I'd agree with Don's point about the sample variance being unbaised under random sampling. Because of the linearity of the estimate, the lack of independence of samples is not a problem here. This should not be confused with the problem of t tests. There the source of the problem is that the variance of the sample mean, var(1/n sum(Z(x_i)) takes the form
sigma**2/n + (1/n**2)sum(C_ij) ( sum over all i,j: i not equal to j)
If the covariance terms for i not equal to j are all zero, then the variance of error reduces to sigma**2/n ....and this is where the number of independent samples n comes into it. If the samples are not independent, then the second term of the above does not necessarily fall away to zero quickly (in particular, in an extreme case, if the covariance falls very slowly we may have C_ij approx equal to sigma**2 and so the total above acts like sigma**2/n + (n-1/n)*sigma**2 = sigma**2. In other words the error does not reduce at all with an increasing number of samples - let alone reduce like 1/n).
So, for this t test business, a crude method of getting a number of 'independent' samples would be to take the lenght of the field divided by range (provided that we have enough sample data to cover the field at a sampling spacing less than the range). This could be used in place of the raw number of samples n - which as said before will give a very poor result.
Colin Daly
-----Original Message-----
From: Donald E. Myers [mailto:[EMAIL PROTECTED]]
Sent: Tue 12/7/2004 6:52 PM
To: [EMAIL PROTECTED]
Cc:
Subject: [ai-geostats] Continuing discussion on "F and t tests"
The sample variance (assuming that you use the "n-1" divisor) is an
unbiased estimator of the population variance provided you use random
sampling. Note the "ing" on the word sampling, it is not quite correct
to talk about "random samples" or "independent samples". or at least it
may be mis-leading. Random sampling pertains to how the data is
collected, not the end result.
Note moreover that one can always compute a sample variance for a given
data set but this does not show that the random variable or random
function has a finite variance.
The sample variance (even when sampling from a normal population) is
relatively speaking more variable as an estimator of the variance than
the sample mean is as an estimator of the population mean. The sampling
distribution in this restricted case is chi-square, the chi-square
distribution has a "fat" tail (as contrasted with a normal distribution).
If correctly (or maybe you would want to say "adequately" ) estimated,
the sill of a second order stationary random function would be the
variance of the random function. In general, the sample variance will
not estimate the sill (because you are not using random sampling).
Donald Myers
http://www.u.arizona.edu/~donaldm
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