I faced the below formula for standard deviation.
Is it correct or no? If it is How can I simplify or expand it.
$$ \sigma(S) \ = \sum_{s \in S} \frac{s^2}{|S|} - avg(S)^2 $$
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$\begingroup$ It is wrong for standard deviation, as it will have squared units. However, you meant variance. What formula do you use for calculating variance? $\endgroup$– DaveCommented Nov 11, 2019 at 5:00
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$\begingroup$ What you have there is a formula for variance. It's okay algebraically but you shouldn't use it that way for any numerical calculation. $\endgroup$– Glen_bCommented Nov 11, 2019 at 6:22
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$\begingroup$ Thanks @Glen_b for clarification.... $\endgroup$– MAACommented Nov 11, 2019 at 11:25
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1 Answer
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No, it's not correct. What are $S$ and $s$ here? We use $X$ and $x$ instead.
The population variance, which uses Greek letter is:
$$\sigma^2=\frac{1}{N}\sum^N_{i=1}(x_i-\mu)^2$$
However, since you probably don't know the true mean, which is $\mu$, you should use the sample variance, in English letter:
$$s^2=\frac{1}{n-1}\sum^n_{i=1}(x_i-\bar{x})^2$$
where $\bar{x}$ is the average of X.
So the sample standard deviation is:
$$s=\sqrt{\frac{1}{n-1}\sum^n_{i=1}(x_i-\bar{x})^2}$$
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1$\begingroup$ This misses the point about an equation for a sample version of $var(X)= \mathbb{E}[X^2] - (\mathbb{E}[X])^2$. Reasoning through it in my head, the given formula is equivalent to the usual MLE for variance. This is why I asked about which formula the OP wants to use for calculating variance. $\endgroup$– DaveCommented Nov 11, 2019 at 6:03
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$\begingroup$ @Dave, I understand what you mean, you want to use the stat formula (I have the book Statistical Inference by George Casella too). However, if the person use avg(). I would assume he or she is not trying to use the theorem. :-) $\endgroup$ Commented Nov 11, 2019 at 6:07
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1$\begingroup$ I would have thought $s$ is $x_i$ and $|S|$ is $n$, with the expression in the question being either $\frac{1}{n}\sum\limits^n_{i=1}x_i^2-\mu^2$ for a population or $\frac{1}{n}\sum\limits^n_{i=1}x_i^2-{\bar x}^2$ for a sample $\endgroup$– HenryCommented Nov 11, 2019 at 8:53