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For a given sample set $S$ with $N$ individual samples $x_i$, I can easily find the average distance from the maximum by doing something like this:

$\sigma_{max_N}:=\sqrt{\frac{1}{N}\sum\limits_{i=1}^N {\left(x_i - \max{\left(S\right)}\right)}^2}$

Trying to do so for larger and larger sample sizes, I (unsurprisingly) end up with different limit values given different distributions.

For instance, for a uniform distribution on a unit interval, I get a value around 0.577 with digits after that varying wildly.

A normal distribution with the same mean $\left(\frac{1}{2}\right)$ and variance $\left(\frac{1}{12}\right)$ gives me something around 1.5, though the exact value shifts around a lot more.

For an exponential distribution with parameter $\lambda=2\sqrt{3}$, I end up at around 4.5.

Is there a way to get an expected value of $\sigma_{max_N} \to \sigma_{max}$ for a given distribution as the sample size $N \to \infty$? It doesn't necessarily have to be analytic.

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2 Answers 2

up vote 4 down vote accepted

I presume we are talking about a collection of i.i.d random variables, with finite moments. Using a bit more convenient notation, let's consider (we can afterwards apply the continuous mapping theorem for the square root),

$$\sigma^2_{max_n}:=\frac{1}{n}\sum\limits_{i=1}^n {\left(X_i - X_{(n)}\right)}^2$$

$$= \frac{1}{n}\sum\limits_{i=1}^n X_i^2 -2X_{(n)}\bar X+X_{(n)}^2$$

where $X_{(n)}$ is the maximum order statistic, and $\bar X$ is the sample mean. We want the probability limit of this expression, which will be, if it exists, the sum of the probability limits of the three components.

We have easily that

$$\frac{1}{n}\sum\limits_{i=1}^n X_i^2 \xrightarrow{p} E(X_1^2)$$

but the main question is

Does the maximum order statistic converges in probability somewhere?

(while the usual issue is whether extreme order statistics, properly transformed, converge in distribution - to some variant of the Extreme Value Distribution).
Applying the definition of convergence in probability, and searching for the value of $k$, we would want that $\exists \;k:$

$$\begin{align}P[|X_{(n)}-k|< \epsilon]=1 &\Rightarrow P[k-\epsilon<X_{(n)}<k+\epsilon]=1\\ &\Rightarrow F_{Y_{(n)}}(k+\epsilon)- F_{Y_{(n)}}(k-\epsilon)=1 \\&\Rightarrow [F_X(k+\epsilon)]^n-[F_X(k-\epsilon)]^n =1 \end{align}$$

since $F_{Y_{(n)}}(y) = [F_X(y)]^n$.

RANDOM VARIABLES with support bounded from above

We see what happens here: for random variables with bounded support from above, denote this upper bound simply $\theta$, if we set $k=\theta$ then, by the properties of the cumulative distribution function, the equality above will hold, and so $$X_{(n)} \xrightarrow{p} \theta \Rightarrow X_{(n)}^2 \xrightarrow{p} \theta^2$$ by the continuous mapping theorem. The sample mean will converge to the expected value so in all, for this class of random variables,

$$\sigma^2_{max_n} \xrightarrow{p} E(X_1^2) -2\theta E(X_1) + \theta^2$$

Verification: For a sample of i.i.d uniforms $U(0,1)$ we have $E(X_1^2) =1/3$, $E(X_1) =1/2$, $\theta=1$.

So (using again the continuous mapping theorem) $$\sigma_{max_n} \xrightarrow{p} \sqrt{\frac 13 -2\cdot1\cdot \frac 12 +1^2} = \sqrt{1/3} \approx 0.577$$ as the OP found.

RANDOM VARIABLES with support unbounded above
Intuitively, here the maximum order statistic will tend to $+\infty$ as $n\rightarrow \infty$. So we have to accept approximations, since our sample will be finite after all. Treating the continuous case the OP treated, for a normal $N(0.5, 1/12)$ a virtual "zero-tail probability" value is $\theta=2$. Moreover, $$E(X_1^2)= \operatorname{Var}(X) + \mu^2 = 1/12 + (1/2)^2 = 1/3$$. So here

$$\sigma_{max_n} \approx \sqrt{\frac 13 -2\cdot2\cdot \frac 12 +2^2} =\sqrt{2.33}\approx 1.52$$

This matches the OP simulations, but of course it is very sensitive to the exact tail probability chosen. Why not choose $\theta=3$?... This is not a robust approximation. Moreover we are silent about the size of $n$ -although in finite samples, we will never observe an actual infinity, with really large $n$, well, there is a whole world in the interval $(2,\infty)$. So again, the above value may match the specific batch of simulations, but see @whuber's comments below for a way out (and also, the OP appears to have found a way...). In any case, this valuet does not represent an upper bound on the $\sigma_{max_n}$.

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combining your guessing approach with @Dougal 's link to "a good approximation" for the normal distribution, I arrived at $\theta$~2.03025 for my used sample size of 10,000,000. Putting that into the formula you found, I get $\sigma_{max}$~1.55724. This seems to match up really well, thank you. –  kram1032 Jul 28 '14 at 8:34
You're welcome. It is always a nice moment when different answers are productively combined. –  Alecos Papadopoulos Jul 28 '14 at 10:01
The solution for the Normal distribution cannot be correct, because as $n$ grows large the expected squared difference between the values and their maximum must diverge. Counterexamples to your claim of a maximum at $1.52$ start at $n\gt 10^7$, approximately, which is not an unheard-of sample size. Not convinced? Take two minutes to simulate it for a range of values of $n$: sim <- function(n, mu=1/2, sigma=1/12) { x <- rnorm(n, mu, sqrt(sigma)); sqrt(mean((x - max(x))^2)) }; sapply(1:7, function(k) mean(replicate(1e2, sim(10^k)))) (in R). –  whuber Jul 28 '14 at 14:13
If you reason that asymptotically $\bar{X}$ and $X_{(n)}$ will be independent, then your initial analysis leads to a simple estimate of $\sigma_{\max n}$. For a Normal distribution with sd $\sigma$ it should approach $\sigma \Phi^{-1}(2^{-1/n})$ from above. –  whuber Jul 28 '14 at 14:22
@whuber Per your first comment, I don't see where we disagree -the first thing I wrote is that the maximum order statistic will diverge. For samples of the size you describe, naturally we should expect to observe values in $[2,\infty]$ (I admit I tend to think about sample sizes in the range of thousands and not of tens of millions, due obviously to my affiliation with econometrics!). The second comment is very interesting, I will try to work it out. –  Alecos Papadopoulos Jul 28 '14 at 15:17

This alternative is easier to analyze:

$$ \newcommand{\E}{\mathbb{E}} d(S) := \frac{1}{N} \sum_{i=1}^N (\max(S) - x_i) $$

Then we have

$$\begin{align*} \E d(S) &= \frac{1}{N} \sum_{i=1}^N (\E \max(S) - \E x_i) = \E \max(S) - \E x \end{align*}$$

$\E \max(S)$ might or might not be easy to find for a given distribution.

If you want to stick to the root-mean-square type distance, I'll call it $d_2$ because I think $\sigma_{\max_N}$ is misleading:

$$\begin{align*} d_2^2 &= \frac{1}{N} \sum_{i=1}^N (x_i - \max(S))^2 \\&= \frac{1}{N} \sum_{i=1}^N \left( x_i^2 + \max(S)^2 - 2 x_i \max(S) \right) \\&= \frac{1}{N} \sum_{i=1}^N x_i^2 + \max(S)^2 - 2 \frac{1}{N} \sum_{i=1}^N x_i \max(S) \end{align*}$$ Then $$\begin{align*} \E d_2^2 &= \E x^2 + \E( \max(S)^2 ) - 2 \frac{1}{N} \sum_{i=1}^N \E \left( x_i \max(S) \right) \end{align*}$$

Now, $$\begin{align*} \E x_i \max(S) &= \E\left[ x_i \max(S) \mid x_i = \max(S) \right] \Pr(x_i = \max(S)) \\&\qquad + \E\left[ x_i \max(S) \mid x_i \ne \max(S) \right] \Pr(x_i \ne \max(S)) \\&= \E\left[ \max(S)^2 \right] \frac{1}{N} + \E\left[ x_i \max(S) \mid x_i \ne \max(S) \right] \frac{N-1}{N} \\&\approx \frac{1}{N} \E\left[ \max(S)^2 \right] + \frac{N-1}{N} \E\left[ x_i \right] \E\left[ \max(S) \right] \end{align*}$$ when $N$ is large.

Then $$\begin{align*} \E d_2^2 &\approx \E x^2 + \E( \max(S)^2 ) - 2 \frac{1}{N} \E\left[ \max(S)^2 \right] - 2 \frac{N-1}{N} \E\left[ x_i \right] \E\left[ \max(S) \right] \\&= \E x^2 + \frac{N - 2}{N} \E\left[ \max(S)^2 \right] - 2 \frac{N-1}{N} \E\left[ x_i \right] \E\left[ \max(S) \right] . \end{align*}$$

Via Jensen's inequality, $\E d_2 = \E \sqrt{d_2^2} \le \sqrt{ \E d_2^2}$.

So you can at least get an approximate upper bound from the first and second moments of $x$ and $\max S$.

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Why don't you push this for $N\rightarrow \infty$? –  Alecos Papadopoulos Jul 28 '14 at 0:03
@AlecosPapadopoulos I don't think the $\lim_{N \to \infty}$ case for distributions with unbounded support is interesting, since $\lim_{N \to \infty} \mathbb{E}[\max S] = \infty$; it's the asymptotic results that matter. I suppose you can just do that for the different cases of the extreme value distribution. –  Dougal Jul 28 '14 at 2:42
I like the alternate metric you found. I'm a little dumbfolded that I didn't think of that myself. Obviously, since $\max{\left(S\right)}$ is the maximum value, subtracting another value from it will never go negative, so it's a valid metric, and I suspsect that, since no squaring is involved, it will also be more robust. Thanks. –  kram1032 Jul 28 '14 at 8:42
The "alternative" is not a distance: you need to use the absolute value of the differences. –  whuber Jul 28 '14 at 14:15
@whuber Luckily, $\max S = \max_i x_i \ge x_i$, so the difference is immaterial. (Which is why I changed the order from the original notation.) –  Dougal Jul 28 '14 at 15:38

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