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I was reading this paper http://comjnl.oxfordjournals.org/content/20/4/359.full.pdf and in the last paragraph of the first page, Rule 1 is given as $\alpha_{j+1} > \bar{\alpha} + k s_{\alpha}$. Here, $\bar{\alpha}$ and $s_{\alpha}$ refer to the mean and standard deviation of a distribution $\alpha$ while $\alpha_{j+1}$ is a clustering criterion value. So far so good. But $k$ is listed as the standard deviate.

What is a standard deviate and how is it different from a standard deviation? Googling standard deviate gives this link http://encyclopedia2.thefreedictionary.com/standard+deviate The link explains that a standard deviate is $\left(x- \bar{x}\right)/ \sigma$. Is this correct? I don't see any references on that site.

Also, is there a relation between the standard deviate and the tail of the distribution?

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I corrected your formula in the first sentence of the third paragraph: the deviance of an observation $x$ from the mean is standardized over $s_{x}$, what you wrote was something different. –  Alexis Jul 1 at 20:29
    
I'd have assumed that 'standard deviate' was something like a z-score, and your dictionary reference suggests the same - but don't assume the definition in the paper precisely matches the one in the dictionary. –  Glen_b Jul 1 at 23:34
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I'd have assumed that a "standard deviate" was someone who browsed not-very-wierd porn sites, but I'll take your word for it. –  keshlam Jul 2 at 1:41

1 Answer 1

up vote 3 down vote accepted

Without getting into technicalities, by what you have written:

Standard deviation of a variable $X$ is a measure of how "dispersed" $X$ is, i.e. on average, how far away are the individual observations of $X$ from its mean.

Standard deviate of a particular observation $x_i$ of $X$ is how many "units" is $x_i$ away from the sample mean $\bar{x}$. The units are centered and scaled by the mean and standard deviation of $X$ (i.e. standardized). So suppose the standard deviate of $x_i$ is $k$, which means $x_i$ is $k$ standardized units away from the mean. This is how it works:

$$ x_i=\bar{x}+k\sigma \Leftrightarrow k=\frac{x_i-\bar{x}}{\sigma} $$

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Please note that the units are not centered at all: they are only scaled by $\sigma$. Your phrase "...away from..." describes the centering by the mean. Thus, in brief, a "standard deviate" is a way of describing the data using a different system of units, much like the relationship between degrees Fahrenheit and degrees Centigrade. Second, the referenced paper is sloppy in its terminology, causing your answer not to reflect what it is actually doing. It is re-expressing the data in terms of (unbiased) estimates of the parameters (based on corrected sample moments). –  whuber Jul 1 at 21:41
    
Thanks ved. Your answer is correct. After reading on, on page2 para1 of the paper, the authors have indeed defined standard deviate, same as yours and same as the link I had referenced. Also, the connection between the standard deviate and the tail is clear, because in the evaluation of the paper, the authors standardized all the possible clustering criterion values and thereafter, used different values of k to determine the criterion value for the clustering iteration which falls beyond that k. This value represents a 'significant change' (debatable imho) used to determine optimal clustering. –  Dhruv Gairola Jul 2 at 0:01

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