# Is it valid to calculate standard deviation for n=2?

My colleagues and I are having discussion whether it's valid to calculate standard deviation for $n=2$ or not? I think it's valid since I can calculate based on the equation, but higher N will give more power in the analysis. Can anyone comment? Thank you!

• Valid, sure; useful, not so much. – André Nicolas Dec 21 '15 at 16:15
• I can't think of a counter-argument. What are they ? – Yves Daoust Dec 21 '15 at 16:52
• @AndréNicolas: for $N=2$, the standard deviation and range coincide (to a constant). Both are a cheap measure of spread that can be quite useful. An example application is the measure of temporal noise in images when you cannot afford taking more than two of them. – Yves Daoust Dec 21 '15 at 16:58

Standard deviation is a measure of spread from the mean, so it is defined even when $N=1$ (although in that case it will always be 0). Certainly when $N=2$, it is a meaningful statistic.
And you are right -- if $N$ is larger, the statistic will be more powerful.
• I am not so sure about the $0$ spread. I'd rather say it's indeterminate. – Yves Daoust Dec 21 '15 at 17:00
• For $N>1$, $\hat\sigma$ is an estimator of $\sigma$. For $N=1$, nothing can be estimated (I wouldn't take $0$ as a good estimator of $\sigma$). – Yves Daoust Dec 21 '15 at 19:13
• @YvesDaoust the first one was my intuition too - no indeterminacy: $$\sigma^2 = \frac{1}{n} \sum_{k = 1}^n (x_k - \mu)^2 = \frac{(x_1 - \mu)^2}{1} = 0$$ since $x_1 = \mu$... – gt6989b Dec 21 '15 at 19:15