $\mu$ and $\sigma$?


3 Answers 3


If we combine two of the answers here (Approximate order statistics for normal random variables), we have for the $r$th $\it{smallest}$ order statistic

$$E[r,n] \approx \mu + \sigma \ \Phi^{-1} \left( \frac{r-\frac{\pi}{8}}{n-\frac{\pi}{4}+1}\right) $$

For the largest value we want $r=n,$ so we have

$$E[Y] \approx \mu + \sigma \ \Phi^{-1} \left( \frac{n-\frac{\pi}{8}}{n-\frac{\pi}{4}+1}\right) $$

  • $\begingroup$ Thanks. This is an interesting result. However, I am not sure how to interpret the case when n tends to infinity. The posted answer gives: E[Y] = \mu + \sigma / \Phi. However, intuitively, I would expect the answer to be E[Y] = \m + 3 \sigma or something along those lines. Am I missing something here? $\endgroup$ Aug 13, 2020 at 21:24
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    $\begingroup$ As $n$ tends to infinity, you should expect the sample maximum to increase. And that is exactly what the formula predicts. The argument in parentheses in the last formula above tends towards 1 as $n$ increases. The inverse cdf will continue to increase as $n$ increases. $\endgroup$
    – soakley
    Oct 6, 2020 at 13:25
  • $\begingroup$ Certainly, as n increases, the sample maximum is expected to increase. No question there. My point though was that the increase seems to be too little. For example, even if you have just a million samples, one would expect that we will see max value at least 3 SDs away from mean. While the posted result only seems to suggest 1/phi SD away. Clearly there is a small disconnect in the result or its interpretation. $\endgroup$ Oct 6, 2020 at 16:29
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    $\begingroup$ That is not 1/phi. It is the inverse of the standard normal cdf. $\endgroup$
    – soakley
    Oct 6, 2020 at 18:56
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    $\begingroup$ For a million observations and using the formula above, I get a value of 4.85 standard deviations above the mean. $\endgroup$
    – soakley
    Oct 6, 2020 at 19:48

First note that\begin{align}Y_n=\max\{X_1,\ldots,X_n\}&=\max\{\sigma\epsilon_1+\mu,\ldots,\sigma\epsilon_n+\mu\}\\&=\sigma\max\{\epsilon_1,\ldots,\epsilon_n\}+\mu\\&=\sigma\xi_n+\mu\end{align} hence that $(\mu,\sigma)$ is also a location-scale parameter for the maximum. Asymptotically, the Normal distribution belongs to the domain of attraction of the Gumbel distribution, meaning that $$\sqrt{2\log(n)}(\xi_n-d_n)\stackrel{{\cal L}}{\longrightarrow} G_0$$with $G_0(x)=\exp\{-\exp(-x)\}$ the Gumbel pdf and $$d_n = \sqrt{2\log(n)}-\dfrac{\log\log n + \log(4\pi)}{2\sqrt{2\log(n)}}$$



I found this paper referenced in a thread on the maths stack exchange (Approximate order statistics for normal random variables), so I had a look. For the maximum, $r=n$.

"In a sample of size n the expected value of the rth largest order statistic is given by


where $\phi(x)=1/\sqrt(2\pi)exp(-\frac{1}{2}x^2)$ and $\Phi(x)=\int^x_{-\infty}\phi(z)dz.$"

  • Royston, J. P. (1982), 'Algorithm AS 177: Expected Normal Order Statistics (Exact and Approximate)', Journal of the Royal Statistical Society. Series C (Applied Statistics), 31(2):161-165.

So $Y$ is an order statistic. Let's label its density function $g_{(n)}(x)$, to indicate that it's the pdf of the variable in the nth position (i.e. its the pdf of the maximum in the sample). Let's also label the normal $N(\mu, \sigma^2)$ density function as $f(x)$. It's a standard result that $$g_{(n)}(x)=n[F(x)]^{n-1}f(x),$$ where $F(x)$ is the cumulative density function of $N(\mu, \sigma^2)$ (as a reference, I suggest Mathematical Statistics (7th ed.) by Wackerly, Mendenhall, and Scheaffer, p.333).

It is at this point that I'm unable to proceed - I don't know how to evaluate the expected value of $Y$, given that it has such a strange pdf. However, I'd advise you to search for "expected value of order statistic" - in particular, I found a thread on this topic on the maths stack exchange site:

EDIT: As pointed out by Khol, thread is for a uniform distribution, not a normal distribution. The uniform is apparently more straightforward to deal with. Apologies for the partial answer!


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    $\begingroup$ Linked thread is for a uniform random sample, which is distinctly simpler than a normal random sample. $\endgroup$
    – khol
    May 2, 2018 at 4:36

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