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Consider 10 values that follow a standard normal distribution. What would you expect to be the lowest value?

I tried to simulate this problem in R. I basically just simulated 100000 standard normal distribution with 10 values and took the mean of the each lowest value.

> mean(replicate(100000,min(rnorm(10))))
[1] -1.536875

This corresponds to a probability to get a lower value of

> pnorm(-1.536875)
[1] 0.06216196

I tried to reach these values analytically but I really have no idea how to approach this.

I thought about it for quite some time now and also tried to look it up. But I can't find a solutions to this simple problem. Probably just overlooking something obvious. Someone that can help me?

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  • $\begingroup$ You're after the expectation of the distribution of the 1st Order statistic, or the expectation of the distribution of the minimum. If you look here it gives the general form for the density of the kth order statistic $\endgroup$
    – Glen_b
    Jan 25, 2014 at 11:30
  • $\begingroup$ See here and here and the answer here. If you use google you can turn up lecture notes on the topic (like this) $\endgroup$
    – Glen_b
    Jan 25, 2014 at 11:40
  • $\begingroup$ OK, if you know what you are looking for, you can indeed find a lot of information. Less straightforward then I thought. Thanks $\endgroup$
    – statastic
    Jan 25, 2014 at 12:50
  • $\begingroup$ Nevertheless the normal case has been well studied. Well, more has been written about the maximum, but that's just a matter of flipping about the mean (which for the standard normal is just taking the negative). There are tables, for example, and asymptotic approximations (which are a bit rough at n=10 but which I think do relatively well at higher n). The tables here give an expected value for the maximum at n=10 of 1.53875. $\endgroup$
    – Glen_b
    Jan 25, 2014 at 13:18

1 Answer 1

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As has been clarified by Steck (1962), results contained in Ruben (1954) lead to the following expression for the expected value of the $1$st (minimum) and $n$th (maximum) order statistic, of a sample of $n$ i.i.d standard normal random variables:

$$E[X_{(n)}] = -E[X_{(1)}] = \frac {n(n-1)}{2\sqrt{\pi}}\mathcal Z_{n-2}(0,...,0;\rho=1/3) $$

where $\mathcal Z_{n-2}() $ represents the multivariate standard normal cumulative distribution function of $n-2$ identical and equicorrelated standard normal random variables with common correlation coefficient $\rho=1/3$, all r.v's evaluated at $0$. This is not an analytical solution, but this integral can be (and has been) evaluated. For $n=10$ the value of the integral, as given in Gupta (1963) is $\mathcal Z_{8}(0,...,0;\rho=1/3) = 0.06061$ and so we obtain

$$E[X_{(10)}] = -E[X_{(1)}] = \frac {10\cdot 9}{2\sqrt{\pi}}\cdot 0.06061 = 1.538799$$

Chen & Tyler (1999) provide the following approximation to these expected values:

$$E[X_{(n)}] = -E[X_{(1)}] \approx \Phi^{-1}(0.5264^{1/n})$$

Where $\Phi^{-1}(\cdot)$ is the inverse (not the reciprocal) of the standard normal cumulative distribution function. They report that for $5<n<10$ this has less than $2$% approximation error, while for $n\ge 10$ the approximation error falls below $1$%. They derived this formula by the following approach: They asked "what would the expected value be if the distribution of the extreme order statistics was symmetric around the mean?" If it was symmetric the expected value would equal the value at the median, which is $\Phi^{-1}(0.5^{1/n})$. But since the distribution is not symmetric around the mean, they searched for a correction term, which lead them to the expression above. For $n=10$, this approximation gives

$$E[X_{(10)}] = -E[X_{(1)}] \approx \Phi^{-1}(0.5264^{1/10}) = \Phi^{-1}(0.937846) =1.536941$$

which has percentage approximation error $-1.2074\cdot10^{-3}$, much less than $1$% $=10^{-2}$.

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