User:NorwegianBlue/kladd

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[edit] This a page that I use for testing before posting an entry

There are no secrets here :-), but there may be nonsense, things in the middle of a cut-and-paste process, stuff that does not parse correctly, or a half-finished entry which I am preparing in response to something you wrote. In case of the latter, you are of course welcome to read it, but save your comments till the time when my entry is actually posted.

[edit] Table of contents

1 This a page that I use for testing before posting an entry
2 Table of contents
3 Evaluating a definite integral
4 Calculating the expected standard deviation
5 N generations
6 Help needed in understanding 1925 Biometrika paper
7 D3 function

NorwegianBlue is trying to take a short wikibreak and will be back on Wikipedia the 12th of July. Possibly, however, NorwegianBlue will not be able to keep away from Wikipedia for that long, and might sneak away to an internet café, and make some small edits once in a while anyway.

[edit] Evaluating a definite integral

I'm curious about the function

$w(n)=\int_{-\infty}^{\infty} (1-(1-\alpha)^n-\alpha^n) \, dx$ ,

where $\, \alpha=\Phi(x)$ , i.e. the cumulative standard normal distribution, and n is an integer greater than one. I suspect that the usage of α without any indication that it is a function of x may be non-standard (at least, it had me confused). However, I wanted to reproduce it exactly as it appears in the paper in which I found it: Tippet, LHC. On the Extreme Individuals and the Range of Samples Taken From a Normal Population. Biometrika vol 17, 364-387, 1925. It is my understanding that this function gives the expected value of the range of a sample of size n, taken from a standard normal distribution. In the field of Statistical process control, the function w is known as d2.

These are my questions:

For n = 2, it is easy to show that the integral is numerically equal to 2/sqrt(π) within machine precision, and I feel reasonably certain that 2/sqrt(π) is indeed the exact value. I would like to know how one determines whether this is the case. As I am neither a statistician nor a mathematician, I would need the details spelled out.
Can this integral be expressed in terms of simple functions for values of n greater than 2? If so, how?
Is my suspicion avove, that the notation today would be considered non-standard, correct? If so, what would standard notation be?

Thanks. --NorwegianBlue^talk 14:45, 9 February 2008 (UTC)

[edit] Calculating the expected standard deviation

The expected value of the standard deviation of the range of samples from a standard normal distribution was tabulated in a paper by Egon S. Pearson, The Percentage Limits for the Distribution of Range in Samples from a Normal Population, Biometrika Vol 24, No 3/4 (Nov 1932), pp. 404-417. The value 0.8525 is given for samples of size 2. I would like to calculate this quantity with higher precision, for various sample sizes. This function is known as d3 in the field of statistical process control. The expected value of the range itself is given in my previous question. Since Pearson used numerical methods (I did not understand the paper well enough to understand the calculations), I suspect that a closed-form solution may not exist, although I'm not at all sure of this.

As stated in my previous question, I am neither a statistician nor a mathematician. I am, however, capable of writing a program for evaluating an integral, if I only knew what calculations would be required. I would be very grateful if someone would enlighten me. --NorwegianBlue^talk 14:45, 9 February 2008 (UTC)

[edit] N generations

Flavius Zeno (c. 425–491)

[edit] Help needed in understanding 1925 Biometrika paper

$d_2(n)=\int_{-\infty}^{+\infty} (1-(1-\Phi(x))^n-\Phi(x)^n) \, dx$

Tippett then defines as equation (10) the following:

$\, \mu_2 = \sigma^2 = 2 \int_{-\infty}^{+\infty} \int_{-\infty}^{x_1}(1 - \alpha_1^n - (1 - \alpha_n)^n - (\alpha_1 - \alpha_n)^n)dx_1 dx_n - \overline{w}^2$

The square root of $\, \mu_2$ should correspond to the control chart constant d3, and Table IV confirms that this is indeed the case. My problem is that I haven't been able to evaluate $\, \mu_2$ .

I don't see a problem with the way you are calculating the double integral. However, there is something terribly wrong about the function you are trying to integrate. To explain, I'll introduce some notation:

$a(n,x_1,x_n)=1-\Phi(x_1)^n-(1-\Phi(x_n))^n-(\Phi(x_1)-\Phi(x_n))^n\;\!$

$b(n,x_1)=\int_{-\infty}^{x_1}a(n,x_1,x_n)\ dx_n$

$c(n)=\int_{-\infty}^{\infty}b(n,x_1)\ dx_1$

$d_3(n)=\sqrt{2c(n)-d_2(n)^2}$

In order for any of this to be meaningful, b must exist for any

x 1

. For this to happen, at the very least we must have $\lim_{x_n \to - \infty}a(n,x_1,x_n)=0$ for every

x 1

. This is clearly not the case, because $\Phi(-\infty)=0$ and so $a(n,x_1,-\infty)=1-2\Phi(x_1)^n$ . Thus the function is not integrable. Check that you have copied the equations exactly as they appear in the paper, and that the notation means exactly what you think it does; If so, there is possibly a mistake in the paper. -- Meni Rosenfeld (talk) 12:10, 24 February 2008 (UTC)

ooo

$\, d_3(n)=\sqrt{2\int_{-\infty}^{+\infty} \int_{-\infty}^{x_1}(1-\Phi(x_1)^n-(1-\Phi(x_n))^n {\color{red}+} (\Phi(x_1)-\Phi(x_n))^n)dx_n dx_1 -d_2(n)^2}$