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$$\sqrt{\frac{\sum_{i = 1}^N{w_i \left( x_i - \overline{x}_w \right)^2}}{\frac{\left( N' - 1 \right) \sum_{i = 1}^N {w_i}}{N'}}}$$

Is it acceptable to remove the $\frac{N' - 1}{N'}$ from the formula of weighted standard deviation?

Since $\frac{N' - 1}{N'}$ is approximately 1. If it is allowed, can you give me a reference. Thank you very much.

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  • $\begingroup$ What's $N'$ here means? $\endgroup$ – oszkar Apr 16 at 9:54
  • $\begingroup$ the N' prime is the number of weights that are not equal to zero $\endgroup$ – Ji Pa Apr 16 at 9:59
  • $\begingroup$ Are you sure it's the number of non-zero weights, not the sum of them? And why do you want to remove it? $\endgroup$ – oszkar Apr 16 at 10:38
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The N-1 normalization factor comes from the unbiased estimator of the variance. This makes sure that the expectation of your estimator is the true variance. If you change it by N, the expectation will no longer be exactly the true variance.

However, as N increases, this error becomes of course smaller. But just make sure to take this into account (you can compute the expectation and variance of your estimator) if you want to report the quality or confidence intervals of your estimator.

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  • $\begingroup$ So it is alright to remove the (N'-1)/N' from the formula it will still be a weighted standard deviation? but it will be biased? $\endgroup$ – Ji Pa Apr 17 at 3:04
  • $\begingroup$ As @oszkar said, I don't really know why you want to remove it so bad. And as I told you, it would be biased but asymptotically unbiased. $\endgroup$ – Oriol B Apr 17 at 17:50

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