Garage of Code: The value of more options

I took a course on Statistical Modeling of Extreme Values in university. The following results may have been covered elsewhere, but it was missing from the course I took. I think that's a shame because they lend themselves more easily to quick estimations than the GEV. When valuating insurance it may of course be important to use the formal tools, but for most cases in extreme value analysis we just want to know whether something has the chance to be Really Bad.

Suppose we pick N samples from a stochastic variable X that are independent and identically distributed. What is the expected value of the Max of these samples? In particular, how does it grow with respect to N?

1. X ~ U(0, 1)

$\mathbb{E}(\text{Max}(X_1,...,X_N)) = \int x f(x)F(x)^{N-1}dx = \int_{0}^{1}Nx^{N}dx = \frac{N}{N+1}$

Another way to see this is that the expected value of 1 - Max(X) shrinks with the speed of O(1/N). So with N=10, the max will be about 0.9, with N=100, the max will be about 0.99, and so on...

2. X ~ Exp(1)

$\mathbb{E}(\text{Max}(X_1,...,X_N)) = \int xNf(x)F(x)^{N-1}dx = \int_{0}^{\infty} x * Ne^{-x}(1-e^{-x})^{N-1}dx = [\text{Binomial theorem}] = N \sum_{k=0}^{N-1} {N-1 \choose k}(-1)^k \int_{0}^{\infty} x e^{-x(k+1)}dx = \\ N \sum_{k=0}^{N-1} {N-1 \choose k}(-1)^k \frac{1}{(k+1)^2}$

At this point, I had to take help from Wolfram Alpha [1]. The expression evaluates to the Digamma function of N, plus a constant [2]. The constant is about 0.58. The Digamma function grows as ln(N) - 1/(2N). Calculating some values:

N	E(Max({X}))
10	2.8
100	5.2
1000	7.5
1,000,000	14.4

Basically, the expected max grows logarithmically. Every extra order of magnitude adds about 2.3 (ln(10)) to the expected max value.

3. X ~ Pow(a), a > 1

$\mathbb{E}(\text{Max}(X_1,...,X_N)) = \int xNf(x)F(x)^{N-1}dx = aN \int_{1}^{\infty} x^{-a}(1-x^{-a})^{N-1}dx = [\text{Binomial theorem}] = aN \sum_{k=0}^{N-1} {N-1 \choose k} (-1)^{k} \int_{1}^{\infty} x^{-a(k+1)}dx = \\ aN \sum_{k=0}^{N-1} {N-1 \choose k} (-1)^{k} \frac{1}{a(k+1)-1}$

Once again, the waters were too deep for me here, so I had to take help of Wolfram Alpha [3][4]. The answer for the whole expression comes out to:

$\frac{(N-1)!(-\frac{1}{a})}{(N-1-\frac{1}{a})!}$

The value of (-1/a)! can be calculated with a gamma function calculator. For a=2, it comes out to about 1.77. Note that the input to the gamma function should be (1-1/a). For the remainder of the expression, we can use the Stirling approximation, which says that:

$N! \approx e^{N log(N)}$

Also assuming that N is fairly large, so we replace N-1 with N. Now we have:

$\frac{N!}{(N-\frac{1}{a})!} \approx e^{Nlog(N) - (N-\frac{1}{a})log(N-\frac{1}{a})} =...= (1 - \frac{1}{N-\frac{1}{a}})^N * (N-\frac{1}{a})^{\frac{1}{a}}$

The first term goes to 1/e as N becomes large. The second term is approximately N^(1/a). The approximation E(Max({X})) = N^(1/a) is used in the table below.

N	E(Max({X})), a=2	E(Max({X})), a=3
10	3.16	2.15
100	10	4.64
1000	31.6	10
1,000,000	1000	100

References

[1] Wolfram Alpha input

[2] Digamma function (wikipedia)

[3] Wolfram Alpha input

[4] Why, do you ask, did I not just use WA on the original expression? Actually, computation time in the free version times out then.

Garage of Code

codecogs equations

fredag 10 september 2021

The value of more options

1. X ~ U(0, 1)

2. X ~ Exp(1)

3. X ~ Pow(a), a > 1

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